maths.lanark.s-lanark.sch.ukmaths.lanark.s-lanark.sch.uk/.../2018/02/advanced-higher-all.pdf ·...

Outcome 1 – Algebraic Skills

The Binomial Expansion of Expressions like (x + y)n

The Binomial expansion provides a method of expanding (1 + x)n in powers of x where the index n is a positive integer.More generally, the theorem gives an expansion (x + y)n in powers of x and y.

By expanding, using the distributive law, it can be shown that (1 + x)n for n = 1, 2, 3, 4, etc. is as follows :-The coefficients form an array known as Pascal’s Triangle.

(1 + x)0 = 1 1

(1 + x)1 = 1 + x 1 1

(1 + x)2 = 1 + 2x + x2 1 2 1

(1 + x)3 = 1 + 3x + 3x2 + x3 1 3 3 1

(1 + x)4 = 1 + 4x + 6x2 + 4x3 + x4 1 4 6 4 1

(1 + x)5 = 1 + 5x + 10x2 + 10x3 + 5x4 + x5 1 5 10 10 5 1

so (1 + x)6 has its coefficients => 1 6 15 20 15 6 1

=> (1 + x)6 = 1 + 6x + 15x2 + 20x3 + 15x4 + 6x5 + x6

Similarly (1 + x)7 = 1 + 7x + 21x2 + 35x3 + 35x 4 + 21x5 + 7x6 + x7

Similarly for (x + y)n, n = 1, 2, 3, 4, etc.

Example (x + y)4 :–by writing (x + y) as x[1 + y

x ] we can see that the coefficients come also from Pascal’s Triangle.

(x + y)4 = x4[1 + yx ]4 = x4[1 + 4( y

x ) + 6( yx )2 + 4( y

x )3 + ( yx )4 ]

= x4[1 + 4 yx + 6 y2

x2 + 4 y3

x 3 + y4

x4 ]

= x4 + 4x3y + 6x2y2 + 4xy3 + y4

Note (1) The sum of the powers of x and y is 4.(2) The coefficients are from Pascal’s Triangle.(3) As the powers of x decrease, the powers of y increase.

Advanced Higher ©TeeJay Publishers 2000 Maths Support

page 2

Exercise 1

1. By writing x – y as x[1 −yx

], find an expression for (x – y)5

2. Write down expressions for

(a) (x + 1)5 (b) (1 – x)4 (c) (a + 2b)3 (d) (2a – b)5

(e) (2x – 3y)5 (f) (x +1x

)5 (g) (a – 2b)6 (h) (2a + b)7

Expanding (1 + x)n using The Binomial Coefficients.

From Pascal’s Triangle :-

(1 + x)0 = 1 1

(1 + x)1 = 1 + x 1 1

(1 + x)2 = 1 + 2x + x2 1 2 1

(1 + x)3 = 1 + 3x + 3x2 + x3 1 3 3 1

(1 + x)4 = 1 + 4x + 6x2 + 4x3 + x4 1 4 6 4 1

(1 + x)5 = 1 + 5x + 10x2 + 10x3 + 5x4 + x5 1 5 10 10 5 1

What are the entries in the n th line ?Notice that the 5th line has 6 entries, the 4th line has 5 entries, the 3rd line has 4 entries, and so on.So the n th line must have (n + 1) entries, where the first is 1 and the second is n.

The third entry is n(n – 1)1 × 2

, the fourth entry is n(n – 1)(n – 2)1 × 2 × 3

and so on.

So the 1 5 10 10 5 1 comes from

1 5 5 × 41 × 2

5 × 4 × 31 × 2 × 3

5 × 4 × 3 × 21 × 2 × 3 × 4

5 × 4 × 3 × 2 × 11 × 2 × 3 × 4 × 5

In general, the coefficient of xr , where p is any power in the expansion is :–

n(n – 1)(n – 2)(n – 3)....(n – r + 1)1× 2 × 3 × ...... × r

1 × 2 × 3 × ...... × r or r(r – 1)(r – 2)(r – 3)..... × 3 × 2 ×1 is written in shorthand as r! and is called r factorial or factorial r.

n! appears on most calculators. Find it and try using it.


page 3

This factorial notation can be used to simplify the formula for the coefficient

n(n – 1)(n – 2)(n – 3)....(n – r + 1)1 × 2 × 3 × ...... × r

can be written as :-

n(n – 1)(n – 2)(n – 3)....(n – r + 1)(n – r)(n – r – 1)(n – r – 2) × ... × 3 × 2 ×1(1 × 2 ×3 × 4 × ........ × r)(n – r)(n – r –1)(n – r – 2) × ... × 3 × 2 ×1

= n!r!(n – r )!

This coefficient n!r!(n – r )!

can be written as nCr or nr⎛

⎝ ⎜ ⎞

⎠ ⎟

Sonr⎛

⎝ ⎜ ⎞

⎠ ⎟ = nCr = n!

r!(n – r )! and is known as the binomial coefficient.

The word binomial means “ 2 terms ” and the binomial expansion for (1 + x)n is written as

(1 + x)n = nCrr=0

n

∑ xr

=> (1 + x)n = 1 +n1⎛

⎝ ⎜ ⎞

⎠ ⎟ x +

n2⎛

⎝ ⎜ ⎞

⎠ ⎟ x2 +

n3⎛

⎝ ⎜ ⎞

⎠ ⎟ x3 + ... +

nr⎛

⎝ ⎜ ⎞

⎠ ⎟ xr + ..... + xn

Also (x + y)n = xn +n1⎛

⎝ ⎜ ⎞

⎠ ⎟ xn–1y +

n2⎛

⎝ ⎜ ⎞

⎠ ⎟ x n–2y2 + ....... +

nr⎛

⎝ ⎜ ⎞

⎠ ⎟ xn–ryr + ... + yn

=> (x + y)n = xn + n1 xn –1y + n(n –1)

1× 2 xn– 2y2 + n(n –1)(n – 2)1× 2× 3 xn – 3y3 + ...... + yn

Examples:– Expand (a) (1 + x)6 (b) (1 – x)4 (c) (4 + x)5

(d) (2 + x)4 and find the value of (i) (2·1)4 (ii) (1·9)4

Solutions :–

(a) (1 + x)6 = 1 + 61 x + 6× 5

1× 2 x2 + 6× 5 × 41× 2× 3 x3 + 6 × 5× 4 ×3

1× 2× 3× 4 x4 + 6 × 5× 4 ×3 × 21× 2 ×3 × 4× 5 x5 + x6

=> (1 + x)6 = 1 + 6x + 15x2 + 20x3 + 15x4 + 6x5 + x6

(b) (1 – x)4 = (1 + (–x))4 = 1 + 41 (-x) + 4× 3

1× 2 (-x)2 + 4× 3× 21× 2 × 3 (-x)3 + (-x)4

=> (1 – x)4 = 1 – 4x + 6x2 – 4x3 + x4

(c) (4 + x)5 = 45 + 51 44 x + 5× 4

1× 2 43 x2 + 5× 4 × 31× 2× 3 42 x3 + 5× 4 ×3 × 2

1× 2 × 3× 4 4x4 + x5

=> (4 + x)5 = 1024 + 1280x + 640x2 + 160x3 + 20x4 + x5


page 4

this function also appears on some calculators.

Explore its use.

(d) (2 + x)4 = 24 + 41 23x + 4× 3

1× 2 22 x2 + 4× 3× 21× 2× 3 2x3 + x4

=> (2 + x)4 = 16 + 32x + 24x2 + 8x3 + x 4

(i) (2·1)4 = (2 + 0·1)4 = 16 + 32(0·1) + 24(0·1)2 + 8(0·1)3 + (0·1)= 16 + 3·2 + 0·24 + 0·008 + 0·0001 = 19·4481

(ii) (1·9)4 = (2 – 0·1)4 = 16 + 32(–0·1) + 24(–0·1)2 + 8(–0·1)3 + (–0·1)= 16 – 3·2 + 0·24 – 0·008 + 0·0001 = 13·0321

Exercise 2

1. Expand the following using the Binomial Expansion :-

(a) (3 + x)3 (b) (5 + 2x)3 (c) (3 + x)4

(d) (2 – x)4 (e) (x + 2y)3 (f) (2x – 3y)3

(g) (1 + 3x)4 (h) (2 – 3x)5 (i) (2x +3x

)5

2. Expand (2 + x)5 and use your expansion to find (a) (2·1)5 (b) (1·9)5

3. Expand (2 + x)7 in ascending powers of x up to and including the term in x3.Hence evaluate (2·1)7 correct to 6 significant figures.

4. Use the Binomial Theorem to evaluate the following to the stated degree of accuracy.

(a) (1·01)4 correct to 5 decimal places.

(b) (0·998)5 correct to 7 significant figures

(c) (0·99)10 correct to 4 decimal places.

(d) (1·99)10 correct to 4 significant figures.

Further examples can be found in the following resources.

The Complete A level Maths (Orlando Gough)

Page 251 Exercise 5.4:1

Understanding Pure Mathematics (A.J.Sadler/D.W.S.Thorning)

Page 226 Exercise 8D


page 5

Express proper rational functions as sums of partial fractions.

Given the expression 2x – 1

+3

x + 1+

42x – 3

we can express it as a single fraction.

2x – 1

+3

x + 1+

42x – 3

= 2(x + 1)(2x – 3) + 3(x – 1)(2x – 3) + 4(x – 1)(x + 1)(x – 1)(x + 1)(2x – 3)

= 14x 2 – 17x – 1(x – 1)(x + 1)(2x – 3)

14x 2 – 17x – 1(x – 1)(x + 1)(2x – 3)

is called a rational function.

2x – 1

+3

x + 1+

42x – 3

are called the corresponding partial fractions. Our object is to work in reverse and attempt to find the partial fractions for a given rational function.There are a number of different types to consider.

In general terms we shall refer to the rational function as f(x)g(x)

,

where f(x) and g(x) are polynomial functions.

Before we can begin to find the partial fractions of a rational function the degree of f(x) must be less than the degree of g(x).

This will mean carrying out a long division (or similar method) first, if necessary.

e.g. x3

x2 – 4= x +

4xx2 – 4

from

xx2 – 4 x3

x3 – 4x4x

Type 1 Having ensured that the degree of f(x) is less than the degree of g(x), and where g(x) has linear and different factors.

Example 1 x + 162x2 + x – 6

=x + 16

(2x – 3)(x + 2) by factorising

Let x + 162x2 + x – 6

=A

(2x – 3)+

B(x + 2)

cont’d ....


page 6

Multiply both sides by (2x – 3)(x + 2) [to remove the denominators]i.e. x + 16 = A(x + 2) + B(2x – 3)Substitute x = 3/2 35/2 = 7/2 A => A = 5 [since (2x – 3) equals zero]

Substitute x = – 2: 14 = –7B => B = –2 [since (x + 2) equals zero]

Hence x + 162x2 + x – 6

=5

(2x – 3)– 2

(x + 2)

Example 2 2x 2 + 4x(x2 – 1)(2x + 1)

=2x2 + 4x

(x – 1)(x + 1)(2x + 1)

Let 2x2 + 4x(x2 – 1)(2x + 1)

=A

(x – 1)+

B(x + 1)

+C

(2x + 1)

Multiply both sides by (x – 1)(x + 1)(2x + 1)ie. 2x2 + 4x = A(x + 1)(2x + 1) + B(x – 1)(2x + 1) + C(x – 1)(x + 1)Substitute x = 1: 6 = 6A => A = 1

Substitute x = –1: –2 = 2B => B = –1Substitute x = –1/2 : –3/2 = –3/4 C => C = 2

Hence 2x 2 + 4x(x2 – 1)(2x + 1)

=1

(x – 1)– 1

(x + 1)+

2(2x + 1)

Exercise 1

Find Partial Fractions for the following rational functions.

1. (a) 4x – 9(x – 2)(x – 3)

(b) 3 – 8xx(1 – x)

(c) x + 24x2 – x – 12

(d) 2(3x + 4)x2 + 4x


page 7

Type 2 Having ensured that the degree of f(x) is less than the degree of g(x), and where g(x) has linear and repeated factors.

e.g. –8x2 + 14x – 154x3 + 4x2 – 7x + 2

=–8x2 + 14x – 15(2x – 1)2(x + 2)

(use synthetic division to factorise)

Let –8x2 + 14x – 154x3 + 4x2 – 7x + 2

=A

(2x – 1)+

B(2x – 1)2 +

C(x + 2)

Multiply both sides by (2x – 1)2(x + 2)

–8x2 + 14x – 15 = A(2x – 1)(x + 2) + B(x + 2) + C(2x – 1)2

Substitute x = 1/2 : –10 = 5/2 B => B = – 4

Substitute x = –2 : –75 = 25C => C = –3

Note : we have now used all possible values of x from the factors. This will happen in repeated factors.

Therefore substitute x = 0 and use the values of B and C already found.

Substitute x = 0 : –15 = –2A + 2B + C => –15 = –2A – 8 – 3 => A = 2

= > –8x2 + 14x – 154x3 + 4x2 – 7x + 2

=2

(2x – 1)– 4

(2x – 1)2 – 3(x + 2)

Exercise 2


1. (a) 3x2 + 1x(x + 1)2 (b) 3x2 + 2

x(x – 1)2

(c) x 2 – 2x + 10(x + 2)(x – 1)2 (d) 5x2 – 6x – 21

(2x – 3)(x – 4)2


page 8

Type 3 Having ensured that the degree of f(x) is less than the degree of g(x), and where g(x) has a linear factor and an irreducible quadratic factor.

e.g. 5x – 7x3 + 3x2 + 2x + 6

=5x – 7

(x + 3)(x2 + 2) (use synthetic division to factorise)

Let 5x – 7x3 + 3x2 + 2x + 6

=A

(x + 3)+

Bx + C(x2 + 2)

Multiply both sides by (x + 3)(x2 + 2)

5x – 7 = A(x2 + 2) + (Bx + C)(x + 3)Substitute x = –3: –22 = 11A => A = –2

Note : We have now used all possible values of x from the factors. Now put x = 0 and x = 1 (simple values not already used) and use the values

of the constants already found.

Substitute x = 0: –7 = 2A + 3C => –7 = – 4 + 3C => C = –1

Substitute x = 1: –2 = 3A + 4(B + C) => –2 = –6 + 4B – 4 => B = 2

= > 5x – 7x3 + 3x2 + 2x + 6

=–2

(x + 3)+

2x – 1(x2 + 2)

Exercise 3


1. (a) 8x – 1(x – 2)(x2 + 4)

(b) 7x2 – x + 14(x – 2)(x2 + 4)

(c) x(3x + 2)(x – 2)(x2 + 4)

(d) x3 + 2x2 + 61(x + 3)2(x2 + 4)

*careful



Page 98 Exercise 2.3:3


Page 451 Exercise 18A


page 9

Answers

Pascal’s Triangle and the Binomial Theorem

Exercise 1 Page 31. x5 − 5x4y + 10x3y2 − 10x2y3 + 5xy 4 − y5

2. (a) x5 + 5x4 + 10x3 + 10x 2 + 5x + 1 (b) 1 − 4x + 6x2 − 4x3 + x 4

(c) a3 + 6a2b + 12ab2 + 8b3(d) 32a5 − 80a4b + 80a3b2 − 40a2b3 + 10ab4 − b5

(e) 32x5 − 240x4y + 720x3y2 − 1080x2 y3 + 810xy 4 − 243y5

(f) x5 + 5x3 + 10x +10x

+5x3 +

1x 5

(g) a6 − 12a5b + 60a4b2 − 160a3b3 + 240a2b4 − 192ab5 + 64b6

(h) 128a7 + 448a6b + 672a5b2 + 560a4b3 + 280a3b4 + 84a2b5 + 14ab6 + b7

Exercise 2 Page 51. (a) 27 + 27x + 9x 2 + x3 (b) 125 + 150x + 60x2 + 8x3

(c) 81 + 108x + 54x 2 + 12x3 + x4 (d) 16 − 32x + 24x2 − 8x3 + x4

(e) x3 + 6x2y + 12xy 2 + 8y3 (f) 8x3 − 36x2y + 54xy2 − 27y3

(g) 1 + 12x + 54x2 + 108x3 + 81x4

(h) 32 − 240x + 720x2 − 1080x3 + 810x4 − 243x5

(i) 32x5 + 240x3 + 720x +1080

x+

810x3 +

243x5

2. (a) 40·84101 (b) 24·76099

3. 180·109

4. (a) 1·04060 (b) 0·9900399 (c) 0·9044 (d) 973·9

Partial Fractions

Exercise 1 Page 7

1. (a) 1x − 2

+3

x − 3(b) 3

x−

51 − x

(c) 4x − 4

−3

x + 3(d) 2

x+

4x + 4

Exercise 2 Page 8

1. (a) 1x

+2

x + 1– 4

x + 1( )2 (b) 2x

+1

x − 1+

5x − 1( )2

(c) 2x + 2

−1

x − 1+

3x − 1( )2 (d) − 3

2x − 3+

4x − 4

+7

x − 4( )2

Exercise 3

1. (a) 3x − 2

−3x − 2x 2 + 1

(b) 5x − 2

+2x + 3x2 + 4

(c) 2x − 2

+x + 4x2 + 4

(d) 3x + 3

+4

x + 3( )2 −2x − 1x2 + 4


page 10

Outcome 2 – Differentiation

Differentiate functions involving sums, products, quotients and composites of the elementary functions, xn n ε Q, sinx, cosx, tanx, ex and lnx.

Differentiation from First Principles

An approximation for the gradient of the tangent AC to the curve at A can be found by taking a second point B, on the curve and calculating the gradient of the chord AB instead.

If A is the point [x, f(x)] and the other point B is taken as[x+h, f(x+h)] where |h| is small and h can be positive or negative, we can calculate the gradient of the chord AB.

mAB = f x + h( ) − f x( )x + h( ) − x

=f x + h( ) − f x( )

h

mAB tends to a limit as h tends to zero. This limit is denoted by ′ f (x) , the derivative of f x( ) and gives the gradient of the the tangent AC to the curve y = ƒ(x) at A.

i.e. ′ f x( ) = Limh→0

f x+ h( ) − f x( )h

. This is known as Differentiation from First Principles.

ExampleFind the derivative of x2 from first principles.f x( ) = x2 , f x + h( ) = x + h( ) 2 = x2 + 2xh + h2

′ f (x) = Limh→0

f x + h( ) − f x( )h

= Limh→0

x 2 + 2xh + h2( ) − x2

h= Lim

h→02x + h( ) = 2x

Exercise 1

Differentiate the following functions from First Principles :-

1. (a) f x( ) = x3 (b) f x( ) = x2 + 2x (c) f x( ) = 3x2 + 4x – 5


The Complete A level Maths (Orlando Gough) Page 52-53 Exercise 1.3:1 Question 5.

Understanding Pure Mathematics (A.J.Sadler/D.W.S.Thorning)Page 258 Exercise 10A Question 3.


page 11

x

yB[(x+h),f(x+h)]

A[x,f(x)]

C

tangent

Standard derivatives

f x( ) ′ f (x)

xn nxn−1

ax + b( )n an ax + b( ) n−1

sinx cosx

cosx ––sinx

sin(ax + b) acos(ax + b)

cos(ax + b) –asin(ax + b)

All of the above were covered for the Higher course.

Exercise 2Differentiate the following functions with respect to x :-

1. f x( ) = x 3− x 2 + 5 x− 6 2. f x( ) = 3x2 + 7 −4x

3. f x( ) = x +1x

4. f x( ) = x32 − x

12 + x

−12

5. f x( ) =1x 2 −

1x3 6. f x( ) =

xx2 +

x2

x

7. f x( ) = 4x + 5( )5 8. f x( ) = 2x4 − 3( )12

9. f x( ) =3

4 − x2( ) 10. f x( ) =

4

x3 + 3x( )13

11. f x( ) = cos3x 12. f x( ) = sinx


The Complete A level Maths (Orlando Gough) Page 54 Exercise 1.3:2 Questions 1,2,3. and Page 160 Exercise 3.4:3 Question 3.

Understanding Pure Mathematics (A.J.Sadler/D.W.S.Thorning)Page 261 Exercise 10B Questions 1 - 37 and Page 370 Exercise 15F Questions 1-32(only sin/cos)

New Trigonometric functions :-secantθ secθ( ) = 1

cosθ , cosecantθ cosecθ( ) = 1

sinθ , cotangentθ cotθ( ) = 1

tanθ


page 12

Differentiate a Simple Composite function using the Chain Rule

The Chain rule (An alternative form if not already met) If y is a function of u, and u is a function of x , then, if y is now regarded as a function of x

=> dydx

= dydu

×dudx

Example Differentiate y = x2 + 3x − 5( )5

Either dydx

= 5 x 2 + 3x − 5( )42x + 3( )

Or Let u = x2 + 3x − 5 then y = u5 and u = x2 + 3x − 5

dydu

= 5u4 dudx

= 2x + 3

then dydx

= dydu

×dudx

= 5u4 2x + 3( ) = 5 x2 + 3x − 5( )42x + 3( )

This method could be adopted in more complicated functions, but otherwise use the original method of differentiating the outer function and multiplying by the derivative of the inner function.

Exercise 3Differentiate the following functions using the Chain Rule as above :-

1. y = x 2 + 4x − 5( )32. y = x3 + 5

3. y = 1 + 2 x( )44. y =

34 − x 2( )



Page 64 Exercise 1.3:6 Questions 2.


Page 319 Exercise 13A Questions 1 - 20.


page 13

Differentiate a Product

The Product Rule :- Used to differentiate a product of two functions. If u and v are functions of x , ie. u(x) and v(x) then :-

ddx

u x( ) × v x( )[ ] = u x( ) × ddx

v x( )[ ] + v x( ) × ddx

u x( )[ ]

ddx

uv( ) = u dvdx

+ v dudx

Examples

1. y = xcosx Put u = x and v = cosxdudx

= 1 dvdx

= −sinx

dydx

= −xsinx + cosx = cosx – xsinx

2. y = x2sin3x Put u = x2 and v = sin3x

dudx

= 2x dvdx

= 3cos3x

dydx

= 3x2cos3x + 2xsin3x

Exercise 4

Differentiate the following functions using the Product rule as above :-

1. y = x2 x − 3( )2 2. y = x 2x + 3( )3

3. y = x x − 6( ) 4. y = x x − 3( )3

5. y = x + 1( )2 x − 1( )4 6. y = x3 x − 1( )

7. y = xsinx 8. y = x2sinx

9. y = sinxcosx 10. y = sin2xcos5x


The Complete A level Maths (Orlando Gough) Page 67 Exercise 1.3:7 Questions 1(i), (ii), (iii), (iv), (v) and Page 160 Exercise 3.4:3 Question 5(i), (ii), (iii).

Understanding Pure Mathematics (A.J.Sadler/D.W.S.Thorning)Page 330 Exercise 13E Questions 1 - 13.


page 14

Differentiate a quotient

The Quotient Rule :- Used to differentiate a rational function. If u and v are functions of x , ie. u(x) and v(x) then

ddx

u x( )v x( )⎡

⎣ ⎢

⎤

⎦ ⎥ =

v x( ) ddx

u x( )[ ] – u x( ) ddx

v x( )[ ]v x( )[ ]2

ddx

uv

⎛ ⎝ ⎜ ⎞

⎠ ⎟ =

v dudx

– u dvdx

v 2

Examples

1. y = x2 – 1x2 + 1

Put u = x2 – 1 and v = x2 + 1

dudx

= 2x dvdx

= 2x

v2 = x 2 + 1( )2

dydx

= 2x x2 + 1( ) – 2x x 2 – 1( )

x2 + 1( ) 2 = 2x3 + 2x – 2x3 + 2xx2 + 1( )2 = 4x

x2 + 1( )2

2. y = 2xx2 + 1( )

Put u = 2x and v = x2 + 1( )12

dudx

= 2 dvdx

= 12

x2 + 1( )− 12 × 2x

v2 = x 2 + 1( )

dydx

= 2 x2 + 1( )

12 – 2x2 x2 + 1( )−

12

x2 + 1( ) = 2 x2 + 1( )−

12 x2 + 1( ) – x 2[ ]x2 + 1( ) = 2

x2 + 1( )32

Exercise 5Differentiate the following functions using the Quotient rule as above :-

1. y =x2

x + 32. y =

4 – xx2

3. y =4x

1 − x( )3 4. y =2x2

x − 2

5. y =1 − 2x( )3

x3 6. y =x + 1( )x2



Page 67 Exercise 1.3:7 Questions 1(vi), (vii), (viii), 8. and Page 160 Exercise 3.4:3 Questions 5(iv), (v).

Understanding Pure Mathematics (A.J.Sadler/D.W.S.Thorning)Page 330 Exercise 13E Questions 14 - 24.


page 15

Derivatives of New Functions

The derivatives of y = tanx , y = cosecx, y = cotx, y = ex and y = lnx

1. y = tanx = sinxcosx

Put u = sinx and v = cosx

dudx

= cosx dvdx

= – sinx

v2 = cos2 x

dydx

= cos x × cos x – sin x × − sin x( )cos2 x

= cos2 x + sin2 x

cos2 x = 1

cos2x = sec2x

2. y = cosecx = 1sinx

= sinx( )−1

dydx

= − sinx( )−2 × cosx = − cosxsin2x

= − 1sinx

×cosxsinx

= − cosecx cotx

3. y = secx = 1cosx

= cosx( )−1

dydx

= − cosx( )−2 × − sinx = sinxcos2x

= 1cosx

×sinxcosx

= secx tanx

4. y = cotx = 1tanx

= tanx( )−1

dydx

= − tanx( )−2 × sec2x = −1

tan2x× sec2x = −

cos2xsin2x

×1

cos2x= −

1sin2x

= − cosec2x

5. The following result can be obtained. If y = ex

then dydx

= ex

or ddx

ex( ) = ex

6. It follows that we can now find ddx

loge x( )If y = lnx = loge x then x = ey ( by definition)

and so dxdy

= ey

therefore dydx

=1ey

dydx

=1x

or ddx

loge x( ) =1x


page 16

′ f (x) = Limh→0

f x + h( ) − f x( )h

= Limh→0

ex +h − ex

h = Lim

h→0

ex (eh –1)h

= ex Limh→0

(eh – 1)h

= ex

Examples

1. y = tan2x 2. y = tan2x

dydx

= 2sec2 2x dydx

= 2tanx.sec2 x

3. y = 3cosec2x 4. y = 2sec3xdydx

= 3 × −2cosec2xcot2x dydx

= 2 × 3sec3xtan3x

= −6cosec2xcot2x = 6sec3xtan3x

5. y = cot3x dydx

= 3cot 2x × −cosec2 x = − 3cot2xcosec2x

6. y = e3x

dydx

= 3e3x from y = eu where u = 3x

dydx

= eu dudx

= 3 dydx

= dydu

×dudx

= eu × 3 = 3e3x

In general ddx

e f x( )( ) = ′ f x( )e f x( )

7. y = ln3xdydx

= 3 ×1

3x=

1x

from y = lnu where u = 3x

dydu

=1u

dudx

= 3 dydx

= dydu

×dudx

=1u× 3 =

13x

× 3 =1x

In general ddx

ln f x( )[ ]{ } = ′ f x( ) ×1

f x( )

8. y = ex2 9. y = esinx dydx

= 2xex 2 dydx

= cosxesinx

10. y = x2e2x Put u = x2 and v = e2x

dudx

= 2x dvdx

= 2e2x

dydx

= 2x2e2x + 2xe2x = 2xe2x x + 1( ) (The Product Rule)


page 17

cont’d .....

11. y =lnxx

Put u = lnx and v = xdudx

=1x

dvdx

= 1

v2 = x2

dydx

=x ×

1x− lnx × 1

x2 =1 − lnx

x2 (The Quotient Rule)

Exercise 6

Differentiate using the Chain, Product and Quotient Rules as above :-

1. y = tan32x 2. y = − 2cosec4 x

3. y = secxtanx 4. y = x2cotx

5. y = ln 3x + 2( ) 6. y = x + 2( )e−x

7. y =ex

x + 28. y =

x2

lnx

9. y = ln x2 + 1( )⎛ ⎝

⎞ ⎠ 10. y = xe−2x2

11. y = ln 1 + x1 − x⎛ ⎝ ⎜ ⎞

⎠ ⎟ 12. y =

ex − e− x

ex + e− x


The Complete A level Maths (Orlando Gough) Page 105 Exercise 2.5:2 Questions 5, 6 and Page 108 Exercise 2.5:4 Questions 8, 10Page 160 Exercise 3.4:3 Questions 10,11,13 and 16.

Understanding Pure Mathematics (A.J.Sadler/D.W.S.Thorning)Page 370 Exercise 15F Questions 1-34(cosec/sec/cot only) and Page 480 Exercise 19A Questions 1-15 and Page 487 Exercise 19B Questions 1-15, 17-20.


page 18

Higher Derivatives

Let y = x 2 + 3x + 4dydx

= 2x + 3

Here dydx

is defined as a function of x and so can be differentiated with respect to x.

ie. ddx

dydx

⎛ ⎝ ⎜ ⎞

⎠ ⎟ = 2

ddx

dydx

⎛ ⎝ ⎜ ⎞

⎠ ⎟ is usually written as d 2y

dx 2 , ′ ′ f x( ) or ′ ′ y and is called the 2nd derivative of y with

respect to x.

This process can be repeated to get the 3rd, 4th , ------ , nth derivative which are written as d3ydx3 , d4 y

dx 4 , ..............., dn ydx n

Motion in a straight line

Take the x-axis to be the straight line along which the motion takes place.The displacement is defined as the distance from the origin in time t and is denoted by x(t).

0 x(t) x

Velocity is defined as the rate of change of displacement with respect to time andis denoted by v(t).

ie.v t( ) = ddt

x(t)( ) or simply v = dxdt

Acceleration is defined as the rate of change of velocity with respect to time andis denoted by a(t).

ie.a t( ) = ddt

v(t)( ) or simply a = dvdt

= ddt

dxdt

⎛ ⎝ ⎜ ⎞

⎠ ⎟ = d2x

dt2

If this has a positive value, it is called acceleration and if negative it is called a deceleration.

Another notation used is

x - displacement x•

– velocity x••

– acceleration

x•

= dxdt

x• •

= d2xdt2


page 19

Examples

1. A car is travelling along a straight road. The distance, x (metres), travelled in t seconds, is x = 10t – 5t2

Find its velocity when t = 0·5 secs.

x = 10t – 5t2 v = dx

dt = 10 – 10t

at t = 0·5, v = 10 – 5 = 5 m/s.

2. A car is travelling along a straight road. Its velocity, v (metres per second), in t seconds, is v = 10 + 6t2 – t3

Find the acceleration when t = 3 secs.v = 10 + 6t2 – t3

a = dvdt

= 12t – 3t2

at t = 3, a = 36 – 27 = 9 m/s2

3. A car is travelling along a straight road. Its distance, x (metres), travelled in t seconds, is x = 5 + 2t + t3

Find the velocity and acceleration when t = 3 secs.x = 5 + 2t + t3

v = dxdt

= 2 + 3t2 , at t = 3, v = 2 + 27 = 29m/s

a = d2xdt2 = 6t , at t = 3, a = 18 m/s2


page 20

Exercise 7

1. A body moves in a straight line and the motion is such that x, the number of metres from a fixed point after t secs, is given by

x = 3 – 4t + t2.(a) How far is the body from the fixed point at the start ?(b) What is its position after 4 seconds ?(c) What is its velocity after 3 seconds ?(d) What is the initial acceleration ?

2. If x = 4t3– 3t2 – 2t – 1, where x is in metres and t in seconds, find(a) The velocity at the end of the 3rd and 4th seconds.(b) The acceleration at the end of the 3rd and 4th seconds. (c) The average velocity during the 4th second.(d) The average acceleration during the 4th second.



Page 370 Exercise 7.1:1.


Page 312 Exercise 12D.


page 21

3. A motor bike starts from rest and its displacement x m after t secs is given by:-

x =16

t3 +14

t2

Calculate the initial acceleration and the acceleration at the end of the 2nd second.

4. A body is moving in a straight line, so that after t seconds its displacement x metres from a fixed point O, is given by

x = 9t + 3t2 – t3.(a) Find the initial displacement, velocity and acceleration of the body.

(b) Find the time at which the body is instantaneously at rest.

5. A body moves along a straight line so that after t seconds its displacement from a fixed point O on the line is x metres.If x = 3t2(3 – t), find:–

(a) the initial velocity and acceleration.

(b) the velocity and acceleration after 3 seconds.

y

x1 2

(1,5)(2,4)

(3/2,-3/2)

′ f x( )

f x( )

Using the Second Derivative to determine the nature of Stationary Points

Consider the function Now consider the function defined by f x( ) = 2x3 − 9x2 + 12x ′ f x( ) = 6x2 − 18x + 12′ f x( ) = 6x2 − 18x + 12 ′ ′ f x( ) = 12x − 18

For St. values ′ f x( ) = 0 For St. values ′ ′ f x( ) = 0 6x2 − 18x + 12 = 0 12x − 18 = 0 6 x2 − 3x + 2( ) = 0 x =

32

6 x − 1( ) x − 2( ) = 0 y = −32

x = 1 or x = 2 y = 5 y = 4Therefore f x( ) = 2x3 − 9x2 + 12x Therefore ′ f x( ) = 6x2 − 18x + 12

has a

has stationary points at (1,5) and(2,4). stationary point at 32

, −32

⎛ ⎝ ⎜ ⎞

⎠ ⎟

Nature Nature

x <- 1 -> <- 2 -> x <- 32

-> x – 1 – 0 + + + + 12x – 18 – 0 + x – 2 – – – – 0 + ′ ′ f x( ) – 0 +

′ f x( ) + 0 – – 0 + ′ f x( ) f x( )

ie. Maximum T. Pt. at (1,5) ie. Minimum T. Pt. at 32

, − 32

⎛ ⎝ ⎜ ⎞

⎠ ⎟

Minimum T. Pt. at (2,4)

Now compare the graphs of f x( ) and ′ f x( )

The gradient of f x( ) = 2x3 − 9x2 + 12x is given by ′ f x( ) .

The gradient of ′ f x( ) = 6x2 − 18x + 12 is given by ′ ′ f x( ) .

At the maximum turning point (1,5) , ′ ′ f x( ) = negative. ie ′ ′ f 1( ) = – ve.At the minimum turning point (2,4) , ′ ′ f x( ) = positive. ie ′ ′ f 2( ) = +ve.


page 22

(0,2)

(–1,4)

(1,0)(–2,0) x

y

In general If, at x = a,

(1) ′ ′ f a( ) = + ve , then f x( ) has a minimum stationary value.(2) ′ ′ f a( ) = – ve , then f x( ) has a maximum stationary value,(3) ′ ′ f a( ) = 0 , then f x( ) has a possible point of inflexion , but use a table of signs to check.At a point of inflexion , it can be shown that ′ f x( ) = 0 , a necessary but not sufficient condition.

For example; f x( ) = x4 => ′ f x( ) = 4x3 For S.V. ′ f x( ) = 0 ie. 4x3 = 0 =>x = 0 ie. St. Pt at (0,0)

Now ′ ′ f x( ) = 12x 2 and ′ ′ f 0( ) = 0 -- a possible point of inflexion. But consider the sign of ′ f x( ) at x = 0

x <- 0 -> ′ f x( ) – 0 + ie. (0,0) is a minimum T.Pt. f x( )

Examples

1. Sketch the graph of the function f (x) = x + 2( ) x − 1( )2 .′ f x( ) = x − 1( )2 + 2 x + 2( ) x − 1( ) (using the product rule)

For S.V. ′ f x( ) = 0x − 1( ) 2 + 2 x + 2( ) x − 1( ) = 0

x − 1( ) x − 1( ) + 2x + 4[ ] = 0 3 x − 1( ) x + 1( ) = 0 (or 3x 2 − 3 = 0)

i.e. x = −1 and x = 1

Stationary points at (–1,4) and (1,0)′ ′ f x( ) = 6x [the 2nd derivative]′ ′ f −1( ) = negative i.e. (–1,4) is a Maximum Turning point.′ ′ f −1( ) = positive i.e. (1,0) is a Minimum Turning point.

When f (x) = 0 , x + 2( ) x − 1( )2= 0 => x = –2 and x = 1.Curve crosses the x -axis at (–2,0) and (1,0)When x = 0, f (0) = 2 i.e. y = 2.Curve crosses the y -axis at (0,2)


page 23

2. Find the coordinates and nature of the stationary point on the curvef x( ) = x3 − 81 lnx

′ f x( ) = 3x 2 −81x

For S.V. ′ f x( ) = 0

3x 2 −81x

= 0

3x3 − 81 = 0 => x3 = 27 => x = 3, y = 27 – 81 ln 3′ ′ f x( ) = 6x +

81x 2

′ ′ f 3( ) = +ve i.e. (3, 27 – 81 ln 3) is a Minimum Turning point.

3. Find the coordinates and nature of the stationary point on the curve f x( ) = ex − 4x

′ f x( ) = ex − 4

For S.V. ′ f x( ) = 0 e x − 4 = 0 => ex = 4 => x = ln 4, y = 4 – 4 ln 4

′ ′ f x( ) = e x

′ ′ f ln4( ) = +ve i.e. (ln 4, 4 – 4 ln 4) is a Min. Turning point.



Page 61 Exercise 1.3:5. Questions 3,10(i) and Page 105 Exercise 2.5:2 Question 8 andPage 108 Exercise 2.5:4 Question 15


Page 268 Exercise 10D Questions 9 - 14 and Page 480 Exercise 19A Questions 31,32,33.


page 24

Exercise 8Use the second derivative to find the stationary values and their nature for the following functions.

1. y = x – lnx 2. y = xlnx

3. y = xe– x 4. y = 12 sinθ + sin2θ

Optimisation Problems

Example1. An isosceles triangle is inscribed in a circle of radius r.

Show that the area of the triangle isA = r2sinθ 1 + cosθ( )

where θ is the angle between the equal sides.

Find the maximum possible area of the triangle.

From sinθ =xr

=> x = rsinθ and cosθ =yr

=> y = rcosθ

The base is 2rsinθ and the height is r + rcosθ

Area = 12 2rsinθ(r + rcosθ) = r 2sinθ 1 + cosθ( )

A θ( ) = r 2sinθ 1 + cosθ( ) = r2sinθ + r 2sinθcosθ = r 2sinθ +12

r 2sin2θ

′ A θ( ) = r2cosθ + r2cos2θ

For S.V. ′ A θ( ) = 0

r 2cosθ + r 2cos2θ = 0

r 2 2cos2θ + cosθ − 1( ) = 0 => r 2 2cosθ − 1( ) cosθ + 1( ) = 0

cosθ =12

and cosθ = −1 => θ =π3

or θ = π (but θ ≠ π )

S.V. at θ =π3

′ ′ A θ( ) = −r 2sinθ − 2r 2sin2θ

′ ′ A π3

⎛ ⎝ ⎜ ⎞

⎠ ⎟ = −r2sin π

3− 2r 2sin 2π

3= – ve i.e. a maximum S.V. at θ =

π3

The maximum area is

A π3

⎛ ⎝ ⎜ ⎞

⎠ ⎟ = r2sin π

31 + cos π

3⎛ ⎝ ⎜ ⎞

⎠ ⎟ = r 2 3

21 +

12

⎛ ⎝ ⎜ ⎞

⎠ ⎟ =

3 34

r 2


page 25

θ

r r

r

xy θ

Exercise 9

1. Four squares each of side s cm are cut from the corners of a metal square of side 16 cm. The metal is then bent to make an open topped tray of volume, V cm3.

(a) Prove that V = 4s3 – 64s2 + 256s.(b) Find the value of s which makes the volume a maximum.


The Complete A level Maths (Orlando Gough) Examples can be found throughout most differentiation sections of trigonometric,exponential and logarithmic functions.


Page 268 Exercise 10D Questions 21 - 25.


page 26

2. A sector of a circle with radius r cm has an area of 16 cm2.

(a) Show that the perimeter P cm of the sector is given by

P(r) = 2(r +16r

)

(b) Find the minimum value of P.

3. A cylindrical tank has a radius of r metres and a height of h metres.The sum of the radius and the height is 2 metres.

(a) Prove that the volume, in m3, is given by V =πr2(2 – r)

(b) Find the maximum volume.

4. ABCD is a kite which has AC as its axis of symmetry.Angle BAD is right angled and BC and DC are 20 cm.

(a) Show that the area of triangle BCDis given by the expression 200 sin2θ and find an expression for BD2.

(b) Use this expression for BD2 to showthat the area of triangle BAD is given bythe expression 200 – 200 cos2θ andhence show that the area of the kiteis given by the expressionA(θ) = 200(1 – cos2θ + sin2θ)

(c) Find the value of θ which makes the area a maximum and find this maximum area.

A

B

C

D

2020

2θ

Answers

Exercise 1 Page 111. 3x 2 2. 2x + 2 3. 6x + 4Exercise 2 Page 12

1. 3x 2 − 2x + 5 2. 6x +4x2 3. 1

2x12

−1

2x32

4. 32

x12 −

1

2x12

−1

2x32

5. − 2x3 +

3x 4 6. − 3

2x52

+32

x12 7. 20 4x + 5( )4 8. 4x3

2x4 − 3( )12

9. 3x

4 − x2( )32

10. − 4(x 2 + 1)

x3 + 3x( )43

11. −3sinxcos2x 12. cosx2 sinx

Exercise 3 Page 13

1. 6x + 12( ) x2 + 4x − 5( )2 2. 3x2

2 x3 + 5( ) 3.

4 1 + 2 x( )3

x4. 3x

4 − x2( )32

Exercise 4 Page 14

1. 2x x − 3( ) 2x − 3( ) 2. 8x + 3( ) 2x + 3( )2 3. 3 x − 4( )2 x − 6

4. x − 3( )2 7x − 3( )2 x

5. 2 x + 1( ) 3x + 1( ) x − 1( )3 6. x2 7x − 6( )2 x − 1

7. sinx + xcosx 8. x 2sinx + xcosx( )

9. cos2x 10. 2cos2xcos5x − 5sin2xsin5x

Exercise 5 Page 15

1. x2 + 6xx + 3( )2 2. x − 8

x3 3. 4 2x + 1( )1 − x( )4 4. 2x x − 4( )

x − 2( )2

5. − 3 1 − 2x( )2

x 4 6. − 3x + 4( )2x3 x + 1

Exercise 6 Page 18

1. 6 tan2 2x sec2 2x 2. 8cosec4xcotx 3. secx tan 2x + sec2x( )

4. x 2cotx − xcosec2x( ) 5. 33x + 2

6. − x + 1( )e−x

7. x + 1( )e x

x + 2( ) 2 8. x 2lnx − 1( )lnx( )2 9. x

x2 + 1

10. e−2 x21 − 4x2( ) 11. 2

1 − x2 12. 4ex + e−x( )2


page 27


page 28

Exercise 8 Page 24

1. Min at (1, 1) 2. Min at (1/e, 1/e) 3. Max at (1,1/e)

4. Max at (π/3,3√3/4), P of I at (π, 0), Min at (5π/3,–3√3/4)

Exercise 9 Page 25 - 26

1. (a) Proof (b) s = 8/3 cm

2. (a) Proof (b) P = 16 cm

3. (a) Proof (b) V = 32π/27 cm3

4. (a) Proof (b) Proof

(c) θ = 8π/3 , A = 200(1 + √2)

Exercise 7 Page 20 - 21

1. (a) 3 m (b) 3 m (c) 2 m/s (d) 2 m/s2

2. (a) 88 m/s, 166 m/s (b) 66 m/s2, 90 m/s2

(c) 127 m/s (d) 78 m/s2

3. (a) 1/2 m/s2 (b) 21/2 m/s2

4. (a) 0, 9 m/s, 6 m/s2 (b) 3 secs

5. (a) 0 m/s, 18 m/s2 (b) –27m/s, –36 m/s2

Outcome 3 – Integration

Integrate an expression requiring a standard result.

Standard Indefinite Integrals

f x( ) f x( )∫ dx

xn xn +1

n + 1+ c

ax + b( )n ax + b( )n+1

a n + 1( ) + c

sin x − cosx + c

cos x sin x + c sin ax + b( ) −

1a

cos ax + b( ) + c

cos ax + b( ) 1a

sin ax + b( ) + c

All of the above were covered for the Higher course.In addition the following trigonometrical identities were useful.

sin2 x =12

1 – cos 2x( )

cos2 x =12

1+ cos2x( )Exercise 1

Integrate the following functions :-

1. f x( ) = 8x3 2. f x( ) =6x3 3. f x( ) = x

4. f x( ) =1x

5. f x( ) =3x4 + 6

x2 6. f x( ) =1− 3x

x

7. f x( ) = 3x + 4( )4 8. f x( ) = 1− 2x( )4 9. f x( ) =1

2x − 3( ) 2

10. f x( ) =1

4x + 1( )11. f x( ) =

3

3x + 1( )32

12. f x( ) = sin 3x

13. f x( ) = cos 13

x⎛ ⎝ ⎜

⎞ ⎠ ⎟ 14. f x( ) = sin2 x 15. f x( ) = cos2 1

2x

⎛ ⎝ ⎜

⎞ ⎠ ⎟


The Complete A level Maths (Orlando Gough) Page 171 Exercise 4.1:2 Questions 1, 2.

Understanding Pure Mathematics (A.J.Sadler/D.W.S.Thorning)Page 296 Exercise 12A Questions 1, 2, 3 and Page 373 Exercise 15G Questions 1, 3, 4, 8, 9, 10, 11, 12, 13, 14.


page 29

New Integrals

The exponential function ex∫ dx .

(a) If f x( ) = ex then f ′ x( ) = ex and so ex∫ dx = ex + c

(b) If f x( ) = e2 x then ′ f x( ) = 2e2x and so 2e2 x∫ dx = e2x + c

ie e2x∫ dx =12

e2x + c

(c) If f x( ) = e1− 2x then ′ f x( ) = −2e1−2 x and so −2e1− 2x∫ dx = e1−2 x + c

ie e1− 2x∫ dx = −12

e1− 2x + c

In General eax +b∫ dx =1a

eax +b + c

The logarithmic function 1ax + b∫ dx

(a) If f x( ) = ln x then ′ f x( ) =1x

and so 1x∫ dx = ln x + c

(b) If f x( ) = ln 2x then ′ f x( ) =22x

and so 22x∫ dx = ln 2x + c

ie 2 12x∫ dx = ln 2x + c

so 12x∫ dx =

12

ln 2x + c

(c) If f x( ) = ln 3x + 2( ) then ′ f x( ) =3

3x + 2 and so 3

3x + 2∫ dx = ln 3x + 2( ) + c

ie 3 13x + 2∫ dx = ln 3x + 2( ) + c

so 13x + 2∫ dx =

13

ln 3x + 2( ) + c

In General 1ax + b∫ dx =

1a

ln ax + b( ) + c


page 30

The integral of sec2∫ xdx

(a) If f x( ) = tan x then f ′ x( ) = sec2 x and so sec2∫ xdx = tan x + c

(b) If f x( ) = tan 3x then f ′ x( ) = 3sec2 3x and so 3sec2 3∫ xdx = tan 3x + c

ie 3 sec2 3∫ xdx = tan 3x + c

so sec2 3∫ xdx =13

tan 3x + c

In General sec2∫ ax + b( )dx =1a

tan ax + b( ) + c

Exercise 2 Integrate the following functions :-

1. f x( ) = e5 x 2. f x( ) = e−2 x 3. f x( ) = 3e12

x

4. f x( ) =13x

5. f x( ) =1

x + 56. f x( ) =

12x − 3

7. f x( ) = ex + e−x( )28. f x( ) =

1+ ex

e x 9. f x( ) = e2 x +1

e2x

10. f x( ) =6

3x + 211. f x( ) =

31− 2x

12. f x( ) =5

6 − 7x

13. f x( ) = sec2 4x 14. f x( ) = sec2 π + 2x( ) 15. f x( ) = 3sec2 2x


The Complete A level Maths (Orlando Gough) Page 172 Exercise 4.1:2 Questions 15, 18.

Understanding Pure Mathematics (A.J.Sadler/D.W.S.Thorning)Page 480 Exercise 19A Questions 16, 17, 18, 19, 22, 23. and Page 487 Exercise 19B Questions 26, 27, 28, 34, 35, 36.


page 31


page 32

Definite Integrals

1. f x( )a

b

∫ dx = F a( ) − F b( ) where ′ F (x) = f (x)

eg. x2 + 1( )dx =x3

3+ x

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

1

3

∫1

3

=33

3+ 3

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ −

13

3+ 1

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ = 12 −

43

=323

or 102/3

2. f x( )a

b

∫ dx = − f x( ) dxb

a

∫

eg. x2 + 1( )dx =x3

3+ x

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

1

3

∫1

3

=33

3+ 3

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ −

13

3+ 1

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ = 12 −

43

= 102/3

and − x2 + 1( )dx = −x3

3+ x

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

3

1

∫3

1

= −13

3+ 1

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ −

33

3+ 3

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

= −43−12

⎡ ⎣ ⎢

⎤ ⎦ ⎥ = − −

323

⎡ ⎣ ⎢

⎤ ⎦ ⎥ =

323

= 102/3

Examples

1. 1x2

−3

−2

∫ dx = x−2

−3

−2

∫ dx = −x−1[ ]−3

−2= −

1x

⎡ ⎣ ⎢

⎤ ⎦ ⎥ −3

−2

= −1−2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ − −

1−3

⎛ ⎝ ⎜

⎞ ⎠ ⎟ =

12−

13

=16

2. 12x −1( )1

5

∫ dx = 2x −1( )1

5

∫−

12dx =

2x −1( )12

2 × 12

⎡

⎣

⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥

1

5

= 2x −1( )12

⎡ ⎣ ⎢

⎤ ⎦ ⎥

1

5

= 912 −1

12 = 3 −1 = 2

3. sin 2x0

12π

∫ dx = −12

cos2x⎡ ⎣ ⎢

⎤ ⎦ ⎥

0

12π

= −12

cosπ⎛ ⎝ ⎜

⎞ ⎠ ⎟ − −

12

cos0⎛ ⎝ ⎜

⎞ ⎠ ⎟ =

12

+12

= 1

4. sec2

−12π

12π

∫12

x dx = 2 tan 12

x⎡ ⎣ ⎢

⎤ ⎦ ⎥ −

12π

12π

= 2 tan 14π − 2 tan −

14π

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 2 ×1− 2 × −1( ) = 4

5. e−3x

0

2

∫ dx = −13

e−3x⎡ ⎣ ⎢

⎤ ⎦ ⎥

0

2

= −13

e−6⎛ ⎝ ⎜

⎞ ⎠ ⎟ − −

13

e0⎛ ⎝ ⎜

⎞ ⎠ ⎟ = −

13

e−6⎛ ⎝ ⎜

⎞ ⎠ ⎟ − −

13

⎛ ⎝ ⎜

⎞ ⎠ ⎟ =

13

1 − e−6( )

6. 2x − 34

5

∫ dx = 2ln x − 3( )[ ]4

5= 2ln 2 − 2ln1 = 2ln 2 = ln 4


page 33


The Complete A level Maths (Orlando Gough) Page 176 Exercise 4.1:3 Questions 1, 2, 3, 4.

Understanding Pure Mathematics (A.J.Sadler/D.W.S.Thorning)Page 303 Exercise 12B Question 1. and Page 374 Exercise 15G Questions 36, 37, 38, 41.

Exercise 3

Integrate the following functions :-

1. 2x2 −

5x4

⎛ ⎝ ⎜

⎞ ⎠ ⎟

1

2

∫ dx 2. x −1x

⎛ ⎝ ⎜

⎞ ⎠ ⎟ 1

4

1

∫ dx 3. 2x2 + 3x1

4

∫ dx

4. x + 3( )1

6

∫ dx 5. 14 + 5x( ) 20

1

∫ dx 6. x − 4( )13

3

12

∫ dx

7. cos2xdx0

π4∫ 8. cosec 2xdxπ

4

π2∫ 9. sin2 xdx

0

2π

∫

10. e−3xdx0

2

∫ 11. e1−x dx0

1

∫ 12. ex2dx

0

1

∫

13. 1x − 35

9

∫ dx 14. 13x + 20

1

∫ dx 15. 11− 2x−4

0

∫ dx


page 34

Integration by Substitution

It is sometimes difficult to reduce an integral to one of the Standard integrals. Such integrals may be made simpler by changing the variable by means of a substitution of a new variable.

Examples (where the simple substitution is not given)

1. x4∫ 1+ 2x5( )3dx (note that d

dx1 + 2x5( ) = 10x4 )

Let u = 1+ 2x5 then dudx

= 10x4 and so x4 dx =1

10 du

The integral now becomes 1

10∫ u3 du =1

10×

14× u4 + c

=140

1+ 2x5( )4 + c

2. 2xe x2

∫ dx (note that ddx

x2( ) = 2x)Let u = x2 then du

dx= 2x and so 2xdx = du

The integral then becomeseu ∫ du = eu + c

= ex 2

+ c

3. sin2∫ xcosx dx (note that ddx

sinx( ) = cosx )Let u = sinx then du

dx= cosx and so cosxdx = du

The integral then becomesu2∫ du =

13

u3 + c

=13

sin3x + c

4. lnxx∫ dx (note that d

dxlnx( ) =

1x)

Let u = lnx then dudx

=1x

and so 1x

dx = duThe integral then becomes

u∫ du =12

u2 + c

=12

lnx( ) 2 + c


page 35

5. x2 2x3 + 3( )5∫ dx

Let u = 2x3 + 3 then dudx

= 6x2 and so 6x2dx = du and x2 dx =16

du The integral then becomes

16

u5∫ du =136

u6 + c

=136

2x3 + 5( )6 + c

6. 2x x + 2( )5∫ dx

Let u = x + 2 then dudx

= 1 and so dx = du and from u = x + 2 then x = u – 2The integral then becomes

2 u − 2( )u5∫ du = 2u6 − 4u5( )∫ du =27

u7 −23

u6 + c

=27

x + 2( )7 −23

x + 2( )6 + c

7. x4

x5 + 1( )3∫ dx

Let u = x 5 + 1 then dudx

= 5x 4 and so x4dx =15

du The integral then becomes

15u3∫ du =

15

u−3∫ du =15× −

12

u−2 + c = −1

10u2 + c

= −1

10 x5 + 1( )2 + c

8. 2xx2 + 4∫ dx


= 2x and so 2xdx = duThe integral then becomes

duu∫ = lnu + c

= ln x2 + 4( ) + c

9. xx2 + 5∫ dx


= 2x and so 2xdx = du and xdx = 12

duThe integral then becomes

12u∫ du =

12

1u∫ du =

12

lnu + c

=12

ln x 2 + 5( ) + c

contd....


page 36

Exercise 4Integrate the following functions (no substitution given) :-

1. x x2 − 3( )∫5dx 2. x2∫ x3 −1( )2

dx

3. x 1 − x 2( )∫ dx 4. cosxsin4xdx∫5. x

1 − x2( )3∫ dx 6. x3

1 + x4∫ dx

7. sinxcos3x

dx∫ 8. cosxsin6x∫ dx

9. ex

3ex −1∫ dx 10. sec2xtanx∫ dx

Examples (where the substitution will be given)

1. x 3x − 2( )∫ dx given u = 3x – 2

Letu = 3x – 2 then dudx

= 3 and sodx =13

du

and from u = 3x –2 then x =13

u+ 2( ) The integral then becomes

13

u + 2( ) u∫13

du = 19

u32 + u

12

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ ∫ du = 1

925

u52 +

23

u32

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ + c

=245

3x − 2( )52 +

227

3x − 2( )32 + c


The Complete A level Maths (Orlando Gough) Page 189 Exercise 4.2:2 Questions 1,3,5,7.

Understanding Pure Mathematics (A.J.Sadler/D.W.S.Thorning)Page 320 Exercise 13B Questions 1 – 20.

10. tanxdx =sinxcosx∫∫ dx

Let u = cosx then dudx

= −sinx and sosinxdx = −du The integral then becomes−

1u∫ du = −lnu + c

= −ln cosx( ) + c Questions 9, 10, and 11 demonstrated the special integral of the form

′ f (x)f x( )∫ dx = ln f x( )[ ] + c

Simple substitution will not be given whereas more difficult ones will be given.

contd....


page 37

Exercise 5Integrate the following functions (substitution given) :-

1. 9x∫ 3x + 2( )3dx where u = 3x + 2 2. 7x∫ 2x + 3( )5dx where u = 2x + 3

3. 3x 1+ x2( )∫ dx where u = 1 + x2 4.3x

2x + 3( )∫ dx where u = 2x + 3

Examples (where the substitution will be given – Special substitutions)

In integrals which contain the expression √(a2 – x2), one of the following could be given : either u = √(a2 – x2) or x = asinθ

1. 1 − x2( )∫ dx given x = sinθ

From x = sinθ, dxdθ

= cosθ , dx = cosθ dθ

and √(1 – x2) = √(1 – sin2θ ) =√(cos2θ ) = cosθ The integral becomes

cosθcosθdθ∫ = cos2θdθ =12∫∫ 1+ cos2θ( )dθ =

12

+ 12

cos2θ⎛ ⎝ ⎜

⎞ ⎠ ⎟ ∫ dθ

=12θ + 1

4sin2θ + c

2. x(x + 2)5∫ dx given u = x + 2


= 1 and so du = dx and from u = x + 2 then x = u – 2The integral then becomes

u − 2( )∫ u5du = u6 − 2u5( )∫ du = 17

u7 −13

u6 + c = 17

x + 2( )7 −13

x + 2( )6 + c

3. xx + 4( )∫ dx given u = x + 4


= 1 and so du = dx and from u = x + 4 then x = u – 4The integral then becomes

u − 4u∫ du = u

12∫ − 4u

−12 du = 2

3u

32 − 2u

12 + c = 2

3x + 4( )

32 − 2 x + 4( )

12 + c


The Complete A level Maths (Orlando Gough) Page 186 Exercise 4.2:1 Question 1. and Page 189 Exercise 4.2:2 Question 1.

Understanding Pure Mathematics (A.J.Sadler/D.W.S.Thorning)Page 500 Exercise 20A Questions 3 – 9.

contd....


page 38

We must return to the original variable.Since sinθ = x then θ = sin-1xcosθ = √(1 – x2) and sin2θ = 2sinθ cosθ = 2x√(1 – x2)

1 − x2( )∫ dx = 12 sin-1x +

14 2x√(1 – x2) + c

= 12 sin-1x +

12 x√(1 – x2) + c

2. x2

9 − x2( )∫ dx given x = 3sinθ

From x = 3sinθ, dxdθ

= 3cosθ , dx = 3cosθ dθ

and √(9 – x2) = √(9 – 9sin2θ ) =√(9cos2θ ) = 3cosθ The integral becomes

9sin2θ.3cosθ3cosθ∫ dθ = 9sin2∫ θ dθ = 9. 1

2∫ 1 − cos2θ( )dθ =92−

92

cos2θ⎛ ⎝ ⎜

⎞ ⎠ ⎟ ∫ dθ

=92θ −

94

sin2θ + cWe must return to the original variable.

Since sinθ = x3

then θ = sin-1 x3

⎛ ⎝ ⎜

⎞ ⎠ ⎟

cosθ = √(1 – sin2x) = 9 −x3

⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

= 9 − x2

9=

9 − x2

3=

9 − x2

3

and sin2θ = 2sinθ cosθ = 2

x3

⎛ ⎝

⎞ ⎠

9 − x 2

3 =29

x 9 − x2( )x2

9 − x2( )∫ dx = 92

sin-1 x3

⎛ ⎝ ⎜

⎞ ⎠ ⎟ −

x2

9 − x 2( ) + c

Exercise 6Integrate the following functions (substitution given) :-

1. x1 − x2( )∫ dx given x = sinθ 2. 4 − x2( )∫ dx given x = 2sinθ

3. x9 − x2( )∫ dx given x = 3sinθ 4. x 2

4 − x2( )∫ dx given x = 2sinθ


The Complete A level Maths (Orlando Gough) Page 193 Exercise 4.2:3 Question 1,2,3,5,7.

Understanding Pure Mathematics (A.J.Sadler/D.W.S.Thorning)Page 500 Exercise 20A Questions 10 – 16.

A1A3

A2

1 2 3

y = x2 – 3x + 2

–1 5

y = x2 – 4x – 5

x

y


page 39

The Area enclosed by a curve and the x axis

The Area enclosed by a curve and the x axis has been already met at Higher level.A = f x( )

a

b

∫ dx or ydxa

b

∫ = F b( ) − F a( ) where F(x) is the integral of f(x) or y and F(b) and F(a) are the values of F(x) at x = b and x = a.

Examples1. Find the area enclosed by the curve f(x) = x2 + 2 between x = 1 and x = 3.

The area is above the x -axis

ydx = x2 + 1( )dx1

3

∫1

3

∫ =x3

3+ 1

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥ 1

3

= (9 + 1) – (1/3 + 1)

=> Area = 263

2. Find the area enclosed by the curve f(x) = x2 – 4x – 5 and the x -axis. The area is below the x -axis.

ydx =−1

5

∫ x2 − 4x − 5( )−1

5

∫ dx

= x3

3− 2x2 − 5x

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥ −1

5

= ( 125/3 – 50 – 25) – (–1/3 –2 –5) = (–100/3) – (–22/3) = –78/3

The negative indicates that the area enclosed is below the x -axis as shown.Area = 78/3

3. Find the area enclosed by the curve f(x) = x2 – 3x + 2 and the x -axis,from x = 0 to x = 3.The area lies partly above and partly below the x -axis.A1 = ydx =

0

1

∫ x2 − 3x + 2( )0

1

∫ dx

= x3

3−

32

x2 + 2x⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

0

1

= 5/6

A2 = f x( )dx =1

2

∫ x 2 − 3x + 2( )1

2

∫ dx

= x3

3−

32

x2 + 2x⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

1

2

=–1/6

A3 = f x( )dx =2

3

∫ x2 − 3x + 2( )2

3

∫ dx

=> The total area = 5/6 + 1/6 + 5/6 = 15/6

contd....

x

y

1 3

y = x2 + 2

x

y


page 40

Exercise 7

1. Find the area enclosed by the following curves and the x -axis between the lines given.

(a) y = 3x2 + 2, x = 0, x = 2

(b) y = x3– x, x = 1, x = 2

2. Find the area enclosed by the following curves and the x -axis.(A rough sketch could help)

(a) y = 6 + x – x2

(b) y = x(x –2)(x – 3)

The Area enclosed between two curves

The Area enclosed between two curves has been already met at Higher level.

A = f x( ) − g x( )[ ]a

b

∫ dx where the curvesf(x) andg(x) intersect at x = a and x = b

Note Provided the graph of f(x) is above the graph of g(x), their positions relative to the x -axis does not matter.

Examples

1. Find the area enclosed between the curve y = x(4 – x) and the straight line y = x.

The curve and straight line meet where x = x(4 – x)

i.e. x2 – 3x = 0 x(x – 3) = 0i.e. x = 0 and x = 3

A = x 4 − x( ) − x[ ]0

3

∫ dx = 3x − x2( )0

3

∫ dx

= 3x 2

2−

x3

2

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

0

3

= ( 272

– 9) – 0 = 92


The Complete A level Maths (Orlando Gough) Page 176 Exercise 4.1:3 Question 5,6,7,8,9,25.

Understanding Pure Mathematics (A.J.Sadler/D.W.S.Thorning)Page 304 Exercise 12B Questions 2 - 14.

contd....

y = xy = x(4 – x)

(3, 3)

x

y


page 41

2. Find the area enclosed between the curves y =(x – 1)2 + 4 and y = 5 + 4x – x2.

The curves meet where(x – 1)2 + 4 = 5 + 4x – x2

i.e. 2x2 – 6x = 0 =>2x(x – 3) = 0i.e. x = 0 and x = 3

A = 5 + 4x − x2( ) − x2 − 2x + 5( )[ ]0

3

∫ dx

= 6x − 2x2( )0

3

∫ dx = 3x 2 −23

x3⎡ ⎣ ⎢

⎤ ⎦ ⎥

0

3

= (27 – 18) – 0

= 9

Exercise 8

Find the areas enclosed by the following curves.

1. y = x(10 – x) and y = 4x. 2. y = 4x – x2 and y = x2– 4x + 6.

3. y = 2√x and y =x2

4 4. y = x3 + x2 – 5x and y = x2 – x.

5. y = sinx and y = cosx 6. y 2= 4ax and x2 = 4ay (both are parabolas)


The Complete A level Maths (Orlando Gough) Page 176 Exercise 4.1:3 Questions 48,50,51,52.

Understanding Pure Mathematics (A.J.Sadler/D.W.S.Thorning)Page 303 Exercise 12B Questions 16 – 22.

y

xO

(3,8)

y =(x –1)2+4

–1 5

(0,5)

y =5+4x - x2


page 42

y = x2

x

y

1

4

x

y

1

8y = x3

Exercise 9

Find the area enclosed by the following curves and the y -axis between the lines given.

1. x = y 2, y = 3. 2. y = x 3, y = 1, y = 8.

3. x =1y , y = 2, y = 3. 4. y 2= 1 – x, y = 0, y = 1.

5. y = 1/x 3, y = 8, y = 27. 6. y =lnx, y =2, y =5


The Complete A level Maths (Orlando Gough) Page 181 Exercise 4.1:4 Questions 16(Area only).

Understanding Pure Mathematics (A.J.Sadler/D.W.S.Thorning)Page 304 Exercise 12B Question 15.

The Area enclosed by a curve and the y-axis

The Area enclosed by a curve and the y -axis is given by A = f y( )

a

b

∫ dy or xdya

b

∫ = F b( ) − F a( ) where F(y) is the integral of f(y) or y and F(b) and F(a) are the values of F(y) at y = b and y = a.

Examples

1. Find the area enclosed by the curve y = x2 between y = 1 and y = 4, x ≥ 0.

From y = x2, x = √y

A = y12dy =

23

y32

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥ 1

4

∫

=23

4( )32

⎛ ⎝ ⎜

⎞ ⎠ ⎟ −

23

1( )32

⎛ ⎝ ⎜

⎞ ⎠ ⎟

= 16/3 – 2/3 = 42/3

2. Find the area enclosed by the curve y = x3 between y = 1 and y = 8.From y = x3, x = 3√y

A = y13

1

8

∫ dy =34

y43

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

1

8

=34

8( )43

⎛ ⎝ ⎜

⎞ ⎠ ⎟ −

34

1( )43

⎛ ⎝ ⎜

⎞ ⎠ ⎟

= 12 – 3/4 = 111/4

y

x2

y = x3 – 2x2

a b

y = f(x)

x

y


page 43

The Volume of Revolution

The volume of revolution is formed when the area bounded by a curve y = f(x), the x - axis, x = a and x = b is rotated completely about the x – axis.

V = πy2dx

Example

Find the volume generated, by rotating about the y - axis, the area enclosed by the curve y = x2 + 1, x > 0, the x – axis and the line y = 2.

From y = x2 + 1, x2 = y – 1

V = π(y – 1)dy = π12

y2 − y⎡ ⎣ ⎢

⎤ ⎦ ⎥

1

2

= π[(2 – 2) – (1/2 – 1)]

= π2

The volume of revolution can also be formed when the area bounded by a curve y = f(x), the y – axis, y = a and y = b is rotated completely about the y – axis.

V = πx2dy

a

b

∫

1

2

∫

a

b

∫

y

xa

b

y = f(x)

y = x2 + 1y

x

1

2

1

2

ExampleFind the volume generated, by rotating about the x – axis, the area enclosed by the curve y = x3 – 2x2 and the x – axis.

The curve crosses the x axis where x3 – 2x2 = 0

x2(x – 2) = 0 => x = 0 and x = 2

∴ V = 0

2

∫ π(x3 – 2x2)2dx

= 0

2

∫ π(x6 – 4x5 + 4x4) dx

= π17

x7 −23

x6 +45

x5⎡ ⎣ ⎢

⎤ ⎦ ⎥

0

2

= π27

7−

27

3+

27

5

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ − 0 = π27 1

7−

13

+15

⎛ ⎝ ⎜

⎞ ⎠ ⎟ =

128105

π


page 44

Exercise 10

1. Find the volumes of solids of revolution formed when the regions bounded by the following curves and the x – axis are rotated through one revolution about the x – axis.Sketch the curves.

(a) y =4x

, x = 1 and x = 4 (b) y = √x, x = 0 and x = 4

(c) x + 2y = 2, x = 0 and x = 2 (d) y =1x2 , x = 1

3 and x = 1

2

(e) y = x(x – 1) (f) y = √(9 – x2)

(g) y2= 8x, x = 0 and x = 4 (h) y = sinx, x = 0 and x =π

2. Find the volumes of solids of revolution formed when the regions in the first quadrant bounded by the following curves and the y – axis are rotated through one revolution about the x – axis.Sketch the curves.

(a) x = √y and y = 4 (b) x = y2 and y = 1

(c) xy = 1, y = 3 and y = 6 (d) y = 4 – x2, y = – 4 and y = 4

(e) xy2 = 2, y = 2 and y = 4 (f) x = y2 + 1, y = – 1 and y = 1

(g) y = lnx, y = 2 and y = 5 (h) y = cosx


The Complete A level Maths (Orlando Gough) Page 181 Exercise 4.1:4 Questions 16(Area only).

Understanding Pure Mathematics (A.J.Sadler/D.W.S.Thorning)Page 304 Exercise 12B Question 15.


page 45

Answers

Exercise 1 Page 29

1. 2x4 + c 2. −3x2 + c 3. 2

3x

32 + c (etc.)

4. 2 x 5. x3 −6x

6. 2x12 − 2x

32

7. 3x + 4( )5

158. −

1 − 2x( )5

109. −

12 2x − 3( )

10. 4x + 12

11. −2

3x + 112. −

13

cos3x

13. 3sin 13

x 14. 12

x −14

sin2x 15. 12

x +12

sin x

Exercise 2 Page 31

1. 15

e5x 2. −12

e−2 x 3. 6e12

x

4. 13

ln 3x 5. ln x + 5( ) 6. 12

ln 2x −3( )

7. 12

e2 x + 2x − 12

e−2x 8. x − e− x 9. 12

e2 x −12

e−2x

10. 2 ln 3x + 2( ) 11. −32

ln 1− 2x( ) 12. −57

ln 6 − 7x( )

13. 14

tan 4x 14. 12

tan π + 2x( ) 15. 32

tan 2x

Exercise 3 Page 33

1. −11

242. −

5

123. 154

5

4. 38

35. 1

366. 9

7. 1

28. 1 9. π

10. 2

51−

1e

⎛ ⎝

⎞ ⎠ 11. e – 1 12. 2e – 2

13. ln3 14. 1

3ln 5

215. ln3

Exercise 4 Page 36

1. 112

x2 −3( )62. 1

9x3 −1( )3

3. −13

1 − x 2( )32

4. −15

sin5 x 5. 14 1− x2( )2 6. 1

4ln 1+ x4( )

7. 12

sec2 x 8. −1

5sin5 x9. 1

3ln 3ex −1( )

10. ln tan x( )


page 46

Exercise 5 Page 37

1.15 (3x + 2)5 –

12 (3x + 2)4 + c 2.

14 (2x + 3)7 –

78 (2x + 3)6 + c

3. (1 + x)32 + c 4.12

(2x + 3)32 –92

(2x + 3)12 + c

Exercise 6 Page 38

1. – √(1 – x2) + c 2. 2sin–1(x/2) + x √(4 – x2) + c

3. – √(9 – x2) + c 4. 2sin–1(x/2) – x2 √(4 – x2) + c

Exercise 7 Page 40

1. (a) 12 (b) 21/4 2. (a) 205/6 (b) 31/12

Exercise 8 Page 41

1. 36 2. 22/3 3. 51/3

4. 4 5. √2 6.16a2

3

Exercise 9 Page 42

1. 9 2. 111/4 3. 2√3 – 2√2

4. 2/3 5. 71/2 6. e2(e3 – 1)

Exercise 10 Page 44

1. (a) 12π (b) 8π (c) 2π (d) 19π3 3

(e) 1π (f) 18π (g) 64π (h) 1π2

30 2

2. (a) 8π (b) 1π (c) 1π (d) 32π5 6

(e) 7π (f) 56π (g) 1(e10– e4)π (h) π(π – 2)48 15 2

Outcome 4 – Properties of Functions

Find the vertical asymptote(s) of a rational function.

Asymptotes

An Asymptote is a straight line to which a curve approaches more and more closely as x becomes larger or smaller, or approaches a certain value.

Use the computer or a graphics calculator to obtain the graphs of the following rational functions.

Show the asymptotes clearly and write down their equations.

1. f x( ) =1x

2. f x( ) =1

x − 3

3. f x( ) =1

x + 24. f x( ) =

2x + 33 − x

5. f x( ) =1

x −1( ) x + 3( ) 6. f x( ) =x

x −1

7. f x( ) =x

x + 2( ) x − 3( ) 8. f x( ) =x2

x + 2( ) x − 3( )

9. f x( ) =x −1x2 10. f x( ) =

x2

x −1

11. f x( ) =x3

x 2 −112. f x( ) =

1x 2 + 1

When using a computer or a graphics calculator, it is not always possible to be sure that you have identified the equation of the asymptote correctly.A little algebra will identify the equations.You will have seen that there are vertical and non - vertical asymptotes.

Vertical asymptotes are found from the zeros of the denominator ie. are in the form x = k The way the curve approaches the asymptotes must also be detemined.

Examples 1. (If the degree of the numerator < the degree of the denominator) :-

e.g. f x( ) =2x + 3

x 2 + 5x + 4=

2x + 3x + 4( ) x + 1( )

Vertical asymptotes occur at the zeros of the denominator set x + 4( ) x + 1( ) = 0

=> x = −4 and x = −1 are asymptotes. (vertical lines)

We now have to determine how the curve approaches these asymptotes.As x → −4− (means x tends to –4 from a negative direction ie. x < −4)

y → −( )−( ) −( ) → −∞ (assign a sign to each part of the rational function and find the

net sign. , −∞ means negative infinity)


page 47cont’d ....

(a function of the form )P(x)Q(x)

As x → −4+ (means x tends to – 4 from a positive direction ie. x > −4)

y →−( )

+( ) −( ) → +∞ (assign a sign to each part of the rational function and

find the net sign. +∞ means positive infinity) We can show this in an asymptote diagram.

x = −4 ← ⎯ ⎯ (indicates y → +∞)

(indicates y → −∞) ⎯ → ⎯

As x → −1− (means x tends to –1 from a negative direction ie. x < −1)

y →+( )

+( ) −( ) → −∞ (assign a sign to each part of the rational function and

find the net sign. −∞ means negative infinity)

As x → −1+ (means x tends to –1 from a positive direction ie. x > −1)

y → −( )+( ) −( ) → +∞ (assign a sign to each part of the rational function and

find the net sign. +∞ means positive infinity)

We can show this in an asymptote diagram.

x = −1 ← ⎯ ⎯ (indicates y → +∞)

(indicates y → −∞)⎯ → ⎯

2. (If the degree of the numerator = the degree of the denominator) :-

f x( ) =x2 + 2x + 1x 2 + 5x + 4

=x + 1( )2

x + 4( ) x + 1( )Vertical asymptotes occur at the zeros of the denominator

x + 4( ) x + 1( ) = 0 x = −4 and x = −1 are asymptotes.

We now have to determine how the curve approaches these asymptotes.


y → +( )−( ) −( ) → +∞ (assign a sign to each part of the rational function and





y →+( )

+( ) −( ) → −∞ (put a sign to each part of the rational function and


We can show this in an asymptote diagram.x = −4

(indicates y → +∞)⎯ → ⎯

← ⎯ ⎯ (indicates y → −∞)


y →+( )

+( ) −( ) → −∞ (put a sign to each part of the rational function and


As x → −1+ (means x tends to -1 from a positive direction ie. x > −1)

y → +( )+( ) +( ) → +∞ (put a sign to each part of the rational function and



← ⎯ ⎯ (indicates y → +∞)

(indicates y → −∞)⎯ → ⎯

3. (If the degree of the numerator > the degree of the denominator) :-

f x( ) =x 2 + 4x + 3

x + 2=

x + 1( ) x + 3( )x + 2

Vertical asymptotes occur at the zeros of the denominator x + 2 = 0

x = −2 is an asymptote. We now have to determine how the curve approaches this asymptotes.


y → −( ) +( )−( ) → +∞ (put a sign to each part of the rational function and

find the net sign. +∞ means negative infinity)




y →−( ) +( )

+( ) → −∞ ( put a sign to each part of the rational function and

find the net sign. −∞ means positive infinity)


(indicates y → +∞)⎯ → ⎯

← ⎯ ⎯ (indicates y → −∞)Exercise 1

Find all the vertical asymptotes of the following rational functions.

1. f x( ) =4

x − 22. f x( ) =

3x −1x 2 + 2x − 3

3. f x( ) =12

x 2 − 2x − 34. f x( ) =

x + 4x − 2

5. f x( ) =x2

4 − x2 6. f x( ) =x x + 1( )

x −1( ) x + 2( )

7. f x( ) =x −1( ) x − 4( )

x8. f x( ) =

x 2 + 3x −1

9. f x( ) =x3

x 2 + 310. f x( ) =

xx 2 + 4

11. f x( ) =x2

x −112. f x( ) =

2x 2

x 2 −1

Find the non-vertical asymptote of a rational function.

The non - vertical asymptote is in the form y = c or y = mx + c.The way the curve approaches the asymptotes must also be determined.

Examples

1. (Note the degree of the numerator < the degree of the denominator)f x( ) =

2x + 3x 2 + 5x + 4

=2x + 3

x + 4( ) x + 1( )Divide each term of the numerator and denominator by the highest power of x .

ie. divide by x2 f x( ) =

2x

+3x2

1+ 5x

+ 4x2

As x → ±∞ (ie large positive and negative values of x, 2x

, 3x2 , 5

x and 4

x2 , all tend to zero.)As a result, f x( ) → 0 and so y = 0 is a horizontal asymptote. (the x - axis !)Now we must find out if the curve approaches this asymptote from above or below.



Proceed as follows.As x → +∞ ,y → 0+ (above the value of 0, since the fraction will be very small

and positive)As x → −∞ , y → 0− (below the value of 0, since the fraction 2

x is negative and

larger than the fraction 3x2 , making the numerator negative.

At the same time, the denominator becomes 1 – fraction + fraction making the net sign positive.)


Indicates y → 0− Indicates y → 0+

(from below) (from above)

2. (If the degree of the numerator = the degree of the denominator) :-

f x( ) =x2 + 2x + 1x 2 + 5x + 4

=x + 1( )2

x + 4( ) x + 1( )

Divide the numerator by the denominator :- 1

Write as f x( ) = 1− 3x + 3( )x2 + 5x + 4

using long division x2 + 5x + 4 x2 + 2x + 1

x2 + 5x + 4 – 3x – 3

Now divide each term of 3x + 3( )x2 + 5x + 4

by the highest power of x

ie. divide by x2 f x( ) = 1−

3x

+3x2

⎛ ⎝ ⎜

⎞ ⎠ ⎟

1 + 5x

+ 4x2

⎛ ⎝ ⎜

⎞ ⎠ ⎟

As x → ±∞ (ie large positive and negative values of x, 3x

, 3x2 , 5

x and 4

x2 ,all tend to zero.)As a result, f x( ) →1 and so y = 1 is a horizontal asymptote.

Now we must find out if the curve approaches this asymptote from above or below.

Proceed as follows.As x → +∞ ,y →1− (below the value of 1, since the fraction will be very small

and positive)As x → −∞ , y →1+ (above the value of 1, since the fraction 3

x is negative and

larger than the fraction 3x2 , making the numerator negative.

At the same time, the denominator becomes approximately 1.f(x) = 1 – (–/1) => 1+ => slightly bigger than 1.


Indicates y →1+ Indicates y →1−

(from above) (from below)


page 51

3. (If the degree of the numerator > the degree of the denominator) :-

f x( ) =x 2 + 4x + 3

x + 2=

x + 1( ) x + 3( )x + 2

Divide the numerator by the denominator :- x + 2 Write as f x( ) = x + 2 − 1

x + 2 using long division x + 2 x2 + 4x + 3

x2 + 2x 2x + 3 2x + 4

–1Now divide each term of 1

x + 2 by the highest power of x .

ie. divide by x f x( ) = x + 2 −

1x

1 + 2x

As x → ±∞(ie large positive and negative values of x, 1x

and 2x

tend to zero.)

As a result, f x( ) → x + 2 and so f x( ) = x + 2 is a oblique asymptote. (slanting)Now we must find out if the curve approaches this asymptote from above or below.Proceed as follows.As x → +∞ ,y → x + 2( )− (below the line f x( ) = x + 2, since the fraction will be

very small and positive)As x → −∞ , y → x + 2( )+ (above the line f x( ) = x + 2, since the fraction 1

x is

negative and the denominator becomes 1 – fraction making the net sign positive.)

We can show this in an asymptote diagram. (Indicates y → x + 2( )− )

(Asymptote) (Indicates y → x + 2( )+ )

Exercise 2Find the non - vertical asymptote of the following rational functions.

1. f x( ) =4

x − 22. f x( ) =

3x −1x 2 + 2x − 3

3. f x( ) =12

x 2 − 2x − 34. f x( ) =

x + 4x − 2

5. f x( ) =x2

4 − x2 6. f x( ) =x x + 1( )

x −1( ) x + 2( )

7. f x( ) =x −1( ) x − 4( )

x − 28. f x( ) =

x 2 + 3x −1

9. f x( ) =x3

x 2 + 310. f x( ) =

xx 2 + 4

11. f x( ) =x2

x −112. f x( ) =

2x 2

x 2 −1


page 52

Sketch the graph of a rational function including appropriate analysis of stationary points.Curve Sketching

1 Sketching the graph of a rational function using calculus

(a) Find all the stationary points and their nature.

(b) Find all the asymptotes and investigate the approach of the curve to each.

(c) Find all the crossings of the y axis. and the x axis (if easily found)

Note The table of signs may be easier than finding the second derivative in order to determine the nature of the stationary points.

Examples Sketch the following curves.

1. f x( ) =2x2 + x −1

x −1=

x + 1( ) 2x −1( )x −1

(a) Stationary points and their nature.

′ f x( ) =x −1( ) 4x + 1( ) − 2x2 + x −1( ) ×1

x −1( ) 2 =4x2 − 3x −1 − 2x2 − x + 1

x − 1( )2 =2x2 − 4x

x −1( )2

For S.V. ′ f x( ) = 02x2 − 4x = 0 2x x − 2( ) = 0x = 0 and x = 2y = 1 y = 9Stationary points at (0,1) and (2,9)

′ ′ f x( ) =x −1( )2 × 4x − 4( ) − 2x2 − 4x( ) × 2 x −1( )

x −1( ) 4 =4 x −1( )3 − 4x x −1( ) x − 2( )

x −1( )4

=4 x −1( )2 − 4x x − 2( )

x −1( )3

′ ′ f 0( ) =+( ) − 0−( ) = −( ) ie. (0, 1) is a Maximum Turning point.

′ ′ f 2( ) =+( ) − 0

+( ) = +( ) ie. (2, 9) is a Minimum Turning point.

(b) Asymptotes.

( i ) Vertical.Vertical asymptotes occur at the zeros of the denominator

x −1 = 0 x = 1 is an asymptote. x = 1

As x → 1− , y → +( ) +( )−( ) → −∞

As x → 1+ , y → +( ) +( )+( ) → +∞



Outcome 5 – Systems of Linear Equations

Use matrix techniques to solve systems of linear equations.

Solve a 3 x 3 system of linear equations using Gaussian Elimination on an augmented matrix.

MATRICES

A matrix (matrices) is a rectangular array of numbers arranged in rows and columns, the array being enclosed in round (or square) brackets.

e.g. xy

⎛

⎝ ⎜ ⎞

⎠ ⎟

3 10 5⎛

⎝ ⎜

⎞

⎠ ⎟

6 8 103 4 5⎛

⎝ ⎜

⎞

⎠ ⎟ 4 -2 5( )

2 rows 2 rows 2 rows 1 row 1 column 2 columns 3 columns 3 columns

Each number in the array is called an entry or an element of the matrix and is identified by first stating the row and then the column in which it appears.

A matrix is often denoted by a capital letter. The order of a matrix is given by stating the number of rows followed by the number of columns.

eg. A = 4 6 82 3 4⎛

⎝ ⎜

⎞

⎠ ⎟ B =

3 42 −6⎛

⎝ ⎜

⎞

⎠ ⎟

A has 2 rows and 3 columns B has the same number of rows and and is said to be of order columns and is called a square 2 x 3 (read 2 by 3). matrix of order 2.

In General , a matrix A, with m rows and n columns (order m x n), can be represented asfollows, where aij denotes the row and column of each element.

A =

a1 1 a1 2 a1 3 - - - a1n

a2 1 a2 2 a2 3 - - - a2n

a3 1 a3 2 a3 3 - - - a3n

’ ’ ’ - - - ’’ ’ ’ - - - ’’ ’ ’ - - - ’

am1 am2 am 3 - - - amn

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟


page 74

The solution of a system of linear equations in three variables.

The general form of a system of equations is :

a1x + b1y + c1z = d1

a2 x + b2y + c2 z = d2

a3x + b3 y + c3z = d3

or in matrix form A X = B

where A = a1 b1 c1

a2 b2 c2

a3 b3 c3

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

X = xyz

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

and B = d1

d2

d3

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

A is an 3 x 3 matrix , X is an 3 x 1 matrix and B is an 3 x 1 matrix.

Gaussian Elimination is best demonstrated by example.

The process is as follows:-Solve :-

x + y + 2z = 3 ....... row1 2x – y – z = 2 ....... row2

3x – 2y + 2z = –2 ....... row3

Eliminate x from row1ie. row2 – 2 x row1 2x – y – z = 2

2x + 2y + 4z = 6 ie – 3y – 5z = – 4 ....... row4 Eliminate x from row3ie. row3 – 3 x row1 3x – 2y + 2z = –2

3x + 3y + 6z = 9 ie – 5y – 4z = – 11 ....... row5

Eliminate y from row5

ie. row5 – 53

x row4 – 5y – 4z = – 11

– 5y – 253

z = – 203

ie 133

z = –133

z = –1

Substitute in row4 – 3y + 5 = – 4 y = 3Substitute in row1 x + 3 – 2 = 3 x = 2

Notice that we arranged the system of equations until we arrived at

x + y + 2z = 3 – 3y – 5z = – 4

133

z = –133

This working can be set out more neatly as shown below


page 75

x y z b1 1 2 32 – 1 – 1 23 – 2 2 – 20 – 3 – 5 – 40 – 5 – 4 – 11

0 0 133

–133

This whole process is called Gaussian Elimination.

For the following general 3 x 3 system

a1x + b1y + c1z = d1 ............ (1) a2x + b2y + c2z = d2 ............ (2) a3x + b3y + c3z = d3 ............ (3)

The 3 x 3 matrix of the coefficients of x, y and z, a1 b1 c1

a2 b2 c2

a3 b3 c3

d1

d2

d3

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

is called the augmented matrix.

Leave (1) alone

Eliminate x from equation (2) by subtracting m1 (= a2

a1

) times equation (1) from equation (2)

ie. in the above example m1 = 2 giving –3y – 5z = – 4 ............ (4)

Similarly, eliminate x from equation (3) by subtracting m2 = ( a3

a1

) times equation (1) from equation (3)

ie. in the above example m2 = 3 giving – 5y – 4z = –11 ............ (5)

Thus we have x + y + 2z = 3 2x – y – z = 2 ⎯ → ⎯ – 3y – 5z = – 43x – 2y + 2z = – 2 ⎯ → ⎯ – 5y – 4z = –11

Leave (4) alone

Eliminate y from equation (5) by subtracting m3 = ( 5

3 ) times equation (4) from equation (5)

ie. in the above example m3 = 53 giving 13

3 z = –13

3

The last equation gives the value of z and we then back-substitute into the equation above to obtain



the value of y and use both values to obtain x from the first equation.

This process is called Gaussian Elimination.The final equations are called the Pivotal Equations. m1, m2 etc are called the Multipliers. The above example will look like this in tabular form.

m x y z b1 1 1 2 32 2 –1 –1 23 3 –2 2 –21 –3 –5 –4

53

–5 –4 –11

133

–133

Checks on your working can be added to the above table.This column is the sum of the coefficients of x, y, z and b.We then perform the same operations on this and check that the sum agrees.

m x y z b check1 1 1 2 3 72 2 –1 –1 2 23 3 –2 2 –2 11 –3 –5 –4 –12

53

–5 –4 –11 –20

133

–133

0

check 6 –2 3from (1 + 2 + 3) (1 – 1 – 2) (2 – 1 + 2)

Note (i) You can re-order the set of original equations so that the smallest coefficient of x , y or z is in the first column before the first set of eliminations.

(ii) At the second set of eliminations you can reorder the 2 equations in the same way.


page 77

Example Use Gaussian Elimination to solve the following system of equations.

x + y + 2z = 3 ............ R1 4x + 2y + z = 13 ............ R2 2x + y – 2z = 9 ............ R3

m x y z b check1 1 1 2 3 7 R1

4 4 2 1 13 20 R2

2 2 1 –2 9 10 R3

R2 – 4R1 –2 –7 1 –8 R4

R3 – 2R1 –1 –6 3 –4 R5

Now change the order of the last 2 lines.

m x y z b check1 1 1 2 3 7 R1

4 4 2 1 13 20 R2

2 2 1 –2 9 10 R3

–1 –6 3 –4 R5

–2 –7 1 –8 R4

R4 – 2R5 5 –5 R6

From 5z = – 5 => z = –1Substitute in – y – 6z = 3 => y = 3Substitute in x + y + 2z = 3 => x = 2

Exercise 1Solve the following equations using Gaussian Elimination (You may have to interchange rows before you start) :-

1. x + y + z = 1 2. x – 2y + z = 6 3x + 3y + z = 4 3x + y – 2z = 4 3x + 2y + 2z = 7 7x – 6y – z = 10

3. 5x – y + 2z = 25 4. x + y + z = 2 3x + 2y – 3z = 16 3x – y + 2z = 4 2x – y + z = 9 2x + 3y + z = 7

5. 5x – 3y + 6z = 0x + 5y + 2z = 0

– x + 2y + 5z = 0


page 78

Ill - conditioning

A system of equations is said to be ill-conditioned when small changes in the coefficients of x, y and z produce relatively large changes in the solution.

Therefore any uncertainty in the coefficients or round-off in the elimination could drastically affect our answers. This could occur if the coefficients result from experiment.

eg. x + 1000y = 1x + 999y = 2 has solutions x = 1001 and y = –1

However x + 999y = 1x + 1000y = 2 has solutions x = –998 and y = 1

A change of 1 in 1000 has brought drastic changes in our solutions.

One sign of ill-conditioning is that the pivots decrease in magnitude at each stage, tending towards zero.

Example Show, by calculating the exact values, that the following system demonstrates ill-conditioning.

1x + 1y = 3 0·1y = 1 ie. y = 10 and x = –71x + 0·9y = 2

1x + 0·95y = 3 0·05y = 1 ie y = 20 and x = 0.21x + 0·9y = 2

This system is said to be ill-conditioned since relatively small changes in y produces a large change in the solution.

Exercise 2 Show, by calculating the exact values, that the following system demonstrates ill-conditioning.

(a) (i) 1x + 1 y = 2 (ii) 1x + 1y = 2 1x + 1·000y = 2·0001 1x + 1·0001y = 1·9999

(b) (i) 1x + 0·99y = 1.99 (ii) 1x + 0·99y = 2·00 0·99x + 0·98y = 1.97 0·99x + 0·98y = 1·97


page 79

Answers

Exercise 1 Page 78

1. x = 5, y = –7/2, z = –1/2 . 2. x = 5, y = 3, z = 7.

3. x = 5, y = 2, z = 1 . 4. x = 3, y = 1, z = –2.

5. x = 0, y = 0, z = 0.

Exercise 2 Page 79

1. (a) (i) x = 1, y = 1 (ii) x = – 1, y = 3.

(b) (i) x = 0·01, y = 2 (ii) x = –97, y = 100


page 80


page 1

Outcome 1 – Use further differentiation techniquesDifferentiate an Inverse Function

The Derivatives of Inverse Trigonometric Functions

1. y = sin– 1x If y = sin–1x, then by definition, x = siny .

=> dxdy

= cosy [assume that dydx

=1dxdy

, dxdy

≠ 0]

=> dydx

=1

cos y But, since cos2y = 1 – sin2y = 1 – x2

=> cosy = ±√(1 – x2)

Since y ∈ [–π2

, π2

] , then cosy > 0

and so cosy = √(1 – x2)

dydx

=1

1− x2

y = sin–1x => dydx

=1

1− x2

2. y = cos– 1x If y = cos–1x then by definition => x = cosy

dxdy

= − sin y [assuming dydx

=1dxdy

, dxdy

≠ 0]

=> dydx

=−1

sin y

But sin2y = 1 – cos2y = 1 – x2

so siny = ±√(1 – x2)

Since y ∈ [0,π ], then siny > 0

and so siny = √(1 – x2)

dydx

=−1

1− x2

y = cos–1x => dydx

=−1

1− x2


page 2

3. y= tan– 1x

If y = tan–1x then by definition => x = tany

dxdy

= sec2 y

= 1 + tan2y = 1 + x2

dydx

=1

1+ x2

y = tan–1x => dydx

=1

1 + x2

Examples:–1. y = sin–1 3x

y = sin–1u where u = 3x

dydu

=1

1 − u2 and du

dx= 3

dydx

=dydu

×dudx

= 11 − u2

× 3

= 3

1 − 9x2

2. y = cos–1( x2

) (we can use the chain rule here).

dydx

=−1

1− x2

⎛ ⎝ ⎜

⎞ ⎠ ⎟

2×

12

= −1

1 − x4

2×

12

= −14 − x2

4

×12

= −14 − x2

2

×12 =

−14 − x2

3. y = tan–1 4x

dydx

=1

1+ (4x)2 × 4 (by using the chain rule) = 41 + 16x2

4. y = cos−1( 11 + x

)

Let u =1

1+ x = (1 + x)–1

=> y = cos–1u where u = (1 + x)–1

dydu

=−1

(1 − u2) and

dudx

= −(1 + x)−2 =−1

(1 + x)2

cont’d ....


page 3

4.(contd)

u2 =1

(1 + x)2

=> 1 − u2 = 1 − 1(1+ x)2 =

(1 + x)2 –1(1 + x)2

=> 1 − u2 =x2 + 2x1+ x

=> −11 − u2

=−(1 + x)x2 + 2x

dydx

=dydu

×dudx

= −(1 + x)x2 + 2x

×−1

(1 + x)2 = 1(1 + x) x 2 + 2x

Exercise 1

Find the derivatives of the following, using the Chain , Product and Quotient Rules.

1. y = sin–1(√x) 2. y = tan–1(√x)

3. y = x tan–1x 4. y = x tan–1( x2 )

5. y = x sin–1x + √(1 – x2) 6. y = cos–1(2x – 1)

7. y = sin–1( x −1x + 1 ) 8. y = tan–1( x −1

x + 1 )

9. y = tan–1( 2x(1 − x2) ) 10. y = sin–1( 2x

(1 − x2) )

11. y = tan–1(secx) 12. y = cos(sin–1x)

13. y = tan−1(ex ) 14. y = sin−1(x2 ) − xex



Page 164 Exercise 3.5:2 Questions 2, 3, 4, 5.


Page 376 Exercise 15H Questions 1 – 12


page 4

Differentiate a Function defined Implicitly

Consider a function y, defined by the equation :- 3x2 + 7xy + 9y2 = 6It is difficult or even impossible to write y in terms of x .

Such a function y, is called an implicit function.These can be differentiated term by term with respect to x, assuming that y is a function of x.

Examples

1. In the above example :

(i) d(3x 2)dx

= 6x

(ii) 7xy is a product and is differentiated using the product rule.d(7xy)

dx=

d(7x)dx

y + 7xd(y)dx

= 7y + 7xdydx

(iii) 9y2 is a function of y and y is a function of xand so differentiate it using the chain rule.

d(9y2)dx

=d(9y2)

dx×

d(y)dx

= 18y dydx

.

(iv) 6 is a constant and so dydx

= 0 .

Therefore differentiating 3x2 + 7xy + 9y2 = 6 gives

=> 6x + 7y + 7x dydx

+ 18y dydx

= 0

=> (7x + 18y) dydx

= −(6x + 7y) => dydx

= −(6x + 7y)(7x + 18y)

2. Find the equation of the tangent at the point (2, 1) on the curve, 2x2 – 3xy – y2 = 1.Firstly, differentiate with respect to x.

4x − (3y + 3x dydx

) – 2y dydx

= 0

4x – 3y – 3x dydx

– 2y dydx

= 0

(3x + 2y) dydx

= 4x − 3y => dydx

=4x − 3y3x + 2y

When x = 2 and y = 1, => dydx

=58

The gradient of the tangent at (2, 1) is 58

The equation of the tangent is y −1 =58

(x − 2) => 5x – 8y = 2


page 5

3. Find the derivative of sinx + 2cosy = 1

=> cos x − 2sin ydydx

= 0

=> dydx

=cos x2sin y

4. Show that (1, 2) is a stationary point on the curve x2 – xy + y3 = 7 and find its nature.

=> (1, 2) does lie on the curve since 12 −1 × 2 + 23 = 7

Differentiating with respect to x gives:-

2x – (y + x

dydx

) + 3y2 dydx

= 0

=> 2x − y − xdydx

+ 3y2 dydx

= 0

=> dydx

=y − 2x3y2 − x

At (1, 2), dydx

= 0, hence (1, 2) is a stationary point on the curve.

To find d2ydx 2 , differentiate 2x − y − x dy

dx+ 3y2 dy

dx= 0

x dydx is a product and so

ddx

(x dydx

) =dydx

+ x d2ydx2

3y2 dydx is a product and so

ddx

(3y2 dydx

) = 6y dydx

×dydx

+ 3y2 d2ydx2

= 6y( dydx

)2 + 3y2 d2 ydx2

So the derivative of 2x − y − x dydx

+ 3y2 dydx

= 0

is 2 −dydx

−dydx

− x d2ydx2 + 6y( dy

dx)2 + 3y2 d2y

dx2 = 0

At the point (1, 2), dydx

= 0

and so 2 −d2ydx2 + 12 d2y

dx2 = 0 => d2ydx 2 =

−211 (i.e. < 0)

=> (1, 2) is a Maximum Turning Point.


page 6

Exercise 2

1. Find dydx

in each of these relations :

(a) x2 – y2 = 0 (b) y2 = 2x + 2y

(c) xy2 = 9 (d) 4x2 – y3 + 2x + 3y = 0

(e) x3y + xy3 = x –y (f) sinxcosy = 1

(g) exy = 2 (h) ex lny = x

2. Find the equations of the tangents at the given points on the following curves:-

(a) x3 – 2y3 = 3xy at (2, 1) .

(b) x2y2 = x 2 + 5y2 at (3, 32).

(c) y(x + y)2 = 3(x3 – 5) at (2,1).

3. For the curve xy(x + y) = 84 , find dydx

at (3, 4).

4. Find the gradient of the curve x2 + 3xy + y2 = x + y + 8 at the point (1, 2).

5. For the curve, 4x2 +y3= 2x + 7y, find the values of :-dydx and

d2ydx 2 at the point (–1, 2).

6. Show that (–1, 3) and (0, 0) are stationary points on the curve:-3x2 + 2xy – 5y2 + 16y = 0 and find the nature of each.



Page 69 Exercise 1.3:8 Questions 1 – 13.


Page 332 Exercise 13F Questions 1 – 14.


page 7

Related Rates of Change in problems where the Functional Relationship is given Explicitly and Implicitly.

Examples

1. Air is blown into a spherical ballon at a rate of 160 cm3/sec.Find the rate of increase of the radius when the radius is 5 cm.

We require to find drdt , given that

dVdt = 160

(i) By the chain rule, drdt

=drdV

. dVdt

V =43πr 3 = >

dVdr

= 4πr2 = >drdV

=1

4πr2

drdt

=drdV

. dVdt =

14πr 2 .160 =

40πr2

When r = 5, drdt =

4025π = 0·51 cm/sec.

(ii) By implicit differentiation

V =43πr 3 =>

dVdt

=ddt

43πr3⎛

⎝ ⎜ ⎞ ⎠ ⎟ =

ddr

43πr3⎛

⎝ ⎜ ⎞ ⎠ ⎟

drdt = 4πr 2 dr

dt

Since dVdt = 160 ,

drdt =

14πr 2 .160 = 0·51 cm/sec.

2. A cylinder, of radius r and height h, is closed at both ends. Its total surface area is 15 units2.

Find an expression for drdh

The total surface area of a cylinder closed at both ends, A = 2πr2 + 2πrh . => 2πr 2 + 2πrh = 15

Differentiate r with respect to h implicitly.

4πr drdh

+ 2πh drdh

+ 2πr = 0

=> drdh

(4πr + 2πh) = –2πr

=> drdh

=–2πr

4πr + 2πh = –r

2r + h


page 8

Exercise 3

1. If P = (2m + 3)4, find dmdt

when m = 1, given that dPdt

= 2

2. If r =1+ p1 + p2 , find dp

dt when p = 2, given that dr

dt = 14

3. The radius of a circular oil slick is increasing at a rate of 0·2 metres/second.Find the rate at which the area is increasing when the radius is 10 m.

4. A rectangle has dimensions x cm and y cm. Both x and y are changing but in such a way that the area of the rectangle remains constant at 40cm2 .

(a) Show that dydt

= −40x 2 . dx

dt(b) If x increases at a rate of 0·2 cm/sec, find the rate at which y

is changing when x = 8.

5. The volume of a cylinder is constant at 50 cm3, but both height h and radius r are changing.

(a) Show that dhdt

= −100πr3 . dr

dt(b) At an instant when the radius is 5 cm, the height is decreasing by 3 cm/sec.

Find the rate of change of radius at this instant.

6. A particle moves along a straight line so that its velocity, v m/sec, when it is s metres from a fixed point, is given by v = s 2 + 3 .

Find an expression for its acceleration, a metres/sec2, in terms of s.

7. A spherical balloon is inflated so that its volume increases at a constant rate of 200 cm3 /second. Find the rate of increase of the surface area of the balloon when the radius is 100 cm.

8. The volume of a sphere is increasing at a rate of 6 cm3/sec.Find the rate at which the surface area is increasing when the radius is 3 cm.



No reference


Page 324 Exercise 13C Questions 1 – 11


page 9

Logarithmic Differentiation.

The differentiation of complicated functions, particularly those that have powers consisting of functions of x, or contain products and quotients, can be made simpler by taking logarithms (to the base e).

Examples

1. Differentiate y = ax Take ln of both sides

lny = lnax => lny = x lna

Differentiate implicitly:-1y

dydx

= lna (lna is a constant).

=> dydx

= y ln a = ax ln a

3. Differentiate y = 2sin x

Take ln of both sideslny = ln 2sin x

lny = sinx ln2Differentiate implicitly:-1y

dydx

= cos x ln 2

dydx

= y cos x ln 2 = 2sin x cos x ln 2

2. Differentiate y = xx

Take ln of both sides lny = ln xx => lny = xlnxDifferentiate implicitly:-1y

dydx

= ln x + x ×1x

= ln x + 1

=> dydx

= y(ln x + 1) = xx (ln x + 1)

4. Differentiate y = (sin x)x

Take ln of both sideslny = ln(sin x)x

lny = x ln(sin x)Differentiate implicitly:-1y

dydx

= ln(sin x) + x × cos xsin x

= ln(sin x) + x cot x

=> dydx

= (sin x)x (ln(sin x) + x cot x)

5. Differentiate y = 1 + x1 − x

Take ln of both sides:-

lny = ln 1 + x1 − x

= ln 1+ x1− x⎛ ⎝ ⎜

⎞ ⎠ ⎟

12

=> lny =ln 1+ x1− x⎛ ⎝ ⎜

⎞ ⎠ ⎟

12 = 1

2{ln(1 + x) – ln(1 – x)}

Differentiate implicitly :-

=> 1y

dydx

=12

11 + x

+1

1 − x⎛ ⎝ ⎜

⎞ ⎠ ⎟ =

12

2(1 + x)(1 – x)

⎛

⎝ ⎜

⎞

⎠ ⎟ =

1(1 + x)(1− x)

=> dydx

= y 1(1+ x)(1 − x)⎛

⎝ ⎜

⎞

⎠ ⎟ = 1 + x

1 − x⎛ ⎝ ⎜

⎞ ⎠ ⎟

12 1

(1+ x)(1 − x)⎛

⎝ ⎜

⎞

⎠ ⎟ = 1

(1 − x)32 (1 + x)

12


page 10

Exercise 4

Differentiate using logarithmic differentiation.

1. y = 10x 2. y = 2x 2

3. y = x− x 4. y = xsin x

5. y = x1x 6. y = x ln x

7. y = (ln x)x 8. y = (ln x) ln x

9. y = x5

(3x + 5)10. y = x3(2x – 1)5

(x + 1)2





Page 487 Exercise 19B Questions 1 – 24.


page 11

Parametric Differentiation.

It is sometimes convenient to define the coordinates of a moving point by means of a two equations, expressing x and y separately in terms of a third variable, say t, called a parameter.

e.g. The position of a ball, relative to the x and y axes, may be plotted at 1 second intervals. Here, clearly the x and y positions of the ball depend on time t.

It can be shown that the path of the ball is given by the equationsx = at2 and y = 2at where “a” is a constant.

These equations are called parametric equations of the curve, t being the parameter.By replacing t by various values in the two equations, we obtain the coordinates of various points on the curve.

e.g. At point P(t) ; x = at2, y = 2at. Hence P is the point (at2, 2at)P(0) ; x =0, y = 0. Hence P is the point (0, 0)

P(1) ; x = a, y = 2a. Hence P is the point (a, 2a) P(2) ; x = 4a, y = 4a. Hence P is the point (4a, 4a) P(3) ; x = 9a, y = 6a. Hence P is the point (9a, 6a) and so on.

x = at2, y = 2at are the parametric equations of a parabola.It is often useful to eliminate t from the two equations to obtain an equation involving x and y but not t.

This equation is called the constraint equation. e.g. x = at2 and y = 2at

From y = 2at => t =y

2a

From x = at2 => x = ay2

4a2 => x = y2

4a

=> y2 = 4ax

which is the general equation of a parabola whose axis of symmetry is the x – axis.

Other common parametric equations are :

(a) x = acosθ, y = asinθ , are the parametric equations of a circle x2 + y2 = r2

(b) x = acosθ, y = bsinθ , are the parametric equations of an ellipse x2

a2 +y2

b2 = 1

(c) x = asecθ, y = btanθ , are the parametric equations of a hyperbola x2

a2 −y2

b2 = 1

(d) x = ct, y =ct

, are the parametric equations of a rectangular hyperbola xy = c2


page 12

Finding the first derivative of a parametric function

If a function is defined as x = x(t) , y = y(t), (i.e. x and y are functions of t), then :-

since dydx

=dydt

×dtdx

and dxdt

=1dtdx

, then dydx

=

dydtdxdt

.

Examples

1. Find dydx

on the curve x = at2, y = 2at .

y = 2at => dydt

= 2a, x = at2 =>dxdt

= 2at

And so :– dydx

=2a2at

=1t

2. Find a formula for the gradient of the tangent to the curve whose parametric equations are x = a(θ – sinθ), y = a(1 – cosθ)

y = a(1 – cosθ) => dydθ

= asinθ

x = a(θ – sinθ) => dxdθ

= a(1– cosθ)

and so :- dydx

=asinθ

a(1− cosθ)=

sinθ1− cosθ

3. Find the coordinates of the points on the curve x = 1 – t2, y = t3 + t at which the gradient is 2.

y = t3 + t => dydt

= 3t2 + 1, x = 1− t2 => dxdt

= −2t

and so dydx

=3t2 + 1−2t

=> mtan =3t2 + 1−2t

= 2

i.e. –4t = 3t2 + 1 => 3t2 + 4t + 1 = 0 => (t + 1)(3t + 1) = 0

=> t = – 1, t = –13

At t = – 1, x = 0, y = –2. i.e. at the point (0, –2).

At t = –13

, x =89

, y =−1027

, i.e. at the point ( 89

,−1027

).


page 13

4. Find the equation of the tangent to the curve x = at2, y = 2at, at the point P(t).

y = 2at => dydt

= 2a , x = at2 => dxdt

= 2at

and so dydx

=2a2at

=1t

=> mtan =1t

=> Equation of the tangent at P(t) = P(x,y) = P(at2, 2at ) is:-

y− 2at =1t

(x− at2 )

=> ty – 2at2 = x – at2

=> x – ty = – at2

Exercise 5

1. Find dydx

in terms of the parameter.

(a) x = t3 + t2, y = t2 + t (b) x = 4cosθ, y = 3sinθ

(c) x =1

1+ t, y =

t1− t

(d) x =t −1t + 1

, y =2t− 1t− 2

(e) x = (t + 1)2, y = t2 – 1 (f) x =t

1− t , y =t2

t + 3

(g) x = cos 2θ, y = 4sin θ (h) x = acos2θ, y = asin3θ

(i) x = etcos t, y = etsin t (j) x = a(t – cost), y = a(1 + sint)

2. Find the equations of the tangents to these curves at the point P(x,y) :-

(a) x = ct, y =ct (b) x = at2, y =at(t2 – 1)

(c) x =a2

(t +1t) , y =

a2

(t − 1t) (d) x = secθ, y = tanθ



Page 322 Exercise 6.3:2 Questions 1 – 12, selected others.


Page 327 Exercise 13D Questions 1 – 224.


page 14

Finding the Second Derivative of a Parametric Function

If a function is defined as x = x(t) , y = y(t), then

since dydx

=dydt

×dtdx

and dxdt

=1dtdx

then dydx

=

dydtdxdt

Also, d2ydx 2 =

ddx

dydx

⎛ ⎝ ⎜

⎞ ⎠ ⎟ =

ddt

dydx

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ×

dtdx

or ddt

dydx

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ÷

dxdt

Examples

1. For the curve x = at2 , y = 2at , find dydx

and d2ydx 2

y = 2at => dydt

= 2a , x = at2 => dxdt

= 2at

And so dydx

=2a2at

=1t

d2ydx 2 =

ddx

dydx

⎛ ⎝ ⎜

⎞ ⎠ ⎟ =

ddt

dydx

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ×

dtdx

=ddt

1t

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ×

12at

= −1t2 ×

12at

= −1

2at3

2. For the curve x = acosθ, y = asinθ , find dydx

and d2ydx 2

y = asinθ => dydθ

= acosθ x = acosθ => dxdθ

= −asinθ

And so dydx

=a cosθ−asinθ

= – cosθsinθ

= –cotθ

d2ydx 2 =

ddx

dydx

⎛ ⎝ ⎜

⎞ ⎠ ⎟ =

ddθ

dydx

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ×

dθdx

= ddθ

– cosθsinθ

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ×

1–asinθ

= sinθ(sinθ ) + cosθ(cosθ )

sin2 θ×

1−asinθ

= sin2θ + cos2 θsin2 θ

×1

−asinθ = –1

asin3θ


page 15

3. Find the turning points on the curve x = t, y = t3 – 3t and determine their nature.

y = t3 − 3t => dydt

= 3t2 − 3 , x = t => dxdt

= 1

and so dydx

= 3t2 − 3

For stationary values, dydx

= 0 => 3t2 − 3 = 0 => t = –1 or t = 1

When t = –1, x = –1 and y = 2 i.e. (–1, 2)When t = 1, x = 1 and y = –2 i.e. (1, –2)

d2ydx 2 =

ddx

dydx

⎛ ⎝ ⎜

⎞ ⎠ ⎟ =

ddt

dydx

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ×

dtdx

=ddt

3t2 − 3( ) × dtdx

= 6t x 1 = 6t

When t = –1, d2ydx 2 is negative, and so (–1, 2) is a Maximum stationary point.

When t = 1, d2ydx 2 is positive, and so (1, –2) is a Minimum stationary point.

Exercise 6

1. Find dydx

and d2ydx 2 in terms of t :-

(a) x =1t2 , y = 1 + t (b) x = (t + 1)2 , y = t 2 −1

(c) x = 4cost, y = 3sint (d) x = cos2t, y = sint

(e) x = 2cost – cos2t, y = 2sint + sin2t

2. A curve has parametric equations x = t – cost, y = sint.Find the coordinates of the points at which the gradient of the curve is zero.

3. Find the coordinates of the stationary points on the curves and determine their nature:-(a) x = 4 – t2, y = 4t – t3 (b) x = (5 – 3t)2, y = 6t – t2

(c) x = t2 + 1, y = t(t – 3)2

4. Given that x = t – sint, y = 1 – cost, show that y2 d2ydx 2 + 1= 0


The Complete A level Maths (Orlando Gough) Page 321 Exercise 6.3:2 .Understanding Pure Mathematics (A.J.Sadler/D.W.S.Thorning)Page 327 Exercise 13D Question 25 Page 334 Exam question 10.


page 16

Application of Parametric Differentiation to Motion in a Plane

Velocity

Suppose the point P moves along a curve in the x-y – plane and suppose we know its position at any time t is given by x = f(t) and y = g(t). (its parametric equations).

=> dxdt

is the velocity in the x – direction and dydt

is the velocity in the y – direction

The magnitude of the velocity is therefore v = dxdt

⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

+dydt

⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

which gives the instantaneous speed of the particle at P.

Examples (1) The position of a golf ball, t seconds after being hit, is given by :-x = 10t, y = 30t – 5t2.

Find the speed of the golf ball when it first hits the ground.The golf ball hits the ground when y = 0.

30t – 5t2 = 0 => 5t(6 – t) = 0 => t = 0 and t = 6dxdt

= 10 and dydt

= 30 – 10t

At t = 6, dxdt

= 10 and dydt

= –30

The speed when it first hits the ground is v = [(10)2 + (30)2 ] = 31·6 m/s.

(2) A cannon ball is fired and its position t seconds later, is given by

x = 10t, y = 2 + 9t – 5t2 .

The target is 2 metres above the ground.

(i) Find how far away the target should be if the cannon ball is to hit it.(ii) What is the speed of the cannon ball when it hits the target ?

(i) When y = 2, => 2 + 9t – 5t2 = 2=> 9t – 5t2 = 0 => t(9 – 5t) = 0

=> t = 0 or t = 95

When t = 95

, x = 10t = 18The target should be placed with its centre 18 metres horizontally from the cannon and at a height of 2 metres.

cont’d ......


page 17

(ii) dxdt

= 10 and dydt

= 9 – 10t

At t = 95

=> dxdt

= 10 and dydt

= –9

The speed when it hits the target is v = [(10)2 + (−9)2] = 13·5 m/s




Understanding Pure Mathematics (A.J.Sadler/D.W.S.Thorning). No reference

Exercise 71. At time t, the position of a moving point is given by

x = t(2 – t), y = t(3 – t).Find the speed when t = 0 and when t = 2.

2. At time t, the position of a moving point is given byx = t + 1, y = t2 – 1.

Find the speed when t = 2.

3. At time t, the position of a moving point is given byx = cos2t, y = 2sint.


4. At time t, the position of a moving point is given byx = et, y = e–2t.

Find the speed when t = ln3.

5. At time t, the position of a moving point is given byx = sect, y = tant .

Find the speed when t = π6

.

6. At time t, the position of a moving point is given byx = ln(t + 1) , y = t2 .


7. A particle moves so that its position at time t is given by:- x = 4cost, y =3sint.

Show that its speed is 9 + 7sin2 t .Hence find the maximum and minimum speeds and the corresponding positions of the particle.


page 18

Answers

Exercise 1

1. 1

2 x(1− x) 2.

12(1− x) x 3. tan−1 x +

x1+ x 2

4. tan−1 x2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ +

2x4 + x2 5. sin−1 x 6.

−1(x − x2 )

7. 1(x + 1) x

8. 11 + x2 9. 2

(1 + 3x2 ) (1− x2 )

10.2

(1 − x2) (1 − 5x 2)11.

sec x tan x1+ sec2 x 12.

−x1 − x2

13.ex

1 + e2x 14.2x

1 − x4− ex2

[2x2 + 1]

Exercise 2

1. (a) xy

(b) 1y− 1

(c) −y2x

(d) 2(4x + 1)3(y2 – 1)

(e) 1− 3x2y – y3

1+ 3xy2 + x3 (f) cotx coty (g) −yx

(h) ye−x − yln y

2. (a) 3x – 4y = 2 (b) 5x + 8y = 27 (c) 2x – y = 3

3. −4033

4. −76

5. 2, −565

6. (–1, 3) Minimum, (0,0) Maximum.

Exercise 4

1. 10x ln10 2. 2x 2

× 2x ln x 3. –x–x(lnx + 1)

4. xsin x (cos x ln x +sin x

x) 5. x

1x− 2

⎛ ⎝ ⎜

⎞ ⎠ ⎟

(1− ln x) 6. 2xlnx – 1lnx

7. (ln x)x [ln(ln x) +1

ln x] 8.

(ln x) ln x

x[ln(ln x) + 1]

9.x4(27x + 50)2(3x + 5)32 10.

3x 2(2x −1)4(4x 2 + 5x – 1)(x + 1)3

Exercise 3

1. 2000 2. –50 3. 4π

4. (a) Proof (b) –0·125 cm/sec 5. 11·8 cm/sec

6. 2s(s2 + 3) 7. 0·2 cm/sec2 8. 4 cm/sec2.


page 19

Exercise 5

1.(a) 2t + 1

t(3t + 2) (b) −34

cot θ (c) −(1 + t)2

(1− t)2 (d) −3(t + 1)2

2(t − 2)2

(e) t

t + 1 (f) (t + 6)(1− t)2

(t + 3)2 (g) –cosecθ (h) −32

sin θ( )

(i) cos t + sintcos t − sint

(j) cos t1+ sint

2.(a) x + t2y = 2ct (b) (3t2 −1)x − 2ty = at2(t2 + 1)

(c) (t2 + 1)x – (t2 – 1)y = 2at (d) xsecθ – ytanθ = 1

Exercise 6

1.(a) −12

t3, 34

t5 (b)t

1 + t, 12(1 + t)3 (c) −

34

cot t,− 316

cosec3 t

(d) −12

cosec t,− 14

cos ec3 t (e)cos t + 1

sint, −1sin3 t(2cos t −1)

2. π2

,1⎛ ⎝ ⎜

⎞ ⎠ ⎟ , 3π

2,–1

⎛ ⎝ ⎜

⎞ ⎠ ⎟

3.(a) Maximum S.Pt. at 83

,169

3⎛ ⎝ ⎜

⎞ ⎠ ⎟ , Minimum S.Pt. at 8

3, −16

93

⎛ ⎝ ⎜

⎞ ⎠ ⎟

(b) Maximum S.Pt. at (16, 9)

(c) Maximum S.Pt. at (2, 4), Minimum S. Pt. at (10, 0)

4. Proof

Exercise 7

1. √13, √37 2. √17 3. 2

4. 19

730 5. 13

20 6. 12

17

7. Max of 4 at (0,3) and (10,–3). Min of 3 at (4,0) and (–4,0).


page 20

Outcome 2 – Use further integration techniques

(a) Integrals of the form 11− x 2∫ dx and 1

a2 − x2∫ dx

(i) If f (x) = sin−1 x => ′ f (x) =1

1 − x2

then 11− x 2∫ dx = sin−1 x + c

(ii) If f (x) = sin−1 x2

⎛ ⎝ ⎜

⎞ ⎠ ⎟

=> ′ f (x) = 1

1− x2

⎛ ⎝ ⎜

⎞ ⎠ ⎟

2⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

×12

= 1

1− x2

4

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

×12

= 14 − x2

4

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

×12

= 14− x2

2

×12

= 14− x2

then 14 − x2

∫ dx = sin−1 x2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + c

Alternatively, integrate 14 − x2

∫ dx by substitution.

Let x = 2t, dx = 2dt and 4 – x2 = 4 – 4t2 = 4(1 – t2) and √(4 – x2) = √4(1 – t2) = 2√(1 – t2)

The integral 14 − x2

∫ dx becomes 12 1− t2∫ 2dt

= 11− t2∫ dt

= sin-1t + c

But t = x2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ and so 1

4 − x2 ∫ dx = sin−1 x

2⎛ ⎝ ⎜

⎞ ⎠ ⎟ + c

In general 1a2 − x2∫ dx = sin−1 x

2⎛ ⎝ ⎜

⎞ ⎠ ⎟ + c


page 21

Examples Integrate the following :-

(1) 116 − x 2∫ dx = sin−1 x

4⎛ ⎝ ⎜

⎞ ⎠ ⎟ + c

(2) 116 + x2∫ dx = 1

4tan−1 x

4⎛ ⎝ ⎜

⎞ ⎠ ⎟ + c

(b) Integrals of the form 11+ x 2∫ dx and 1

a2 + x2∫ dx

(i) If f(x) = tan −1 x => ′ f (x) = 11+ x2

then 11+ x 2∫ dx = tan −1 x + c

(ii) If f(x) = tan −1 x3

⎛ ⎝ ⎜

⎞ ⎠ ⎟

and ′ f (x) =1

1+x3

⎛ ⎝ ⎜

⎞ ⎠ ⎟

2⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

×13 =

1

1+x2

9⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

×13 =

19 + x2

9⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

×13

= 3

9 + x2

then 39 + x2∫ dx = 3 1

9 + x2∫ dx = tan −1 x3

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + c

so 19 + x2∫ dx = 1

3tan −1 x

3⎛ ⎝ ⎜

⎞ ⎠ ⎟ + c

Alternatively, integrate 19 + x2∫ dx by substitution.

Let x = 3t, dx = 3dt and 9 + x2 = 9 + 9t2 = 9(1 + t2)

The integral 19 + x2∫ dx becomes 1

9 1+ t2( )∫ 3dt

= 13

11+ t2∫ dt = tan −1 t + c

But t = x3

and so 1

9 + x2∫ dx = 13

tan −1 x3

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + c

In general 1a2 + x2∫ dx = 1

atan−1 x

a⎛ ⎝ ⎜

⎞ ⎠ ⎟ + c


page 22

(3)1

25 − 9x2∫ dx =1

9 259− x2⎛

⎝ ⎜

⎞ ⎠ ⎟

∫ dx = 13

1259− x2

∫ dx

= 13

sin−1 x53

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ + c

= 13

sin−1 3x5

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + c

(4)1

9 + 4x2∫ dx = 1

4 94

+ x2⎛ ⎝ ⎜

⎞ ⎠ ⎟

∫ dx = 14

194

+ x2∫ dx

= 14×

132

tan−1 x32

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ + c

= 16

tan−1 2x3

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + c

Exercise 1Integrate

1. 149 − x2∫ dx 2. 1

49 + x2∫ dx

3. 19 − x2∫ dx 4. 1

100 + x2∫ dx

5. 136 − 25x 2∫ dx 6. 1

36 + 25x2∫ dx

7. 225 + 4x2∫ dx 8. 3

36 − 9x2∫ dx

Evaluate

9. 21+ x21

3

∫ dx 10. 14 − x20

2

∫ dx

11. 31− x21

2

1

∫ dx 12. 19 + x 20

3

∫ dx



Page 193 Exercise 4.2:3 Questions 16, 17, 19, 20.


Page 376 Exercise 15H Questions 13 – 24.


page 23

(c) Integrals by means of Partial Fractions.

We already know how to express a rational function as a sum of partial fractions from Unit 1, Outcome 1.We will apply these techniques to integrating rational functions.

Examples (where the denominator has linear and different factors).

(1) x + 162x2 + x − 6∫ dx

Let x + 162x2 + x − 6

= x + 162x − 3( ) x + 2( ) = Α

2x − 3+

Βx + 2

Multiply both sides by 2x − 3( ) x + 2( ) => x + 16 = A(x + 2) + B(2x – 3)

Replace with x = 32 :

352

= 72

A => A = 5 Replace with x = –2 : 14 = –7B => B = –2

The integral becomes:

x + 16

2x2 + x − 6∫ dx = 5

2x − 3−

2x + 2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ∫ dx

=5 12x − 3∫ dx – 2 1

x + 2 dx∫

= 52

ln(2x – 3) − 2 ln(x + 2) + c

(2) 2x2 + 4xx −1( ) x + 1( ) 2x + 1( )∫ dx

Let 2x2 + 4xx −1( ) x + 1( ) 2x + 1( ) = A

x −1+

Bx + 1

+C

2x + 1

Multiply both sides by (x – 1)(x + 1)(2x – 1)

=> 2x2 + 4x = A(x + 1)(2x + 1) + B(x – 1)(2x + 1) + C(x – 1)(x + 1)

Put x = 1: 6 = 6A => A = 1 Put x = –1: –2 = 2B => B = –1

Put x = – 12 : –

32

= – 34

C => C = 2

The integral becomes:–

2x2 + 4xx −1( ) x + 1( ) 2x + 1( )∫ dx = 1

x −1−

1x + 1

+2

2x + 1∫ dx

contd...


page 24

1x −1

−1

x + 1+

22x + 1∫ dx = ln(x –1) – ln(x + 1) + ln(2x + 1) + c

Note: You may find the following method of simplifying the right -hand side useful when you come to first order differential equations in a later outcome.When all integrals are logarithms, c can be replaced by lnK since they are both constants in different forms.

and so ln(x –1) – ln(x + 1) + ln(2x + 1) + c = ln(x –1) – ln(x + 1) + ln(2x + 1) + lnK

= ln K x −1( ) 2x + 1( )x + 1( )

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ by the laws of logarithms

(3) 2x3 + 7x2 − 2x − 22x2 + x − 6∫ dx

Use long division 2x2 + x − 6 2x3 + 7x 2 − 2x − 2x + 3

) 2x3 + x2 − 6x

6x2 + 4x − 2

6x2 + 3x − 18 x + 16

The integral can now be written as

2x3 + 7x2 − 2x − 22x2 + x − 6∫ dx

= x + 3 +x + 16

2x2 + x − 6⎛ ⎝ ⎜

⎞ ⎠ ⎟ ∫ dx = x + 3 +

x + 162x − 3( ) x + 2( )

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ ∫ dx

Now express x + 162x − 3( ) x + 2( ) in its partial fractions:-

Let x + 162x − 3( ) x + 2( ) = Α

2x − 3+

Βx + 2

Multiply both sides by (2x – 3)(x + 2)

=> x + 16 = A(x + 2) + B(2x – 3)

Put x = 32

: 352 = 7

2A => A = 5

Put x = –2: 14 = –7B => B = –2The integral becomes:

2x3 + 7x2 − 2x − 22x2 + x − 6∫ dx = x + 3 +

52x − 3

−2

x + 2⎛ ⎝ ⎜

⎞ ⎠ ⎟ ∫ dx

= 12 x2 + 3x +

52 ln(2x –3) – 2ln(x + 2) + c


page 25

Exercise 2

Integrate

1. x + 8x + 2( ) x + 4( )∫ dx 2. x2

x2 − 4∫ dx

3. x2 − 6x − 7x −1( ) x − 2( ) x + 3( )∫ dx 4. x3 − 2x −13

x 2 − 2x − 3∫ dx

Example where the denominator has linear and repeated factors.

(1) −8x2 + 14x −152x −1( )2 x + 2( )∫ dx

Let −8x2 + 14x −152x − 1( )2 x + 2( ) = Α

2x −1+

Β2x −1( )2 +

Cx + 2

Multiply both sides by 2x −1( )2 x + 2( ) => −8x2 + 14x −15 = A(2x −1)(x + 2) + B(x + 2) + C(2x −1)2

Put x = 12 : –10 = 5

2B => B = – 4

Put x = –2: –75 = 25C => C = –3

Put x = 0: –15 = –2A + 2B + C

–15 = –2A – 8 – 3 => A = 2


−8x2 + 14x −152x −1( )2 x + 2( )∫ dx = 2

2x −1−

42x −1( )2 −

3x + 2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ ∫ dx

= ln(2x – 1) + 22x −1

– 3ln(x + 2) + c

Exercise 3

Integrate

1. 3x2 + x + 1x x + 1( )2∫ dx 2. x 2 − 2x + 10

x + 2( ) x −1( )2∫ dx

3. 252x − 1( )2 x + 2( )∫ dx 4. 5x + 2

x − 2( )2 x + 1( )∫ dx


page 26

Examples where the denominator has a linear factor and an irreducible quadratic factor of the form x2 + a2 .

(1) x −1x + 1( ) x2 + 1( )∫ dx

Let x −1x + 1( ) x2 + 1( ) = Α

x + 1+Βx + Cx2 + 1

Multiply both sides by (x + 1)(x2 + 1)

=> x – 1 = A(x2 + 1) + (x + 1)(Bx + C)Put x = –1: –2 = 2A => A = –1Put x = 0: –1 = A + C => C = 0Put x = 1: 0 = 2A + 2(B + C) => B = 1


x −1x + 1( ) x2 + 1( )∫ dx = −1

x + 1+

xx2 + 1

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ∫ dx

= – ln(x + 1) + 12 ln(x2 + 1) + c

(2)x2 + 2x + 1x x2 + 1( )∫ dx

Let x2 + 2x + 1x x 2 + 1( ) = A

x+

Bx + Cx 2 + 1

Multiply both sides by x(x2 + 1)

x2 + 2x + 1 = A(x2 + 1) + x(Bx + C)Put x = 0: 1 = A => A = 1Put x = 1: 4 = 2A + B + C => B + C = 2Put x = –1: 0 = 2A + B – C => B – C = –2The integral becomes:

x2 + 2x + 1x x2 + 1( )∫ dx = 1

x+

2x2 + 1

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ∫ dx

= lnx + 2tan–1x + c

}Solve givingB = 0, C = 2


page 27

(3) 1x − 2( ) x2 + 1( )∫ dx

Let 1x − 2( ) x2 + 1( ) = A

x − 2+

Bx + Cx 2 + 1

=> Multiply both sides by (x – 2)(x2 + 1)

1 = A(x2 + 1) + (x – 2)(Bx + C)

Put x = 2: 1 = 5A => A = 15

Put x = 0: 1 = A – 2C => C = −25

Put x = 1: 1 = 2A – B – C => B = −15


1x − 2( ) x2 + 1( )∫ dx =

15

x − 2+−

15

x −25

x2 + 1

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

∫ dx

= 15 x − 2( )∫ −

15

x + 2x2 + 1

⎛ ⎝ ⎜

⎞ ⎠ ⎟ dx

= 15 x − 2( )∫ −

15

xx2 + 1

+2

x2 + 1⎛ ⎝ ⎜

⎞ ⎠ ⎟ dx

= 15 x − 2( )∫ −

15

xx2 + 1

⎛ ⎝ ⎜

⎞ ⎠ ⎟ −

15

2x 2 + 1

⎛ ⎝ ⎜

⎞ ⎠ ⎟ dx

= 15 x − 2( )∫ −

110

2xx 2 + 1

⎛ ⎝ ⎜

⎞ ⎠ ⎟ −

25

1x 2 + 1

⎛ ⎝ ⎜

⎞ ⎠ ⎟ dx

=

15 ln(x – 2) –

110 ln(x2 + 1) –

25 tan–1x + c

Exercise 4Integrate :-

1. 3x + 1x −1( ) x2 + 1( )∫ dx 2. 3x 2 + 92x

x + 6( ) x2 + 1( )∫ dx

3. xx4 −1∫ dx 4. x

x + 1( ) x2 + 4( )∫ dx



Page 196 Exercise 4.2:5 Questions 7, 8(i),(ii),(iii) only, 9, 10.


Page 488 Exercise 19B Questions 46, 47, 48.Page 494 Exercise 19D Question 6.


page 28

Integration by Parts

We have already used a formula to differentiate a product of two functions u(x) and v(x).

i.e. ddx

uv( ) = udvdx

+ vdudx

Integrating both sides of the above formula with respect to x

=> ddx

uv( )⎛ ⎝ ⎜

⎞ ⎠ ⎟ ∫ dx = u

dvdx

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ∫ dx + v

dudx

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ∫ dx

i.e. uv = udvdx

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ∫ dx + v

dudx

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ∫ dx

Rearranging gives

udvdx

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ∫ dx = uv − v

dudx

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ∫ dx or simply udv = uv − vdu∫∫

This formula enables us to integrate a product of two functions and the method is known as integration by parts .

N.B. When choosing which function is u and which is dvdx

, care must be taken to

ensure that v dudx

is simple enough to integrate.

Examples

(1) x cos xdx∫ Let u = x and dv = cosxdx du = 1dx and v = sinx

∴ x cos xdx∫ = xsinx – sin xdx∫ = xsinx + cosx + c

Note if we try to let u = cosx and dv = xdx instead, this will lead to :-

du = –sinxdx and v = 12 x2

∴ x cos xdx∫ = 12

x2 cos x +12∫ x2 sin xdx

This time, the second integral is not any simpler than the original.

(2) xe xdx∫ Let u = x and dv = ex dx du = 1dx and v = ex

∴ xe xdx∫ = xex – ex dx∫= xex – ex + c

(3) x ln xdx∫ Let u = lnx and dv = xdx

du = 1x

dx and v = 12

x2

∴ x ln xdx∫ = 12

x2 ln x – 1x×

12

x2dx∫ = 1

2x2 ln x – x

2dx∫ = 1

2x2 ln x – 1

4x 2 + c


page 29


1. x sin xdx∫ 2. x sin3xdx∫

3. x ln xdx∫ 4. 1x3 ln xdx∫

5. xe xdx∫ 6. x cos4xdx∫

Integration by Parts involving Repeated Applications.

Examples

(1) x2 sin xdx∫ Let u = x2 and dv = sinxdx du = 2xdx and v = –cosx

∴ x2 sin xdx∫ = – x2 cosx + 2x cos xdx∫Let u = 2x and dv = cosxdx

du = 2dx and v = sinx

= −x2 cos x + 2x sin x − 2sin xdx∫ = −x2 cos x + 2x sin x + 2 cosx + c

(2) x2exdx∫ Let u = x2 and dv = exdx du = 2xdx and v = ex

∴ x2exdx∫ = x2e x − 2xe xdx∫Let u = 2x and dv = exdx

du = 2dx and v = ex

= x2e x − 2xex − 2ex dx∫[ ] = x2e x − 2xe x + 2ex + c


1. x2 cos xdx∫ 2. x2 sin 3xdx∫

3. x2e2x dx∫ 4. x2 cos2xdx∫

5. x2e− xdx∫ 6. x3ex dx∫


page 30

Integration by Parts involving a “Dummy” Function.

Functions like lnx, sin–1x, cos–1x and tan–1x do not have a standard integral but have a standard derivative.In order to integrate them, we introduce a “dummy” function, namely the number 1 (one).

i.e. for u = lnx, u = sin–1x or u = tan–1x, let dv = 1 each time.Examples

(1) ln xdx∫ Let u = lnx and dv = 1dx

du = 1x

dx and v = x

∴ ln xdx∫ = xlnx – x ×1x

dx∫ = xlnx – 1dx∫

= xlnx – x + c

(2) sin−1 dx∫ Let u = sin–1x and dv = 1dx

du = 11 − x2

dx and v = x

∴ sin−1 dx∫ = x sin−1 x −x

1 − x2dx∫

For x1− x2

dx∫ , use the substitution t = 1 – x2 => dt = –2xdx

i.e. xdx = − 12

dt

The integral becomes:x

1− x2dx∫ = −

12 t

dt = −12

t−

12 dt∫∫ = −

12

2t12 + c

= – 1 − x2 + c

∴ sin−1 dx∫ = x sin−1 x + 1− x2 + cExercise 7

Integrate :-

1. tan−1 xdx∫ 2. sin−1 3xdx∫

3. tan−1 2xdx∫ 4. sin−1 12

xdx∫

5. cos−1 xdx∫ 6. tan−1 12

xdx∫

7. ln 2xdx∫ 8. (ln x)2dx∫


page 31

Exercise 8Integrate

1. ex sin 2xdx∫ 2. e− x sin xdx∫3. e−2x cos3xdx∫ 4. ex cos2 xdx∫

Further examples can be found in the following resources.The Complete A level Maths (Orlando Gough) Page 199 Exercise 4.2:6 Questions 1, 2, 4, 5, 8, 12 .Understanding Pure Mathematics (A.J.Sadler/D.W.S.Thorning)Page 505 Exercise 20B Questions 1 – 25.

Integration by Parts involving an Integral that returns to its original form.

There are some integrals which return to themselves when integration by parts is used. We have a special way of dealing with them.

Examples(1) ex sin xdx∫ Let u = ex and dv = sinxdx

du = exdx and v = –cosx

∴ ex sin xdx∫ = –ex cosx + ex cos xdx∫Let u = ex and dv = cosx

du = ex and v = sinx

= −ex cos x + ex sin x − ex sin xdx∫

∴ ex sin xdx∫ + e x sin xdx∫ = e x sin x − ex cos x

i.e. => 2 ex sin xdx∫ = ex sin x − ex cosx

=> ex sin xdx∫ =12

(ex sin x − ex cosx) + c

(2) e2x cos xdx∫ Let u = e2x and dv = cosx dx du = 2e2xdx and v = sinx

∴ e2x cos xdx∫ = e2x sin x − 2 e2x sin xdx∫Let u = e2x and dv = sinx dx

du = 2e2x dx and v = –cosx

= e2x sin x − 2[−e2x cos x + 2 e2x cos xdx]∫ = e2x sin x + 2e2x cosx − 4 e2x cos xdx∫

i.e. => 5 e2x cos xdx∫ = e2x sin x + 2e2x cos x

=> e2x cos xdx∫ =15

(e2x sin x + 2e2x cos x) + c

which is theoriginal integral

which is theoriginal integral


page 32

Solve First Order Differential Equations (Variable Separable)

Differential Equations

A differential equation is an equation connecting x, y and the differential coefficients dydx and

d2ydx 2 .

For example (1) ydydx

+ 2xy = x and (2) d2ydx 2 − 3 dy

dx+ 4 = sin x .

The order of the differential equation is the value of the largest differential coefficient in the equation .

i.e. (1) above is of the first order and (2) is of second order.

In order to solve a differential equation in the variables x and y, it is necessary to find some function y = f(x) which satisfies the equation.

Let us consider the first order differential equation dydx

= 2x + 3

We know, by integration, that solutions to this differential equation will be of the form y = x2 + 3x + c

We say that y = x2 + 3x + c is the general solution of dydx

= 2x + 3

If we were to draw the graphs of y = x2 + 3x + c , for various values of c, we would obtain a family

of curves, each of which has the property that dydx

= 2x + 3

A particular solution could be found by choosing values of x and y.

First Order Differential Equations

Any first order differential equation that can be expressed in the form

f (y) dydx

= g(x)

can be solved by separating the variables (variable separable)

Suppose f (y) dydx

= g(x)

Integrating both sides with respect to x

f y( )∫dydx

dx = g x( )∫ dx

f y( )∫ dy = g x( )∫ dx


page 33

Examples Find the general solutions of :-

(1) ydydx

=1x2 => ydy =

1x2 dx

=> ydy =1x2∫∫ dx

=> y2

2= −

1x

+ c

=> y2 = −2x

+ C

(2) x2 dydx

= y + 3 => dyy + 3

=dxx2

=> dyy + 3∫ =

dxx2∫

=> ln(y + 3) = −1x

+ c

(3) (x + 2) dydx

= 1 => dydx

=1

x + 2

=> dy∫ =1

x + 2∫ dx

=> y = ln(x + 2) + c

(4) dydx

=1 + y2

1+ x2 => dy1 + y2 =

dx1 + x2

=> dy1 + y2∫ =

dx1+ x2∫

=> tan −1 y = tan−1 x + c

Exercise 9

Find the general solutions of :-

1. (1 + x) dydx

= xy 2.dydx

= x(1 − y)2

3.dydx

= ex y2 4. x(y −1) dydx

= 2y

5. sin x cos y = sin y cosx dydx 6. y − x dy

dx= 1 + x 2 dy

dx


page 34

Examples Find the particular solutions of :-

(1) dydx

= x(y − 2) , given x = 0 when y = 5.

dyy − 2

= xdx

=> dyy − 2∫ = xdx∫

ln(y − 2) =x2

2+ c

When x = 0 and y = 5, c = ln3

i.e. ln(y − 2) =x2

2+ ln 3

=> ln(y − 2) − ln 3 =x2

2

=> ln (y − 2)3

=x2

2

=> (y − 2)3

= ex 2

2 => y − 2 = 3ex2

2

=> y = 3ex 2

2 + 2

(2) (1 + x2) dydx

= 4ey , given x = 0 when y = 0.

dyey =

41 + x2 dx

dyey∫ =

41+ x2 dx∫

−1e y = 4 tan−1 x + c

When x = 0 and y = 0 , –1 = 0 + c => c = –1

i.e. − 1e y = 4 tan−1 x −1

=> ey =1

1− 4 tan−1 x

y = ln 11 − 4 tan−1 x⎛ ⎝ ⎜

⎞ ⎠ ⎟ = − ln 1 − 4 tan−1 x( )


page 35

Exercise 10 Find the particular solutions of :-

1. (1 − cos2x) dydx

= 2sin2x when x = π4

and y = 1.

2. (1 + x2) dydx

= 1+ y2 when x = 0 and y = 1.

3.dydx

= x(y − 2) when x = 0 and y = 5.

4. dydx

= (1− y2) when x = π6

and y = 0.

5. dydx

= y cos x when x = 0 and y = 1.

6. dydx

= tan x tan y when x = π4 and y = π

4



Page 206 Exercise 4.4:2 Questions 2, 3, 4.



Problems involving First Order Differential Equations , solved by the Variable Separablemethod.

Examples (1) At t hours after noon, there are x bacteria in a culture.

The growth of the bacteria is modelled by the differential equation

dxdt

=k(n − x)x

n where n and k are positive constants.

(a) Show that the general solution of this equation is x =n

1 + Ae−kt where A is an arbitrary constant.

(b) Given that there are 200 bacteria at noon, find the relation between n and A.(c) Given also that x tends to 600 as t tends to ∞ , find n and A.

cont’d .....


page 36

Solution (a) dxdt

=k(n − x)x

n

=> nx(n − x)

dx = kdt

nx(n − x)

dx∫ = kdt∫ Let nx(n − x)

=Ax

+B

n − x

n = A(n – x) + Bx

Put x = 0: n = An => A = 1 Put x = n: n = Bn => B = 1

Integral becomes => n

x(n − x)dx∫ = 1

xdx +

1n − x

dx∫∫ = kdt∫

ln x – ln (n – x) = kt + c

i.e. ln xn − x

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = kt + c

i.e. xn − x

= ekt+ c = ektec

=> xn − x

= Aekt

=> n − xx

= Ae−kt

=> xAe−kt = n − x

=> xAe−kt + x = n

=> x(Ae−kt + 1) = n

=> x =n

(Ae−kt + 1)

(b) At t = 0, x = 200 i.e. 200 =n

A + 1 => n = 200(A + 1)

(c) As x –> 600, t –> ∞ i.e. 600 =n

0 + 1=> n = 600

From n = 200(A + 1), => A = 2

(2) An infectious disease spreads at a rate which is proportional to the product of the number infected and the number uninfected.

Initially, one half of the population is infected and the rate of spread is such that, were it to remain constant, the whole population would become infected in 24 days.Calculate the proportion of the population which is infected after 12 days.

cont’d ......


page 37

Solution Let x be the fraction infected and so 1 – x is the fraction unaffected.dxdt

is the rate at which the the disease spreads.

Hence dxdt

∝ x(1 – x) => dxdt

= kx(1 – x) where k is a constant of variation.

Initially, x = 12

and dxdt

is equal to the constant rate at which the remaining half would become infected in 24 days.

i.e. dxdt

=12

24=

148

Substituting x = 12

and dxdt

= 148

into dxdt

= kx(1 – x)

=> 148

= k. 12

. 12

=> k = 112

i.e. dxdt

=112

x(1− x)

dxx(1− x)

=112

dt => Let 1x(1− x)

=Ax

+B

1 − x

1 = A(1 – x) + BxPut x = 0: 1 = A => A = 1Put x = 1: 1 = B => B = 1

Integral becomes

1x(1 − x)

dx =1x∫∫ dx +

11− x

dx∫ = 112

dt∫

lnx – ln(1 – x) = 112

t + c

When t = 0, x = 12

, => ln( 12

)– ln( 12

) = 0 + c => c = 0

∴ lnx – ln(1 – x) = 112

t

ln x1− x⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 1

12t => x

1 − x = e

t1 2

When t = 12, x1 − x

= e => x =e

1 + e≈ 0 ⋅73

Hence 73% of the population is infected after 12 days.

Note These situations arise where the rate of change of a quantity Q is proportional to Q. The rate of change is often taken with respect to time.

i.e. dQdt

varies as Q .

(i) If Q is increasing with respect to time, then dQdt

= kQ

(ii) If Q is decreasing with respect to time, then dQdt

= – kQ


page 38

Exercise 11

1. The rate at which a radioactive material decays at any instant is proportional to the mass remaining at that instant. Given that there are x grams present after t days, the differential equation in x is modelled by :- dx

dt= –kx.

The half–life if a radioactive material is roughly 25 days.(This means that a given mass will decay to half its mass in 25 days)

Find the time taken for 100 g of the material to decay to 20 g.

2. A pan of milk is heated in a kitchen where the temperature is 15°. When the milk reaches boiling point it is left to cool and after t minutesthe temperature of the milk is φ degrees.

The rate of cooling is propertional to φ – 15.

The differential equation modelling this situation is dφdt

= –k(φ – 15).

(a) Find the general solution of the differential equation.(b) After 10 minutes the temperature of the milk was 50°.

(i) Calculate the temperature of the milk 5 minutes after it boiled.(ii) The milk is required when the temperature is 45°.

Calculate how long this takes after the milk boiled.

3. The population of a small town was 468 in 1980 and 534 in 1990.Assuming that the rate of increase of the population p, is proportional to p,

(a) write down a differential equation representing the above information and find the solution to the equation.

(b) calculate the population in 2000.

4. The surface area of a pool is 10000 m2 and is partially covered with weeds. At any instant, the weeds are increasing in area at a rate proportional to its area at that instant.(a) If the area of the weed is x m2 formed in t days, form a differential equation.(b) Initially, the area is 100 m2 and after 7 days, the area is 1000 m2. Show that :-

lnx

100⎛ ⎝ ⎜

⎞ ⎠ ⎟ =

17

t ln(10)

(c) Find the area of the pool not covered by weeds after 10·5 days.

(d) Find the time t, when the weeds cover half the surface of the pool.

5. A man is given a drug which causes an initial level of 2 mg of the drug per litre of his blood. After t hours there are x mg of the drug per litre and it is known that the rate of decrease of x is proportional to x. (i.e. dx

dt= kx ). After 1 hour, x = 1·6.

(a) Calculate the value of x after 3 hours.(b) Calculate the time after which x = 0·5.

cont’d ......


page 39





Page 515 Exercise 20D Questions 41, 42.

6. A container is shaped so that when the depth of water is x cm, the volume of water in the container is (x2 + 3x) cm3. Water is poured into the container so that, when the depth of water is x cm, the rate of increase is (x2 + 4) cm3/sec.

(a) Show that the differential equation for this situation can be modelled by:-dxdt

=x 2 + 42x + 3 , where t is the time in seconds.

(b) Solve the differential equation to obtain t in terms of x, given that initially the container is empty.

7. The motion of a particle is such that its speed at time t is given by:-dvdt

=12

v − v2( )When t = 0, v = 0·2.Express v in terms of t.

8. Heat is supplied to an electric kettle at a constant rate of 2000 watts, but heat is lost to the surroundings at a rate of 20 watts for every °C difference between the kettle and that of the surroundings. One watt causes the temperature of the kettle to rise at a rate of 0·02°C per minute.If the temperature of the surroundings is 15°C and θ°C is the temperature of the the kettle after t minutes, the differential equation modelling this is :-

dθdt

= 40 −25θ −15( )

How long will it take for the temperature to rise from 15°C to 100°C?


page 40

Homogeneous First Order Differential Equations.

A homogeneous differential equation is one in which the sum of the powers of x and y in each term is the same.To solve these equations we use a substitution y = vx and then treat the new differential equation as a variable separable.

Substituting y = vx gives dydx

= v + xdvdx

by the product rule.

Examples

1. Solve dydx

=2x − yx + y

by using the substitution y = vx

Let y = vx and so dydx

= v + xdvdx

.

Substituting, the equation becomes

v + x dvdx

=2x − vxx + vx

=2 − v1+ v

x dvdx

=2 − v1 + v

− v

= 2 − v − v(1+ v)1 + v

= 2 − 2v − v2

1+ v

i.e. => 1+ v2 − 2v − v2 dv =

1x

dx

=> 1+ v2 − 2v − v2 dv∫ =

1x

dx∫ – 1

2 ln(2 – 2v – v2) = lnx + lnc

i.e. ln(2 − 2v − v2)− 1

2 = lncx

∴ 1

(2 − 2v − v2 )= cx

2 − 2v − v2 =Ax 2 (where A = c-2)

2 − 2 yx−

y2

x2 = Ax2

=> 2x2 – 2xy – y2 = A

This section is optional


page 41

2. Find the general solution of x tan yx

⎛ ⎝ ⎜

⎞ ⎠ ⎟

dydx

+ x − ytan yx

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 0

Let y = vx and so dydx

= v + xdvdx

Substituting, the equation becomes

x tan v v + xdvdx

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + x − vx tan v = 0

v + xdvdx

⎛ ⎝ ⎜

⎞ ⎠ ⎟ +

1tan v

⎛ ⎝ ⎜

⎞ ⎠ ⎟ − v = 0

xdvdx

= −1

tan v

tan vdv = −1x∫ dx∫

sin vcos v

dv = −1x∫ dx∫

− ln(cosv) = − ln x + c (or ln C)

ln secv( ) = ln Cx

⎛ ⎝ ⎜

⎞ ⎠ ⎟ => secv = C

x x secv = C

x sec yx

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = C

Exercise 12

Solve the following homogeneous differential equations :-

(a) xy dydx

= x2 + y2 (b) xy3 dydx

= x4 + y4

(c) (x2 + y2 ) dydx

= xy (d) (x − y) dydx

= y − 4x

(e) (x2 – xy) dydx

+ 2y2 = 0 (f) x dydx

− tan yx

⎛ ⎝ ⎜

⎞ ⎠ ⎟

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = y



Page 515 Exercise 20D Questions 31, 33, 36, 37, 38.


page 42

Reduction Formulae – An application of integration by parts.

It is sometimes possible to express an integral in terms of a simpler one of the same form. This is usually done by integrating the function by parts.

Examples(1) If In

= sinn xdx0

π2∫ express In in terms of In−2

and

hence evaluate I6 and I7.

Firstly express sinn x as sinn−1 x.sin x i.e. a product.Now use integration by parts.In = sinn xdx

0

π2∫

= sinn−1 x.sin xdx0

π2∫ Let u = sinn−1 x and dv = sinxdx

du = (n −1)sinn−2 x cosxdx and v = – cosx

In = − sinn -1 cos x[ ]0

π2 + (n – 1) sinn−2 x.cos2 xdx

0

π2∫

= 0 + (n – 1) sinn−2 x.(1− sin2 x)dx0

π2∫

= (n – 1) sinn−2 xdx0

π2∫ – (n – 1) sinn xdx

0

π2∫

In = (n – 1) In−2 – (n – 1) In

=> In + (n – 1) In = (n – 1) In−2

=> n In = (n – 1) In−2

=> In =(n −1)

nIn−2

This is a Reduction formula since it reduces finding In to finding In−2.

Now for I6 = 56

I4 and I4 = 34

I2 and I2 = 12

I0 .

and so I6 = 56

. 34

I2 = 56

. 34

. 12

I0 .

But I0 = sin00

π2∫ x dx = 1

0

π2∫ dx = x[ ]0

π2 = π

2 => I6 = 5

6. 34

. 12

. π2

= 5π32

Similarly I7 = 67

. 45

. 23

I1

But I1 = sin10

π2∫ x dx = −cos x[ ]0

π2 = −cos π

2+ cos0 = 1

=> I7 = 67

. 45

. 23

.1 = 1635

This section is optional


page 43

Exercise 13

1. If Ι n = x n∫ e x dx , prove that Ι n = xnex − nΙ n−1.

Show that I3 = ex x3 − 3x2 + 6x − 6( ) + c .

2. If In = xn cos0

π2∫ x dx , prove that In =

π2

⎛ ⎝ ⎜

⎞ ⎠ ⎟

n

− n n −1( )In−2.

Show that I6 =π2

⎛ ⎝ ⎜

⎞ ⎠ ⎟

6

− 30 π2

⎛ ⎝ ⎜

⎞ ⎠ ⎟

4

+ 360 π2

⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

− 720 .

(2) If In = x ln x( )n1

e

∫ dx , where n is a positive integer,

prove that In =12

e2 −12

nIn−1 and evaluate I3.

In = x ln x( )n1

e

∫ dx Let u = (ln x)n and dv = xdx

du = n(ln x)n−1. 1x

dx and v = 12

x2

In = 12

x2 ln x( )n⎡ ⎣ ⎢

⎤ ⎦ ⎥ 1

e

– n2

x1

e

∫ ln x( )n−1dx

= 12

e2 lne( )n −12

ln1( )n −n2

In−1

= 12

e2 −n2

In−1

This is a Reduction formula since it reduces finding In to finding In−1.

Now for I3 = 12

e2 −32

I2 = 12

e2 −32

( 12

e2 −22

I1)

= 12

e2 −34

e2 +32

I1

= − 14

e2 +32

(12

e2 −12

I0 )

= 12

e2 −34

I0

but I0 = x ln x( )01

e

∫ dx = x1

e

∫ dx =12

x2⎡ ⎣ ⎢

⎤ ⎦ ⎥ 1

e

=12

e2 −12

=> I3 = 12

e2 −34

12

e2 −12

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 1

8e2 +

38


page 44

Answers

Exercise 1

1. sin–1(x7 ) + c 2.

17

tan−1 x7

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + c 3. sin−1 x

3⎛ ⎝ ⎜

⎞ ⎠ ⎟ + c

4. 110

tan−1 x10

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + c 5.

15

sin−1 5x6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + c 6.

130

tan−1 5x6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + c

7.15

tan −1 2x5

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + c 8. sin−1 x

2⎛ ⎝ ⎜

⎞ ⎠ ⎟ + c 9.

π6

10. π4

11. π 12. π12

Exercise 2

1. 3ln(x + 2) – 2ln(x +4) + c 2. x – ln(x + 2) + ln(x –2) + c

3. 3ln(x – 1) – 3ln(x – 2) + ln(x + 1) + c 4. 12 x2 + 2x + 2ln(x – 3) +3ln(x + 1) + c

Exercise 3

1. lnx + 2ln(x + 1) + 3x + 1

+ c 2. 2ln(x + 2) – ln(x –1) – 3x −1

+ c

3. ln(x + 2) – ln(2x – 1) – 52x −1

+ c 4. 13

ln(x – 2) – 13

ln(x + 1) – 4x − 2

+ c

Exercise 4

1. 2ln(x – 1) – ln( x2 + 1) + tan −1 x + c 2. 152

ln( x2 + 1) – 12ln(x + 6) + 2tan −1 x + c

3. 14

ln(x + 1) + 14

ln(x – 1) – 14

ln( x2 + 1) + c

4. 110

ln( x2 + 4 ) – 15

ln(x + 1) + 25

tan −1 x2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + c

Exercise 5

1. –xcosx + sinx + c 2. −x3

cos3x +19

sin3x + c 3. 23

x32 ln x − 4

9x

32 + c

4. −1

2x 2 ln x − 14x2 + c 5. xex − ex + c 6.

14

x sin 4x +1

16cos4x + c

Exercise 6

1. x2 sin x + 2x cos x – 2sin x + c 2. −13

x2 cos3x +29

x sin3x +227

cos3x + c

3. 12

x2e2x −12

xe2x +14

e2x + c 4. 12

x2 sin2x +12

x cos2x − 14

sin 2x + c

5. −x2e− x − 2xe −x − 2e−x + c 6. x3ex −3x2e x + 6xex − 6ex + c


page 45

Exercise 7

1. x tan −1 x −12

ln(1+ x2 ) + c 2. x sin−13x −13

(1 − 9x 2) + c

3. x tan −1 2x −14

ln(1 + 4x2 ) + c 4. x sin−1 12

x + (4 − x2) + c

5. x cos−1 x − (1− x2 ) + c 6. x tan −1 12

x − ln(4 + x2 ) + c

7. x ln 2x − x + c 8. x(ln x)2 − 2x ln x + 2x + c

Exercise 8

1.15

ex sin 2x −25

ex cos2x + c 2. −12

e−x cos x −12

e−x sin x + c

3.3

13e−2x sin3x −

213

e−2x cos3x + c 4. ex cos2 x +15

ex sin 2x −25

e x cos2x + c

Exercise 9

1. y =ex+ c

x + 1 =Ae x

x + 1 2. y = 1− 2x2 + C 3. y =

−1ex + c

4.ey

y= cx2 or y =

ey

cx2 5. cos xcos y

= c 6. y =(c + 1)x + 1

(x + 1)

Exercise 10

1. y = ln(1 – cos2x) + 1 2. tan −1 y = tan−1 x +π4

3. y = 3e12 x2

+ 2

4. y = sin x −π6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ 5. y = esin x 6. sin y =

12cos x

Exercise 11

1. 58 days 2. (a) θ = Ae−kt + 15, (b) (i) 69·5° (ii) 11·7 minutes

3. (a) p = 468e0⋅0132 t (b) 609 4. (a) dxdt

= kx (b) Proof (c) 6837 m2 (d) 12 days

5. (a) 1·024 mg (b) 6·22 hours 6. Proof and t = ln x2 + 44

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ +

32

tan−1 x2

⎛ ⎝ ⎜

⎞ ⎠ ⎟

7. v =1

1 + 4e− 12 t 8. 4·74 mins

Exercise 12

(a) y2 = 2x 2(ln x + c) (b) y4 = 4x4(ln x + c) (c) ln y =x 2

2y2 + c

(d) yx− 2

⎛ ⎝ ⎜

⎞ ⎠ ⎟

yx

+ 2⎛ ⎝ ⎜

⎞ ⎠ ⎟

3

= cx 4 (e) x + y( ) 2

x2y= c

Exercise 13

1. Proof, Proof 2. Proof, Proof


Page 46

Outcome 3 – Complex Numbers

contd....

Introduction

We are in a position where we can solve a range of quadratic equations, (a) by factorisation (b) by quadratic formula.

When the roots are real, (b2 – 4ac > 0) we obtain one or more solutions.Some equations however have no real roots (b2 – 4ac < 0).We must therefore extend our notation of number.eg. where x2 – 49 = 0, x2 = 49 => x = –7 or 7.

but x2 + 49 = 0 has no real roots since x2 = – 49 has no solution in terms of rational or irrational numbers.

We introduce therefore the concept of an IMAGINARY NUMBER :-

Suppose there exists a non-zero number i such that i2 = – 1.Then in the equation x2 = – 49, we can write x2 = 49(–1)

=> x2 = 49i2 => x = –7i or 7i.A number of the form bi , where b is a real number, is called an IMAGINARY NUMBER and x + yi where x and y are real is called a COMPLEX NUMBER.

z is commonly used to denote a complex number and so if z = x + yi , then the real part of the complex number z is written as Re(z) = x and the imaginary part of the complex number is written as Im(z) = y .

This can now be used to solve any quadratic equation ax2 + bx + c = 0, which has no real roots.

Examples

1. Find the roots of the equation x2 – 2x + 5 = 0

x =−b ± b2 − 4ac

2a

=2 ± 4 − 20

2

=2 ± −16

2

=2 ± 16( ) −1( )

2

=2 ± 16i2

2 =

2 ± 4i2

= 1 ± 2i

The roots are 1 + 2i and 1 – 2i

Note:- that if one root of a quadratic equation is x = a + bi then the other root is x = a – bi.These are called the :- “complex conjugates” of each other.


Page 47

Exercise 11. Accepting that i2 = –1, what would you expect to be the values of the following ?

(a) (2i)2 (b) (3i)2 (c) (4i)2 (d) (–2i)2 (e) (–3i)2

2. What would you expect the roots of the following equations to be in terms of i ?(a) x2 + 4 = 0 (b) x2 + 9 = 0 (c) x2 + 3 = 0

3. Assuming that the normal formula for solving a quadratic equation can be used, solve the following quadratic equations:-(a) x2 – 2x + 2 = 0 (b) x2 – 4x + 5 = 0

(c) x2 – 4x + 13 = 0 (d) x2 + 2x + 2 = 0

(e) 4x2 – 4x + 5 = 0 (f) x2 + 6x +10 = 0

(g) 2x2 – 2x + 1= 0 (h) 9x2 – 6x + 2 = 0

4. In Question 3 we assumed that the formula would still apply. Check, by substitution, that the solutions you found do, in fact, satisfy the appropriate equations.

5. Look back at your solutions to Question 3. For each equation, evaluate:-

(a) the sum of the roots (b) the product of the roots.

Describe what you find .

6. Solve the following equations as fully as you can:-(a) x3 – 1 = 0 (b) x4 – 1 = 0(c) x3 – x2 – x – 2 = 0 (d) (x2 + 4)(x2 + 9) = 0

To check the roots, substitute into the equation x2 – 2x + 5 = 0

x = 1 + 2i => (1 + 2i)2 – 2(1 + 2i) + 5 = 1 + 4i + 4i2 – 2 – 4i + 5 = 0.

x = 1 – 2i => (1 – 2i)2 – 2(1 – 2i) + 5 = 1 – 4i + 4i2 – 2 + 4i + 5 = 0.

Note also that the :-

(i) sum of the roots = 1 + 2i +(1 – 2i) = 2 (which is real)i.e. the negative of the x coefficient, b of the quadratic.

(ii) product of the roots = (1 + 2i)(1 – 2i) = 1 – 4i2 = 5 (real)i.e. the constant c of the quadratic equation.

(iii) 1 + 2i and 1 – 2i are called complex conjugates.If z = x + yi, then the complex conjugate of z, is denoted by => z = x – yi. (reads as “z bar”).


Page 48

From Exercise 1 , you should have the following results:

Addition of Complex Numbers:- (a + bi) + (c + di) = (a + c) + (b + d)i (a complex number)

Subtraction of Complex Numbers must therefore be”- (a + bi) – (c + di) = (a – c) + (b – d)i (a complex number)

Multiplication of Complex Numbers :- (a + bi)(c + di) = ac + adi + bci + bdi2 = ac + adi + bci – bd

= (ac – bd) + (ad + bc)i (a complex number)

Division of complex numbers makes use of the fact that (a + bi)(a – bi) = a2 – b2 (a real number). (a + bi) and (a – bi) are called complex conjugates of each other.

If z = a + bi then z = a – bi and vice versa. We make use of this in division of complex numbers.

Examples1. (3 + i) + (1 + 2i) = (3 + 1) + (1 + 2)i = 4 + 3i

2. (2 – 3i) – (1 + 2i) = (2 – 1) + (–3 –2)i = 1 – 5i

3. (2 – i)(3 + 2i) = 6 + i – 2i2 = 6 + i + 2 = 8 + i

4. 5 + 3i1 −3i

= 5 + 3i( ) 1 + 3i( )1− 3i( ) 1+ 3i( ) = 5 + 18i − 9

1+ 9

= − 4 + 18i10

= − 25

+95

i

contd....



Page 344 Exercise 6.5:1 Questions 1 – 5


Page 463 Exercise 18D Questions 6, 8, 10


Page 49

Exercise 21. Express each of the following in the form x + yi .

(a) (3 + 7i) + (2 + i) (b) (9 – 2i) – (3 + i)

(c) (–2 + i) + (7 – 4i) (d) (3 + 2i) + (3 – 2i)

(e) (–2 + i) – (–2 – i) (f) (a + bi) + (a – bi)

(g) (a + bi) – (a – bi)

2. Using the fact that i2 = –1, express i3, i4, i5, i6, i7, i8, i9 and i10 in their simplest form.

3. Simplify (a) 2i x 4i (b) –2i2

(c) i(3 + 2i) (d) –i(1– 4i)

(e) (2 + i)(3 + i) (f) (6 – 5i)(2 + 3i)

(g) (2 + 3i)(2 – 3i) (h) (a + bi)(a – bi)

(i) (a + bi)(c + di) (j) (a + bi)(c – di)

contd....

5. (1 – i)3 = 1 – 3i + 3i2 – i3 (by Pascal’s Triangle or the Binomial Theorem)

= 1 – 3i – 3 + i = – 2 – 2i

6. √(15 + 8i)

Let √(15 + 8i) = x + yi => 15 + 8i = (x + yi)2

= x2 – y2 + 2xyiEquating real and imaginary parts:–

=> x2 – y2 = 15 [1] and 2xy = 8 [2]From [2] => y = 4

x

Substituting in [1] => x2 – 16x2 = 15

=> x4 – 15x2 – 16 = 0=> (x2 – 16)(x2 + 1) = 0=> x2 – 16 = 0 or x2 + 1 = 0But x is real, and so x2 + 1 = 0 gives no suitable values.=> x = – 4 or 4 and y = –1 or 1therefore, √(15 + 8i) = – 4 – i or 4 + i

= ±(4 + i)


Page 50

...cont’d 3(k) (1 + i)3 (l) (1 + i)4

(m) (1 + i)4(1 – i)5 (n) (3 + i)2 + (3 – i)2

(o) (cost + isint)2

(p) (cosA + isinA)(cosB + isinB)

4. Simplify

(a) (2 + i)(2 – i) (b) (1 – 2i)(1 + 2i) (c) (5 – i)(5+ i)

5. Simplify and express in the form x + yi

(a) 4 + ii

(b) 12 + i

(c) 2 − i1 − 2i

(d) 5 + i

5 − i(e) a + bi

a − bi (f) a + bi

c + di

(g) 10 + 5i2 − i

(h) 1cos A + isin A

(i) cos A + isin Acos A − isin A

6. Simplify (x – 1 – i)(x – 1 + i).

Hence state an equation which has (1 + i) and (1 – i) as its roots.

7. Find the square root of each of these :-

(a) 3 – 4i (b) 21 – 20i (c) 2i (d) – 24 + 10i





Page 463 Exercise 18D Questions 1, 2, 3, 4, 5, 7


Page 51

The Geometrical Representation of Complex Numbers

The Argand Diagram

Complex numbers can be represented geometrically using the x and y axes as the Real (Re) and the Imaginary (Im) axes respectively.

A diagram showing complex numbers is called an ARGAND DIAGRAM.

If z = x + yi, z can be represented geometrically as shown below:-

Examples

1. The points A(1,0), B(0,1), C(–1,0) and D(0,–1) are shown below:-

Corresponding to:-OA, there is a complex number :- 1 + 0i i.e. 1OB, there is a complex number :- 0 + 1i i.e. iOC, there is a complex number :- –1 + 0i i.e. –1OD, there is a complex number :- 0 + (–1)i i.e. – i

2. z = 5 + 3i can be representedas shown opposite.

y(Im)

x(Re)

y

x

z = x + yi

z

x(Re)

y(Im)

B(0,1)

A(1,0)C(–1,0)

D(0,–1)

y(Im)

x(Re)

3

5

(5,3)z = 5 + 3i


Page 52

The Modulus and Argument of a Complex number.

On occasions it is convenient to express complex numbers in another form, particularly when we wish to find or illustrate the product, quotient or powers of complex numbers.

e.g. z = x + yi

where x = r cosθ y = r sinθ

r2 = x2 + y2

z = x + yi = r cosθ + ri sinθ = r(cosθ + i sinθ).

r is known as the Modulus (magnitude) of z and θ the Argument (amplitude) of z and are written as Mod z or |z| and arg(z) respectfully.

ie. |z| = x2 + y2 and arg(z) = θ = tan −1 yx

⎛ ⎝ ⎜

⎞ ⎠ ⎟ , but only for –π < θ ≤ π.

3. If zA = xA + yAi and zB = xB + yB i, then :-=> zA + zB = (xA + xB) + (yA + yB)i.

4. If zA = xA + yAi and zB = xB + yB i, then :-=> zA – zB = (xA – xB) + (yA – yB)i.

zB zA+zB

zB

zA

y(Im)

x(Re)

zB

zA– zB–zB

zA

y(Im)

x(Re)

y(Im)

x(Re)

y

x

(x,y)

r

θ


Page 53

Examples

Find the modulus and arguments of

1. z = 1 + i|z| = √(12 + 12) = √2

arg(z) = tan–1(1) = π4

2. z = –1 + i |z| = √((–1)2 + 12) = √2

arg(z) = tan −1 1−1

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = π – π

4

= 3π4

3. z = –√3 – i

|z| = √((–√3)2 + (–1)2) = 2

arg(z) = tan −1 −1− 3

⎛ ⎝ ⎜

⎞ ⎠ ⎟

= tan −1 13

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = –

5π6

(since –π < θ < π)

4. z = 1 – i|z| = √(12 + (–1)2) = √2

arg(z) = tan −1 −11

⎛ ⎝ ⎜

⎞ ⎠ ⎟

= tan–1(–1) = – π4

(since –π < θ ≤ π)

Exercise 3

1. Find the modulus and argument of :-

(a) 1 + √3i (b) 2 – 2i (c) –√2 – √2i

(d) 2i (e) 3 (f) –√3 + i

(g) –3i (h) –5 (i) –3 – 3i

contd....

(1,1)y(Im)

x(Re)θ

(–1,1)y(Im)

x(Re)

θ

(–√3,–1)

y(Im)

x(Re)θ

(1,–1)

y(Im)

x(Re)θ


Page 54

3. Given that z1 = – 3 + 3√3i and z2 = √3 + i,

(a) (i) find |z1|, |z2|, z1

z2

.

(ii) find arg(z1), arg(z2), arg z1

z2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ .

(b) Repeat for z1 = 3i and z2 = √2 – √2i . What do you notice?

2. Given that z1 = – 3 + 3√3i and z2 = √3 + i,

(a) (i) find |z1|, |z2|, |z1z2| (ii) find arg(z1), arg(z2), arg(z1z2)

(b) Repeat for z1 = 3i and z2 = √2 – √2i . What do you notice?

6. Given that z = 1 + i

(a) (i) find z z (ii) find |z z | and arg(z z )

(b) Repeat for z = –√3 – i. What do you notice?

7. Summarise your findings for the results in this exercise.



Page 353 Exercise 6.5:4 Questions 1 – 5Page 354 Exercise 6.5:5 Questions 1, 2.Page 356 Exercise 6.5:6 Questions 1, 2, 3, 5, 7, 8.


Page 467 Exercise 18F Questions 1 – 7


(a) (i) find |z|, |iz| (ii) find arg(z), arg(iz)



(a) (i) find |z|, | z | (ii) find arg(z), arg( z )



Page 55

Exercise 41. (i) For each of the following, find |z1|, |z2| and |z1z2|

(ii) Find also arg(z1), arg(z2) and arg(z1z2)(iii) Hence, find the product of z1z2 in its simplest form each time.

(a) z1 = 2[cos π4

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin π

4⎛ ⎝ ⎜

⎞ ⎠ ⎟ ] z2 = 3[cos π

4⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin π

4⎛ ⎝ ⎜

⎞ ⎠ ⎟ ]

(b) z1 = √2[cos π6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin π

6⎛ ⎝ ⎜

⎞ ⎠ ⎟ ] z2 = 2[cos π

3⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin π

3⎛ ⎝ ⎜

⎞ ⎠ ⎟ ]

(c) z1 = r1[cosθ1 + isinθ1] z2 = r2[cosθ2 + isin θ2]This is the general result.

Expressing z = x + yi in the form r[cosθ + isinθ]

We know z = x + yi can be express in the form z = r[cosθ + isinθ] where r = |z| and θ = arg(z). This known as the Polar Form or the Modulus and Argument form.

Examples We saw earlier :-

(i) z = 1 + i can be written z = √2[cos π4

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin π

4⎛ ⎝ ⎜

⎞ ⎠ ⎟ ]

(ii) z = –1 + i can be written as z = √2[cos 3π4

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin 3π

4⎛ ⎝ ⎜

⎞ ⎠ ⎟ ]

(iii) z = –√3 – i can be written as z = 2[cos −5π6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin −5π

6⎛ ⎝ ⎜

⎞ ⎠ ⎟ ]

(iv) z = 1 – i can be written as z = √2[cos −π4

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin −π

4⎛ ⎝ ⎜

⎞ ⎠ ⎟ ]

2. Write down, using the results from above, |z1z2|, arg(z1z2) and z1z2 :-

(a) z1 = 3[cos 3π4

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin 3π

4⎛ ⎝ ⎜

⎞ ⎠ ⎟ ] z2 = 2[cos 5π

4⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin 5π

4⎛ ⎝ ⎜

⎞ ⎠ ⎟ ]

(b) z1 = 12

[cos 2π15

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin 2π

15⎛ ⎝ ⎜

⎞ ⎠ ⎟ ] z2 = 2[cos 13π

15⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin 13π

15⎛ ⎝ ⎜

⎞ ⎠ ⎟ ]

(c) z1 = 10[cos 5π9

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin 5π

9⎛ ⎝ ⎜

⎞ ⎠ ⎟ ] z2 = 10[cos 13π

9⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin 13π

9⎛ ⎝ ⎜

⎞ ⎠ ⎟ ]

3. (i) Find |z1|, |z2| and z1

z2

each time.

(ii) Find arg(z1), arg(z2) and arg z1

z2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ .

(iii) Find the quotient of z1

z2 in its simplest form.

(a) z1 = 2[cos π2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin π

2⎛ ⎝ ⎜

⎞ ⎠ ⎟ ] z2 = 3[cos π

4⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin π

4⎛ ⎝ ⎜

⎞ ⎠ ⎟ ]

(b) z1 = √2[cos π6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin π

6⎛ ⎝ ⎜

⎞ ⎠ ⎟ ] z2 = 2[cos π

3⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin π

3⎛ ⎝ ⎜

⎞ ⎠ ⎟ ]

(c) z1 = r1[cosθ1 + isinθ1] z2 = r2[cosθ2 + isin θ2]This is the general result.


Page 56

5. Express each of the following in its simplest form, using earlier results.

(a) [cos π3

⎛ ⎝ ⎜

⎞ ⎠ ⎟ +isin π

3⎛ ⎝ ⎜

⎞ ⎠ ⎟ ] x [cos π

6⎛ ⎝ ⎜

⎞ ⎠ ⎟ +isin π

6⎛ ⎝ ⎜

⎞ ⎠ ⎟ ] x [cos π

4⎛ ⎝ ⎜

⎞ ⎠ ⎟ +isin π

4⎛ ⎝ ⎜

⎞ ⎠ ⎟ ]

(b) 4[cos π3

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin π

3⎛ ⎝ ⎜

⎞ ⎠ ⎟ ] x 2[cos π

6⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin π

6⎛ ⎝ ⎜

⎞ ⎠ ⎟ ]

(c) 20[cos π4

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin π

4⎛ ⎝ ⎜

⎞ ⎠ ⎟ ] ÷ [cos π

6⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin π

6⎛ ⎝ ⎜

⎞ ⎠ ⎟ ]

(d) [cos 2π3

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin 2π

3⎛ ⎝ ⎜

⎞ ⎠ ⎟ ] ÷ 3[cos π

2⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin π

2⎛ ⎝ ⎜

⎞ ⎠ ⎟ ]

6. Summarise your results from this exercise.






4. Write down, using the results from earlier, z1

z2

, arg z1

z2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ and z1

z2

each time.

(a) z1 = 3[cos 3π4

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin 3π

4⎛ ⎝ ⎜

⎞ ⎠ ⎟ ] z2 = 2[cos 5π

4⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin 5π

4⎛ ⎝ ⎜

⎞ ⎠ ⎟ ]

(b) z1 = 2[cos 13π15

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin 13π

15⎛ ⎝ ⎜

⎞ ⎠ ⎟ ] z2 = 1

2[cos 2π

15⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin 2π

15⎛ ⎝ ⎜

⎞ ⎠ ⎟ ]

(c) z1 = 10[cos 13π9

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin 13π

9⎛ ⎝ ⎜

⎞ ⎠ ⎟ ] z2 = 10[cos 5π

9⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin 5π

9⎛ ⎝ ⎜

⎞ ⎠ ⎟ ]


Page 57

De Moivre’s Theorem.

(cosθ + isinθ)n = cos(nθ) + isin(nθ)

Exercise 5

1. Using the results of Exercise 4, write down the value of :-

[cosπ6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin(

π6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ )]n for n = 0, 1, 2, 3, 4, 5, 6.

2. Repeat question 1 for :-

[cosπ4

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin

π4

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ]n for n = 0, 1, 2, 3, 4.

For positive integers n :- (cosθ + isinθ)n = cos(nθ) + isin(nθ)

This is known as De Moivre’s Theorem.

3. Does this result still hold when n is negative ?

Consider [cosπ3

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin

π3

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ]–2

Use De Moivre’s Theorem to prove that :-

[cosπ3

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin

π3

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ]–2 = cos

2π3

⎛ ⎝ ⎜

⎞ ⎠ ⎟ – isin

2π3

⎛ ⎝ ⎜

⎞ ⎠ ⎟

4. Simplify :-

(a) [cos5π24

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin

5π24

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ]4 (b) [cos

2π5

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin

2π5

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ]5

(c)cos2θ + i sin2θ( )5

cos3θ + i sin3θ( )3 (d) (cosθ + isinθ)8(cosθ – isinθ)4

5. Since (cosθ + isinθ)3 = cos3θ + isin3θ, deduce expressions forcos3θ and sin3θ in terms of cosθ and sinθ.

6. If z = cosθ + isinθ, show that :-

(i) z + 1z = 2cosθ and (ii) z2 +

1z2 = 2cos2θ

7. If z = cosθ + isinθ, then :- zn = (cosθ + isinθ )n = cos(nθ) + isin(nθ) and

z−n = (cosθ + isinθ )−n = cos(–nθ) + isin(–nθ).

Show that (i) cos(nθ ) = 12

zn + z −n( ) and (ii) sin(nθ) = 12i

z n − z−n( )

(note)


Page 58

SUMMARY of Exercises 3, 4 and 5

|z1z2| = |z1| x |z2| arg(z1z2) = arg(z1) + arg(z2)

z1

z2 =

z1

z2 arg z1

z2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ = arg(z1) – arg(z2)

|iz| = |z| arg(iz) = arg(z) + π2

| z | = |z| arg( z ) = – arg(z)

|z z | = |z|2 arg(z z ) = 0

De Moivre’s Theorem


proofIf z = cosθ + i sinθ and |z| = 1, arg(z) = θ Then z2 = (cosθ + i sinθ)2

|z2| = |z| x |z| = 1, arg(z2) = arg[(z) x (z)] = arg(z) + arg(z) = θ + θ = 2θ

and so z2 = (cosθ + isinθ)2 = cos2θ + isin2θ Similarly for z3

If z = cosθ + isinθ and |z| = 1, arg(z) = θ Then z3 = (cosθ + isinθ)3

|z3| = |z2| x |z| = 1, arg(z3) = arg(z2 x z) = arg (z2) + arg(z)

= 2θ + θ = 3θ and so z3 = (cosθ + isinθ)3 = cos3θ + isin3θ For positive integers n,








Page 59

The Fundamental Theorem of Algebra.

All complex polynomial equations have at least one complex root.

This is called the Fundamental Theorem of Algebra.

Consequently, if a polynomial is of degree n, then there are precisely n roots in the set of complex numbers. Some or all may be real, some or all may be complex.

If z = x + yi is a root of a polynomial equation then z = x – yi is also a root.(x – z) and (x – z ) are therefore factors of the polynomial.

More precisely (x – z)(x – z ) = x2 – (z + z )x + z z is a quadratic factor.

Dividing the polynomial by this quadratic factor will reveal the remaining factor(s) which may be real or complex.

Examples

1. Find the roots of the equation z3 – 2z2 – 8z + 21 = 0=> Let f(z) = z3 – 2z2 – 8z + 21.

f(1) ≠ 0, => z – 1 is not a factor.f(–1) ≠ 0, => z + 1 is not a factor.f(3) ≠ 0, => z – 3 is not a factor.f(–3) = 0, => z + 3 is a factor.

Using synthetic division:-–3 1 –2 –8 21

–3 15 –211 –5 7 0

i.e. z3 – 2z2 – 8z + 21 = (z + 3)(z2 – 5z + 7)

z2 – 5z + 7 = 0 => z = 5 ± 25 − 28

2 = 5 ± 3i

2 = 52

±3

2i

The roots are –3, 52

+3

2i and 5

2−

32

i .

2. Verify that z = 1 + i is a root of the equation, z4 – 3z3 + 5z2 – 4z + 2 = 0.Hence find all the roots of the form p + qi.

If z = 1 + i is a root ,then :-(1 + i)4 – 3(1 + i)3 + 5(1 + i)2 – 4(1 + i) + 2 must = 0

Expanding :-

1 + 4i + 6i2 + 4i3 + i4 – 3(1 + 3i + 3i2 + i3) + 5(1 + 2i + i2) – 4(1 + i) + 2 = 1 + 4i – 6 – 4i + 1 – 3 – 9i + 9 + 3i + 5 + 10i – 5 – 4 – 4i + 2

= 0 ∴ z = 1 + i is a root .

cont’d....


Page 60

2. If z = 1 + i is a root then z = 1 – i is also a root.

=> the quadratic factor formed by these roots must be :-[z – (1 + i)][z – (1 – i)] = [z – 1 – i][z – 1 + i] = [(z – 1) – i][(z – 1) + i] = (z – 1)2 – i2 = z2 – 2z + 1 + 1 = z2 – 2z + 2

For the remaining quadratic factor divide z4 – 3z3 + 5z2 – 4z + 2 by z2 – 2z + 2

∴ z4 – 3z3 + 5z2 – 4z + 2 = (z2 – 2z + 2)(z2 – z + 1)

z2 – z + 1 = 0,

=> z = 1 ± 1 − 4

2

= 1 ± 3i

2

= 12

±3

2i

∴ The roots are z = 1 + i, 1 – i, 12

+3

2i and 1

2−

32

i

z2 – z + 1z2 – 2z + 2 z4 – 3z3 + 5z2 – 4z + 2

z4 – 2z3 + 2z2 – z3 + 3z2 – 4z – z3 + 2z2 – 2z

z2 – 2z + 2 z2 – 2z + 2

0

.... cont’d


Page 61

Exercise 6

1. Find all the roots of the equation z3 – 11z + 20 = 0.

2. Verify that z = 1 + i is a root of the equation z4 + 3z2 – 6z + 10 = 0.Hence find all the roots of the equation.

3. Verify that z = –2 + 3i is a root of the equation z4 + 7z2 – 12z + 130 = 0.Hence find all the roots of the equation.

4. Given that 2 – i is root of the equation 3z3 – 10z2 + 7z + 10 = 0,find the other roots.

5. Given that 1 – 2i is root of the equation z3 + z + 10 = 0,find the other roots.

6. Given that 3 + i is root of the equation z3 – 3z2 – 8z + 30 = 0,find the other roots.

7. Show that – 1 + i is root of the equation z4 – 2z3 – z2 + 2z + 10 = 0,find the other roots.



Page 351 Exercise 6.5:3 Questions 8 – 12, 15, 16, 24 – 27.


Page 463 Exercise 18D Questions 6, 10.


Page 62

Geometric Interpretation of Equations and Inequalities in the Complex Plane

The solution set of complex equations and inequalities can be represented by sets of points in an Argand diagram.

e.g. |z| = a, |z – a| = b, |z – 1| = |z – i|, |z – a| > b

Examples

1. Solve :- |z| = 1

If z = x + yi and |z| = 1 then √(x2 + y2) = 1i.e. x2 + y2 = 1All points representing complex numbers z with modulus of 1 lie on the circumference of a circle, centre the origin and radius 1.

2. Solve :- |z – 1| = 2

If z = x + yi then z – 1 = (x – 1) + yi.Therefore |z – 1| = 2 becomes√[(x – 1)2 + y2] = 2.i.e. (x – 1)2 + y2 = 4All points representing complex numbers z,for which |z – 1| = 2 lie on the circumferenceof a circle, centre (1,0) and radius 2.

3. Solve :- |z – 2| = |z + 2i|

If z = x + yi then z – 2 = (x – 2) + yiand z + 2i = x + (y + 2)iTherefore |z – 2| = |z + 2i| becomes :-√[(x – 2)2 + y2] = √[x2 + (y + 2)2]=> (x – 2)2 + y2 = x2 + (y + 2)2

=> x2 – 4x + 4 + y2 = x2 + y2 + 4y + 4=> y = – x

Since 2 is represented by (2,0) and –2i is represented by (0,–2), all points representing complex numbers z for which |z – 2| = |z + 2i| lie on the perpendicular bisector of the line joining (2,0) and (0,–2).

contd....

Im

Re

z1

Im

Re

z – 12

(1,0)

Im

Re2

y = –x

–2i


Page 63

4. |z + 2| > 4

If z = x + yi then

=> z + 2 = (x + 2) + yi.

Therefore |z + 2| > 4 becomes:- √[(x + 2)2 + y2] > 4.=> (x + 2)2 + y2 > 16All points representing complex numbers z for which |z +2| > 4 lie outside the circumference of a circle, centre (–2,0) and radius 4.

Exercise 7

1. Find the set of points z, where |z| = 2.

2. Find the set of points z, where |z – 3| = 5.

3. Find the set of points z, where |z + 3| = |z – 4i|.

4. Find the set of points z, where |z – 2| > 3.

5. Find the set of points z, where |z – (3 + 2i)| = 4.

6. Find the set of points z, where |z – 1| ≤ 4.

7. Find the set of points z, where |z – 2| = |z + 1 – i|.



Page 472 Exercise 18F Questions 1 - 6, 12,13.

Im

Re

z + 24

(–2,0)


Page 64

Answers Exercise 1

1. (a) – 4 (b) – 9 (c) – 16 (d) – 4 (e) – 9

2. (a) ± 2i (b) ± 3i (c) ± √3i

3. (a) 1 ± i (b) 2 ± i (c) 2 ± 3i (d) – 1 ± i (e) 1

2 ± i (f) – 3 ± i (g) 1

2 ± 1

2i (h) 1

3 ± 1

3i

4. Proofs.

5. (a) Sum = 2 (b) Sum = 4 (c) Sum = 4 (d) Sum = – 2Prod = 2 Prod = 5 Prod = 13 Prod = 2

(e) Sum = 1 (f) Sum = – 6 (g) Sum = 1 (h) Sum = 23

Prod = 54

Prod = 10 Prod = 12

Prod = 29

For a quadratic equation in the form ax2 + bx + c = 0,

the sum of the roots = – ba and the product of the roots =

ca

6. (a) x = 1 (b) x = ± 1, ± i (c) x = 2, – 12

± 32

i (d) x = ± 2i, ± 3i

Exercise 2

1. (a) 5 + 8i (b) 6 – 3i (c) 5 – 3i (d) 6(e) 2i (f) 2a (g) 2bi

2. i 3 = – i, i 4 = 1, i 5 = i, i 6 = – 1, i 7 = – i, i 8 = 1, i 9 = i, i 10 = – 1.

3. (a) – 8 (b) 2 (c) – 2 + 3i (d) – 4 – i (e) 5 + 5i (f) 27 + 8i (g) 13 (h) a2 + b2

(i) (ac – bd) + (bc – ad)i (j) (ac + bd) + (bc – ad)i

(k) – 2 + 2i (l) – 4 (m) 16 – 16i (n) 16

(o) cos2t + isin2t (p) cos(A + B) + isin(A + B)

4. (a) 5 (b) 5 (c) 265. (a) 1 – 4i (b) 2

5−

15

i (c) 45

+35

i

(d) 1213

+513

i (e) a2 − b2

a2 + b2 +2ab

a2 + b2 i (f) ac + bdc2 + d2 +

bc − adc2 + d2 i

(g) 3 + 4i (h) cosA – isinA (i) cos2A + isin2A

6. x2 – 2x + 2 and x2 – 2x + 2 = 07. (a) ±(– 2 + i) (b) ±(5 – 2i) (c) ±(1 + i) (d) ±(1 + 5i)


Page 65

Exercise 3

1. (a) |z| = 2, arg(z) = π3

(b) |z| = 2√2, arg(z) = −π4

(c) |z| = 2, arg(z) = −3π4

(d) |z| = 2, arg(z) = π2

(e) |z| = 3, arg(z) = 0 (f) |z| = 2, arg(z) = 5π6

(g) |z| = 3, arg(z) = – π2

(h) |z| = 5, arg(z) = π

(i) |z| =3√2, arg(z) = −3π4

2. (a) (i) |z1| = 6, |z2| = 2, |z1z2| = 12

(ii) arg(z1) = 2π3

, arg(z2) = π6

, arg(z1z2) = 5π6

(b) (i) |z1| = 3, |z2| = 2, |z1z2| = 6

(ii) arg(z1) = π2

, arg(z2) = – π4

, arg(z1i2) = π4

∴ |z1z2| = |z1| x |z2| and arg(z1z2) = arg(z1) + arg(z2)

3. (a) (i) |z1| = 6, |z2| = 2, z1

z2

= 3

(ii) arg(z1) = 2π3

, arg(z2) = π6

, arg z1

z2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ = π

2

(b) (i) |z1| = 3, |z2| = 2, z1

z2

= 32

(ii) arg(z1) = π2

, arg(z2) = – π4

, arg z1

z2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ = 3π

4

∴ z1

z2

= z1

z2

and arg z1

z2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ = arg(z1) – arg(z2)

4. (a) (i) |z| = √2, |iz| =√2(ii) arg(z) = π

4 , arg(iz) = 3π

4

(b) (i) |z| = 2, |iz| = 2(ii) arg(z) = – 5π

6, arg(iz) = – π

3∴ |iz| = |z| and arg(iz) = arg(z) + π

2

or multiplying by i rotates z by π2

about the origin


Page 66

5. (a) (i) |z| = √2, | z | =√2(ii) arg(z) = π

4 , arg( z ) = – π

4

(b) (i) |z| = 2, | z | = 2(ii) arg(z) = – 5π

6, arg( z ) = 5π

6

∴ | z | = |z| and arg( z ) = – arg(z)or z is a reflection of z in the real axis.

6. (a) (i) z z = 2

(ii) |z z | = 2, arg(z z ) = 0

(b) (i) z z = 4

(ii) |z z | = 4, arg(z z ) = 0

∴ z z is always real, |z z | = |z|2 and arg(z z ) = 0

Exercise 4

1. (a) (i) |z1| = 2, |z2| = 3, |z1z2| = 6 (ii) arg(z1) = π4

, arg(z2) = π4

, arg(z1z2) = π2

(iii) 6[cos π2

+isin π2

]

(b) (i) |z1| = √2, |z2| = 2, |z1z2| = 2√2 (ii) arg(z1) = π6

, arg(z2) = π3

, arg(z1z2) = π2

(iii) 2√2[cos π2

+isin π2

]

(c) (i) |z1| = r1, |z2| = r2, |z1z2| = r1r2

(ii) arg(z1) = θ1, arg(z2) = θ2 , arg(z1z2) = θ1 + θ2

(iii) r1r2[cos(θ1 + θ2) + isin(θ1 + θ2)]

2. (a) |z1z2| = 6, arg(z1z2) = 0, z1z2 = 6[cos0 + isin0].

(b) |z1z2| = , arg(z1z2) = π, z1z2 = cosπ + isinπ.

(c) |z1z2| = 100, arg(z1z2) = 0, z1z2 = 100[cos0 + isin0],


Page 67

3. (a) (i) |z1| = 2, |z2| = 3, z1

z2

= 23

(ii) arg(z1) = π2

, arg(z2) = π4

, arg z1

z2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ = π

4 (iii) 2

3[cos π

4⎛ ⎝ ⎜

⎞ ⎠ ⎟ +isin π

4⎛ ⎝ ⎜

⎞ ⎠ ⎟ ]

(b) (i) |z1| = √2, |z2| = 2, z1

z2

= 22

(ii) arg(z1) = π6

, arg(z2) = π3

, arg z1

z2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ = −

π6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ (iii) 2

2[cos −

π6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ +isin −

π6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ]

(c) (i) |z1| = r1, |z2| = r2, z1

z2

= r1r2

(ii) arg(z1) = θ1, arg(z2) = θ2 , arg z1

z2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ = θ1 – θ2

(iii) r1r2[cos(θ1 – θ2) +isin(θ1 – θ2)]

4. (a) 32

, − π2

, 32

[cos −π2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin −

π2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ].

(b) 4 , 11π15

, 4[cos 11π15

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin 11π

15⎛ ⎝ ⎜

⎞ ⎠ ⎟ ].

(b) 1 , 8π9

, cos 8π9

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin 8π

9⎛ ⎝ ⎜

⎞ ⎠ ⎟ .

5. (a) cos 3π4

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin 3π

4⎛ ⎝ ⎜

⎞ ⎠ ⎟ (b) 8[cos π

2⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin π

2⎛ ⎝ ⎜

⎞ ⎠ ⎟ ]

(c) 20[cos π12⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin π

12⎛ ⎝ ⎜

⎞ ⎠ ⎟ ] (d) 1

3[cos π

6⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin π

6⎛ ⎝ ⎜

⎞ ⎠ ⎟ ]

Exercise 5

1. (cosπ6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin

π6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ )0 = 1

(cosπ6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin

π6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ )1 = cos

π6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin

π6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ =

32 +

12 i

(cosπ6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin

π6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ )2 = cos

π3

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin

π3

⎛ ⎝ ⎜

⎞ ⎠ ⎟ =

12 +

32 i

(cosπ6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin

π6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ )3 = cos

π2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin

π2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = i

(cosπ6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin

π6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ )4 = cos

2π3

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin

2π3

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = –

12 +

32 i

(cosπ6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin

π6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ )5 = cos

5π6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin

5π6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = –

32 +

12 i

(cosπ6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin

π6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ )6 = cos(π) + isin(π) = – 1


Page 68

2. (cosπ4

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin

π4

⎛ ⎝ ⎜

⎞ ⎠ ⎟ )0 = 1

(cosπ4

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin

π4

⎛ ⎝ ⎜

⎞ ⎠ ⎟ )1 = cos

π4

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin

π4

⎛ ⎝ ⎜

⎞ ⎠ ⎟ =

12 +

12 i

(cosπ4

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin

π4

⎛ ⎝ ⎜

⎞ ⎠ ⎟ )2 = cos

π2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin

π2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = i

(cosπ4

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin

π4

⎛ ⎝ ⎜

⎞ ⎠ ⎟ )3 = cos

3π4

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin

3π4

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = –

12 +

12 i

(cosπ4

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin

π4

⎛ ⎝ ⎜

⎞ ⎠ ⎟ )4 = cos(π) + isin(π) = –1

3. Proof

4. (a) cos 5π6

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + isin 5π

6⎛ ⎝ ⎜

⎞ ⎠ ⎟ = – 3

2 + 1

2i

(b) cos2π + isin2π = 1(c) cosθ + isinθ(d) cos4θ + isin4θ

5. cos3θ = 4cos3θ – 3cosθ, sin3θ = 3sinθ – 4sin3θ

6. Proofs

7. Proofs

Exercise 6 1. – 4, 2 ± i 2. 1± i, – 1 ± 2i 3. – 2 ± 3i, 2 ± √6i4. −

23

, 2 ± i 5. – 2, 1 ± 2i 6. – 3, 3 ± i7. – 1 ± i, 2 ± i

Exercise 7 1. x2 + y2 = 4, a circle centre O, radius 2.

2. (x – 3)2 + y2 = 25, a circle centre (3, 0), radius 5.

3. 6x – 8y = 7, the perpendicular bisector of (– 3, 0) and (0, 4).

4. (x – 2)2 + y2 > 9, outside the circle centre (2, 0), radius 3.

5. (x – 3)2 + (y – 2)2 = 16, a circle centre (3, 2), radius 4.

6. (x – 1)2 + y2 ≤ 16, on or inside the circle centre (1, 0), radius 4.

7. y = 3x – 1, the perpendicular bisector of (2, 0) and (– 1, 1).


Page 69

Outcome 4 – Understand and use Sequences and Series

Sequences and Series

Introduction

A set of numbers, stated in a definite order, such that each number can be obtained from the previous number according to some rule, is called a Sequence.Each number of the sequence is called a Term.

3, 5, 7, 9, 11, - - - - - - - - - is a sequence. This sequence is Infinite. 3, 5, 7, 9, - - - - - - - , 47 is a sequence. This sequence is Finite.

An expression for the nth term (un ) of a sequence is useful since any specific term can be obtained from it.1, 4, 9, 16, 25, - - - - - - - - has an nth term un = n2 .

The Sum of the terms of a sequence is called a Series and is denoted by Sn .

Arithmetic Sequences and Series

An Arithmetic sequence is one in which each term differs from the previous term by a constant called the common difference ( d ) .

i.e.. un +1 − un = d or un +1 = un + d

If the first term is denoted by a and the common difference by d, then u1 = a u2 = a + d u3 = a + 2d u4 = a + 3d un = a + (n – 1)d

The nth term of an Arithmetic Sequence is denoted by the formula :-

un = a + (n −1)d

Example Find a formula for the nth term of the sequence 8, 11, 14, 17, ...... .Hence, find the 20 th term.

a = 8 , d = 11 – 8 = 3 un = a + (n −1)d = 8 + (n – 1) x 3 = 3n + 5

=> un = 3n + 5 => u20 = 3 x 20 + 5 = 65


Page 70

Examples1. For 2 + 5 + 8 + 11 + ....., find u15 and S8 .

a = 2 , d = 3 un = a + (n −1)d

= 2 + 3(n – 1) = 3n – 1

u15 = 3(15) – 1 = 44

Sn =n2

[2a + (n −1)d] S8 =

82

[2(2) + (8 −1)3] = 4[4 + 21] = 100

The Sum of the first n terms of an Arithmetic Series

The sum of the first n terms of the series is :-

Sn = a + (a + d) + (a + 2d) + ..... + [a + (n − 2)d] + [a + (n −1)d] [1]

Rewriting these terms in reverse order gives

Sn = [a + (n −1)d] + [a + (n − 2)d] + ....... + (a + 2d) + (a + d) + a [2]

Adding corresponding pairs of terms from [1] and [2] :-

2Sn = [2a + (n −1)d]+ [2a + (n −1)d] + .... + [2a + (n −1)d] + [2a + (n −1)d] + [2a + (n −1)d]

=> 2Sn = n[2a + (n −1)d] since there are n terms

=> Sn =n2

[2a + (n −1)d]

The Sum of the first n terms of an Arithmetic Series is found using :- Sn =

n2

[2a + (n −1)d]

Note : If the series is finite, then

Sn =n2

[2a + (n −1)d] => Sn =n2

[a + (a + (n −1)d)]

= n2

[first term + last term] = n [first term + last term]2

= n x (the average of the first and last terms)

= n2

(a + l)


Page 71

2. If the first term is 37 and the common difference is – 4, find u15 and S8. un = a + (n −1)d

u15 = 37 + 14(– 4) = –19

Sn =n2

[2a + (n −1)d] S8 =

82

[2(37) + (8 − 1) × −4] = 4[74 – 28] = 184

3. Find the number of terms in the series 5 + 8 + 11 + 14 +.......+ 62.

a = 5 , d = 3 (this time we do not know the value of n).To find n, set un = a + (n −1)d = 62 and solve for n.

=> 5 + 3(n – 1) = 62 => 3n + 2 = 62 => n = 20 (i.e. the number of terms is 20.)

4. Find the sum of 2 + 4 + 6 + 8 + 10 +...........+ 146. a = 2, d = 2. We must determine the value of n first.

Set un = a + (n −1)d = 146 and solve for n. => 2 + 2(n – 1) = 146 => n = 73

Sn = n[a + l]2

= 73[2 + 146]2

= 5402

5. The second term of an Arithmetic sequence is 18 and the fifth term is 21.Find the common difference, the first term and the sum of the first 10 terms.

Use simultaneous equations => Second term = a + 1d = 18 => Fifth term = a + 4d = 21

These equations solve to give a = 17, d = 1

=> Common difference, d = 1 and the first term a = 17

Sum of first ten terms S10 =102

[2(17) + (10 −1)1] = 215


Page 72

Exercise 1 1. Find the formula for the nth term of each of the following sequences and

find the requested term.(a) 3, 11, 19,........... ; u19 (b) 8, 5, 2, ............. ; u15

(c) 7, 6·5, 6,............ ; u12

2. Find the number of terms in each of the following sequences. (a) 2, 4, 6, ..........., 46.

(b) 50, 47, 44, ......., 14.

(c) 2, –9, –20, ......, –130.

3. State the values of a and d in each of the following series and find the requested Sn . (a) 4 + 10 + 16 +.........., S12 . (b) 15 + 13 + 11 +........, S20 . (c) 20 + 13 + 6 +.........., S16 .

4. For each of the Arithmetic Sequences, find Sn indicated. (Find a and d first). (a) u2 = 15, u5 = 21, S10 . (b) u4 = 18, d = –5, S16 . (c) u3 = 7, u12 = 61, S15 .

5. S10 = 120 , S20 = 840, find S30 .

6. u15 = 7, S9 = 18, find a, d and u20 .

7. How many terms of the Arithmetic series28 + 24 + 20 + ........

does it take to give a sum of zero ?

8. The sixth term of an Arithmetic sequence is twice the third term.If the first term is 3, find d and the tenth term.

9. How many terms of the Arithmetic series 1 + 3 + 5 +........ will give a sum of 1521 ?



Page 218 Exercise 5.1:3 Questions 1 ,2, 3, 4, 5, 6, 7, 9,10,14, 23, 24, 26, 27, 29, 30.




Page 73

The Sum of the first n terms of a Geometric Series

Let Sn denote the sum of n terms, a the first term and r the common ratio.

Sn = a + ar + ar2 + ar3 + ar4 + ......ar(n−2 ) + ar( n−1) [1]

rSn = ar + ar2 + ar3 + ar4 + ......ar (n−2) + ar(n−1) + arn [2]

[1] x r => Sn – rSn = a − arn

[1] – [2] => (1 − r)Sn = a − arn

=> Sn =a(1 − rn )(1− r )

, r ≠ 1

Note If r > 1, it is more convenient to use this result in the form :-

Sn =a(rn −1)

(r −1)

The Sum of the first n terms of a Geometric Series

Sn =a(1 − rn )(1− r )

r ≠ 1

Geometric Sequences and Series

Introduction

A Geometric sequence is one in which the ratio of each term to the previous term is a constant called the common ratio (r).

i.e. un +1

un= r or un +1 = unr

If the first term is denoted by a and the common ratio by r, then u1 = a u2 = u1r = ar u3 = u2 r = ar 2

u4 = u3r = ar 3

un = un−1r = ar (n−1)

The nth term of a Geometric Sequence is denoted by the formula

un = ar(n−1)


Page 74

3. A Geometric Series has the first term 27 and common ratio 43

.Find the least number of terms the series can have if its sum exceeds 550.

If Sn = 550, then a(rn −1)(r −1)

= 550 => 27( 4

3( )n− 1)

( 43( ) −1)

= 550

=> 81 43

⎛ ⎝ ⎜

⎞ ⎠ ⎟

n

−1⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ = 550

=> 43

⎛ ⎝ ⎜

⎞ ⎠ ⎟

n

−1⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ =

55081

=> 43

⎛ ⎝ ⎜

⎞ ⎠ ⎟

n

=63181

=> n = 7·136..

(done by taking logs of both sides).

For Sn > 550 => n > 7·136 => n ≥ 8 => n = 8

4. Given u3 = 32, u6 = 4. Find a, r, and S8.

From u3 = ar2 = 32 and u6 = ar5 = 4 we can divide the two

expressions to give ar5

ar2 =4

32 => r 3 =

18

=> r =12

Substituting for r in u3 gives a 12⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

= 32 => a = 128

S8 = 1281 − 1

2⎛ ⎝ ⎜

⎞ ⎠ ⎟

8⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

1 − 12

⎛ ⎝ ⎜

⎞ ⎠ ⎟

= 128

255256

⎛ ⎝ ⎜

⎞ ⎠ ⎟

12

= 256 ×255256

= 255

Examples1. Find u10 for the Geometric Sequence :- 144, 108, 81, 60 3

4, ....

a = 144, r = 108144

=912

=34

=> u10 = arn−1 = 144 ×34

⎛ ⎝ ⎜

⎞ ⎠ ⎟

9

= 10 ⋅ 812...

2. Find S19 = 3 – 6 + 12 – 24 + ......

a = 3, r = −63

= −2

=> S19 =a r n − 1( )

r − 1=

3 × (−2)19 −1( )−2 −1

= −1× −524289 = 524289


Page 75

Exercise 2A

1. Find the common ratio for each of these Geometric Series.

(a) 1, 3, 9, 27, ..... (b) 12, 6, 3, 1·5, .....

(c) 7, 0·7, 0·07, ..... (d) 18, 54, 162, .....

(e) 2·25, 1·5, 1, ..... (f) 14

, 18

, 116

, .....

(g) 1, –1, 1, –1, ..... (h) 1, –2, 4, –8, .....

2. Write down the first 4 terms of these Geometric Sequences :- (a) un = 3(n−1) (b) un = 3(–2)(n−1)

(c) un = 6 12⎛ ⎝ ⎜

⎞ ⎠ ⎟

( n−1)

3. Find the required term in the following Geometric Sequences. (Find un first)

(a) 1, 2, 4,... ; u5 (b) 2, 6, 18,... ; u6 (c) 4, 12, 36,... ; u6 (d) 2, 20, 200,... ; u5

(e) 1, –2, 4,... ; u6 (f) 6, 3, 3/2,... ; u7

4. Find the formula for the nth term of these Geometric Sequences :- (i.e. find un ).

(a) 1, 2, 4, .... (b) 3, 6, 12, ....

(c) 2, –6, 18, .... (d) 9, 3, 1, ....

(e) 4, 2, 1, .... (f) 12

, 14

, 18

, 116

, ...

5. Find the common ratio and the 5th term of these Geometric Sequences :-

(a) a = 6, u3 = 24 (b) a = 50, u4 = 400 (c) a = 36, u2 = –12

6. Find the sum of each of the following Geometric Series and simplify the answer as far as possible.

(a) 1 + 2 + 4 + ... to 8 terms (b) 2 + 6 + 18 + ... to 6 terms.

(c) 2 – 4 + 8 – ... to 5 terms (d) 2 – 6 + 18 – ... to 5 terms

(e) 1 + 12

+ 14

+ ... to 6 terms (f) 1 + 13

+ 19

+ ... to 5 terms

(g) 1 + x + x2 + ... to n terms (h) 1 – y + y2 – ... to n terms.

7. Find n if :- (a) 3 + 32 + 33 + 34 + .... + 3n = 363 (b) 2 + 22 + 23 + 24 + ....... + 2n = 510


Page 76

cont’d...

ExampleFind the sum to infinity of the Geometric Series 16 + 12 + 9 + .....

a = 16, r = 1216

= 34

Since –1 < r < 1, => S∞ exists. (This check should always be made)

and so S∞ =a

1 − r = 16

1 − 34

= 64

The Sum to Infinity of a Geometric Series

Consider the infinite series 8 + 4 + 2 + 1 + 12

+ 14

+ 18

+ 116

+ ......

S1 = 8 S2 = 12 S3 = 14 S4 = 15

S5 = 15 12

S6 = 15 34

S7 = 15 78

S8 = 151516

S9 = 15 3132

S10 = 15 6364

S11 = 15127128

S12 = 15 255256

The sum Sn looks as if it is approaching a value of 16.By taking a sufficiently large value of n, we can make Sn as near to 16 as we wish. We say that Sn tends to a limit of 16 as n approaches infinity and we write :-

Sn –> 16 (tends to 16) as n –> ∞ (tends to infinity) i.e. Lim

n→∞Sn = 16 or S∞ = 16

In general, Sn =a(1 − rn )(1− r )

, and if – 1 < r < 1, then r n → 0 for large values of n.

i.e. r n –> 0 as n –> ∞ and this means S∞ =a

1 − r

The Sum to Infinity of a Geometric Series

S∞ =a

1 − r , iff –1 < r < 1

(“iff” means “if and only if ”)


Page 77

Exercise 2B.

Find the common ratio and hence state whether the sum to infinity exists. If the sum to infinity exists find it.

(a) 1 + 13

+ 19

+.... (b) 1 + 2 + 4 +....

(c) 4 + 1 + 14

+.... (d) 8 + 4 + 2 +....

(e) 1 – 5 + 25 -.... (f) 10 – 9 + 8·1 –.

(g) 1 – 12

+ 14

–.... (h) 2 + 43

+ 89

+....



Page 221 Exercise 5.1:4 Questions 1,2,3,4,5,6,7.8,10,12,26,27.


Page 216 Exercise 8B Questions 1 – 26.

Expansion of 11 − r

as a Geometric Series with Common Ratio r

Reminder of the Binomial Theorem

(x + y)n = xn +n1

xn−1y +n(n −1)1 × 2

xn−2y2 +n(n −1)(n − 2)

1× 2 ×3xn−3y3 + ....... + yn

(for all values of n)

but 11 − r

=1

1+ (−r)= (1 + (−r ))−1

=> (1 + (−r))−1 = 1 +(−1)

1(−r ) +

(−1)(−2)1× 2

(−r)2 +(−1)(−2)(−3)

1× 2 × 3(−r )3 + .......

=> 11 − r

= 1 + r + r2 + r3 +........

which is a Geometric Series with Common Ratio r.


Page 78

Expansion of 1a + b

as a Geometric Series with common Ratio −ab

or −ba

.

1a + b

=1

b ab

+ 1⎛ ⎝ ⎜

⎞ ⎠ ⎟

=1

b 1 + ab

⎛ ⎝ ⎜

⎞ ⎠ ⎟

=1b

1+ab

⎛ ⎝ ⎜

⎞ ⎠ ⎟ −1

expands to

= 1b

1+ (–1)1

ab

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + (–1)(-2)

1×2ab

⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

+ (–1)(−2)(−3)1×2×3

ab

⎛ ⎝ ⎜

⎞ ⎠ ⎟

3

+ (–1)(−2)(−3)(−4)1×2×3×4

ab

⎛ ⎝ ⎜

⎞ ⎠ ⎟

4

......⎧ ⎨ ⎪ ⎩ ⎪

⎫ ⎬ ⎪ ⎭ ⎪

=1b

1 − ab

+a2

b2 −a3

b3 +a4

b4 ......⎧ ⎨ ⎪ ⎩ ⎪

⎫ ⎬ ⎪ ⎭ ⎪ a Geometric Series with a Common Ratio of −a

b

Similarly,

1a + b

=1

a 1 + ba

⎛ ⎝ ⎜

⎞ ⎠ ⎟

=1a

1 +ba

⎛ ⎝ ⎜

⎞ ⎠ ⎟ −1

expands to

= 1a

1+ (–1)1

ba

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + (–1)(–2)

1×2ba

⎛ ⎝ ⎜

⎞ ⎠ ⎟

2

+ (-1)(−2)(−3)1×2×3

ba

⎛ ⎝ ⎜

⎞ ⎠ ⎟

3

+ (–1)(−2)(−3)(−4)1×2×3×4

ba

⎛ ⎝ ⎜

⎞ ⎠ ⎟

4⎧ ⎨ ⎪ ⎩ ⎪

⎫ ⎬ ⎪ ⎭ ⎪

=1a

1− ba

+b2

a2 −b3

a3 +b4

a4 ......⎧ ⎨ ⎪ ⎩ ⎪

⎫ ⎬ ⎪ ⎭ ⎪ a Geometric Series with a Common Ratio of −b

a

Examples

1. Expand 11 − 2x

in ascending powers of x giving the first 4 terms.

From 1

1 − r= 1+ r + r 2 + r 3 + ......

Then1

1 − 2x= 1 + (2x) + (2x)2 + (2x)3 + ....

= 1 + 2x + 4x2 + 8x3+ .....

2. Expand 11 + 3x


1

1 + 3x=

11− (−3x)

= 1+ (−3x) + (−3x)2 + (−3x)3 + ....

= 1 – 3x + 9x2 – 27x3 + ....

3. Evaluate 1

0 ⋅ 9 to 4 decimal places

10 ⋅ 9

=1

1 − 0 ⋅1= 1 + (0 ⋅1) + (0 ⋅1)2 + (0 ⋅1)3 + (0 ⋅1)4 + ....

= 1 + 0·1 + 0·01 + 0·001 + 0·0001 + .... = 1·1111


Page 79

Exercise 3 1. Expand the following in ascending powers of x, giving the first 4 terms.

(a) 11 + 2x

(b) 11 − 3x

(c) 1

1 + x2

2. Expand the following in ascending powers of x giving the first 4 terms.

(a) 12 + 4x

(b) 13 − x

(c) 12 − 3x

3. Find the first 4 terms in the expansion of √(1 – x) in ascending powers of x.Deduce the value of √(0·9) to 3 decimal places.

4. Expand 1− 3x1 + 4x

in ascending powers of x as far as x3. ((1 − 3x) × 1(1 + 4x)

).

4. Expand 12 + 3x


. From 1a + b

=1a

1− ba

+b2

a2 −b3

a3 +b4

a4 ......⎧ ⎨ ⎪ ⎩ ⎪

⎫ ⎬ ⎪ ⎭ ⎪

12 + 3x

= 1

2(1 + 3x2

) which expands to

= 12

1 − 32

x +94

x2 −278

x3 + .........⎛ ⎝ ⎜

⎞ ⎠ ⎟

= 12−

34

x +98

x2 −2716

x3 + ....

The Sequence 1 +1n

⎛ ⎝ ⎜

⎞ ⎠ ⎟

n

and its Limit.

Using the Binomial Theorem

x + y( ) n = 1n xn + n1 xn -1y + n(n−1)

1×2 xn−2 y2 + n(n−1)(n−2)1×2×3 x n−3y3 + ..... .

If x is replaced by 1 and y is replaced by 1n

then :-

1 +1n

⎛ ⎝ ⎜

⎞ ⎠ ⎟

n

= 1n + n11n-1 1n( ) + n(n−1)

1×2 1n−2 1n( )2+ n(n−1)(n−2)

1×2×3 1n−3 1n( )3+ .....

1 +1n

⎛ ⎝ ⎜

⎞ ⎠ ⎟

n

= 1 + 1+ (n−1)2!n + (n−1)(n−2)

3!n2 + (n−1)(n−2)(n−3)4!n3 + ....

= 2 +12!

1 − 1n

⎛ ⎝ ⎜

⎞ ⎠ ⎟ +

13!

1− 1n

⎛ ⎝ ⎜

⎞ ⎠ ⎟ 1 − 2

n⎛ ⎝ ⎜

⎞ ⎠ ⎟ +

14!

1− 1n

⎛ ⎝ ⎜

⎞ ⎠ ⎟ 1 − 2

n⎛ ⎝ ⎜

⎞ ⎠ ⎟ 1− 3

n⎛ ⎝ ⎜

⎞ ⎠ ⎟ + ........

Limn→∞

1 +1n

⎛ ⎝ ⎜

⎞ ⎠ ⎟

n

= 1 +11!

+12!

+13!

+14!

+ .... = 2 ⋅ 718 = e , to 3 decimal places.


Page 80

Summation of other Series using the Sigma Notation

The Sigma notation ∑

The Sigma notation is used as a more concise way of writing a series.

e.g. 12 + 22 + 32 + 42 + ..... + n2 can be written more concisely as k2

k= 1

n

∑ (i.e. the sum of all k 2 for k = 1 to k = n).

In general f k( )k= 1

n

∑ is the series with the first term f(1), second term f(2),

third term f(3).....and last term f(n).

Examples 1. Write the following series in full :-

(a) k k + 1( )k =5

10

∑For k = 5, k(k + 1) = 5(6) = 30

For k = 6, k(k + 1) = 6(7) = 42 and so on to k = 10

Thus k k + 1( )k =5

10

∑ = 30 + 42 +56 + 72 + 90 + 110.

(b) (2k2 −1k= 1

4

∑ )

For k = 1, 2k2 −1 = 1 For k = 2 , 2k2 −1 = 7

and so on to k = 4

Thus (2k2 −1k= 1

4

∑ ) = 1 + 7 + 17 + 31.



Page 254 Exercise 5.4:2 Questions 4,5.


Page 230 Exercise 8F Questions 1, 2.


Page 81

Exercise 4

1. Write each of the following series in full

(a) k2

k= 1

5

∑ (b) 2k −1( )k= 1

9

∑ (c) 2520kk= 1

10

∑

2. Express each of the following series in the form f k( )k= 1

n

∑

(a) 1 + 2 + 3 + 4 +.......+ 50 (b) 5 + 10 + 15 +.......+ 30

(c) 3 + 5 + 7 + 9 + 11 + 13

Summation of a Series

The sums of certain finite series can be found or proved by a number of methods.For example

kk= 1

n

∑ = 1 + 2 + 3 + 4 +.........+ n = n(n + 1)2

can be proved using Sn =n2

[2a + (n −1)d], since it is an Arithmetic Series with a = 1 and d = 1.

i.e. Sn =n2

[2(1) + (n −1)(1)] => kk= 1

n

∑ =n(n + 1)

2

Examples Evaluate :-

1. k + 2( )k=1

10

∑ = k + 2k=1

10

∑k=1

10

∑ =12× (10 ×11) + (10 × 2) = 75

2. 4k + 5( )k=1

20

∑ = 4 k + 5k =1

20

∑k =1

20

∑ = 4 12× 20 × 21

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + (20 × 5) = 940

2. Express the following in Σ notation:- 1 + 4 + 7 + 10 + 13 +.......+ 298 Here f(k) is the expression for the kth term of the series. This is an Arithmetic Series, hence a = 1 and d = 3. The kth term = a + (k – 1)d = 1 + (k – 1)3 = 3k – 2. The final term is 298 and so 3k – 2 = 298 i.e.. k = 100

Thus 1 + 4 + 7 + 10 + 13 +.......+ 298 = 3k − 2( )k= 1

100

∑


Page 82

Exercise 5

Evaluate:-

1. kk=1

10

∑ 2. 2kk=1

20

∑ 3. 2k + 3( )k=1

8

∑

4. 4k + 5( )k=1

20

∑ 5. 3k —1( )k=1

10

∑





Page 220 Exercise 8C Questions 1 – 2.


Page 83

Answers

Exercise 1 1. (a) un = 8n − 5, u19 = 147 (b) un = 11− 3n, u15 = −34

(c) un = 7 ⋅ 5 − 0 ⋅ 5n, u12 = 1 ⋅52. (a) n = 23 (b) n = 13 (c) n = 133. (a) a = 4, d = 6, S12 = 444 (b) a = 15, d = –2, S20 = –80

(c) a = 20, d = –7, S16 = –5204. (a) a = 13, d = 2, S10 = 220 (b) a = 133, d = –5, S16 = –72

(c) a = –5, d = 6, S15 = 555

5. a = –15, d = 6, S30 = 2160 6. a = 0, d = 1/2, u20 = 9·5

7. n = 0 or 15 8. d = 3, u10 = 30 9. 39 terms

Exercise 2A

1. (a) r = 3 (b) r = 1/2 (c) r = 1/10 (d) r = 3

(e) r = 2/3 (f) r = 1/2 (g) r = –1 (h) r = –2

2. (a) 1, 3, 9, 27 (b) 3, –6, 12, –24, (c) 6, 3, 3/2, 3/4,

3. (a) un = 2n−1, u5 = 16 (b) 2 × 3n−1, u6 = 486

(c) un = 4 × 3n−1, u6 = 972 (d) 2 ×10n−1, u5 = 20000

(e) un = (−2)n−1, u6 = −32 (f) 6 ×12

⎛ ⎝ ⎜

⎞ ⎠ ⎟

n−1

, u7 =3

32

4. (a) un = 2n−1 (b) un = 3 × 2n−1

(c) un = 2 × (−3)n−1 (d) un = 9 × 13⎛ ⎝ ⎜

⎞ ⎠ ⎟

n−1

=13

⎛ ⎝ ⎜

⎞ ⎠ ⎟

n−3

(e) un = 4 × 12

⎛ ⎝ ⎜

⎞ ⎠ ⎟

n−1

=12

⎛ ⎝ ⎜

⎞ ⎠ ⎟

n−3

(f) un =12×

12

⎛ ⎝ ⎜

⎞ ⎠ ⎟

n−1

=12

⎛ ⎝ ⎜

⎞ ⎠ ⎟

n

5. (a) r = ±2, u5 = 96 (b) r = 2, u5 = 800 (c) r = – 13

, u5 = 49

6. (a) 255 (b) 728 (c) 22 (d) 122

(e) 1 3132

(f) 1 4180

(g) 1 – x n

1 − x(h) 1 – (–y)n

1 + y

7. (a) n = 5 (b) n = 8


Page 84

Exercise 2B

(a) r =13

, S∞ = 112

(b) r = 2, S∞ does not exist

(c) r =14

, S∞ = 5 13

(d) r =12

, S∞ = 16

(e) r = −5, S∞ does not exist (f) r = −9

10, S∞ =

10019

(g) r = −12

, S∞ =23

(h) r =23

, S∞ = 6

Exercise 3

1. (a) 11 + 2x

= 1 − 2x + 4x 2 − 8x3 + ... (b) 11 − 3x

= 1+ 3x + 9x2 + 27x3 + ...

(c) 1

1 + x2

= 1 − x2

+x 2

4−

x3

8+ ...

2. (a) 12 + 4x

=1

2(1+ 2x)=

12− x + 2x 2 − 4x3 + ...

(b) 13 − x

=1

3 1 − 13

x⎛ ⎝ ⎜

⎞ ⎠ ⎟

=13

+x9

+x2

27+

x3

81+ ...

(c) 12 − 3x

=1

2 1 − 32

x⎛ ⎝ ⎜

⎞ ⎠ ⎟

=12

+3x4

+9x2

8+

27x3

16+ ...

3. (1 − x)1

2 = (1 + (–x))1

2 = 11

2 +12

1!(−x) +

12(− 12)2!

(−x)2 +12(– 12)(− 32)

3!(– x)3 + ...

(0 ⋅ 9)1

2 = (1 − 0 ⋅1)12 = 1 – (0·5 x 0·1) – (0·125 x 0·001) – (0·0625 x 0·001) – ....

= 0·949...

4. 1− 3x1 + 4x

= 1 − 7x + 28x2 −112x3 + ....

Exercise 4

1. (a) 12 + 22 + 32 + 42 + 52 (b) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17

(c) 25201

+2520

2+

25203

+ ...... +252010

2. (a) k1

50

∑ (b) 5k1

6

∑ (c) 2k1

6

∑ + 1

Exercise 5

1. 55 2. 420 3. 96 4. 940 5. 155


Page 85

Outcome 5 – Elementary Number Theory - Methods of Proof

Mathematical Proof

Introduction

There are several methods of mathematical proof but all depend on a series of logical steps.All proofs start with an implication or proposition and so it is necessary to establish whether the implication is true or false.

Examples

1. If x > 5 then 2x > 10.This is obviously true and can be shown by solving the inequality 2x > 10

2. If x2 = 25 then x = 5This is false and can be shown by solving the equation x2 = 25.

These are examples of extremely simple direct proofs.

Exercise 1 Assuming the variables are on the set of real numbers, which of the following implications are true or false.

1. If x = 5 then x2 = 25.

2. If a > 0 then a2 > 0.

3. If ax2 + bx + c = 0 has real roots then b2 – 4ac < 0.

4. If f '(a) = 0 then f(a) is a maximum stationary value of f.

5. If u.v = 0 and u ≠ 0 and v ≠ 0, then u is perpendicular to v.

6. If g(x) = x2 and f(x) = x – 1 then g(f(x)) = x2 – 1.

Implication statements are often called “If ... then ...” statements but the notation for this is to use the implication symbol “ => ”, which makes it much less cumbersome.

Example 1. above becomes x > 5 => 2x > 10

Example 2. above becomes x2 = 25 => x = 5.


Page 86

The Converse of an Implication

If “statement p ⇒ statement q”, then the converse will be “statement q ⇒ statement p”. Examples

1. Statement :- x > 5 ⇒ 2x > 10. True.Converse :- 2x > 10 ⇒ x > 5. True.

2. Statement :- x2 = 25 ⇒ x = 5. False.Converse :- x = 5 ⇒ x2 = 25. True.

To prove an implication false it is often only necessary to give a single counter example.In Example 2. x2 = 25 ⇒ x = 5 is false.

Proof :- Counter example x = – 5.

Exercise 2State in words the converse of each statement and say if true or false. If the converse is false, give a reason.

1. If a number ends in zero then it is divisible by 5.2. If n is a prime number greater than 2 then n is an odd number.3. x = 3 ⇒ x2 = 94. If a and b are odd numbers then a + b is even.5. If 3 is a root of x2 + x – k = 0 then k is a multiple of 3.

Two Way implications – Equivalence.

If p ⇒ q and q ⇒ p are both true, then we can say p is equivalent to q or more commonly “ p iff q ”, (p, if and only if q). We use the notation p ⇔ q.

In Example 1, we could write x > 5 ⇔ 2x > 10 but we could not say this in Example 2.

Exercise 3In this exercise, state whether the implication can be replaced by the two way implication.

1. a = b ⇒ a + c = b + c. 2. x = y ⇒ –x = –y.3. n = – 3 ⇒ n2 = 9 . 4. If n is odd then n2 is odd, (n ε Z).5. If a triangle ABC is right-angled at A ⇒ a2 = b2 + c2 .

6. If y = 1 – x2 then dydx

= −2x .



Page 1 Exercise 1.1:1Questions 1 – 12


Page 87

Methods of Proof.

1. Direct proof.This method is one which has been used at various points in the course.

Examples1. Prove that the equation of a straight line is y – b = m(x – a).2. Prove Pythagoras’ Theorem.

2. Proof by Contradiction.To prove an implication p ⇒ q, it is sometimes difficult, if not impossible, to prove it directly.In this case, we prove this by contradiction (or by contrapositive).The contrapositive of p ⇒ q is ~ q ⇒ ~p (which is read as not q implies not p.)When ~ q ⇒ ~p is true so is p ⇒ q

Examples

1. Prove that the set of primes is infinite. This is impossible to prove directly.Suppose that the set of primes is finite having n numbers, p1, p , p3, ....., pn ,Here we can make a composite number which is a product of all n primes.

N = p1 × p2 × p3 × p4 × p5 × .....pn

Let us consider the next number after N, namely N + 1.(N + 1) = p1 × p2 × p3 × p4 × p5 × .....pn + 1

If you attempt to divide (N + 1) by any of the known primes, p1 to pn ,there will always be a remainder of “1”.Hence N + 1 is not divisible by any prime number greater than 1 apart from (N + 1) and so (N + 1) must be prime, and it is greater than the largest prime pn, meaning our supposition is false. Hence the set of primes is infinite.

2. Prove that √2 is irrational. Proof is by contradiction.

Begin by supposing that √2 is rational. (i.e. it can be written as a fraction).

Let √2 = mn

, where m and n have no common factor.

⇒ 2 = m 2

n2 ⇒ m 2 = 2n2 ⇒ m 2 is even

⇒ m is even = 2k, (k ∈ Z).

⇒ 2n2 = 4k2 ⇒ n2 = 2k2 ⇒ n2 is even ⇒ n is even. = 2h, (h ∈ Z).

=> m and n have a common factor of 2.

But we stated that m and n had no common factor ! —> This is a contradiction. Therefore √2 cannot be expressed in the form m

n and must be irrational.


Page 88

Exercise 4Prove by contradiction:-

1. √3 is irrational.

2. If n3 is odd where n is an integer then n is odd.

3. If x, y ε R such that x + y is irrational, then at least one of x and y is irrational.

3. Proof by InductionInduction is a method of proof in which we establish the truth of the implication for some starting value and then we prove it for the succeeding value.This method is used in proving Summation examples, the Binomial Theorem and de Moivre’s Theorem.

Examples

1. Prove that rr =1

n

∑ = 12

n n + 1( ) , where rr =1

n

∑ = 1 + 2 + 3 + ....+ (n – 1) + n.

Proof :-

Part A for n = 1, rr =1

n

∑ = rr =1

1

∑ = 1

12

n n + 1( ) = 12×1× 2 = 1 ⇒ True for n = 1.

Part B Assume it is true for n = k, where k ≥ 1.

i.e. rr =1

k

∑ = 12

k k + 1( )

Now show that it is true for n = k + 1.

i.e. rr =1

k+1

∑ = rr =1

k

∑ + (k + 1)

= 12

k k + 1( ) + (k + 1)

= 12

k + 1( ) k + 2( )

= 12

k + 1( ) k + 1( ) + 1( )

Hence, if true for n = k, it is true also for n = k + 1.

Since it is true for n = 1 ⇒ True for n = 2

Since it is true for n = 2 ⇒ True for n = 3 etc.

and so on for all values of n.

Therefore, it is true by Induction.


Page 89

2. Prove that −1( ) r−1

r =1

n

∑ r2 = 12

−1( )n−1n n + 1( )

Proof :-

Part A for n = 1, −1( ) r−1

r =1

n

∑ r2 = −1( ) r−1

r =1

1

∑ r2 = (–1)012 = 1

12

−1( )n−1n n + 1( ) = 12

−1( )1−11 1 + 1( ) = 12×1×1 × 2 = 1

True for n = 1

Part B Assume true for n = k, k ≥ 1

i.e. −1( ) r−1

r =1

k

∑ r2 = 12

−1( )k−1k k + 1( )

Prove true for n = k + 1

−1( ) r−1

r =1

k+1

∑ r2 = −1( ) r−1

r =1

k

∑ r2 + −1( ) k +1-1( ) k +1( )2

= 12

−1( )k−1k k + 1( ) + −1( ) k( ) k +1( )2

= 12

−1( )k k + 1( ) −k + 2 k +1( )[ ] = 1

2−1( )k k + 1( ) k + 2( )

= 12

−1( )k k + 1( ) k + 1( ) +1( )Hence true also for n = k + 1.True for n = 1 ⇒ True for n = 2 since k ≥ 1True for n = 2 ⇒ True for n = 3and so on for all values of n.

Therefore, it is true by Induction.

Exercise 5Prove by Induction:-

1. 1r r + 1( )r =1

n

∑ =n

n + 12. 2r −1( ) = n − 2( )

r =1

n

∑ n + 2( ) , n ≥ 3

3. 3n > n3, for all n ≥ 4 4. (cosθ + sinθ )n = cosnθ + sinnθ

5. The sum of the first n odd numbers is a perfect square. (i.e. 2k −1 = p2

k= 1

n

∑ )

Further examples can be found in the following resources.The Complete A level Maths (Orlando Gough) Page 235 Exercise 5.2 SelectionUnderstanding Pure Mathematics (A.J.Sadler/D.W.S.Thorning)Page 220 Exercise 8C Questions 3,9(a).


Page 90

Number Theory

The Fundamental Theorem of Arithmetic states that any integer, n > 1, can be expressed uniquely as a product of prime numbers apart from the order of the primes.

i.e. 6 = 2 x 3 12 = 2 x 2 x 321 = 3 x 7 36 = 2 x 2 x 3 x 3 etc.

ExampleExpress 430 as a product of its prime numbers.

430 ÷ 2 = 215 ÷ 5 = 43 ÷ 43 = 1 ⇒ 430 = 2 x 5 x 43Repeated division by prime numbers produced the result.

Exercise 6Express the following numbers as a product of primes. (in increasing order.)

1. 490 2. 1125 3. 2728

The Binomial Theorem

Reminder : Factorial notation n! = n(n – 1)(n – 2)(n – 3).......3.2.1

Combination notation nCr or nr⎛

⎝ ⎜ ⎞

⎠ ⎟ = n!

r!(n – r )!

Useful relationships (i)nr⎛

⎝ ⎜ ⎞

⎠ ⎟ =

nn − r⎛

⎝ ⎜

⎞

⎠ ⎟

Proof nn − r⎛

⎝ ⎜

⎞

⎠ ⎟ = n!

(n − r)!(n − (n − r))! = n!

r!(n – r )! =

nr⎛

⎝ ⎜ ⎞

⎠ ⎟

(ii)n

r − 1⎛

⎝ ⎜

⎞

⎠ ⎟ +

nr⎛

⎝ ⎜ ⎞

⎠ ⎟ =

n + 1r

⎛

⎝ ⎜

⎞

⎠ ⎟

Proofn

r − 1⎛

⎝ ⎜

⎞

⎠ ⎟ +

nr⎛

⎝ ⎜ ⎞

⎠ ⎟ = n!

(r −1)!(n − r + 1)! + n!

r!(n – r )!

= n!rr!(n − r + 1)!

+ n!(n − r + 1)r!(n − r + 1)!

= n!r + n!(n − r + 1)r!(n − r + 1)!

= n!(r + n − r + 1)r!(n − r + 1)!

= n!(n + 1)r!((n + 1) − r )!

= (n + 1)!r!((n + 1) − r )!

= n + 1

r⎛

⎝ ⎜

⎞

⎠ ⎟

(this is the rule that states that any number in Pascal’s triangle is the sum

of the 2 numbers directly above it)


Page 91

The Binomial Theorem (x + y)n =nr⎛

⎝ ⎜ ⎞

⎠ ⎟

r =0

n

∑ x n−r yr

This is an alternative notation to that covered already in Unit 1, Learning Outcome 1.

Proof by Induction (x + y)n =nr⎛

⎝ ⎜ ⎞

⎠ ⎟

r =0

n

∑ x n−r yr

Proof :- Part A For n = 1, (x + y)1 = x + y

and 1r⎛

⎝ ⎜ ⎞

⎠ ⎟

r =0

1

∑ x1−ryr = 10⎛

⎝ ⎜ ⎞

⎠ ⎟ x1−0y0 +

11⎛

⎝ ⎜ ⎞

⎠ ⎟ x1−1y1 = x + y

True for n = 1

Part B Assume true for n = k, k ≥ 1 i.e. (x + y)k = kr⎛

⎝ ⎜ ⎞

⎠ ⎟

r =0

k

∑ xk −ryr

Now prove true for n = k + 1 (x + y)(k + 1) = (x + y)k(x + y)

= (x + y)kr⎛

⎝ ⎜ ⎞

⎠ ⎟

r =0

k

∑ xk −ryr

= xkr⎛

⎝ ⎜ ⎞

⎠ ⎟

r =0

k

∑ xk −ryr + ykr⎛

⎝ ⎜ ⎞

⎠ ⎟

r =0

k

∑ xk −ryr

Expanding this expression gives :-

xkr⎛

⎝ ⎜ ⎞

⎠ ⎟

r =0

k


⎝ ⎜ ⎞

⎠ ⎟

r =0

k

∑ xk −ryr

= xk0⎛

⎝ ⎜ ⎞

⎠ ⎟ xk y0 + x

k1⎛

⎝ ⎜ ⎞

⎠ ⎟ xk –1y1 + x

k2⎛

⎝ ⎜ ⎞

⎠ ⎟ xk −2y2 + .......+ y

k0⎛

⎝ ⎜ ⎞

⎠ ⎟ xk y0 + y

k1⎛

⎝ ⎜ ⎞

⎠ ⎟ x k–1y1 + y

k2⎛

⎝ ⎜ ⎞

⎠ ⎟ xk −2y2 + .......

=k0⎛

⎝ ⎜ ⎞

⎠ ⎟ x k+1 +

k1⎛

⎝ ⎜ ⎞

⎠ ⎟ xk y +

k2⎛

⎝ ⎜ ⎞

⎠ ⎟ xk −1y2 + ....... +

k0⎛

⎝ ⎜ ⎞

⎠ ⎟ xk y + y

k1⎛

⎝ ⎜ ⎞

⎠ ⎟ x k— 1y2 +

k2⎛

⎝ ⎜ ⎞

⎠ ⎟ xk −2y3 + .......

= k0⎛

⎝ ⎜ ⎞

⎠ ⎟ xk+1 +

k0⎛

⎝ ⎜ ⎞

⎠ ⎟ +

k1⎛

⎝ ⎜ ⎞

⎠ ⎟

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥ xk y +

k1⎛

⎝ ⎜ ⎞

⎠ ⎟ +

k2⎛

⎝ ⎜ ⎞

⎠ ⎟

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥ x k-1y2 + ...... +

kk⎛

⎝ ⎜ ⎞

⎠ ⎟ yk +1

From n

r − 1⎛

⎝ ⎜

⎞

⎠ ⎟ +

nr⎛

⎝ ⎜ ⎞

⎠ ⎟ =

n + 1r

⎛

⎝ ⎜

⎞

⎠ ⎟ ,

k0⎛

⎝ ⎜ ⎞

⎠ ⎟ +

k1⎛

⎝ ⎜ ⎞

⎠ ⎟ =

k + 11

⎛

⎝ ⎜

⎞

⎠ ⎟ and

k1⎛

⎝ ⎜ ⎞

⎠ ⎟ +

k2⎛

⎝ ⎜ ⎞

⎠ ⎟ =

k + 12

⎛

⎝ ⎜

⎞

⎠ ⎟ etc.

And so xkr⎛

⎝ ⎜ ⎞

⎠ ⎟

r =0

k


⎝ ⎜ ⎞

⎠ ⎟

r =0

k

∑ xk −ryr

= k + 1

0⎛

⎝ ⎜

⎞

⎠ ⎟ xk +1 +

k + 11

⎛

⎝ ⎜

⎞

⎠ ⎟ xk y +

k + 12

⎛

⎝ ⎜

⎞

⎠ ⎟ xk-1y2 + ...... +

k + 1k + 1⎛

⎝ ⎜

⎞

⎠ ⎟ yk +1

= k + 1

r⎛

⎝ ⎜

⎞

⎠ ⎟

r =0

k+ 1

∑ x(k +1)−r yr = (x + y)(k+ 1) . Hence, true also for n = k + 1.

True for n = 1 ⇒ True for n = 2 since k ≥ 1True for n = 2 ⇒ True for n = 3 and so on for all values of n.

hence, true for all values of n, by induction.

note

note


Page 92

Exercise 7

1. Show that if n ε N nr⎛

⎝ ⎜ ⎞

⎠ ⎟

r =0

n

∑ = 2n [Think of (x + y)n]

2. Show that if n ε N n0⎛

⎝ ⎜ ⎞

⎠ ⎟ +

n2⎛

⎝ ⎜ ⎞

⎠ ⎟ +

n4⎛

⎝ ⎜ ⎞

⎠ ⎟ + ...... = 2n−1

AnswersExercise 1 Page 1

1. T 2. T 3. F 4. F 5. T 6. F

Exercise 2 Page 2

1. If a number is divisible by 5, then it ends in 0. False. e.g. 152. If n is an odd number greater than 2, then it is prime. False. e.g. 153. x2 = 9 => x = 3. False. e.g. x = –34. If a + b is even then a and b are odd. False. e.g. 2 and 45. If k is a multiple of 3 then 3 is a root of x2 + x – k = 0. False e.g. x2 + x – 3 = 0

Exercise 3 Page 2

1. Yes 2. Yes 3. No 4 Yes 5. Yes 6. No.

Exercise 4 Page 4

1. Proof 2. Proof 3. Proof

Exercise 5 Page 5

1. Proof 2. Proof 3. Proof 4. Proof 5. Proof

Exercise 6 Page 6

1. 2 x 5 x 72 2. 32 x 53 3. 23 x 11 x 31

Exercise 7 Page 8

1. Proof 2. Proof


Page 1

Outcome 1 – Vectors Revision of Vectors in 3 dimensions from Higher Level

1. Position Vector of a point P.

Relative to an origin O, P is the point (x, y, z) with position vector:-

OP→

or p = xyz

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

[or (x, y, z) is sometimes used]

2. Basic laws(a) The Commutative Law.

a + b = b + a

Proof :- OQ→

= OP→

+ PQ→

and OQ→

= OR→

+ RQ→

= a + b = b + ai.e. a + b = b + a

(b) The Associative Law. a + b( ) + c = a + b + c( )

Proof: OQ→

= a + b and PR→

= b + c

OR→

= OQ→

+ QR→

and OR→

= OP→

+ PR→

= (a + b )+c = a + (b + c )i.e. a + b( ) + c = a + b + c( )

(c) The Zero Vector or Identity Vector a + 0 = 0 + a = a

(d) The Negative of a Vector. a + (–a ) = 0

(e) Multiplication by a Scalar, k.If a is a non-zero vector and k a non-zero number, then(i) |ka | is k times |a | .(ii) if k > 0, ka is parallel to a and in the same direction.(iii) if k < 0, ka is parallel to a and in the opposite direction.

(f) Unit vectorA unit vector is one whose magnitude (length) is one unit.i is a unit vector in the direction of Ox.

j is a unit vector in the direction of Oy.k is a unit vector in the direction of Oz.The position vector of any point can be given in terms of i, j and k.e.g. If P is the point (1, 2, 3), then

p = i + 2 j + 3k

or p = xyz

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

= 123

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

a

b

a

b b + a

P

QR

O

a + b

Q

ab

c

a +b

b + c

OP

R (a + b ) +c or a + b + c )

i

k

x

yz

(1,0,0)

(0,1,0)(0,0,1)

O

j


Page 2

3. Formulae:-(a) The Section Formula:-

If P divides AB in the ratio m:n,

then AP→

PB→ =

mn

giving p = mb + nam+ n

If P is the mid-point of AB, then p =a + b

2

(b) The Scalar (Dot) Product(i) a.b = |a||b|cosθ

where θ is the angle between a and b.a.a = |a|2 since cosθ = 1.

(ii) If a =x1

y1

z1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ and b =

x2

y2

z2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

then a.b = x1x2 + y1y2 + z1z2

(iii) cosθ = a . b| a || b |

= x1x2 + y1y2 + z1z2

x12 + y1

2 + z12( ) x2

2 + y22 + z2

2( )(iv) a and b are perpendicular <=> a.b = 0.

Note also that :-Parallel vectors i.i = j . j = k.k = 1 butPerpendicular vectors i. j = j .k = k.i = 0.

(v) The Distributive Law a.(b + c) = a.b + a.c

(g) The length, or Magnitude, of a vector.

If AB→

=xyz

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ , then | AB

→| = x2 + y2 + z2

The distance between A(x1, y1, z1) and B(x2, y2, z2).

AB→

= b − a =x2

y2

z2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ −

x1

y1

z1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

=x2 − x1

y2 − y1

z2 − z1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

| AB→

|= [(x2 − x1)2 + (y2 − y1)2 + (z2 – z1)2]

N.B. The unit vector in the direction of a vector u is given

by the formula u| u |

(h) Subtraction of vectors:-If the position vector of P is p and the position vector of Q is q

then PQ→

= q − p

a

b

θ

A

BP

n

m


Page 3

Examples1. If a = 5i + 3 j + 7k and b = 2i – 8 j + 4k,

find the angle between a and b.

a.b = 537

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ .

2−84

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

= 14

|a| = √(25 + 9 + 49) = √83 |b| = √(4 + 64 + 16) = √84

cosθ = a . b| a || b |

= 1483 84

=> θ = 80·3°

2. Find the unit vectors which make an angle of 45° with thevector a = 2i + 2 j – k and an angle of 60° with the vector

b = j – k.

Let the unit vector be u = xi + y j + zku.a = |u||a|cosθ

i.e. 2x + 2y – z = 1x3xcos45° => 2x + 2y – z = 32

.....(1)

u.b = |u||b|cosθ

i.e. y – z = 1x√2xcos60° => y – z = 22

.....(2)

Since u is a unit vector then x2 + y2 + z2 = 1 . . . . . . (3)

From (1) and (2) 2x + 2y – z = 32

= 3 22

. . . . . . (1)

y – z = 22

. . . . . . (2)Subtract 2x + y = √2 or y = √2 – 2x

From (1) and (2) 2x + 2y – z = 3 22

. . . . . . (1) 2y – 2z = √2 . . . . . . (2)

Subtract 2x + z = 22

or z = 22

– 2xSubstitute in (3)

x2 + 2 − 2x( )2+

22

− 2x⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

2

= 1 => 9x2 – 6√2x + 32 = 0

6x2 – 4√2x + 1 = 0 => x = 1

3 2 or 12 (quadratic formula)

=> y = 4

3 2 or 0

=> z = 1

3 2 or −12

Hence u = 13 2

, 43 2

, 13 2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ or 1

2,0, − 1

2⎛ ⎝ ⎜

⎞ ⎠ ⎟


Page 4

Exercise 1

1. Which of the following expressions represent vectors and which represent scalars?

(a) b.c + c.a + a.b (b) (b.c)a + (c.a)b + (a.b)c

(c) ((b.c)(c.a))a (d) [(b.c)c + (b.a)a].(b + 2a)

Evaluate these when a = i + k, b = i + j + 2k and c = 2 j + k.

2. A, B and C have coordinates A(1, 6, –2), B(2, 5, 4), C(4, 6, –3).Find the lengths of the sides, cosines of the angles and the areaof triangle ABC.

3. Prove, that in any parallelogram, the sum of the squares on the diagonals is equal to twice the sum of the squares on two adjacent sides.

i.e. Prove that | BD

→|2 + | AC

→|2 = 2 | AB

→|2 + 2 | AD

→|2 by expressing

BD→

as b – d and AC→

as b + d

4. Find the two unit vectors which make an angle of 45° with the vectors a = i and b = k.

5. (a) Find the unit vectors which make an angle of 45° with thevector a = –i + k and an angle of 60° with the vector

b = –2i + 2 j + k.

(b) Show that these two unit vectors are perpendicular.

A

B C

D

d

b


The Complete A level Maths (Orlando Gough) (No reference)

Understanding Pure Mathematics (A.J.Sadler/D.W.S.Thorning)Page 62 Exercise 2C Question 1 – 18Page 412 Exercise 17A Question 1 – 14


Page 5

The Vector Product.

Introduction

Right-Hand systems and Left-Hand systems.

Let a, b and c be three, non-zero, non-coplanar vectors.

[a, b, c] is said to be a right-hand system if a observer at C sees an anti-clockwise rotation that takes OA

→ on to OB

→.

Imagine c as a screw. Seen from below, a right-hand rotation of the screw takes Cinto the page.

[b, a, c] is said to be a left-hand system if a observer at C sees a clockwise rotation that takes OB

→ on to OA

→.

Imagine c as a screw. Seen from below, a right-hand rotation of the screw takes Cout of the page.

Note : The order in which the vectors are written plays a part.[a, b, c] is a R.H. system.Cyclic interchange of a, b and c is also a R.H. system.

[a, b, c], [b, c, a], [c, a, b] are R.H. systems.but [a, c, b], [b, a, c], [c, b, a] are L.H. systems.

a

c b

c

AO

B

a

b

C

c

AO

B

a

b

C

a

c

b

ac

b


Page 6

The Vector Product.

Definition.The Vector Product is denoted by a x b which reads as “a cross b” and is defined by:-(i) magnitude (length) |a x b| = |a||b|sinθ where θ is the angle

between a and b.

(ii) a x b is perpendicular to both a and b.

(iii) [a, b, a x b] form a R.H. system.

Note (a) a x b = 0 <=> a is parallel to b or either a or b is zero.

(b) |a x b| is the area of a parallelogram with sides determined by a and b.

Area = base x height = |a||b|sinθ.

(c) If a and b are non-zero vectors, the following statementsare equivalent :(i) a x b = 0.(ii) a is parallel to b.(iii) a = kb. (a linear multiple of b.)

(d) a x b = – (b x a)Proof :If [a, b, a x b] is a R.H. systemthen [b, a, b x a] is a L.H. system.

So [b, a, – (b x a)] is a R.H. system.i.e. a x b = – (b x a).

(e) ka x b = k(a x b), ka x lb = kl(a x b)

(f) The distributive law:-a x (b + c) = (a x b) + (a x c)(b + c) x a = (b x a) + (c x a)

a a x bb

b x a

a–(b x a)b

b x a

θ

a

b|b|sinθ


Page 7

The Vector Product in component form

Let i, j and k be unit vectors, mutually perpendicular to form a R.H. system. Therefore i x j = k Also i x i = 0

i x k = – j j x j = 0j x i = –k k x k = 0j x k = ik x i = jk x j = –i

These results can be or If moving in a clockwisesummarised in a table. direction, the vector product

is positive.If moving in an anti - clockwisedirection, the vector productis negative.

Let a = x1i + y1 j + z1k and b = x2i + y2 j + z2kthen a x b = (x1i + y1 j + z1k) x (x2i + y2 j + z2k)by the distributive law = x1ix(x2i + y2 j + z2k) + y1 j x(x2i + y2 j + z2k) + z1kx(x2i + y2 j + z2k)= x1ixx2i+x1ixy2 j +x1ixz2k+y1 j xx2i+y1 j xy2 j +y1 j xz2k+z1kxx2i+z1kxy2 j +z1kxz2k = x1x2ixi+x1y2ix j +x1z2ixk+x2y1 j xi+y1y2 j x j +y1z2 j xk+x2z1kxi+y2z1kx j +z1z2kxk= x1x2(0)+x1y2(k)+x1z2(– j )+x2y1(–k)+y1y2(0)+y1z2(i)+x2z1( j )+y2z1(–i)+z1z2(0)= x1y2(k)+x1z2(– j )+x2y1(–k)+y1z2i+x2z1 j +y2z1(–i)= (y1z2 – y2z1)i + (x2z1 – x1z2) j + (x1y2 – x2y1)k

These components can be found more easily by rearranging the components of a and b under the unit vectors.

The component in the direction i is found by evaluating thedeterminant formed by the components of j and k.Similarly, the component in the direction j is found by evaluating the determinant formed by the components of i and k and the component in the direction k is found by

evaluating the determinant formed by the components of i and j .

Alternatively, write the components of the two vectors one under the other,starting with the second and ending with the second, the components being written in cyclic order.i.e. y1 z1 x1 y1 Downward arrows positive.

y2 z2 x2 y2 Upward arrows negative.

i

k

j

i

k j

× i j ki 0 k − jj −k 0 ik j –i 0

i j kx1 y1 z1

x2 y2 z2


Page 8

Examples 1. If a = i + 2 j + 3k and b = 2i – j + k, find (i) a x b (ii) b x a

(i) a x b = i j k1 2 32 –1 1

=(2x1 – 3x(–1))i – (1x1 – 3x2) j + (1x(–1) – 2x2)k = 5i + 5 j – 5k

or a 2 3 1 2

b –1 1 2 –1 a x b = (2 – (–3))i + (6 – 1) j + (–1 – 4)k

= 5i + 5 j – 5k

(ii) b x a = i j k2 −1 11 2 3

=(–1x3 – 2x1)i –(2x3 – 1x1) j + (2x2 – 1x(–1))k = –5i –5 j + 5k

2. If a = 3i + 5 j + 7k, b = i + k and c = 2i – j + 3k, verify that a x (b x c) = (a.c)b – (a.b)c ( This is known as the Vector triple product).

b 0 1 1 0 b x c = i – j – k

c –1 3 2 –1

a 5 7 3 5 i x (b x c) = 2i + 10 j – 8k

b x c –1 –1 1 –1

(a. c)b – (a.b)c

= (6 – 5 + 21)101

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ – (3 + 0 + 7)

2−13

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ = 22

101

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ – 10

2−13

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ =

210−8

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

i.e. a x (b x c) = (a.c)b – (a.b)c

3. Find the area of the triangle with verticesA(1,3,–2), B(4,3,0) and C(2,1,1).Area of a triangle = 1

2(Area of a parallelogram) = 1

2| CA

→ x CB

→|

CA→

= a – c = −12−3

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ , CB

→ = b – c =

22−1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ , CA

→ x CB

→ =

4−7−6

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

Area of a triangle = 12

√(16 + 49 + 36) = 12

√101


Page 9

Exercise 2

1. If a = 3i + 2 j – k, b = i – j – 2k and c = 4i – 3 j + 4k, evaluate(a) a x (b x c) (b) (a x b) x c (c) (a x b).(a x c)(d) (a x b).(b x c) (e) [a x (b x c)].c

2. If a = 3i + 2 j + 5k, b = 4i + 3 j + 2k and c = 2i + j + 10k, find (a) a x b (b) (a x b).c (c) b.(a x c)

3. If a = 3i + j + 2k, b = 2 j – k and c = i + j + k and d = b x (c x a) + (a.c)a, show that b is perpendicular to d.

4. Find a vector perpendicular to each of the vectors a = 4i – 2 j + 3kand b = 5i+ j – 4k.

5. If a = i + j – k and b = 2i – j + k, find(a) a x b (b) a x (a + b) and show that (c) a.(a x b) = 0

6. Find the unit vectors perpendicular to both a = 4i – k and b = 4i + 3 j – 2k.

7. Find the area of triangle ABC where A(4, –8, –13), B(5, –2, –3)and C(5, 4, 10).

8. Prove algebraically the vector triple product a x (b x c) = (a.c)b – (a.b)c

4. Find a unit vector perpendicular to both a = 2i + j – k and b = i – j + 2k .The vector n = a x b is perpendicular to both a and b. Divide n by |n| to obtain a unit vector that has the same direction as n.

Then u = nn

= a × ba × b

a x b = i – 5 j – 3k from

|a x b| = √(12 + (–5)2 + (–3)2) = ±√35

u = ±135

(i − 5 j − 3k)

a 1 –1 2 1

b –1 2 1 –1


The Complete A level Maths (Orlando Gough) - No reference.

Understanding Pure Mathematics (A.J.Sadler/D.W.S.Thorning) - No reference


Page 10

The Equation of a Straight Line.

( i ) In Vector Form.Let d ≠ 0 be a fixed vector. Let A, with position vector a, be a fixed pointand R, with position vector r be a variable point.Let R lie on a straight line L which passes through A and is parallel to d.

Since R lies on L, AR→

is parallel to d (or is zero).

AR→

= td (where t is some parameter).r – a = td

i.e. r = a + td

This is the vector equation of the line L through A parallel to d.The scalar t is a parameter and may take any real value including zero.The vector d is called a direction vector of the line. The cosines of the angles between the vector d and the unit vectors i, j , kare called the Direction Cosines of d.

( i i ) In Parametric Form.

If r = xyz

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

, a = abc

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ and d =

lmn

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ then the vector equation becomes

=> xyz

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

= abc

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ + t

lmn

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

, equating components gives x = a + tly = b + tmz = c + tn

These are the Parametric Equations of the line.

( i i i ) In Symmetrical or Cartesian Form.We can eliminate the parameter, t, to obtain the following:-

From x − al

=y − b

m=

z − cn

(= t) => x − al

=y − b

m=

z − cn

is called the

Symmetrical Form of the equn of a line through (a,b,c) in the dirn lmn

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

Note : If any of the direction vectors is zero, the parametric equations should be used.

Finding the equation of a line given 2 points on the line.

A particular straight line in space can be precisely specified in a variety of ways :

(a) by means of two points,

(b) by means of one point on the line and the direction vector of the line,

(c) as the intersection of two planes. (See later)

A

Rd

L Oa

r


Page 11

Examples

1. Find the equation of the line joining the points A(1, 0, 2) and B(2, 1, 0)

AB→

= b – a = 11−2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ i.e. the direction vector

lmn

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ of the line.

A(1, 0, 2) lies on the line.

Substitute in x − al

=y − b

m=

z − cn

x −11

=y− 0

1=

z − 2− 2

in symmetrical form ........(1)

x = 1+ t, y = t, z = 2 – 2t in parametric form.

Note B(2, 1, 0) also lies on the line, givingx − 2

1=

y −11

=z

− 2 in symmetrical form ...........(2)

x = 2 + t, y = 1 + t, z = –2t in parametric form.

This illustrates that the equation of a line is not unique.However, equation (1) can be reduced to equation (2)by subtracting 1 from each part.

x −11

−1 =y − 0

1−1 =

z − 2− 2

−1 => x − 21

=y −1

1=

z− 2

An alternative check, to show that the lines are the same, is(i) a point on one line lies on the other and(ii) their direction vectors are parallel.

e.g. A(1, 0, 2) satisfies both equations andDirection Vectors are the same in both equations.

2. Find the symmetrical form of the equation of the line throughthe point (6, 3, –5).

(a) in direction 4−87

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

(b) parallel to the line x3

=y −10−2

=z + 813

=> (a) by inspection x − 64

=y − 3−8

=z + 5

7

=> (b) both lines must have the same direction vector 3−213

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

i.e. x − 63

=y − 3−2

=z + 513


Page 12

The Equation of a Plane

The Normal Vector.The normal vector, n, of a plane is perpendicular to the plane if it is perpendicular to every vector which lies in the plane, Π.

The equation of a plane:-( i ) in the Scalar Product Form of the Vector Equation.

Let n ≠ 0 be a fixed vector and let P with position vector p be a fixed point.

Let Π denote the plane through Pnormal to n.Let X with position vector x be a variable point.

Then the following statements are equivalent : (i) X lies on Π. (ii) n is perpendicular to PX.

Hence (iii) n.(x – p ) = 0

i.e. n.x – n. p = 0 => n.x = n. p

This equation is true if and only if X lies on Π.Since n and p are fixed, n. p = k where k is a constant.Therefore the equation of a plane can be written in the form n.x = kwhere k is a constant, n is the normal vector to the plane and x is the position vector of any point on the plane.

n

xpPX

OΠ

Exercise 3

1. Find the symmetrical and vector equation of the line through

the point (5, –2, 6) in the direction 314

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ .

2. Find the symmetrical and vector equation of the line through the point (2, –3, –7) and parallel to the line x + 5

7=

y −133

=z − 4−2

.

3. Find the symmetrical equations of the lines joining the followingpairs of points:(a) (3, 2, –7), (5, –13, –4) (b) (–8, –13, –9), (12, 7, 1)(c) (3, 0, 0), (0, 0, 5) (d) (0, 0, 0), (–10, 4, –6)


The Complete A level Maths (Orlando Gough) (no reference)


Page 420 Exercise 17C Questions 1 – 11


Page 13

( i i i ) in Symmetrical or Cartesian form.In coordinate terms, if n = ai + bj + ck and x = xi + y j + zkthen the equation of the plane, with the aid of the scalar product is written

abc

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ .

xyz

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

= k or ax + by + cz = k

Noteabc

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ is called the normal vector

or its components are called the direction ratiosor if equal to the unit vector, direction cosines.

Examples

1. Find a parametric equation of the plane containing the three points A, B and C whose coordinates are (2, 1, 3), (7, 2, 3) and (5, 3, 5) respectively.

The position vectors of A, B and C are:-a = 2i + j + 3k, b = 7i + 2 j + 3k and c = 5i + 3 j + 5k

AB→

= (7i + 2 j + 3k) – (2i + j + 3k) = 5i + j

AC→

= (5i + 3 j + 5k) – (2i + j + 3k) = 3i + 2 j + 2k

The parametric equation is thereforer = 2i + j + 3k + λ(5i + j ) + µ(3i + 2 j + 2k)

( i i ) in Parametric Form for the Vector Equation.

Consider the plane Π which is parallel to vectors u and v (where u is not parallel to v) and which also contains the point A whose position vector is a.

Let R be any point on Π with position vector r and

so AR→

lies on Π.

If R is any point on this plane,AR→

= λu+µvwhere λ and µ are parameters.

If r is the position vector of R, r = a + AR→

i.e. r = a + λu + µv

Thus any equation of the form r = a + λu + µv , where λ and µ areparameters, represents the plane parallel to the vectors u and v andcontaining the point a.

O

A

R

r

a

v

u

Π

v


Page 14

2. Find the Cartesian equation of the plane containing the pointsA(0, 1, – 1), B(1, 1, 0) and C(1, 2, 0).

AB→

= b – a = 110

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ –

01−1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ =

101

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ , AC

→ = c – a =

120

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ –

01−1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ =

111

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

AB→

x AC→

= −101

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ from

Hence the normal vector has components −101

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

The equation of the plane is of the form – x + z = kSince A(0, 1, – 1) lies on the plane, substitute in the aboveequation to find k. i.e. k = – 1=> The equation of the plane is – x + z = – 1 or x – z = 1

3. Find the equation of the plane containing the pointsA(1, 2, 3), B(0, 3, 2) and C(3, 0, 5).

AB→

= b – a = 032

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ –

123

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ =

−11−1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ,AC→

= c – a = 305

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ –

123

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ =

2−22

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

AB→

x AC→

= 000

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ from

The normal vector has components 000

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ which is impossible

Therefore the points A, B, C must be collinear and an infinity of planes must pass through A, B and C.

Notice that AB→

= −11−1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ and AC

→ =

2−22

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ = 2

−11−1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

4. Find the Cartesian equation of the plane through (–1, 2, 3)containing the direction vectors 8i + 5 j + k and – 4i + 5 j + 7k.

The normal vector n is perpendicular to both the given vectors.

Let n = abc

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ and so

abc

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ .

851

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

= 0 and abc

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ .−457

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ = 0

i.e. 8a + 5b + c = 0 and –4a + 5b + 7c = 0Solve simultaneously to find b and c in terms of a, we obtainb = –2a and c = 2a.

i.e. n = a

2a−2a

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ or

12−2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

AB→

0 1 1 0

AC→

1 1 1 1

AB→

1 –1 –1 –1

AC→

–2 2 2 –2


Page 15

Exercise 4

1. Find the Cartesian equation of the plane:-

(a) with norm 3−27

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ and through the point (3, –1, 1)

(b) with norm 2i – j – k and through (4, 1, –2)

2. Find the Cartesian equation of the plane through the given point, containing the stated directions :

(a) (1, 2, –3), 4−23

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ,

110

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

(b) (1, –1, 1), 3i – 5 j – 7k, – 4i – j + 6k

3. Find the Cartesian equation of the plane through the following setsof three points :-(a) ((2, 1, 3), (4, 1, 4), (2, 3, 6)(b) (3, 0, 0), (0, 5, 0),(0, 0, 7)

The equation of the plane is therefore in the form x + 2y – 2z = k

Since A(–1, 2, 3) lies on the plane, substitute in the aboveequation to find k. i.e. k = – 3The equation of the plane is x + 2y –2z = – 3

5. Find the Cartesian equation of the plane containing the pointP(3, –2, –7) and the line x − 5

3=

y1

=z + 6

4.

The point Q(5, 0, –6) lies on the line and the point P(3, –2, –7) lies on the plane.

A direction vector on the plane PQ→

= 506

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ –

3−2−7

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ =

221

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

The direction vector of the line = 314

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

The normal vector to the plane is221

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ x

314

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ =

7−5−4

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ from

The equation of the plane is therefore in the form 7x – 5y – 4z = k

Since P(3, –2, –7) lies on the plane, substitute in the aboveequation to find k. i.e. k = 59The equation of the plane is 7x – 5y – 4z = 59

2 1 2 2

1 4 3 1


Page 16


The Complete A level Maths (Orlando Gough). No reference


Page 429 Exercise 17D Questions 1 – 12

4. Find the Cartesian equation of the plane containing both the point and the line given :

(a) (5, 8, – 4), x−4

=y − 5

1=

z + 10

(b) (2, –5, 3), x −1−3

=y + 7

5=

z − 32

5. A plane is parallel to both linesx −1

2=

y3

=z −1

4 and x + 1

−1=

y2

=z1

and passes through the point (1, 0, –1). Find its equation.

6. A plane passes through the points (0, 1, 2) and (1, –1, 0) and is

parallel to the direction 1−11

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ .

Find its equation.

Angles between Two lines, Two Planes, or a Line and Plane

( i ) The angle between Two Lines.The angle between two lines is the angle between their directionvectors and can be found using the Scalar Product.

Example Find the size of the angle between the linesx – 1 = y = z – 1 and x = 1 + t , y = 5t, z = –t.

The lines can be expressed in the symmetrical form i.e. x −1

1=

y1

=z −1

1 and x −1

1=

y5

=z−1

Their direction vectors are a = 111

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ and b =

15−1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

Using cosθ° = a . b| a | | b |

= 53.3 3

=59

=> θ = 56·3


Page 17

( i i ) The angle between Two Planes:-

The angle between two planes is defined to be the angle between theirnormal vectors.

Example. Find the angle between the planes x + 2y + z = 0 and x + y = 0.

Their normal vectors are a = 121

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ and b =

110

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ .

cosθ = a . b| a | | b |

= 36 2

= 32 3

= 32

=> θ = π/6

( i i i ) The angle between a Line and a Plane:-

If θ° is the angle between a line and the normal vector to the plane, then (90 – θ)° is the angle between the line and the plane.

Note (i) (90 – θ)° is the angle between the line and its projection on the plane.

(ii) (90 – θ)° is the smallest angle between the line and the plane.

Example. Find the angle between the line x

1=

y1

=z0

and the plane x + z = 0

The normal vector of the plane = 101

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

The direction vector of the line = 110

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

By the scalar product cosθ° = 1

2 2 = 1

2 => θ = 60

The angle between the line and plane = (90 – θ)° = 30°

LineNormalVector

Plane

θ° (90 – θ)°


Page 18

The Intersection of Two Lines, Two or Three Planes, or a Line and Plane.

( i ) The Intersection of Two Lines.

Two distinct lines in a plane are either parallel or intersecting.In three dimensions there are three possibilities : they may be parallel, or intersecting or skew (neither parallel nor intersecting).

The following example demonstrates how to find whether two lines intersect and if they do, how to find the point of intersection.

Examplex + 9

4=

y + 51

=z + 1−2

. . 1( ) x − 8−5

=y − 2−4

=z − 5

8 . . 2( )

x − 8−5

=y − 2−4

=z + 15

8 . . 3( )

(a) Find where lines (1) and (2) intersect.(b) Find where lines (1) and (3) intersect.

=> (a) If there is a point (p, q, r) lying on both lines,

then p + 94

=q+ 5

1=

r + 1−2

= λ and p − 8−5

=q − 2−4

=r − 5

8= µ

Therefore p = 4λ – 9 = –5µ + 8 ......(4)q = λ – 5 = – 4µ + 2 ......(5)

r = –2λ – 1 = 8µ + 5 ......(6)

Solving (4) and (5) => λ = 3 and µ = 1.Substitute these values in (6) gives –7 = 13.

Since these values do not satisfy (6), we conclude

that the lines (1) and (2) do not intersect.

(b) If there is a point (p, q, r) lying on both lines,

then p + 94

=q+ 5

1=

r + 1−2

= λ and p − 8−5

=q − 2−4

=r + 15

8= µ

Therefore p = 4λ – 9 = –5µ + 8 ......(7)q = λ – 5 = –4µ + 2 ......(8)r = –2λ – 1 = 8µ – 15 .....(9)

Solving (7) and (8) => λ = 3 and µ = 1.Substitute these values in (9) gives –7 = –7.

Since these values satisfy (9), we conclude that the lines (1) and (3) intersect.

Using λ = 3 (or µ =1) in (7), (8) and (9),=> p = 3, q = –2 and r = –7.Therefore lines (1) and (3) intersect at (3, –2, –7).


Page 19

( i i ) The Intersection of Two Planes.

Two non-parallel planes will always meet in a straight line.Given the equation of two planes, we can proceed as follows:-

Example Find the equation of the line of intersection of the planes3x – 5y + z = 8 and 2x – 3y + z = 3.

Method 1.For any point (x, y, z) which lies on both planes, the values of x, y and z fit both equations simultaneously.Hence eliminating z from both equations (by subtractionin this case) gives x – 2y = 5.

There are infinitely many pairs of values of x and y which satisfy this equation, but if we choose a value of x then the value of y is fixed and vice-versa.

Let y = t, then x = 5 + 2t and substituting these expressions for x and y into 3x – 5y + z = 8 gives

3(5 + 2t ) – 5t + z = 8 => z = –7 – t

i.e. x = 5 + 2t, y = t, z = –7 – t are the parametricequations of the line

or in Cartesian form x − 52

=y1

=z + 7−1

= t( )

Method 2.To find the equation of a line, we require its directionvector and a point on the line.The direction vector of the line is perpendicular to both normal vectors of the planes.

The normal vectors of the planes are a =3−51

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

, b = 2−31

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ .

The direction vector of the line of intersection must beparallel to a x b.

a x b = −2−11

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

Let z = 0, hence 3x – 5y = 8, 2x – 3y = 3.Solving these gives x = –9, y = –7.i.e. (–9, –7, 0) is a point on the line.

The equation of the line in Cartesian form isx + 9−2

=y + 7−1

=z1

or x + 92

=y + 7

1=

z−1

which can be turned into x − 52

=y1

=z + 7−1

by subtracting 7 from each part.


Page 20

( i i i ) The Intersection of Three Planes.

The solution of a system of 3 planes is a point common to the 3 planes.If the matrix of the coefficients is singular (see Matrices) then the equations do not have a unique solution.The equations could either have:-

(a) a unique solution, intersecting at one point.

(b) a linear solution,in which case oneequation is a linear combination of theother two.(or the 3 planes are identical)There are an infinite set of points which are common to all 3 planes.

(c) no solution, in which casesthe three planes are parallel,

or

two are parallel

or

one plane is parallel to the lineof intersection of the other two.


Page 21

ExampleShow that the planes:-

A : 2x – y + 5z = –4B : 3x – y + 2z = –1C : 4x – y – z = 2

intersect on a line and find its equation.

The normal vector of A : a = 2−15

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

The normal vector of B : b = 3−12

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

The normal vector of C : c = 4−1−1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

The directon vector of the line of intersection of A and Bis parallel to a x b.

i.e. a x b = 3111

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ from

If (a x b).c = 0, the direction vector of the line of intersection is perpendicular to the normal vector of plane C.=> the line of intersection of planes A and B must be

parallel to the plane C.i.e. the three planes intersect on a line.

Let z = 0, A : 2x – y = –4B : 3x – y = –1C : 4x – y = 2

From A and B, x = 3 and y = 10.A point on the line of intersection of A and B is therefore (3,10, 0).

This point also satisfies C, since 12 – 10 – 0 = 2.i.e. (3,10, 0) lies on all three planes.

The equation of the line is x − 33

=y −10

11=

z1

Note If the planes intersect at a point, the soltion can be found by Gaussian elimination or Matrices (next outcome).

a –1 5 2 –1

b –1 2 3 –1


Page 22

Exercise 5 1. Calculate the acute angle between the following pairs of lines :

(a) x −11

=y + 2−2

=z2

, x + 33

=y + 1

0=

z − 2−4

(b) x – 1 = y = z –1, x = 1 + t, y = 5t, z = –t

(c) x + 43

=y −1

5=

z + 34

, x + 11

=y − 4

1=

z − 52

(d) r = 4i – j + λ(i + 2 j – 2k), r = i – j + 2k – µ(2i + 4 j – 4k)

2. Find the acute angle between the following pairs of planes :

(a) 2x + 2y – 3z = 3, x + 3y – 4z = 6

(b) 5x – 14y + 2z = 13, 6x + 7y + 6z = –23

(c) r .(i – j ) = 4, r .( j + k) = 1

(d) r .(i + j + k) = 1, r .(i – j + k) = 0

3. Find the acute angle between the following lines and planes :

(a) x4

=y −1−1

=z + 3−5

, x – 2y + 4z = –3

(b) x − 22

=y + 1

6=

z + 33

, 2x – y – 2z = 4

(c) x + 12

=y3

=z − 3

6, 10x + 2y –11z = 3

(d) r = i – j + λ(i + j + k), r .(i – 2 j + 2k) = 4

4. Find the coordinates of the point of intersection of these lines :

(a) x − 41

=y −8

2=

z − 31

, x − 76

=y − 6

4=

z − 55

(b) x − 21

=y − 9

2=

z −133

, x + 3−1

=y − 7

2=

z + 2−3

5. Find the equation of the line of intersection of the following two planes :

(a) x + y + 2z = 2, x – y – z = 5

(b) 2x – y = 3, x + y + 4z = 1

(c) 2x + 3y + z = 8, x + y + z = 10, 3x + 5y + z = 6.

6. Find the coordinates of the point of intersection of the following line and plane :

(a) x = 4 + t, y = 1 – t, z = 3t, 2x + 4y + z = 9

(b) x −13

=y− 2

1=

z −14

, x – 2y + 3z = 26


Page 23

7. A plane contains the line x −11

=y− 2−1

=z −1

2 and the plane is parallel to

the line x1

=y + 7

2=

z3

.

Find the equation of the plane.

8. (a) Find the coordinates of the point A in which the line L with equation x + 1

2=

y − 2−1

=z + 3

2 meets the plane Π with equation

3x – y + z = 10.(b) Hence find the equation in cartesian form for the line through A

lying wholly in the plane and perpendicular to the line.

9. (a) Find parametric equations for the line L joining the points A(2, – 4, 3) and B(4, 0 –5).

(b) Verify that L is perpendicular to the line M joining the point Ato C(–2, –6, 1).

(c) Determine the equation of the plane containing L and M.

10. (a) Show that the line L with parametric equationsx = 2 – t, y = –3 + 2t , z = –1 – 4tlies on the plane Π with equation 2x + 3y + z = – 6.

(b) Find parametric equations for the line M through the point (3, 2, –4) perpendicular to Π.

(c) Prove that M meets Π at a point lying on L.

11. (a) Show that the line L joining the points A(2, 1, –1) and B(3, –2, 1) is perpendicular to the line M with parametric equations x = 11 + 4t, y = 3 + 2t, z = 1 + t.

(b) Find the equation of the plane Π through L perpendicular to Mand prove that Π meets M at a point C equidistant from A and B

12. (a) Find paramtetric equations for the line joining A(1, –1, 2) andB(4, 5, –7).

(b) Prove that AB intersects the line with parametric equations x = 6 + 4t, y = 2 + t, z = 1 + 2t at right angles.

(c) Find the coordinates of the point of intersection of lines.



No reference


Page 420 Exercise 17C Questions 13 – 18Page 429 Exercise 17D Questions 16 – 28


Page 24

Answers

Exercise 1

1.(a) Number, 8 (b) Vector, 5i + 7 j + 9k (c) Vector, 4i + 4k (d) Number, 45.2. AB = √38, BC = 3√6, CA = √10,

cosA = − 32 95

, cosB = 416 57

, cosC = 136 15

, Area = 12

371

3. Proof 4. 12

i + 12

j , 12

i – 12

j .

5. (a) −3 + 2 26

i + 26

j + 3 + 2 26

k, −3 − 2 26

i – 26

j + 3 − 2 26

k (b) Proof

Exercise 2

1.(a) –10i + 7 j – 16k (b) 5i – 5k (c) –20 (d) –15 (e) –1252.(a) –11i + 14 j + k (b) 2 (c) –2 3. Proof4. 5i + 31 j + 14k 5. (a) – 3 j – 3k (b) – 3 j – 3k (c) Proof

6. −1

133i + 4 j + 12k( ),

113

3i + 4 j + 12k( ) 7. 111/2 8. Proof

Exercise 3

1. x − 53

=y + 2

1=

z − 64

, x =5−26

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

+ t314

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

2. x − 27

=y + 3

3=

z + 7−2

, x =2−3−7

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

+ t73−2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

3.(a) x − 32

=y − 2−15

=z + 7

3(b) x −12

2=

y− 72

=z −1

1

(c) x − 33

=y0

=z5

(d) x–5

=y2

=z–3

Exercise 4 Pages

1. (a) 3x – 2y + 7z = 18 (b) 2x – y – z = 9

2. (a) x – y – 2z = 5 (b) x + 2y + z = 0

3. (a) x + 3y – 2z = –1 (b) 35x + 21y + 15z = 105

4. (a) 3x + 12y + 17z = 43 (b) 4x – 2y + 11z = 51

5. 5x + 6y – 7z = 12

6. 4x + 3y – z = 1

cont’d .....


Page 25

Exercise 5

1. (a) 70·5° (b) 56·3° (c) 22·5° (d) 0

2. (a) 25·2° (b) 70·2° (c) 60° (d) 70·5°

3. (a) 28·5° (b) 22·4° (c) 22·4° (d) 11·1°

4. (a) (1, 2, 0) (b) (–1, 3, 4)

5. (a) x − 315

=y −127

=z7

(b) x + 15

=y + 123

=z + 122

(c) x − 222

=y + 12−1

=z−1

6. (a) (1, 4, –9) (b) (7, 4, 9)

7. 7x + y – 3z = 6

8. (a) (3, 0, 1) (b) x − 31

=y4

=z −1

1

9. (a) x = 2 + t, y = – 4 + 2t, z = 3 – 4t (b) Proof

(c) 2x – 3y – z = 13

10. (a) Proof (b) x = 3 + 2t, y = 2 + 3t, z = –4 + t (c) Proof

11. (a) Proof (b) 4x + 2y + z = 9, C(3, –1, –1)

12. (a) x = 1 + t, y = –1 + 2t, z = 2 – 3t (b) Proof(c) (2, 1, –1)


Page 26

A matrix (matrices) is a rectangular array of numbers arranged in rows and columns, the array being enclosed in round (or square) brackets.

e.g. xy

⎛

⎝ ⎜ ⎞

⎠ ⎟

3 10 5⎛

⎝ ⎜

⎞

⎠ ⎟

6 8 103 4 5⎛

⎝ ⎜

⎞

⎠ ⎟ 4 −2 5( )

2 rows 2 rows 2 rows 1 row 1 column 2 columns 3 columns 3 columns

Each number in the array is called an entry or an element of the matrix and is identified by first stating the row and then the column in which it appears.

A matrix is often denoted by a capital letter. The order of a matrix is given by stating the number of rows followed by the number of columns.

A = 4 6 82 3 4⎛

⎝ ⎜

⎞

⎠ ⎟ B =

3 42 −6⎛

⎝ ⎜

⎞

⎠ ⎟

Matrix A has 2 rows B has the same number of rows and 3 columns and as columns and is called a is said to be of order square matrix of order 2. 2 x 3 , (reads 2 by 3).

In general, a matrix A, of order (m x n), has m rows and n columns and can be represented as follows,where aij denotes element in the ith row and jth column.

A =

a11 a12 a13 − − − a1n

a21 a22 a23 − − − a2n

a31 a32 a33 − − − a3n

" " " " " " "" " " " " " "" " " " " " "

am1 am2 am 3 − − − amn

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟

Equal Matrices :- Two matrices A and B are equal when :-

(a) they are of the same order and (b) their corresponding elements are equal.

The Zero Matrix is a matrix whose elements are all zero.

Transpose of a Matrix -: A new matrix can be formed from A by writing row 1 as column 1, row 2 as column 2, row 3 as column 3, etc. This new matrix is called the Transpose of A, and is

denoted by ′ A (reads as “A dashed” or A transpose).

e.g. A = 3 1 4 25 4 0 7⎛

⎝ ⎜

⎞

⎠ ⎟ , => transpose ′ A =

3 51 44 02 7

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟

Outcome 2 – Use Matrix Algebra


Page 27

Exercise 1

1. State the order of each of the following matrices:-

(a)3 1 4 25 4 0 7⎛

⎝ ⎜

⎞

⎠ ⎟ (b)

2 −14 81 −2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

(c)

1 23 45 67 8

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟

2. List the order of these matrices and say if any pairs are equal.

A = 1 2 3( ) B = 3 2 1( ) C = 1 2 3( )

D = 2−1⎛

⎝ ⎜

⎞

⎠ ⎟ E =

12⎛

⎝ ⎜ ⎞

⎠ ⎟ F =

21⎛

⎝ ⎜ ⎞

⎠ ⎟ G =

21⎛

⎝ ⎜ ⎞

⎠ ⎟

H = 1 23 4⎛

⎝ ⎜

⎞

⎠ ⎟ J =

−1 −2−3 −4⎛

⎝ ⎜

⎞

⎠ ⎟ K =

1 32 4⎛

⎝ ⎜

⎞

⎠ ⎟ L =

1 23 4⎛

⎝ ⎜

⎞

⎠ ⎟

3. Determine the values of x and y in each of the following:-

(a) 3x −y( ) = 12 3( ) (b)x + 34 − y⎛

⎝ ⎜

⎞

⎠ ⎟ =

75⎛

⎝ ⎜ ⎞

⎠ ⎟

(c)x + 2y2x − y⎛

⎝ ⎜

⎞

⎠ ⎟ =

98⎛

⎝ ⎜ ⎞

⎠ ⎟ (d) x2 y2

y3 x3

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ =

4 9−27 8⎛

⎝ ⎜

⎞

⎠ ⎟

4. Write down the transpose of each matrix in question 1 and state the order of each transpose.

5. For the matrix A = 3 1 4 25 4 0 7⎛

⎝ ⎜

⎞

⎠ ⎟ , show that ( ′ A ′ ) = A

6. P = x 9−3 y⎛

⎝ ⎜

⎞

⎠ ⎟ and Q =

5 −39 −4⎛

⎝ ⎜

⎞

⎠ ⎟ . Find x and y, given that ′ P = Q

Addition of matrices -: If two matrices A and B are of the same order, they can be added to make a new matrix A + B, formed by adding each element of A to the corresponding element of B.

In general If A = a bc d⎛

⎝ ⎜

⎞

⎠ ⎟ and B =

p qr s

⎛

⎝ ⎜

⎞

⎠ ⎟ => A + B =

a + p b + qc + r d + s

⎛

⎝ ⎜

⎞

⎠ ⎟


Page 28

Exercise 2

1. Find the sum of the following matrices:-

(a)31⎛

⎝ ⎜ ⎞

⎠ ⎟ +

45

⎛

⎝ ⎜ ⎞

⎠ ⎟ (b)

24⎛

⎝ ⎜ ⎞

⎠ ⎟ +

−13

⎛

⎝ ⎜

⎞

⎠ ⎟ (c)

2ab

⎛

⎝ ⎜

⎞

⎠ ⎟ +

7a−3b⎛

⎝ ⎜

⎞

⎠ ⎟

(d)2u−3v⎛

⎝ ⎜

⎞

⎠ ⎟ +

−2u3v

⎛

⎝ ⎜

⎞

⎠ ⎟ (e) 2 5( ) + 1 4( ) (f) 2 −3( ) + −5 8( )

(g) 1 00 1⎛

⎝ ⎜

⎞

⎠ ⎟ +

2 34 5⎛

⎝ ⎜

⎞

⎠ ⎟ (h)

2 3 15 1 0⎛

⎝ ⎜

⎞

⎠ ⎟ +

1 −2 3−7 1 −4⎛

⎝ ⎜

⎞

⎠ ⎟

2. A = 3 21 −1⎛

⎝ ⎜

⎞

⎠ ⎟ , B =

1 45 2⎛

⎝ ⎜

⎞

⎠ ⎟ and C =

−2 31 −4

⎛

⎝ ⎜

⎞

⎠ ⎟

Find the matrices:-

(a) A + B (b) B + C (c) (A + B) + C (d) A + (B + C)

Is it true that (A + B) + C = A + (B + C) ?

3. Given that A = 3 −4−5 1⎛

⎝ ⎜

⎞

⎠ ⎟ and B =

−3 45 −1

⎛

⎝ ⎜

⎞

⎠ ⎟ , find the matrices:-

(a) A + B (b) B + A

Comment on your results.

4. For the matrices A = 3 1 4 25 4 0 7⎛

⎝ ⎜

⎞

⎠ ⎟ and B =

2 −1 3 −23 8 1 0⎛

⎝ ⎜

⎞

⎠ ⎟

prove that (A + B ′ ) = ′ A + ′ B ?

Subtraction of matrices -: If two matrices A and B are of the same order, they can be subtracted to form a matrix A – B, formed by subtracting each element of B from the corresponding element of A.


⎝ ⎜

⎞

⎠ ⎟ and B =

p qr s

⎛

⎝ ⎜

⎞

⎠ ⎟ => A – B =

a − p b − qc − r d − s

⎛

⎝ ⎜

⎞

⎠ ⎟


Page 29

Exercise 3

1. Subtract the following matrices.

(a) 31⎛

⎝ ⎜ ⎞

⎠ ⎟ −

45⎛

⎝ ⎜ ⎞

⎠ ⎟ (b)

24⎛

⎝ ⎜ ⎞

⎠ ⎟ −

−13

⎛

⎝ ⎜

⎞

⎠ ⎟ (c)

2ab

⎛

⎝ ⎜

⎞

⎠ ⎟ −

7a−3b⎛

⎝ ⎜

⎞

⎠ ⎟

(d) 2u−3v⎛

⎝ ⎜

⎞

⎠ ⎟ −

−2u3v

⎛

⎝ ⎜

⎞

⎠ ⎟ (e) 2 5( ) − 1 4( ) (f) 2 −3( ) − −5 8( )

(g) 1 00 1⎛

⎝ ⎜

⎞

⎠ ⎟ −

2 34 5⎛

⎝ ⎜

⎞

⎠ ⎟ (h)

4 72 3⎛

⎝ ⎜

⎞

⎠ ⎟ −

8 03 9⎛

⎝ ⎜

⎞

⎠ ⎟ (i)

2 3 15 1 0⎛

⎝ ⎜

⎞

⎠ ⎟ −

1 −2 3−7 1 −4⎛

⎝ ⎜

⎞

⎠ ⎟

2. A = 1 23 4⎛

⎝ ⎜

⎞

⎠ ⎟ , B =

−2 30 1

⎛

⎝ ⎜

⎞

⎠ ⎟ and C =

5 2−1 0⎛

⎝ ⎜

⎞

⎠ ⎟ . Find in simplest form :-

(a) A + B (b) A + C (c) A + B + C(d) A – B (e) C – B (f) C – A(g) (A + C) + (A + B) (h) (A + C) – (A + B)

3. Solve each of the following equations for the 2 x 2 matrix X:-

(a) X + 1 00 1⎛

⎝ ⎜

⎞

⎠ ⎟ =

0 22 1⎛

⎝ ⎜

⎞

⎠ ⎟ (b) X +

2 34 5⎛

⎝ ⎜

⎞

⎠ ⎟ =

4 −13 2⎛

⎝ ⎜

⎞

⎠ ⎟

4. If A = 1 23 45 6

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

, B = 3 −21 54 6

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

and C = 4 21 0−3 5

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

, simplify:-

(a) A + B (b) B – C (c) (A + B) – C (d) A + (B – C)

Multiplication of a matrix by a real number –: If k is a real number and A is a matrix,then kA is a new matrix obtained by multiplying each element of A by k.


⎝ ⎜

⎞

⎠ ⎟ then kA =

ka kbkc kd⎛

⎝ ⎜

⎞

⎠ ⎟

Example If A = 2 1 3−1 0 4⎛

⎝ ⎜

⎞

⎠ ⎟ and B =

2 −1 03 4 5⎛

⎝ ⎜

⎞

⎠ ⎟ , show that :-

=> 3A – 2B = 2 5 9−9 −8 2⎛

⎝ ⎜

⎞

⎠ ⎟

3A – 2B = 32 1 3−1 0 4⎛

⎝ ⎜

⎞

⎠ ⎟ – 2

2 −1 03 4 5⎛

⎝ ⎜

⎞

⎠ ⎟ =

6 3 9−3 0 12⎛

⎝ ⎜

⎞

⎠ ⎟ –

4 −2 06 8 10⎛

⎝ ⎜

⎞

⎠ ⎟

= 2 5 9−9 −8 2⎛

⎝ ⎜

⎞

⎠ ⎟


Page 30

Exercise 4

1. If A = 3 4−1 −2⎛

⎝ ⎜

⎞

⎠ ⎟ , find (a) 2A (b) 3A (c) –5A (d) –A

2. If A = 3 4 12 0 3⎛

⎝ ⎜

⎞

⎠ ⎟ and B =

2 −1 34 5 0⎛

⎝ ⎜

⎞

⎠ ⎟ , find in their simplest form:-

(a) A – B (b) 2(A + B) (c) 2A

(d) 2B (e) 2A + 2B (f) 6A

(g) 3(2A) (h) 8B (i) 4(2B)

This question demonstrates the following properties:_(i) k(A ± B) = kA ± kB and (ii) k(tA) = (kt)A

where k and t are real numbers.

3. If A = 2 −34 1⎛

⎝ ⎜

⎞

⎠ ⎟ and B =

−4 13 −2

⎛

⎝ ⎜

⎞

⎠ ⎟ , simplify:-

(a) 3A + 2B (b) 4A – 3B (c) 5A – 4B (d) 2(A – 5B)

4. Solve each of the following equations for the matrix X :-

(a) 3X = 6 −312 9⎛

⎝ ⎜

⎞

⎠ ⎟ (b) 2X +

3 14 2⎛

⎝ ⎜

⎞

⎠ ⎟ =

9 52 8⎛

⎝ ⎜

⎞

⎠ ⎟

(c) 4X – 3 14 7⎛

⎝ ⎜

⎞

⎠ ⎟ =

5 30 13⎛

⎝ ⎜

⎞

⎠ ⎟ (d)

7 1−4 3⎛

⎝ ⎜

⎞

⎠ ⎟ – 3X =

−5 108 9

⎛

⎝ ⎜

⎞

⎠ ⎟

5. Find the matrix X in each of the following:-

(a) 21 −1 32 −7 5⎛

⎝ ⎜

⎞

⎠ ⎟ + X = 3

1 2 −43 −5 1⎛

⎝ ⎜

⎞

⎠ ⎟ (b) 5

1 23 4⎛

⎝ ⎜

⎞

⎠ ⎟ – 3X = 4

−4 73 8

⎛

⎝ ⎜

⎞

⎠ ⎟

6. Given that 2p qr s

⎛

⎝ ⎜

⎞

⎠ ⎟ +

7 −2−4 5

⎛

⎝ ⎜

⎞

⎠ ⎟ =

5 62 1⎛

⎝ ⎜

⎞

⎠ ⎟ , find p, q, r and s.

Multiplication of matrices :- Consider the two linear expressions ax + bycx + dy⎛

⎝ ⎜

⎞

⎠ ⎟

We can write these in the form a bc d⎛

⎝ ⎜

⎞

⎠ ⎟

xy

⎛

⎝ ⎜ ⎞

⎠ ⎟

We can think of this as a product of a 2 x 2 matrix and a 2 x 1 matrix.To obtain the linear expressions from this product, proceed as follows :-

multiply each element of a row in the first matrix by the corresponding element in the column of the second matrix and then add the products to give a (2 x 1) matrix.


Page 31

Examples.

1.3 2 15 −3 7⎛

⎝ ⎜

⎞

⎠ ⎟

xyz

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

= 3x + 2y + z5x − 3y + 7z⎛

⎝ ⎜

⎞

⎠ ⎟ this is a 2 x 1 matrix.

2. 4 3 2( )31−5

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

=(4 x 3 + 3 x 1 + 2 x (–5)) = (12 + 3 – 10) = (5)

Note:- in 1. The product is of a (2 x 3) matrix and a (3 x 1) matrix giving a (2 x 1) matrix.

Note:- in 2. The product is of a (1 x 3) matrix and a (3 x 1) matrix giving a (1 x 1) matrix.

In general The product of an (m x p) matrix and a (p x 1) matrix produces an (m x 1) matrix.

The product of an (m x p) matrix and an (p x n) matrix can be looked upon as an extension of the product of an (m x p) matrix and a (p x 1) matrix .

In the examples above, the number of columns in the first matrix is the same as the number of rows in the second matrix.

The product of of an (m x p) and an (p x n) should therefore produce an (m x n) matrix.

Matrices must be CONFORMABLE for Multiplication.

e.g.3 −2 15 0 6⎛

⎝ ⎜

⎞

⎠ ⎟ −1 58 −31 4

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

these matrices can be multiplied

Order (2 x 3) x (3 x 2) = (2 x 2) matrix.

Think of dominoes.

The above example would then give :-

3 × −1( ) + −2( ) × 8 + 1 ×1( ) 3 × 5 + −2( ) × −3( ) + 1× 4( )5 × −1( ) + 0 × 8 + 6 ×1( ) 5 × 5 + 0 × −3( ) + 6 × 4( )

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ =

−18 251 49

⎛

⎝ ⎜

⎞

⎠ ⎟

First in row x First in column

Third in row x Third in column

Second in row x Second in column

same these numbers must match


Page 32

Exercise 5

1. Find the following matrix products by first considering the order of the product.

(a) 1 2( )34⎛

⎝ ⎜ ⎞

⎠ ⎟ (b) 3 4( )

25⎛

⎝ ⎜ ⎞

⎠ ⎟ (c) 5 −2( )

1−2⎛

⎝ ⎜

⎞

⎠ ⎟

(d) 3 1 2( )213

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

(e) 2 −3 4( )512

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

(f) 8 −5 −1( )xyz

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

(g)3 11 2⎛

⎝ ⎜

⎞

⎠ ⎟

23⎛

⎝ ⎜ ⎞

⎠ ⎟ (h)

2 10 1⎛

⎝ ⎜

⎞

⎠ ⎟

34⎛

⎝ ⎜ ⎞

⎠ ⎟ (i)

2 −11 2⎛

⎝ ⎜

⎞

⎠ ⎟

23⎛

⎝ ⎜ ⎞

⎠ ⎟

(j)1 −11 0⎛

⎝ ⎜

⎞

⎠ ⎟

5−2⎛

⎝ ⎜

⎞

⎠ ⎟ (k)

2 31 −2⎛

⎝ ⎜

⎞

⎠ ⎟

1−3⎛

⎝ ⎜

⎞

⎠ ⎟ (l)

0 −11 0⎛

⎝ ⎜

⎞

⎠ ⎟

6−4

⎛

⎝ ⎜

⎞

⎠ ⎟

(m)1 −12 3⎛

⎝ ⎜

⎞

⎠ ⎟

54⎛

⎝ ⎜ ⎞

⎠ ⎟ (n)

54⎛

⎝ ⎜ ⎞

⎠ ⎟

1 −12 3⎛

⎝ ⎜

⎞

⎠ ⎟ (o) 4 3( )

52⎛

⎝ ⎜ ⎞

⎠ ⎟

(p)2 1 33 0 11 2 3

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

401

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

(q) 2 13 01 2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

401

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

(r)1 −2 3−1 4 23 1 0

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

213

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

(s) cosa sina( )cosa – sinasin a cosa

⎛

⎝ ⎜

⎞

⎠ ⎟

2. By forming a system of simultaneous equations, find x and y.

(a)3 00 −2⎛

⎝ ⎜

⎞

⎠ ⎟

xy

⎛

⎝ ⎜ ⎞

⎠ ⎟ =

128

⎛

⎝ ⎜

⎞

⎠ ⎟ (b)

x 01 y⎛

⎝ ⎜

⎞

⎠ ⎟

23⎛

⎝ ⎜ ⎞

⎠ ⎟ =

6−1⎛

⎝ ⎜

⎞

⎠ ⎟

(c)2 −11 −3⎛

⎝ ⎜

⎞

⎠ ⎟

xy

⎛

⎝ ⎜ ⎞

⎠ ⎟ =

11−7⎛

⎝ ⎜

⎞

⎠ ⎟ (d)

x yy x

⎛

⎝ ⎜

⎞

⎠ ⎟

31⎛

⎝ ⎜ ⎞

⎠ ⎟ =

5−1⎛

⎝ ⎜

⎞

⎠ ⎟

3. If A = 1 23 1⎛

⎝ ⎜

⎞

⎠ ⎟ , B =

4 52 0⎛

⎝ ⎜

⎞

⎠ ⎟ , C =

4−3⎛

⎝ ⎜

⎞

⎠ ⎟

(a) find (i) AB (ii) BA and comment on the result.

(b) find (i) A(BC) (ii) (AB)C and comment on the result.

(c) Show that (AB ′ ) = ′ B ′ A .

4. Given that A = 3 −15 2⎛

⎝ ⎜

⎞

⎠ ⎟ , find A2 and A3 . (Note :- A2 ≠

9 125 4⎛

⎝ ⎜

⎞

⎠ ⎟ )

5. If P = 1 −2−3 6⎛

⎝ ⎜

⎞

⎠ ⎟ and Q =

2 21 1⎛

⎝ ⎜

⎞

⎠ ⎟ , find PQ and QP.


Page 33

6. Find the following products.

(a)1 −2−3 6⎛

⎝ ⎜

⎞

⎠ ⎟

1 00 1⎛

⎝ ⎜

⎞

⎠ ⎟ (b)

1 00 1⎛

⎝ ⎜

⎞

⎠ ⎟

1 −2−3 6

⎛

⎝ ⎜

⎞

⎠ ⎟ (c)

a bc d⎛

⎝ ⎜

⎞

⎠ ⎟

1 00 1⎛

⎝ ⎜

⎞

⎠ ⎟

(d)1 00 1⎛

⎝ ⎜

⎞

⎠ ⎟

a bc d⎛

⎝ ⎜

⎞

⎠ ⎟ (e)

1 00 1⎛

⎝ ⎜

⎞

⎠ ⎟

1 00 1⎛

⎝ ⎜

⎞

⎠ ⎟ (f)

1 00 1⎛

⎝ ⎜

⎞

⎠ ⎟

0 11 0⎛

⎝ ⎜

⎞

⎠ ⎟

The 2 x 2 matrix 1 00 1⎛

⎝ ⎜

⎞

⎠ ⎟ is called the unit matrix of order 2 and is denoted by I.

It behaves like unity in the real number system. If A is a 2 x 2 matrix, then IA = AI = A.

7. If 3 12 1⎛

⎝ ⎜

⎞

⎠ ⎟

a bc d⎛

⎝ ⎜

⎞

⎠ ⎟ =

1 00 1⎛

⎝ ⎜

⎞

⎠ ⎟ , find a, b, c and d .

8. If p qr s

⎛

⎝ ⎜

⎞

⎠ ⎟

2 14 3⎛

⎝ ⎜

⎞

⎠ ⎟ =

1 00 1⎛

⎝ ⎜

⎞

⎠ ⎟ , find p, q, r and s.

9. If A = 1 23 4⎛

⎝ ⎜

⎞

⎠ ⎟ , find p and q such that A2 = pA + q I.

10. If A = 3 −12 −5⎛

⎝ ⎜

⎞

⎠ ⎟ , find p and q such that A2 = pA + q I.

11. If A = 1 13 21 2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

and B = 2 1 33 −1 2⎛

⎝ ⎜

⎞

⎠ ⎟ , find AB and BA.

12. If A = 1 0 21 −1 12 1 3

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

and B = 2 −1 30 1 −11 1 1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

, find AB and BA

13. Calculate M2 and M3 when θ = 60°, given that M = cosθ sinθ– sinθ cosθ⎛

⎝ ⎜

⎞

⎠ ⎟

14. A matrix B is such that B2 = 6B – 9I, where I is the 2 x 2 unit matrix Find integers p and q such that B3 = pB + qI.

15. A matrix A is such that A2 = 3A – 4I, where I is the 2 x 2 unit matrix.Find rational numbers p and q such A3 = pA + qI.

Further examples covering Exs 1, 2, 3, 4, 5 can be found in the following resources :-


Page 161 Exercise 6A Omit 1(m) and (n), 5, 6, 9 and 10


Page 34

The determinant of a 2 x 2 matrix.

If A = a bc d⎛

⎝ ⎜

⎞

⎠ ⎟ , then the determinant of the matrix A is denoted by det A or |A|

and defined as:- detA = ad – bc (a scalar quantity)

Example If A = 4 3−1 −2⎛

⎝ ⎜

⎞

⎠ ⎟ then |A| or det A = 4 x (–2) – (–1) x 3 = – 5.

If I = 1 00 1⎛

⎝ ⎜

⎞

⎠ ⎟ then |I| or det I = 1

Singular and Non-Singular Matrices

A 2 x 2 matrix P, is called a singular matrix iff |P| = 0 and non-singular iff |P| ≠ 0.

The determinant of a 3 x 3 matrix.

If A = a b cd e fg h i

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

, then det A is defined as follows:-

det A = a e fh i – b

d fg i + c

d eg h

In this expression for det A, the factor multiplying a given element is the determinant of the matrix obtained from A by omitting the row and column which contains the given element,together with the sign, according to the chess-board pattern shown below :-

+ − +− + −+ − +

Example If A = 3 1 −22 1 10 1 −2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

then det A = 3 1 11 −2 – 1

2 10 −2 + (–2)

2 10 1

= 3 x (–2 – 1) – 1 x (–4 – 0) + (–2) x (2 – 0)

= –9 + 4 – 4

= –9

(“iff” reads “if and only if”)


Page 35

Exercise 6

1. Evaluate

(a) 3 21 5 (b)

−1 32 −1 (c)

2 −1−4 −1 (d)

2 −14 −2

2. If A = 1 3−2 1⎛

⎝ ⎜

⎞

⎠ ⎟ and B =

2 4−1 5⎛

⎝ ⎜

⎞

⎠ ⎟ ,

find (i) AB and show that det(AB) = detAdetB.

(ii) BA and show that det(AB) = det(BA).

3. Evaluate

(a) cosθ sinθ

–sinθ cosθ (b) cos2θ sin 2θsin 2θ cos2θ (c)

ln2 ln4ln5 ln6

4. Find the determinants of the following matrices.

(a) −2 1 43 −2 50 1 3

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

(b)0 1 30 −1 42 6 −2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

(c)−2 0 13 −4 5−7 −3 2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

(d)1 2 34 5 67 8 9

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

(e)−7 14 72 −8 69 −3 12

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

(f)1 8 −102 4 151 12 5

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

5. If A = 1 2 30 1 32 −1 5

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

and B = 1 0 12 1 34 2 1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

find (i) AB and show that det(AB) = detAdetB.

(ii) BA and show that det(AB) = det(BA).

6. Show that 1 1 1x y z

yz zx xy = (x – y)(y – z)(z – x).

Further examples can be found in the following resources :-


Page 438 Exercise 17E Q 8


Page 36

The Inverse of a 2 x 2 Matrix. If A = a bc d⎛

⎝ ⎜

⎞

⎠ ⎟ and I =

1 00 1⎛

⎝ ⎜

⎞

⎠ ⎟ ,

then AI = a bc d⎛

⎝ ⎜

⎞

⎠ ⎟

1 00 1⎛

⎝ ⎜

⎞

⎠ ⎟ =

a bc d⎛

⎝ ⎜

⎞

⎠ ⎟

and IA = 1 00 1⎛

⎝ ⎜

⎞

⎠ ⎟

a bc d⎛

⎝ ⎜

⎞

⎠ ⎟ =

a bc d⎛

⎝ ⎜

⎞

⎠ ⎟

∴ AI = IA = A.

For this reason the 2 x 2 unit matrix is called the identity matrix for multiplication of 2 x 2 matrices.

Example Consider A = 3 27 5⎛

⎝ ⎜

⎞

⎠ ⎟ and B =

5 −2−7 3⎛

⎝ ⎜

⎞

⎠ ⎟ ,

AB = 3 27 5⎛

⎝ ⎜

⎞

⎠ ⎟

5 −2−7 3⎛

⎝ ⎜

⎞

⎠ ⎟ =

1 00 1⎛

⎝ ⎜

⎞

⎠ ⎟ and BA =

5 −2−7 3⎛

⎝ ⎜

⎞

⎠ ⎟

3 27 5⎛

⎝ ⎜

⎞

⎠ ⎟ =

1 00 1⎛

⎝ ⎜

⎞

⎠ ⎟

∴ AB = BA = I

For this reason B is called the multiplicative inverse of A and is denoted by A–1. In the same way, we can say that A is the (multiplicative) inverse of B and is denoted by B–1.

We generally use the word inverse of a matrix to refer to the multiplicative inverse since the additive inverse is generally called its negative.

In general If A and B are square matrices of the same order such that AB = BA = I, then B is the inverse of A and A is the inverse of B.It can be shown that if these inverses exist, then they are unique and so we can talk about the inverse of A or the inverse of B.

Exercise 7

1. Show that each matrix is the inverse of the other :

(a)3 21 1⎛

⎝ ⎜

⎞

⎠ ⎟ and

1 −2−1 3⎛

⎝ ⎜

⎞

⎠ ⎟ (b)

5 −7−2 3⎛

⎝ ⎜

⎞

⎠ ⎟ and

3 72 5⎛

⎝ ⎜

⎞

⎠ ⎟

(c)2 53 8⎛

⎝ ⎜

⎞

⎠ ⎟ and

8 −5−3 2⎛

⎝ ⎜

⎞

⎠ ⎟ (d)

3 −2−4 3

⎛

⎝ ⎜

⎞

⎠ ⎟ and

3 24 3⎛

⎝ ⎜

⎞

⎠ ⎟

2. Study the pattern in the entries of the pairs of matrices above.Use this pattern to write down the inverse of each of the followingmatrices and check by multiplication.

(a)2 11 1⎛

⎝ ⎜

⎞

⎠ ⎟ (b)

2 33 5⎛

⎝ ⎜

⎞

⎠ ⎟ (c)

4 39 7⎛

⎝ ⎜

⎞

⎠ ⎟

(d)9 −5−7 4⎛

⎝ ⎜

⎞

⎠ ⎟ (e)

5 −73 −4⎛

⎝ ⎜

⎞

⎠ ⎟ (f)

cosθ sinθ– sinθ cosθ⎛

⎝ ⎜

⎞

⎠ ⎟


Page 37

You should have noticed in the matrices in question 2 that :-(i) the difference in the cross product of the entries is always 1. (i.e. det A = 1)

In 4 39 7⎛

⎝ ⎜

⎞

⎠ ⎟ , (4 x 7) – (9 x 3) = 1

(ii) the inverse can be found by interchanging the entries in the main diagonal and changing the signs in the other diagonal.

Does every 2 x 2 Matrix have an Inverse ?

If A = a bc d⎛

⎝ ⎜

⎞

⎠ ⎟ then its inverse, A–1 would appear to be

d −b−c a⎛

⎝ ⎜

⎞

⎠ ⎟ .

If this inverse is correct then A–1A = AA–1 = I

A–1A = d −b−c a⎛

⎝ ⎜

⎞

⎠ ⎟

a bc d⎛

⎝ ⎜

⎞

⎠ ⎟ =

ad − bc 00 ad − bc

⎛

⎝ ⎜

⎞

⎠ ⎟ = (ad – bc)

1 00 1⎛

⎝ ⎜

⎞

⎠ ⎟

∴1

ad − bcd −b−c a

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

a bc d⎛

⎝ ⎜

⎞

⎠ ⎟ =

1 00 1⎛

⎝ ⎜

⎞

⎠ ⎟

Similarly

AA–1 = a bc d⎛

⎝ ⎜

⎞

⎠ ⎟

d −b−c a⎛

⎝ ⎜

⎞

⎠ ⎟ =

ad − bc 00 ad − bc

⎛

⎝ ⎜

⎞

⎠ ⎟ = (ad – bc)

1 00 1⎛

⎝ ⎜

⎞

⎠ ⎟

∴a bc d⎛

⎝ ⎜

⎞

⎠ ⎟ 1

ad − bcd −b−c a⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥ =

1 00 1⎛

⎝ ⎜

⎞

⎠ ⎟

It follows that if ad – bc ≠ 0 :-

the inverse of matrix A = a bc d⎛

⎝ ⎜

⎞

⎠ ⎟ is A–1 =

1ad − bc

d −b−c a⎛

⎝ ⎜

⎞

⎠ ⎟

where ad – bc is the determinant of the matrix A.If detA = 0, => A does not have an inverse and is called a singular matrix.If detA ≠ 0, => A has an inverse and is said to be non–singular.

Examples

1. Given that A = 4 32 2⎛

⎝ ⎜

⎞

⎠ ⎟ , find A–1, if it exists.

detA = 4 x 2 – 3 x 2 = 2 ≠ 0, so A–1 exists. => A–1 = 12

2 −3−2 4⎛

⎝ ⎜

⎞

⎠ ⎟ =

1 − 32

−1 2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

2. Given that P = 3 −1−2 2⎛

⎝ ⎜

⎞

⎠ ⎟ , find P–1, if it exists.

detP = 3 x 2 – (–2) x (–1) = 4 ≠ 0, so P–1 exists. => P–1 = 14

2 12 3⎛

⎝ ⎜

⎞

⎠ ⎟ =

12

14

12

34

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

Note the order of the cross productsThe main diagonal (4 x 7) first.


Page 38

Exercise 8

1. Find the inverse of the following 2 x 2 matrices, if they exist :

(a) A = 2 14 3⎛

⎝ ⎜

⎞

⎠ ⎟ (b) B =

7 416 9⎛

⎝ ⎜

⎞

⎠ ⎟ (c) C =

4 210 5⎛

⎝ ⎜

⎞

⎠ ⎟

(d) D = 5 76 9⎛

⎝ ⎜

⎞

⎠ ⎟ (e) E =

−2 41 −1

⎛

⎝ ⎜

⎞

⎠ ⎟ (f) F =

1 11 0⎛

⎝ ⎜

⎞

⎠ ⎟

2. Given that P = 2 30 1⎛

⎝ ⎜

⎞

⎠ ⎟ and Q =

2 51 3⎛

⎝ ⎜

⎞

⎠ ⎟ , find

(a) P–1 (b) Q–1 (c) (PQ)–1

(d) P–1Q–1 (e) Q–1P –1 (f) (QP)–1

The Inverse of a 3 x 3 Matrix. If A is a 3 x 3 matrix, provided A–1 exists, then :-

=> AA–1 = A–1A = I where I = 1 0 00 1 00 0 1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

Finding the inverse of a 3 x 3 matrix by elementary row operations.

The method used is called “finding the inverse” if it exists by elementary row operations.

For a 3 x 3 matrix A, we somehow change the A into the matrix 1 0 00 1 00 0 1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

whilst also

changing 1 0 00 1 00 0 1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

into the inverse of A by using row operations similar to elimination.

We begin with A I and turn this into I A–1.

We start to operate with I on the right hand side and by operating on A row by row we turn A into the identity matrix I, at the same time turning I into A–1.

If A has no inverse, this becomes apparent because the reduction of A produces a row of 0’s.It is recommended that the product of A and A–1 should be found as a check i.e. that AA–1 or A–1A = I.

(If det A = 0, A–1 does not exist.)

Further examples can be found in the following resources:-


Page 161 Exercise 6A Q 1(m) and (n), 5, 6


Page 39

Examples 1. Find the inverse of the 3 x 3 matrix A = 1 1 11 −1 22 1 −1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

Set out as follows:- row 1row 2row 3

1 1 1 1 0 01 −1 2 0 1 02 1 −1 0 0 1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

row 3 – 2 x row 11 1 1 1 0 01 −1 2 0 1 00 −1 −3 −2 0 1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

row 2 – row 11 1 1 1 0 00 −2 1 −1 1 00 −1 −3 −2 0 1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

2 x row 3 – row 21 1 1 1 0 00 −2 1 −1 1 00 0 −7 −3 −1 2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

7 x row 1 + row 37 7 0 4 −1 20 −2 1 −1 1 00 0 −7 −3 −1 2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

7 x row 2 + row 37 7 0 4 −1 20 −14 0 −10 6 20 0 −7 −3 −1 2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

2 x row 1 + row 214 0 0 −2 4 60 −14 0 −10 6 20 0 −7 −3 −1 2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

114

× row 1

−1

14× row 2

− 17× row 3

1 0 0 −17

27

37

0 1 0 57

−37

−17

0 0 1 37

17

− 27

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟

The inverse of A = 1 1 11 −1 22 1 −1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

is therefore A–1 = 17

−1 2 35 −3 −13 1 −2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

Check your answer using AA–1 = I or A–1A = I

(since det A = 7 => Inverse exists)

||||

||||

||||

||||

||||

||||

||||

|||||||

A I

A– 1I

note :- we have created a triangle of

zeros

note :- we have created a 2nd triangle of

zeros

we work on the bottom triangular

half of A first

we work on the bottom triangular

half of A first

we now work on the top triangular

half of A


Page 40

2. Find the inverse of the 3 x 3 matrix P = 2 2 12 4 13 2 0

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

row 1row 2row 3

2 2 1 1 0 02 4 1 0 1 03 2 0 0 0 1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

2 x row 3 – 3 x row 12 2 1 1 0 02 4 1 0 1 00 −2 −3 −3 0 2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

row 1 – row 22 2 1 1 0 00 −2 0 1 −1 00 −2 −3 −3 0 2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

row 3 – row 22 2 1 1 0 00 −2 0 1 −1 00 0 −3 −4 1 2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

3 x row 1 + row 36 6 0 −1 1 20 −2 0 1 −1 00 0 −3 −4 1 2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

row 1 + 3 x row 26 0 0 2 −2 20 −2 0 1 −1 00 0 −3 −4 1 2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

16× row 1

−12× row 2

− 13× row 3

1 0 0 13

−13

13

0 1 0 −12

12

0

0 0 1 43

− 13

− 23

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟

The inverse P = 2 2 12 4 13 2 0

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

is therefore P–1 =

13

−13

13

−12

12

043

−13

− 23

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟

or 16

2 −2 2−3 3 08 −2 −4

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

Check your answer using PP–1 = I or P–1P = I

(since det P = – 6 => Inverse exists)

||||

||||

||||

||||

||||

||||

|||||||

A I

A– 1I


Page 41

Exercise 9Find the inverses of the following 3 x 3 matrices, if they exist :–

1.3 4 54 3 111 0 3

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

2.4 8 33 5 11 4 3

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

3.0 2 32 0 01 −1 0

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

4.4 2 13 1 23 5 1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

5.1 2 32 3 13 2 1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

6.1 8 52 10 79 7 3

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

Using The Inverse of a 2 x 2 Matrix to Solve a System of Linear Equations

Consider the system of equations : 3x – y = 52x + y = 15

Since 3x − y2x + y⎛

⎝ ⎜

⎞

⎠ ⎟ =

3 −12 1⎛

⎝ ⎜

⎞

⎠ ⎟

xy

⎛

⎝ ⎜ ⎞

⎠ ⎟ , the set of equations can be written as

3 −12 1⎛

⎝ ⎜

⎞

⎠ ⎟

xy

⎛

⎝ ⎜ ⎞

⎠ ⎟ =

515⎛

⎝ ⎜

⎞

⎠ ⎟

which is in the form AX = B where A = 3 −12 1⎛

⎝ ⎜

⎞

⎠ ⎟ , X =

xy

⎛

⎝ ⎜ ⎞

⎠ ⎟ and B =

515⎛

⎝ ⎜

⎞

⎠ ⎟

If the inverse A–1 of A exists => A–1AX = A–1Bi.e. IX = A–1B or simply X = A–1B

In the above example, A–1 = 15

1 1−2 3⎛

⎝ ⎜

⎞

⎠ ⎟

i.e.15

1 1−2 3⎛

⎝ ⎜

⎞

⎠ ⎟

3 −12 1⎛

⎝ ⎜

⎞

⎠ ⎟

xy

⎛

⎝ ⎜ ⎞

⎠ ⎟ =

15

1 1−2 3⎛

⎝ ⎜

⎞

⎠ ⎟

515⎛

⎝ ⎜

⎞

⎠ ⎟

1 00 1⎛

⎝ ⎜

⎞

⎠ ⎟

xy

⎛

⎝ ⎜ ⎞

⎠ ⎟ =

15

2035⎛

⎝ ⎜

⎞

⎠ ⎟ =>

xy

⎛

⎝ ⎜ ⎞

⎠ ⎟ =

47⎛

⎝ ⎜ ⎞

⎠ ⎟ , and so x = 4 and y = 7

Exercise 10Solve the following systems of equations :

1. x – y = 5 2. 3x + y = 7 3. 2x + y = 5x + y = 11 3x + 2y = 5 2x + 3y = –1

4. 3x – 4y = 18 5. 2x + 3y = 5 6. 5x – 3y = 95x + y = 7 4x – 5y = 21 7x – 6y = 9

Further examples can be found in the following resources:-


Page 437 Exercise 17E Q 9


Page 42

Using the Inverse of a 3 x 3 Matrix to Solve a System of Linear Equations.

Consider the system of equations:- x + y + 2z = 3 2x – y – z = 2

3x – 2y + 2z = –2

Since x + y + 2z2x − y − z

3x − 2y + 2z

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

=1 1 22 −1 −13 −2 2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

xyz

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

, we can be re-write our equations as :-

1 1 22 −1 −13 −2 2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

xyz

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

=32−2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

which is in the form AX = B where A = 1 1 22 −1 −13 −2 2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

, X = xyz

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

and B =32−2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

If the inverse A–1 of A exists, A–1AX = A-1B (i.e. IX = A–1B)

Using the elementary row operation method shown on Pages 38 and 39,

A–1 = 113

4 6 −17 4 −51 −5 3

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

i.e.113

4 6 −17 4 −51 −5 3

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

1 1 22 −1 −13 −2 2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

xyz

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

= 113

4 6 −17 4 −51 −5 3

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

32−2

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

=> 1 0 00 1 00 0 1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

xyz

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

= 113

2639−13

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

=> xyz

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

= 23−1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

and so x = 2, y = 3 and z = –1

Exercise 11 Solve the following systems of equations :

1. x – 2y + z = 6 2. 5x – y + 2z = 25 3x + y – 2z = 4 3x + 2y – 3z = 16 7x – 6y – z = 10 2x – y + z = 9

3. x + y + z = 2 4. 2x + 4y + 5z = – 3 3x – y + 2z = 4 4x – y – 7z = 6

2x + 3y + z = 7 6x + 3y – z = 3

Further examples can be found in the following resources:-Understanding Pure Mathematics (A.J.Sadler/D.W.S.Thorning)Page 161 Exercise 6A Q 4(select), 9. Page 437 Exercise 17E Q 11, 12


Page 43

Using 2 x 2 Matrices to Represent Geometrical Transformations in the (x, y) Plane.

2 x 2 matrices can be associated with transformations of all points in a Cartesian plane.Transformations are defined as reflections in the x - axis, y - axis and in the line y = x, rotations of 90° and 180°, and dilatation (enlargement or reduction), or compositions of these.

Examples1. Reflection of the point (x, y) in the x–axis.

Under this reflection (x, y) maps on to ( ′ x , ′ y ) where ( ′ x , ′ y ) is (x, –y).

′ x ′ y

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ =

x−y⎛

⎝ ⎜

⎞

⎠ ⎟ =

1x + 0y0x −1y⎛

⎝ ⎜

⎞

⎠ ⎟ =

1 00 −1⎛

⎝ ⎜

⎞

⎠ ⎟

xy⎛

⎝ ⎜ ⎞

⎠ ⎟

and so 1 00 −1⎛

⎝ ⎜

⎞

⎠ ⎟ is the matrix associated

with reflection in the x -axis.

2. Rotation of π radians (a half-turn) about the origin.

Under this rotation (x, y) maps on to ( ′ x , ′ y ), where ( ′ x , ′ y ) is (–x, –y).

′ x ′ y

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ =

−x−y⎛

⎝ ⎜

⎞

⎠ ⎟ =

−1x + 0y0x − 1y

⎛

⎝ ⎜

⎞

⎠ ⎟ =

−1 00 −1

⎛

⎝ ⎜

⎞

⎠ ⎟

xy⎛

⎝ ⎜ ⎞

⎠ ⎟

and so −1 00 −1

⎛

⎝ ⎜

⎞

⎠ ⎟ is the matrix associated

with a half-turn rotation about the origin.

3. A dilatation where the scale factor is 2 and the centre of dilatation is the origin.

Under this dilatation (x, y) maps on to ( ′ x , ′ y ), where ( ′ x , ′ y ) is (2x, 2y).

′ x ′ y

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ =

2x2y⎛

⎝ ⎜

⎞

⎠ ⎟ =

2x + 0y0x + 2y⎛

⎝ ⎜

⎞

⎠ ⎟ =

2 00 2⎛

⎝ ⎜

⎞

⎠ ⎟

xy

⎛

⎝ ⎜

⎞

⎠ ⎟

and so 2 00 2⎛

⎝ ⎜

⎞

⎠ ⎟ is the matrix associated with

a dilatation, centre the origin and with a scale factor of 2.

(x, y)

(x, –y)

x

y

(x, y)

(–x, –y)

x

y

(x, y)

x

y(2x, 2y)


Page 44

Exercise 12

1. Find the matrices associated with the following transformations.

(a) Reflection in the y – axis.

(b) Reflection in the line y = x.

(c) Reflection in the line y = –x.

(d) A rotation of π2

radians clockwise.

(e) A rotation of π2

radians anti–clockwise.

(f) A dilatation, about O, where the scale factor is k.

2. Prove that the matrix associated with a general rotation of θ radians

about the origin is cosθ −sinθsinθ cosθ

⎛

⎝ ⎜

⎞

⎠ ⎟ .

The diagram opposite may be helpful.

OP, which makes an angle of α radianswith the x–axis, is rotated throughθ radians to OQ.

Hence, show that the matrix associated with a rotation of:-

(i) π radians is −1 00 −1

⎛

⎝ ⎜

⎞

⎠ ⎟

(ii) π2

radians anti–clockwise is 0 −11 0⎛

⎝ ⎜

⎞

⎠ ⎟

(iii) π2

radians clockwise is 0 1−1 0⎛

⎝ ⎜

⎞

⎠ ⎟

Further examples can be found in the following resources :-


Page 168 Exercise 6B Select

P(x, y)

Q(x', y')

x

y

r

r

O

θα


Page 45

AnswersExercise 1

1. (a) 2 x 4 (b) 3 x 2 (c) 4 x 2

2. A = C (1 x 3), F = G (2 x 1), H = L (2 x 2)

3. (a) x = 4, y = –3 (b) x = 4, y = –1 (c) x = 5, y = 2 (d) x = 2, y = –3

4. (a)

3 51 44 02 7

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟

, 4 x 2 (b)2 4 1−1 8 −2⎛

⎝ ⎜

⎞

⎠ ⎟ , 2 x 3 (c)

1 3 5 72 4 6 8⎛

⎝ ⎜

⎞

⎠ ⎟ , 2 x 4

5. Proof 6. x = 5, y = – 4

Exercise 2

1. (a)76⎛

⎝ ⎜ ⎞

⎠ ⎟ (b)

17⎛

⎝ ⎜ ⎞

⎠ ⎟ (c)

9a−2b⎛

⎝ ⎜

⎞

⎠ ⎟ (d)

00⎛

⎝ ⎜ ⎞

⎠ ⎟

(e) 3 9( ) (f) −3 5( ) (g)3 34 6⎛

⎝ ⎜

⎞

⎠ ⎟ (h)

3 1 4−2 2 −4⎛

⎝ ⎜

⎞

⎠ ⎟

2. (a)4 66 1⎛

⎝ ⎜

⎞

⎠ ⎟ (b)

−1 76 −2

⎛

⎝ ⎜

⎞

⎠ ⎟ (c)

2 97 −3⎛

⎝ ⎜

⎞

⎠ ⎟ (d)

2 97 −3⎛

⎝ ⎜

⎞

⎠ ⎟ ,true

3. A = –B 4. Proof

Exercise 3

1. (a)−1−4⎛

⎝ ⎜

⎞

⎠ ⎟ (b)

31⎛

⎝ ⎜ ⎞

⎠ ⎟ (c)

−5a4b

⎛

⎝ ⎜

⎞

⎠ ⎟ (d)

4u−6v⎛

⎝ ⎜

⎞

⎠ ⎟

(e) 1 1( ) (f) 7 −11( ) (g)−1 −3−4 −4

⎛

⎝ ⎜

⎞

⎠ ⎟ (h)

−4 7−1 −6

⎛

⎝ ⎜

⎞

⎠ ⎟

(i)1 5 −212 0 4⎛

⎝ ⎜

⎞

⎠ ⎟

2. (a)−1 53 5

⎛

⎝ ⎜

⎞

⎠ ⎟ (b)

6 42 4⎛

⎝ ⎜

⎞

⎠ ⎟ (c)

4 72 5⎛

⎝ ⎜

⎞

⎠ ⎟ (d)

3 −13 3⎛

⎝ ⎜

⎞

⎠ ⎟

(e)7 −1−1 −1⎛

⎝ ⎜

⎞

⎠ ⎟ (f)

4 0−4 −4⎛

⎝ ⎜

⎞

⎠ ⎟ (g)

5 95 9⎛

⎝ ⎜

⎞

⎠ ⎟ (h)

7 −1−1 −1⎛

⎝ ⎜

⎞

⎠ ⎟

3. (a)−1 22 0

⎛

⎝ ⎜

⎞

⎠ ⎟ (b)

2 −4−1 −3⎛

⎝ ⎜

⎞

⎠ ⎟

4. (a)4 04 99 12

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

(b)−1 −40 57 1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

(c)0 −23 9

12 7

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

(d)0 −23 9

12 7

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟


Page 46

Exercise 4

1. (a)6 8−2 −4⎛

⎝ ⎜

⎞

⎠ ⎟ (b)

9 12−3 −6⎛

⎝ ⎜

⎞

⎠ ⎟ (c)

−15 −205 10

⎛

⎝ ⎜

⎞

⎠ ⎟ (d)

−3 −41 2

⎛

⎝ ⎜

⎞

⎠ ⎟

2. (a)1 5 −2−2 −5 3⎛

⎝ ⎜

⎞

⎠ ⎟ (b)

10 6 812 10 6⎛

⎝ ⎜

⎞

⎠ ⎟ (c)

6 8 24 0 6⎛

⎝ ⎜

⎞

⎠ ⎟

(d)4 −2 68 10 0⎛

⎝ ⎜

⎞

⎠ ⎟ (e)

10 6 812 10 6⎛

⎝ ⎜

⎞

⎠ ⎟ (f)

18 24 612 0 18⎛

⎝ ⎜

⎞

⎠ ⎟

(g)18 24 612 0 18⎛

⎝ ⎜

⎞

⎠ ⎟ (h)

16 −8 2432 40 0⎛

⎝ ⎜

⎞

⎠ ⎟ (i)

16 −8 2432 40 0⎛

⎝ ⎜

⎞

⎠ ⎟

3. (a)−2 −718 −1⎛

⎝ ⎜

⎞

⎠ ⎟ (b)

20 −157 10

⎛

⎝ ⎜

⎞

⎠ ⎟ (c)

26 −198 13

⎛

⎝ ⎜

⎞

⎠ ⎟ (d)

44 −16−22 22⎛

⎝ ⎜

⎞

⎠ ⎟

4. (a)2 −14 3⎛

⎝ ⎜

⎞

⎠ ⎟ (b)

3 2−1 3⎛

⎝ ⎜

⎞

⎠ ⎟ (c)

2 11 5⎛

⎝ ⎜

⎞

⎠ ⎟ (d)

4 −3−4 −2

⎛

⎝ ⎜

⎞

⎠ ⎟

5. (a)1 8 −185 −1 −7⎛

⎝ ⎜

⎞

⎠ ⎟ (b)

7 −61 −4⎛

⎝ ⎜

⎞

⎠ ⎟

6. p = –1, q = 4, r = 3, s = –2

Exercise 5

1. (a) (11) (b) (26) (c) (9) (d) (13)

(e) (15) (f) (8x –5y – z) (g)98⎛

⎝ ⎜ ⎞

⎠ ⎟ (h)

104

⎛

⎝ ⎜

⎞

⎠ ⎟

(i)18⎛

⎝ ⎜ ⎞

⎠ ⎟ (j)

75⎛

⎝ ⎜ ⎞

⎠ ⎟ (k)

−77

⎛

⎝ ⎜

⎞

⎠ ⎟ (l)

46⎛

⎝ ⎜ ⎞

⎠ ⎟

(m)122⎛

⎝ ⎜

⎞

⎠ ⎟ (n) not possible (o) (26) (p)

11137

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

(q) not possible (r)987

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

(s)10⎛

⎝ ⎜ ⎞

⎠ ⎟

2. (a) x = 4, y = – 4 (b) x = 3, y = – 1 (c) x = 8, y = 5 (d) x = 2, y = – 1

3. (a) (i)8 514 15⎛

⎝ ⎜

⎞

⎠ ⎟ (ii)

19 132 4

⎛

⎝ ⎜

⎞

⎠ ⎟ AB ≠ BA

(b) (i)1711⎛

⎝ ⎜

⎞

⎠ ⎟ (ii)

1711⎛

⎝ ⎜

⎞

⎠ ⎟ A(BC) = (AB)C

(c) Proof

4.4 −525 −1⎛

⎝ ⎜

⎞

⎠ ⎟ ,

−13 −1470 −27

⎛

⎝ ⎜

⎞

⎠ ⎟

5.0 00 0⎛

⎝ ⎜

⎞

⎠ ⎟ ,

−4 8−2 4⎛

⎝ ⎜

⎞

⎠ ⎟


Page 47

6. (a)1 −2−3 6⎛

⎝ ⎜

⎞

⎠ ⎟ (b)

1 −2−3 6⎛

⎝ ⎜

⎞

⎠ ⎟ (c)

a bc d⎛

⎝ ⎜

⎞

⎠ ⎟ (d)

a bc d⎛

⎝ ⎜

⎞

⎠ ⎟

(e)1 00 1⎛

⎝ ⎜

⎞

⎠ ⎟ (f)

0 11 0⎛

⎝ ⎜

⎞

⎠ ⎟

7. a = 1, b = – 1, c = – 2, d = 3

8. p = 32

, q = – 12

, r = – 2, s = 1

9. p = 5, q = 2

10. p = – 2, q = 13

11.5 0 5

12 1 138 −1 7

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

, 8 102 5⎛

⎝ ⎜

⎞

⎠ ⎟

12.4 1 53 −1 57 2 8

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

, 7 4 12−1 −2 −24 0 6

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

13.− 1

23

2

−3

2−

12

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

, −1 00 −1

⎛

⎝ ⎜

⎞

⎠ ⎟

14. p = 27, q = – 54

15. p = 5, q = – 12

Exercise 6

1. (a) 13 (b) – 5 (c) – 6 (d) 0

2. Proofs3. (a) 1 (b) cos4θ (c) ln2.ln 6

25 (from ln2ln6 – ln5ln4)

4. (a) 25 (b) 14 (c) 51 (d) 0

(e) 1428 (f) – 320

5. Proofs

6. Proof


Page 48

Exercise 7

1. Proofs

2. (a)1 −1−1 2⎛

⎝ ⎜

⎞

⎠ ⎟ (b)

5 −3−3 2⎛

⎝ ⎜

⎞

⎠ ⎟ (c)

7 −3−9 4⎛

⎝ ⎜

⎞

⎠ ⎟ (d)

4 57 9⎛

⎝ ⎜

⎞

⎠ ⎟

(e)−4 7−3 5⎛

⎝ ⎜

⎞

⎠ ⎟ (f)

cosθ −sinθsinθ cosθ

⎛

⎝ ⎜

⎞

⎠ ⎟

Exercise 8

1. (a)32 − 1

2

−2 1

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ (b)

−9 416 −7⎛

⎝ ⎜

⎞

⎠ ⎟ (c) does not exist (d)

3 − 73

−2 53

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

(e)12 212 1

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ (f)

0 11 −1⎛

⎝ ⎜

⎞

⎠ ⎟

2. (a)12 − 3

2

0 1

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ (b)

3 −5−1 2⎛

⎝ ⎜

⎞

⎠ ⎟ (c)

32 − 19

2

− 12

72

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ (d)

3 − 112

−1 2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

(e)32 − 19

2

− 12

72

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ (f)

3 − 112

−1 2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

Exercise 9

1.

98 − 3

2298

− 18

12 − 13

8

− 38

12 − 7

8

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

2.11 −12 −7−8 9 57 −8 −4

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

3.0 1

2 00 1

2 −113 − 1

323

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

4.

12 − 1

6 − 16

− 16 − 1

18518

− 23

79

19

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

5.− 1

12 − 13

712

− 112

23 − 5

125

12 − 13

112

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

6.− 1

31157

219

1 − 1419

119

− 43

6557 − 2

19

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟


Page 49

Exercise 10

1. x = 8, y = 3 2. x = 3, y = – 2 3. x = 4, y = – 3

4. x = 2, y = – 3 5. x = 4, y = – 1 6. x = 3, y = 2

Exercise 11

1. x = 5, y = 3, z = 7 2. x = 5, y = 2, z = 1

3. x = 3, y = 1, z = – 2 4. x = 76

, y = − 43

, z = 0

Exercise 12

1. (a)−1 00 1

⎛

⎝ ⎜

⎞

⎠ ⎟ (b)

0 11 0⎛

⎝ ⎜

⎞

⎠ ⎟ (c)

0 −1−1 0⎛

⎝ ⎜

⎞

⎠ ⎟

(d)0 1−1 0⎛

⎝ ⎜

⎞

⎠ ⎟ (e)

0 −11 0⎛

⎝ ⎜

⎞

⎠ ⎟ (f)

k 00 k⎛

⎝ ⎜

⎞

⎠ ⎟

2. Proofs


Page 50

Outcome 3 – Further Sequences and SeriesPower series

Suppose that f(x) = a + bx + cx2 + dx3

then ′ f x( ) = b + 2cx + 3dx2

′ ′ f x( ) = 2c + 6dx ′ ′ ′ f x( ) = 6d

all other derivatives are zeroFrom above f(0) = a

then ′ f 0( ) = b ′ ′ f 0( ) = 2c => c =

′ ′ f 0( )2

=′ ′ f 0( )2!

′ ′ ′ f 0( ) = 6d => d = ′ ′ ′ f 0( )6

=′ ′ ′ f 0( )3!

Hence f(x) = a + bx + cx2 + dx3 can be written as

f(x) = f(0) + ′ f 0( )1! x +

′ ′ f 0( )2! x2 +

′ ′ ′ f 0( )3! x3

Any series of the form a0 + a1x + a2 x2 + a3 x3 + a4 x4 + ....is called a Power Series.

In many cases, the sum of such series becomes bigger and bigger as you add on each successive term, in which case the series is said to diverge.On the other hand, some series are such that, as more and more terms are added, the sum approaches more and more closely to a particular limit,(i.e. a single function), in which case it is said to converge to this limit.

Maclaurin’s Theorem states that, under certain circumstances, a function f(x) is given by:-

f(x) = f(0) + ′ f 0( )1! x +

′ ′ f 0( )2! x2 +

′ ′ ′ f 0( )3! x3 + f (iv) 0( )

4! x4 + ....+ f (n ) 0( )

n! xn

The series can be found if f (n ) 0( ) exists for all values of n.Some series converge to f(x) for all values of x and others converge to f(x)for a limited range of values of x.In the following examples, the range of values of x for which the series is valid will be given, but not justified.

Examples:-Use Maclaurin’s Theorem to expand as a series of ascending powers of x:-

1. f(x) = ex

f(x) = ex f(0) = 1 ′ f x( ) = ex ′ f 0( ) = 1 ′ ′ f x( ) = ex ′ ′ f 0( ) = 1 ′ ′ ′ f x( ) = ex ′ ′ ′ f 0( ) = 1

Hence ex = 1 +x1!

+x2

2!+

x3

3!+ ........ for all x ∈ R


Page 51

2. f(x) = ln(1 + x)

f(x) = ln(1 + x) f (0) = 0

′ f x( ) = 1

1 + x ′ f 0( ) = 1

′ ′ f x( ) =−1

1 + x( )2 ′ ′ f 0( ) = –1

′ ′ ′ f x( ) = 2

1 + x( ) 3 ′ ′ ′ f 0( ) = 2

f (iv) x( )= −6

1 + x( )4 f (iv) 0( ) = – 6

Hence

ln(1 + x) = 0 +x1!−

x2

2!+

2x3

3!−

6x4

4!........

= x −x2

2+

x3

3−

x4

4+ ........ for –1 < x ≤ 1

3. f(x) = sinx (x in radians)

f(x) = sinx f (0) = 0′ f x( ) = cosx ′ f 0( ) = 1′ ′ f x( ) = –sinx ′ ′ f 0( ) = 0

′ ′ ′ f x( ) = –cosx ′ ′ ′ f 0( ) = –1f (iv) x( )= sinx f (iv) 0( ) = 0

Hence

sinx = x1!−

x3

3!+

x5

5!− ........ for all x

4. f(x) = tan-1x

f(x) = tan-1x f (0) = 0

′ f x( ) = 1

1 + x2 ′ f 0( ) = 1

′ ′ f x( ) = −2x

1 + x2( )2 ′ ′ f 0( ) = 0

′ ′ ′ f x( ) = 6x2 − 21 + x2( )3 ′ ′ ′ f 0( ) = – 2

(Quotient rule)

Hence

tan-1x = x − x3

3+

x5

5−

x 7

7+ ....... for –1 < x < 1

contd...


Page 52

5. f(x) = (1 + x)n i.e. The Binomial Theorem.

f(x) = (1 + x)n, f (0) = 1′ f x( ) = n(1+ x)n−1, ′ f 0( ) = n′ ′ f x( ) = n(n −1)(1+ x)n−2 ′ ′ f 0( ) = 0 = n(n–1)

′ ′ ′ f x( ) = n(n −1)(n − 2)(1+ x)n−3, ′ ′ ′ f 0( ) = n(n–1)(n – 2)

Hence

(1 + x)n = 1 +n1!

x +n n –1( )

2!x 2 +

n n – 1( ) n – 2( )3!

x3 + ... –1 < x ≤ 1

= n0⎛

⎝ ⎜ ⎞

⎠ ⎟ +

n1⎛

⎝ ⎜ ⎞

⎠ ⎟ x +

n2⎛

⎝ ⎜ ⎞

⎠ ⎟ x 2 +

n3⎛

⎝ ⎜ ⎞

⎠ ⎟ x3 + ........

nn⎛

⎝ ⎜ ⎞

⎠ ⎟ xn

6. An alternative method for the expansion of f(x) = ln(1 + x)

f(x) = ln(1 + x), ′ f x( ) = 11 + x

= 1 + x( )−1 Using the Binomial expansion,

1 + x( )−1 = 1 +−1( )1!

x +−1( ) –2( )

2!x2 +

−1( ) −2( ) –3( )3!

x3 + ...

= 1 − x + x 2 − x3 + ...

Integrate both sides with respect to x,

ln(1 + x) = x − x2

2+

x 3

3−

x4

4+ ... + c

But f(0) = 0 and so c = ln1 = 0Hence:-

ln(1 + x) = x − x2

2+

x3

3−

x4

4+ ... for –1 < x ≤ 1

7. An alternative method for the expansion of f(x) = tan–1x

f(x) = tan–1x, ′ f x( ) = 1

1 + x2 = 1 + x2( )−1

Using the Binomial expansion,

1 + x2( )−1= 1+

−1( )1!

x2 +−1( ) –2( )

2!x 4 +

−1( ) −2( ) –3( )3!

x6 + ...

= 1 − x 2 + x4 − x6 + ...Integrate both sides with respect to x,

tan–1x = x − x3

3+

x5

5−

x7

7+ ... + c

But f(0) = 0 and so c = tan–1(0) = 0Hence

tan–1x = x − x3

3+

x5

5−

x 7

7+ ... for –1 < x < 1


Page 53

Exercise 1 Expand the following functions in ascending powers of x as far as the power indicated.

1. f(x) = cosx as far as x6. 2. f(x) = tanx as far as x3.

3. f(x) = sin–1x as far as x3. 4. f(x) = ln(1 – x) as far as x4.

5. f(x) = e3x as far as x4. 6. f(x) = ln(1 + 2x) as far as x5.

7. f(x) = sin3x as far as x5. 8. f(x) = tan2x as far as x5.

9. f(x) = ln(2 + x) as far as x3. (Hint: ln(2 + x) = ln2(1 + x2 )

Using more than one expansion

Examples

1. Expand f(x) = e–2xsin3x in ascending powers of x as far as the termx4 using the expansions for ex and sinx.

ex = 1 +x1!

+x2

2!+

x3

3!+

x4

4!+ ........

e–2x = 1 +−2x( )1!

+−2x( ) 2

2!+

−2x( )3

3!+

−2x( )4

4!+ ........

= 1 − 2x + 2x2 −43

x3 +23

x4 + ........

sinx = x1!−

x3

3!+

x5

5!− ........

sin3x = 3x1!

−3x( )3

3!+ ........ (ignore higher powers)

= 3x −92

x3 + ........

Hence:-

f(x) = e–2xsin3x = 1− 2x + 2x2 −43

x3 +23

x4 + ........⎛ ⎝

⎞ ⎠ 3x − 9

2x3 + ........⎛

⎝ ⎞ ⎠

= 3x − 6x2 + 6x3 − 4x4 + ........ − 92

x3 + 9x4

= 3x − 6x2 +32

x3 + 5x4



No reference


Page 529 Exercise 21B Question 1(e)


Page 54

2. Expand f(x) = ln(cosx) in ascending powers of x as far as the termin x6 using the expansions for ln(1 + x) and cosx.

The only logarithmic expansion available is that of ln(1 + x).We therefore must find an expression for cosx in the form (1+ ...)

One possibility is cosx = 1 – 2sin2 x2

⎛ ⎝

⎞ ⎠ , but it is easier to use the

artificial form [1 + (cosx – 1)]. i.e. ln(cosx) = ln[1 + (cosx – 1)]

ln(1 + x) = x −x2

2+

x3

3−

x4

4+ ........

ln(cosx) = ln[1 + (cosx – 1)]

= cosx −1( ) − 12

cosx −1( )2 +13

cosx −1( )3 −14

cosx −1( )4 + ........

But cosx = 1 − x2

2!+

x4

4!−

x6

6!+ ........

and so cosx – 1 = −x2

2!+

x 4

4!−

x6

6!+ ........

ln(cosx)

= −x2

2!+

x4

4!−

x6

6!+

⎛ ⎝ ⎜ ⎞

⎠ −

12

−x2

2!+

x4

4!−

x6

6!+

⎛ ⎝ ⎜ ⎞

⎠

2

+13

−x2

2!+

x4

4!−

x6

6!+

⎛ ⎝ ⎜ ⎞

⎠

3

− ....

= −x2

2!+

x 4

4!−

x6

6!−

12

x4

4−

x6

48+ ..... − x6

48+ ..

⎛ ⎝ ⎜ ⎞

⎠ +

13

−x6

8+ ..

⎛ ⎝ ⎜ ⎞

⎠

= −12

x 2 +124

x4 −1

720x6 −

18

x 4 +196

x6 +196

x6 −124

x6

= −12

x 2 −112

x4 −145

x6

Exercise 21. Expand esin x as far as the term in x4 .

2. Expand ln(1 + sinx) as far as the term in x4.

3. Expand exsinx as far as the term in x5.

4. Expand ln(1 + ex) as far as the term in x4.



No reference


Page 529 Exercise 21B Question 2


Page 55

ITERATION

The idea of Iteration has been met before.Iteration is defined as successive repetition of a mathematical process using the result at one stage as the input for the next.We know how to solve the equation 3x2 + 4x – 2 = 0 by using the quadratic formula, if it does not factorise.But if the equation is more complex, eg. x3 – 3x2 + 2x + 7 = 0 or 2x + 1 = ex, there is no formula available.We will develop a method, iteration, which will give numerical approximations to the roots to a required degree of accuracy.

Simple IterationConsider the equation x2 – 4x – 8 = 0Let f(x ) = x2 – 4x – 8 = 0From the table of signs:-

x –3 –2 –1 0 1 2 3 4 5 6 f(x) + + – – – – – – – +

We can see that there is a negative root between –2 and –1 and a positive root between 5 and 6 shown by the change of sign.

To investigate the root between –2 and –1 we try a value of x between –2 and –1 and narrow the interval.f(–2) = 4 f(–1) = –3 ∴ the root is slightly closer to –1 than to –2.f(–1·5) = 0·25 ∴ the root lies between –1·5 and –1.f(–1·4) = –0·44 ∴ the root lies between –1·5 and –1·4.f(–1·45) = –0·0975 ∴ the root lies between –1·5 and –1·45f(–1·475) = 0·0756 ∴ the root lies between –1·475 and –1·45f(–1·46) = –0·0284 ∴ the root lies between –1·475 and –1·46 f(–1·465) = 0·0062 ∴ the root lies between –1·465 and –1·46The root to 2 decimal places is – 1·46 The other root between 5 and 6 can be found in the same way, i.e. 5·46

However the equation:- x2 – 4x – 8 = 0 can be re-written in other ways.

(i) From 4x = x2 – 8 => x = 14

x 2 − 2

(ii) From x(x – 4) = 8 => x = 8x − 4

(iii) From x2 = 4x + 8 => x2 = 4(x + 2) => x = 2 x + 2( )12

(iv) From x2 = 4x + 8 => x = 4 +8x

These formulae are all in the form x = g(x), contd...


Page 56

We can express each of these as a recurrence relation(i) xn+1 =

14

xn2 − 2 (ii) xn+1 = 8

xn − 4

(iii) xn+1 = 2 xn + 2( )12 (iv) xn+1 = 4 +

8xn

With starting values x0, we can use them to determine the roots of this equation.

Using (i) xn+1 = 14

xn2 − 2

For the root between –2 and –1, x0 = –1, converges to x = –1·464.For the root between –2 and –1, x0 = –2, converges to x = –1·464.For the root between 5 and 6, x0 = 5, converges to x = –1·464.For the root between 5 and 6, x0 = 6, diverges.

Using (ii) xn+1 = 8xn − 4

For the root between –2 and –1, x0 = –1, converges to x = –1·464.For the root between –2 and –1, x0 = –2, converges to x = –1·464.For the root between 5 and 6, x0 = 5, converges to x = –1·464.

For the root between 5 and 6, x0 = 6, not defined. (since 84 − 4

→ ∞ )

Using (iii) xn+1 = 2 x + 2( )12 , not valid for the negative root

For the root between 5 and 6, x0 = 5, converges to x = 5·464.For the root between 5 and 6, x0 = 6, converges to x = 5·464.

Using (iv) xn+1 = 4 +8xn

For the root between –2 and –1, x0 = –1, converges to x = 5·464.For the root between –2 and –1, x0 = –2, not defined.For the root between 5 and 6, x0 = 5, converges to x = 5·464.For the root between 5 and 6, x0 = 6, converges to x = 5·464.

Clearly one formula does not give all the roots for x = g(x).

If the equation f(x) = 0 is written in the form x = g(x), the recurrence relationxn+1 = g(xn), for some chosen value of x0, may define a convergent sequence.

In this case, if the sequence converges to α, then α satisfies the equation, since xn–> α and xn+1–> α.

Therefore by xn+1 = g(xn), α = g(α).The sequence may instead diverge.


Page 57

The behaviour of the sequence defined by xn +1 = g(xn ) can be illustrated graphically. The graphs of y = x and y = g(x) are drawn on the same axes.The root α is the x–coordinate of the point of intersection of the graphs at P.In each case, let x0 be the first approximation for α. Starting at x0 on thex–axis, lines drawn parallel alternately to the y–axis and the x–axis.The curve converts x0 to g(x0) then the line converts g(x0) to x1 and so on.

This is known as a staircase diagramThe root α is the x–coordinate of the point of intersection of the graphs at P.All starting values converge to P.

This is known as a cobweb diagramThe root α is the x–coordinate of the point of intersection of the graphs at P.All starting values converge to P.

All starting values diverge.

In Graphs 1, the sequence converges to α. Here 0 < ′ g α( ) < 1.In Graphs 2, the sequence converges to α. Here –1 < ′ g α( ) < 0.In Graphs 3, the sequence diverges. Here ′ g α( ) < – 1.In Graphs 4, the sequence diverges. Here ′ g α( ) > 1.

i.e. for convergence of an iterative scheme for finding a fixed point α,

| ′ g a( ) | < 1 (i.e. –1 < ′ g α( ) < 1)

x0x1 x2α

Graph 3

y = x

y

x

y = g(x)

P

x3 x4

x0 x1x2 α

Graph 1

x0

y = x

y = g(x)

y

x

P

x0 x1x2 α

Graph 2

y = xy

x

y = g(x)

P

All starting values diverge.

x0x1x2

Graph 4

y = x

x

y = g(x)

P

x3x0

y

α


Page 58

For the equation x2 – 4x – 8 = 0 :-

(i) rearranged to x = 14

xn2 − 2 where g(x) =

14

xn2 − 2 , the corresponding

rough graphs are as follows:-

(ii) rearranged to x = 8

xn − 4 where g(x) = 8

xn − 4 , the corresponding


In this graph, the roots converge for starting values of x = –1, –2 and 5 to α = –1·464. They diverge for x = 6 i.e. does not converge to β = 5·464.

Here ′ g x( ) =12

x and | ′ g α( ) | < 1

whereas | ′ g β( ) | > 1

In this graph, the roots converge for starting values of x = –1, –2 and 5 to α = –1·464. For x = 6, it is undefined.Here ′ g x( ) = −

8x − 4( )2 and

–1 < ′ g α( ) < 0 and ′ g β( ) < –1

y

xx0 = –1x0 = –2

x0 = 5 x0 = 6

α

β

y = 14 x2 – 2

y = x

y

x

y = x

x = 4

x0 = –1x0 = –2x0 = 5

y = x

y =8

x – 4 β

α


Page 59

(iii) rearranged to x = 2 xn + 2( )12 where g(x) = 2 xn + 2( )1

2 , the corresponding rough graphs are as follows:-

(iv) rearranged to x = 4 +8xn

where g(x) = 4 +8xn

, the corresponding


In this graph, the roots converge for starting values of x = 5 and x = 6 to β = 5·464.

Here ′ g x( ) = −8x2 and

′ g α( ) < –1 and – 1 < ′ g β( ) < 0

–2

y

x

y =x

y = 2√(x + 2)

βx0 = 5 x0 = 6

In this graph, the roots converge for starting values of x = 5 and 6 to β = 5·464.

Here ′ g x( ) =1

x + 2( ) and:-

0 < ′ g β( ) < 1

x0 = 5 x0 = 6β

y = xy = 4 +

8x

y

x

α

x0 = -1x0 = -2


Page 60

Examples1. Show that, for the equation x2 – 5x + 3 = 0, one of the following

iterative formulae:- (a) xn+1 =xn

2 + 3( )5 (b) xn+1 = 5 − 3

xn

produces a convergent series for x ≈ 5 and the other does not.

(a) g x( ) =x2 + 3( )

5=

x2

5+

35

, ′ g x( ) =25

x and hence ′ g 5( ) = 2

Since | ′ g 5( ) | > 1, this formula will not produce a convergent series when x ≈ 5.

(b) g x( ) = 5 −3x

, ′ g x( ) =3x2 and hence ′ g 5( ) =

325

Since | ′ g 5( ) | < 1, this formula will produce a convergent series when x ≈ 5.

2. (a) Show that the equation x – cosx = 0 has only one root in the interval 0 ≤ x ≤ π

2.

(b) Estimate the root to 1 decimal place.

(c) Use the recurrence relation xn+1 = cos xn to find the root to 3 decimal places.

(a)

(b) x0 ≈ π4

= 0·785... radians

(c) Using xn+1 = cos xn , the root converges to 0·739 radians.

contd...

x

y

y = x

y = cosx

π2

x0 α

1


Page 61

3. Use a recurrence relation to find to 5 decimal places the root ofthe equation e–x = 3(x – 1).

Rearranging gives xn+1 =13

e−xn + 1

Using xn+1 =13

e−xn + 1 with x0 = 1, the root converges to 1·10987

Exercise 3 You should use a graphic calculator in this exercise.

1. (a) Show that the equation x3 – 5x + 3 = 0 has a root between x = 0 and x = 1.

(b) Using the iteration formula xn+1 =3 + xn

3

5 , find this root to 2 decimal places.

2. (a) Show that the equation x3 – 2x2 – 3 = 0 has a root between x = 2 and x = 3.

(b) Investigate graphically the behaviour of the sequence

xn+1 =xn

3 − 32 xn

when x0 = 2 and when x0 = 3.

(c) Use the recurrence relation xn+1 =3

xn2 + 2 , with x0 = 2, to

calculate the root to 3 decimal places.Illustrate graphically the convergence of the sequence.

3. Use the recurrence relation xn+1 = e− xn +1( ) , with x0 = 1·15,

to find the positive root of the equation x2 – 1 = e–x correct to 5 decimal places.

4. (a) Show that x = cosx – 3 has a root between – 4 and – 3.

(b) Use an iterative formula to calculate this root to threedecimal places

contd...

y = x y

x x0 = 1

y =13

e− x + 1


Page 62

5. (a) Show that x = ln(8 – x) has a root between 1 and 2.

(b) Using the iterative formula xn+1 = ln 8 − xn( ), calculate thisroot to 3 decimal places.

6. (a) Show graphically that neither

xn+1 =25

4 − xn3( ) nor xn+1 =

8 − 5xn( )2xn

2 can be used to find

the real root of the equation 2x3 +5x – 8 = 0.

(b) Derive a suitable iterative formula and hence obtain the root to 3 decimal places

7. (a) Show that x4 + x2 – x = 0 has two roots.

(b) Using the iterative formula xn+1 =xn

xn3 + xn

, find the larger

root correct to 2 significant figures.

8. The iterative formula xn+1 =12

xn +cxn

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ is used to solve the

equation x2 = c (i.e. x = √c).Estimate √20 to 3 decimal places using x0 = 4 as a starting value.



Page 115 Exercise 2.6:2 Questions 8, 9(ii), 10(ii), 12(ii)


Page 531 Exercise 21C Questions 1, 2, 3.


Page 63

Answers

Exercise 1

1. 1 − x2

2!+

x4

4!−

x6

6!2. x +

13

x3

3. x +16

x3 4. −x −12

x 2 −13

x3 −14

x 4

5. 1 + 3x +92

x2 +92

x3 +278

x4 6. 2x − 2x 2 +83

x3 − 4x 4 +325

x5

7. 3x −92

x3 +8140

x5 8. 2x +83

x3 +6415

x5

9. ln2 +12

x −18

x2 +124

x 3

Exercise 2

1. 1 + x +12

x2 −18

x4 2. x − 12

x2 +16

x3 −112

x4

3. x + x2 +13

x3 −130

x5 4. ln2 +12

x +18

x2 +1

192x4

Exercise 3

1. (a) Proof (b) 0·66

2. (a) Proof (b) Divergent (c) 2·486

3. 1·14776

4. (a) Proof (b) –3·794

5. (a) Proof (b) 1·821

6. (a) Proof (b) 1·087 using xn+1 = 4 −52

xn⎛ ⎝ ⎜

⎞ ⎠ ⎟

13 or xn+1 =

82xn

2 + 5

7. 0·68

8. 4·472


Page 64

Outcome 4 – Further Ordinary Differential Equations First order linear differential equations using the integrating factor.

Linear differential equations take the form dydx

+ P(x)y = Q(x) where P(x) and Q(x) are functions of x.

To solve this type of equation, we have to multiply all terms by a factor calledthe integrating factor, denoted by µ(x), which converts the whole of the left-hand side of the equation into the derivative of one function.

For dydx

+ P(x)y = Q(x), multiply all terms by µ(x).

=> µ(x) dydx

+ P(x)µ( x)y = µ(x)Q(x) Now d

dx[µ (x )y] = µ(x ) dy

dx+

ddx

[ µ( x)]y (using the product rule)

and so we choose µ(x) such that ddx

[µ( x)] = P(x )µ(x)

i.e. 1µ( x)

ddx

[ µ(x)] = P(x)

=> ddx

[ln( µ( x))] = P(x)

so ln(µ(x)) = P x( )∫ dx

i.e. µ(x) = e P x( ) dx∫

The equation µ(x) dydx

+ P(x)µ( x)y = µ(x)Q(x) becomes

µ(x) dydx

+ddx

[µ(x)]y = µ(x)Q( x)

i.e. ddx

[µ(x)y] = µ(x)Q(x)

i.e. µ(x)y = µ(x)Q(x)∫ dx

=> y =1

µ (x)µ (x)Q( x)∫ dx

where µ(x) = e P x( ) dx∫ is the integrating factor.


Page 65

Examples1. Find the general solution of dy

dx + y

x = 1

Here P(x) = 1x

, Q(x) = 1 and µ(x) = e P x( ) dx∫ = e1x

dx∫ = e lnx = x

Thus the equation becomes:-

x dydx

+ y = x [i.e. µ(x) dydx

+ P(x)µ( x)y = µ(x)Q(x) ]

i.e. ddx

[xy] = x [i.e. ddx

[µ(x)y] = µ(x)Q(x)]

xy = xdx∫ = x2

2+ c

=> y =x2

+cx

2. Find the general solution of xdydx + (x – 2)y = x3

xdydx + (x – 2)y = x3 has to be written as

dydx +

x − 2x y = x2

Here P(x) = x − 2

x = 1 −2x , Q(x) = x2 and

µ(x) = e1− 2

x⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ dx∫ = ex− 2lnx = ex− lnx 2

= exe−lnx 2=exe

ln 1x 2 = 1

x2 ex

Multiply both sides by µ(x) = 1x2 ex gives:-

1x2 e x dy

dx + x − 2

x3⎛ ⎝ ⎜

⎞ ⎠ ⎟ ex y = ex [i.e. µ(x) dy

dx+ P(x)µ( x)y = µ(x)Q(x) ]

=> ddx

1x2 ex y⎡ ⎣ ⎢

⎤ ⎦ ⎥ = ex [i.e. d

dx[µ(x)y] = µ(x)Q(x)]

=> 1x2 ex y = ex∫ dx

=> 1x2 ex y = ex + c

=> y = x2 + cx2e–x

3. Find the general solution of dydx

– 2y = 6e–x

Here P(x) = –2, Q(x) = 6e−x and µ(x) = e −2dx∫ = e−2x

Multiply both sides by e−2x

e−2x dy – 2e−2x y = 6e−3x

=> ddx

e−2 xy[ ] = 6e−3x

=> e−2x y = 6e−3x∫ dx = –2e−3x + c=> y = –2e−x + ce2 x


Page 66

4. Find the general solution of (1 + x2)dydx – xy = x(1 + x2)

Rewrite asdydx –

x1 + x2 y = x i.e. P(x) = –

x1 + x2 , Q(x) = x

=> µ(x) = e−

x1+x 2 dx∫

= e−1

2ln 1+x 2( )

= eln 1

1+x 2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

=1

1 + x2

i.e. 11 + x2

dydx

−x

1 + x2( )32

y =x

1+ x2

=> ddx

11+ x2

y⎡ ⎣ ⎢

⎤ ⎦ ⎥ =

x1 + x 2

=>1

1 + x2y =

x1 + x2∫ dx = 1 + x2 + c

=> y = 1 + x2 + c 1 + x2

5. Find the particular solution of dydx

+ y = 2x + 4 given y = 1 when x = 0

Integrating factor µ(x) = e 1dx∫ = ex

i.e. ex dydx + exy = ex(2x + 4)

ddx [exy] = ex(2x + 4)

exy = ex∫ 2x + 4( )dx [ integration by parts required]

exy = ex(2x + 4) – 2ex∫ dx

exy = ex(2x + 4) – 2ex + c

=> y = 2x + 4 – 2 + ce–x

=> y = 2x + 2 + ce–x

When x = 0 and y = 11 = 2 + ce0 => c = –1

The particular solution is y = 2x + 2 – e–x

u = 2x + 4dvdx = ex

dudx = 2 v = ex


Page 67

Exercise 1

1. Find the general solutions of these linear differential equations:

(a) (x + 1)dydx – y = (x + 1)2 (b)

dydx – ytanx = sinxcosx

(c) tanxdydx + 2y = x cosecx (d)

dydx +

2y1 − x2 = 1 – x

(e) x(x + 1)dydx – y = x3ex (f)

dydx + y = 5cos2x

(g) (1 – x)dydx + xy = (1 – x)2e–x (h)

dydx +

x + 1x y = e–x

(i) x(x + 1)dydx + y = x(x + 1)2e–x (j)

dydx + ycotx = cosx

2. Find the particular solutions of these linear differential equations:

(a) xdydx + 2y = x3 when x = 1 and y = 2.

(b) (1 + x)dydx + 2y = x2 when x = 0 and y = 0.

(c) sinx dydx

– ycosx = 1 when x = π2

and y = 3.

(d) x(x + 1)dydx + y = (x + 1)2ex when x = 1 and y = 0.

(e) xdydx = y + x2(sinx + cosx) when x =

π2 and y = 0.



No reference


No reference


Page 68

Second order linear differential.

Simple Equationsd2 ydx 2 = f(x)Solve by 2 successive integrations.

Exampled2 ydx 2 =

1x2

=> dydx =

1x2∫ dx = −

1x

+ C

=> y = −1x

+ C⎛ ⎝ ⎜

⎞ ⎠ ⎟ ∫ dx = −lnx + Cx + D

where C and D are arbitrary constantsLinear Equations.

Second order linear equations have the form

ad2ydx2 + b

dydx

+ cy = f x( )where a, b and c are constants.

We look for solutions of the form y = AeDx

1. Where f(x) = 0.

i.e. a d2ydx2 + b dy

dx+ cy = 0

If we try y = AeDx as a solution of this equation, then

dydx = DAeDx and

d2 ydx 2 = D2AeDx

Substituting into the equation givesaD2AeDx + bDAeDx + cAeDx = 0

i.e. AeDx (aD2 + bD + c) = 0Thus y = AeDx is a solution provided

aD2 + bD + c = 0 solves to give one of the following :-

(i) 2 distinct real roots. (b2 – 4ac > 0)If D = D1 and D2 then the General solution is

y = AeD1x + Be D2 x

(ii) equal roots. (b2 – 4ac = 0)If D = D1(twice) then the General solution is

y = AeD1x + BxeD1 x = A + Bx( )eD1x

(iii) complex roots. (b2 – 4ac < 0)If D = D1 and D2 where D1 and D2 are of the form p ± qi

then the General solution is:-

y = epx(Acosqx + Bsinqx)

The equation aD2 + bD + c = 0 is called theAuxiliary Equation.(A.E.)


Page 69

Examples

1. Solve d2 ydx 2 + dy

dx – 6y = 0

The Auxiliary Equation (A.E.) is D2 + D – 6 = 0 => (D + 3)(D – 2) = 0 => D = – 3 and D = 2The General Solution(G.S.) is

y = Ae–3x + Be2x

2. Solve d2 ydx 2 + 4 dy

dx + 4y = 0

The Auxiliary Equation (A.E.) is D2 + 4D + 4 = 0 => (D + 2)2 = 0 => D = – 2The General Solution(G.S.) is

y = (A+ Bx)e–2x


dx + 5y = 0

The Auxiliary Equation(A.E.) is D2 + 2D + 5 = 0 => Does not factorise.

Using the quadratic formula

D = −2 ± 4 − 20

2=−2 ± −16( )

2=−2 ± 4i

2= −1 ± 2i

The General Solution(G.S.) is

y = e–x(Acos2x + Bsin2x)

Exercise 2 1. Find the general solutions of these linear differential equations:

(a) d2 ydx 2 – 3 dy

dx + 2y = 0 (b) d2 y

dx 2 – 4 dydx

+ 3y = 0

(c) d2 ydx 2 – 3 dy

dx = 0 (d) 2 d2 y

dx 2 – 9 dydx

+ 9y = 0


(a) d2 ydx 2 + 6 dy

dx + 9y = 0 (b) 4 d2 y

dx 2 + 4 dydx

+ y = 0

(c)d2 ydx 2 – 6

dydx + 9y = 0 (d)

d2 ydx 2 + 2n

dydx + n2y = 0


(a) d2 ydx 2 + 2 dy

dx + 2y = 0 (b) d2 y

dx 2 + 4 dydx

+ 8y = 0

(c)d2 ydx 2 + 6

dydx + 13y = 0 (d)

d2 ydx 2 +

dydx + y = 0


Page 70

2. Where f(x) ≠ 0.

Firstly we find the general solution as above.This is called the Complementary Function (C.F.).

We then find a particular solution of the given equation.This is called the Particular Integral (P.I.).

The General Solution (G.S.) is found by adding the Complementary Function (C.F.) and the Particular Integral (P.I.)

For the P.I. we choose a solution of the same form as f(x).

(i) If f(x) is a polynomial function of degree n, i.e. x2 + 2x – 1try y = Cnx n + Cn–1x

n–1 + Cn–2xn–2 + .......... + C1

find dydx

and d2 ydx 2 and then substitute into the given equation.

By comparing coefficients of the left-hand side and the right-hand side, find the arbitrary constants Cn ......C1.

(ii) If f(x) is an exponential function, i.e. 3e2x

try y = Cepx

find dydx


By comparing coefficients of the left-hand side and the right-hand side, find the arbitrary constants C.

(iii) If f(x) is a trigonometric function, i.e. mcospx, nsinpx ormcospx + nsinpxtry y = C1cospx + C2sinpx

find dydx


By comparing coefficients of the left-hand side and the right-hand side, find the arbitrary constants C1 and C2.

Examples


dx + 2y = 4x + 4

The Auxiliary Equation (A.E.) is D2 + 3D + 2 = 0 => (D + 1)(D + 2) = 0 => D = –1 and D = –2The Complementary Function (C.F.) is

y = Ae–x + Be–2x

For the Particular Integral (P.I.), try y = Cx + Ddydx

= C and d2 ydx 2 = 0

Substitute into the given equation gives0 + 3C + 2Cx + 2D = 4x + 4 2Cx + (3C + 2D) = 4x + 4

cont’d....


Page 71

...cont’d => Comparing coefficients2C = 4 => C = 2

3C + 2D = 4 => D = –1i.e. The Particular Integral (P.I.) is y = 2x – 1

The General Solution (G.S.) = C.F. + P.I.i.e. y = Ae–x + Be–2x + 2x – 1

2. Solve d2 ydx 2 – 2

dydx – 3y = x2 + 4x – 5

The Auxiliary Equation (A.E.) is D2 – 2D – 3 = 0 => (D + 1)(D – 3) = 0 => D = –1 and D = 3

The Complementary Function (C.F.) isy = Ae–x + Be3x

For the Particular Integral (P.I.), tryy = Cx2 + Dx + Edydx

= 2Cx + D and d2 ydx 2 = 2C

Substitute into the given equation gives

2C – 2(2Cx + D) – 3(Cx2 + Dx + E) = x2 + 4x – 5–3Cx2 + (– 4C – 3D)x + 2C – 2D – 3E = x2 + 4x – 5Comparing coefficients

– 3C = 1 => C = – 13

– 4C – 3D = 4 => D = – 89

2C – 2D – 3E = –5 => E = 5527

i.e. The Particular Integral (P.I.) isy = – 1

3 x2 – 89 x +

5527

The General Solution (G.S.) = C.F. + P.I.

i.e. y = Ae–x + Be3x –13 x2 –

89 x +

5527


Page 72

3. Solve d2 ydx 2 – y = 2e3x

The Auxiliary Equation (A.E.) is D2 – 1 = 0 => (D + 1)(D – 1) = 0 => D = –1 and D = 1The Complementary Function (C.F.) is

y = Ae–x + Bex

For the Particular Integral (P.I.), try y = Ce3x

dydx = 3Ce3x and d2 y

dx 2 = 9Ce3x

Substitute into the given equation gives9Ce3x – Ce3x = 2e3x 8Ce3x = 2e3x

Comparing coefficients8C = 2 => C = 1

4i.e. The Particular Integral(P.I.) is y = 1

4 e3x


i.e. y = Ae–x + Bex + 14 e3x

4. Solve d2 ydx 2 + 2

dydx + 2y = 3e–2x

The Auxiliary Equation (A.E.) is D2 + 2D + 2 = 0 => Does not factorise.

Using the quadratic formula:-

D = −2 ± 4 −82

=−2 ± −4( )

2=−2 ± 2i

2= −1 ± i

The Complementary Function (C.F.) isy = e–x(Acosx + Bsinx)

For the Particular Integral (P.I.), try y = Ce–2x

dydx = –2Ce–2x and d2 y

dx 2 = 4Ce–2x

Substitute into the given equation gives4Ce–2x – 4Ce–2x + 2Ce–2x = 3e–2x

2Ce–2x = 3e–2x

Comparing coefficients => C = 32

i.e. The Particular Integral (P.I.) is y = 32 e–2x


i.e. y = e–x(Acosx + Bsinx) + 32 e–2x


Page 73

5. Solve d2 ydx 2 + 6

dydx + 10y = 30sin2x

The Auxiliary Equation (A.E.) is D2 + 6D + 10 = 0 (Does not factorise)

Using the quadratic formula

D = −6 ± 36 − 402

=−6 ± −4( )

2=−6 ± 2i

2= −3 ± i

The Complementary Function (C.F.) isy = e–3x(Acosx + Bsinx)

For the Particular Integral (P.I.), try y = Ccos2x + Dsin2xdydx

= –2Csin2x + 2Dcos2x and d2 ydx 2 = – 4Ccos2x – 4Dsin2x

Substitute into the given equation gives:-(6C +12D)cos2x + (– 12C + 6D )sin2x = 30sin2xComparing coefficients

6C +12D = 0 – 12C + 6D = 30=> C = – 2 and D = 1i.e. The Particular Integral (P.I.) is y = –2cos2x + sin2x

The General Solution (G.S.) = C.F. + P.I.i.e. y = e–3x(Acosx + Bsinx) –2cos2x + sin2x

6. Solve d2 ydx 2 – 6

dydx + 9y = ex + sinx

The Auxiliary Equation (A.E.) is D2 – 6D + 9 = 0 => (D – 3)2 = 0 => D = 3

The Complementary Function (C.F.) isy = e3x(A + Bx)

For the Particular Integral (P.I.), try y = Cex + Dcosx + Esinxdydx = Cex – Dsinx + Ecosx and

d2 ydx 2 = Cex – Dcosx – Esinx

Substitute into the given equation gives4Cex + (8D – 6E)cosx + (6D + 8E)sinx = ex + sinxComparing coefficients4C = 1 8D – 6E = 0 C = 1

46D + 8E = 1

i.e. The Particular Integral (P.I.) is y = 1

4 ex + 350

cosx + 225

sinx


i.e. y = e3x(A + Bx) + 14 ex +

350

cosx + 225

sinx

} D = 350

and E = 225


Page 74

Exercise 2


(a)d2 ydx 2 – 4

dydx + 3y = x3 (b)

d2 ydx 2 – y = 2 – 5x

(c)d2 ydx 2 + 5

dydx + 4y = 32x2 (d)

d2 ydx 2 + 2

dydx + 2y = 1 + x2


(a)d2 ydx 2 + 2

dydx – 3y = 10e2x (b) 4

d2 ydx 2 + 13

dydx + 9y = 7e–2x

(c)d2 ydx 2 –

dydx – 2y = ex (d)

d2 ydx 2 – 4

dydx + 4y = 2e–2x


(a) d2 ydx 2 – 3 dy

dx + 2y = sinx (b) d2 y

dx 2 – 2 dydx

+ y = 10cos2x

(c) 4 d2 ydx 2 + y = 4sinx (d) d2 y

dx 2 + 8 dydx

+ 25y = 26cos3x

7. Solve d2 ydx 2 – y = 2ex

The Auxiliary Equation(A.E.) is D2 – 1 = 0 => (D + 1)(D – 1) = 0 => D = ± 1

The Complementary Function (C.F.) isy = Aex + Be–x

Note If the terms in the C.F. also appear in the P.I.then an extra x is required in the P.I. note*

For the Particular Integral (P.I.), try y = Cxex

dydx = Cex + Cxex (by the product rule)

d2 ydx 2 = Cex + Cex + Cxex = 2Cex + Cxex

Substitute into the given equation gives2Cex + Cxex – Cxex = 2ex

C = 1i.e. The Particular Integral (P.I.) is

y = xex

The General Solution (G.S.) = C.F. + P.I.i.e. y = Aex + Be–x + xex

cont’d .....


Page 75



No reference.


Page 519 Exercise 20E Questions 1 – 5.


(a)d2 ydx 2 + 4

dydx + 4y =x – e2 x (b)

d2 ydx 2 +

dydx – 6y = ex + e–x

(c)d2 ydx 2 + 4

dydx + 8y = x – ex (d) 4

d2 ydx 2 + 4

dydx + y = ex – 2cos2x

5. (a)d2 ydx 2 + 5

dydx + 6y = 3e–2x (b)

d2 ydx 2 – 4

dydx + 3y = 2ex

6. Find the particular solution of

d2 ydx 2 + 2 dy

dx + 2y = sinx for which y = 0, dy

dx = 0 when x = 0

7. Find the particular solution of

d2 ydx 2 –

dydx – 6y = 5e3x for which y = 1, dy

dx = – 6 when x = 0


Page 76

AnswersExercise 1

1. (a) y = (x + 1)(x + c) (b) y = –13 cos2x + csecx

(c) y = cosecx(x + cotx + ccosecx) (d) y = 1 − x1 + x⎛ ⎝

⎞ ⎠

12

x2 + x + c⎛ ⎝

⎞ ⎠

(e) y = x

x + 1⎛ ⎝

⎞ ⎠ xex − ex + c( ) (f) y = (2sin2x + cos2x) + ce-x

(g) y = cex(1 – x) – 12 e-x(1 – x) (h) y =e− x x

2+

cx

⎛ ⎝

⎞ ⎠

(i) y = c(x + 1)

x−

x + 1( )2

xe− x (j) y = c cosecx – 1

4cos2xsin x

2. (a) y =1

5x2 x 5 − 9( ) (b) y = x3 3x + 4( )12 1+ x( )2

(c) y = 4sinx – cosecxcotx (d)x + 1

x⎛ ⎝

⎞ ⎠ ex − e( )

(e) y = x2(sinx – cosx) + x(sinx + cosx) – x(π2 + 1)

Exercise 21. (a) y = Ae2x + Bex (b) y = Ae3x + Bex

(c) y = A + Be3x (d) y = Ae32

x+ Be3x

2. (a) y = (A + Bx)e–3x (b) y = A + Bx( )e− 1

2x

(c) y = (A + Bx)e3x (d) y = (A + Bx)e–nx

3. (a) y = e–x(Acosx + Bsinx) (b) y = e–2x(Acos2x + Bsin2x)

(c) y = e–3x(Acos2x + Bsin2x) (d) y = e−1

2x

Acos 32

x + Bsin 32

x⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

Exercise 3

1. (a) y = Aex + Be3x +13

x3 +43

x2 +269

x +8027

(b) y = Aex + Be−x + 5x − 2(c) y = Ae−x + Be−4x + 8x2 − 20x + 21

(d) y = e−x Acos x + Bsin x( ) +12

x2 − x + 1

2. (a) y = Ae−3x + Be x + 2e2x (b) y = Ae−x + Be− 9

4x− 7e−2x

(c) y = Ae−x + Be2x −12

ex (d) y = A + Bx( )e2x +18

e−2x


Page 77

Exercise 3 (contd)

3. (a) y = Aex + Be2x +3

10cos x +

110

sin x

(b) y = A + Bx( )ex −65

cos2x −85

sin 2x

(c) y = Acos 12

x + Bsin 12

x −43

sin x

(d) y = e−4x Acos3x + Bsin3x( ) +12

cos3x +34

sin 3x

4. (a) y = A + Bx( )e−2x +14

x −14−

116

e2x

(b) y = Ae−3x + Be2x −14

ex −16

e−x

(c) y = e−2x Acos 2x + Bsin 2x( ) +18

x −116

−1

13ex

(d) y = A + Bx( )e− 1

2x

+19

ex +30289

cos2x −16289

sin 2x

5. (a) y = Ae–2x + Be–3x + 3xe–2x

(b) y = Aex + Be3x – xex

6. y = e−x 25

cos x + 65

sin x⎛ ⎝ ⎜

⎞ ⎠ ⎟ −

25

cos x +15

sin x

7. y = 2e−2x − e3x + xe3x


Page 78

Outcome 5 – Further Number Theory and Further Methods of Proof.

Further Mathematical ProofNegation of a Statement:-

Statement : A rhombus has four equal sides. (can be true or false)Negation : A rhombus does not have four equal sides.

If a statement is true then the negation is false.If a statement is false then the negation is true.

Given that p represents a statement, the negation is written ~p (reads “not p”), and is such that if p is true, ~p is false; if p is false, ~p is true.

Examples Negate the following statements:-1. “All cats have tails”.

p : All cats have tails.Since the given statement is taken to be true for every cat, thenegation must assert that at least one cat has no tail and so:-~p : Some cats do not have tails.

2. “Some pilots are women”.p : Some pilots are women.Since the given statement asserts that there is at least one pilotwho is a woman, the negation must assert that all pilots are not women and so:-~p : No pilots are women.

Exercise 11. Which of the following is a negation of “All boys are adventurous” ?

(a) No boys are adventurous.(b) All boys are unadventurous.(c) Some boys are not adventurous.(d) No boys are unadventurous.

2. Which of these is a negation of “No visitors may walk on the grass” ?(a) All visitors may walk on the grass.(b) Some visitors may not walk on the grass.(c) All visitors may not walk on the grass.(d) Some visitors may walk on the grass.

3. Write down the negation of each of the following statements:-(a) For all real x, x2 is positive.(b) Some pupils find mathematics difficult.(c) No dogs like cats.(d) There exists a positive integer x such that x + 3 > 0.(e) Every parallelogram has half turn symmetry.(f) No schoolboy lies.(g) A number which has zero in the units place is divisible by five.(h) All numbers of the form 2n – 1, (n an integer), are prime.


Page 79

Equivalent Statements

If a statement and its converse are true then the implication can be replaced by the two-way implication sign <=>.

If p => q and q => p are both true then we can write p <=> q.This is sometimes read as “ p (is true) if and only if q (is true)”, or “p iff q”.

If p => q, we say that p is a sufficient condition for q ; if q => p, we say that p is a necessary condition for q.

Examples In each of the following, say whether the first statement is:- (a) a necessary condition, (b) a sufficient condition,(c) both necessary and sufficient, (d) neither,for the second condition condition.

1. p: John plays the piano. q: John is a concert pianist.

Since q => p, the first statement is a necessary condition for the second as John must be able to play the piano to be concert pianist, but since p ≠> q, it is not a sufficient condition.

2. p: ABCD is a rhombus q: the diagonals of ABCD bisect each other.

Since p => q, the first statement is a sufficient condition for the second as the diagonals of a rhombus bisect each other.Since q ≠> p, it is not a necessary condition.

The Converse of a Statement

Statement : If a triangle is right–angled, then the square of the hypotenuse isequal to the sum of the squares on the other two sides.

Converse : If the square on the longest side of a triangle is equal to the sum of the squares on the other two sides, then the triangle is right angled.

If p => q then the converse is q => p. (=> means “implies”)

In the cases of a statement and its converse:-(i) the statement may be true and the converse false.(ii) the statement may be be false and the converse true.(iii) both may be false or both may be true.

Exercise 2 State the converse of each of the following, and show by a counter example that the converse is false.

1. (a) If a number ends in 0, it is divisible by 5.(b) All primes greater than 2 are odd numbers.(c) If a quadrilateral is a square, its diagonals intersect at right angles.(d) x = 3 => x2 = 9.(e) If two numbers are odd then their sum is even.(f) If two integers are even, then their product is even.


Page 80

Contrapositive of a StatementIf p => q then the contrapositive is ~q => ~p.i.e. The negation of q implies the negation of p.If p => q is true then ~ q => ~p must also be true.Similarly, if p => q is false then ~ q => ~p must also be false. This is an important logical method of proving by indirect proof.

Method of Proof

Proof using Contrapositive

Exercise 3

1. For each of these, say whether the first statement is:- (i) a necessary condition, (ii) a sufficient condition,(iii) both necessary and sufficient, (iv) neither,for the second condition condition.(a) p: there are more than 8 people in this room

q: there are 9 people in this room.(b) p: ABCD is a parallelogram

q: the diagonals of ABCD are perpendicular.2. Which of the following statements are necessary or/and sufficient for

the statement q: “natural number n is divisible by 6” to be correct ?(a) p: n is divisible by 3 (b) p: n is divisible by 9(c) p: n is divisible by 12 (d) p: n2 is divisible by 12(e) p: n = 384 (f) p: n is even and divisible by 3(g) n = m(m + 1)(m + 2), where m is some natural number.

1. Prove that if x and y are integers and xy = 100, then either x ≤ 10 or y ≤ 10. p : x and y are integers and xy = 100, q : either x ≤ 10 or y ≤ 10. ~p : x and y are integers and xy ≠ 100 ~q : x > 10 and y > 10, then Proof of contrapositive:- (i.e. show ~q => ~p) x > 10 and y > 10 => xy > 100 => xy ≠ 100. => The contrapositive is true and hence the statement is true.

2. Prove that if 7 is a factor of n2 then 7 is a factor of n. p : 7 is a factor of n2 q: 7 is a factor of n. ~p : 7 is a not factor of n2 ~q: 7 is a not factor of n.Proof of contrapositive:-7 is a not factor of n => n = 7m + t for some integers m and t.

=> n2 = 49m2 + 14mt + t2 => n2 = 7(7m2 + 2mt) + t2 = 7k + remainder

=> 7 is not a factor of n2.The contrapositive is true and hence the statement is true.


Page 81

Proof by Contradiction

This is an extension of work in Mathematics 2(AH) Learning Outcome 5.To prove a theorem by this method we assume the theorem does not hold, thenshow that this assumption leads to a contradiction.

Examples

1. Prove that √2 is irrational.Suppose the opposite is true. i.e. √2 is rational.

=> 2 =mn where m and n have no common factor

=> 2 =m2

n2 => m 2 = 2n2 => m 2 is even => m is even = 2k (k e Z)

=> 2n2 = 4k2 => n2 = 2k2 => n2 is even => n is even

=> m and n have a common factor.

But m and n have no common factor.

Therefore √2 cannot be written in the form mn

This contradicts our assumption that √2 is rational. Hence √2 is irrational (by the method of contradiction).

2. If x and y are integers and xy is an odd integer, prove that x andy must both be odd.

Assume the opposite.

i.e. assume that xy is odd and at least one of x and y is even.

If x is even => x = 2z, z is an integer.

=> xy = 2zy (an even integer)

=> xy is even.

Since this is a contradiction, => the initial assumption is wrong.If x and y are integers such that xy is an odd integer, then x and y must both be odd.


Page 82

Proof by Induction

This is an extension of work in Mathematics 2(AH) Learning Outcome 5.We are now going to look at further examples of this method of proof.

Examples 1. Prove that r 2

r =1

n

∑ =16

n n + 1( ) 2n + 1( )

We know that r 2

r =1

n

∑ = 12 + 22 + 32 + ....+ (n – 1)2 + n2

Proof :- for n = 1, r 2

r =1

n

∑ = r 2

r =1

1

∑ = 12 = 1

and 16

n n + 1( ) 2n + 1( ) =16×1 × 2 ×3 = 1 => true for n = 1

Assume true for n = k, k ≥ 1 i.e. r 2

r =1

k

∑ = 16

k k + 1( ) 2k + 1( )

=> prove true for n = k +1 i.e. r 2

r =1

k+ 1

∑ = 16

(k + 1) k + 2( ) 2k + 3( )

r 2

r =1

k+1

∑ = r 2

r =1

k

∑ + (k + 1)2

= 16

k k + 1( ) 2k + 1( ) + k + 1( )2

= 16

k + 1( ) k 2k + 1( ) + 6 k + 1( )[ ] by factorisation

= 16

k + 1( ) 2k2 + 7k + 6[ ] = 1

6k + 1( ) k + 2( ) 2k + 3( )

Hence if true for k, it is true for n = k + 1True for n = 1 => True for n = 2 since k ≥ 1True for n = 2 => True for n = 3 and so on for all nHence true for all n, by induction.

Exercise 41. Prove, using the contrapositive,

(a) that if x and y are integers and xy is odd, then both x and y are odd.(b) that every prime number greater than 3 is of the form

6n ± 1, where n is a positive integer.(c) that if n is a natural number such that n2 is even, then n is even.

2. Prove, by contradiction,(a) that if x and y are integers such that x + y is odd, then one of them

must be odd and one must be even.(b) that if x and y are real numbers such that x + y is irrational, then at

least one of x, y is irrational.(c) that if m and n are integers such that mn2 is even, then at

least one of m or n is even.(d) that if sinθ ≠ 0, then θ ≠ kπ for any integer k.


Page 83

2. Prove that r 3

r =1

n

∑ =14

n2 n + 1( )2 = rr =1

n

∑⎡ ⎣ ⎢

⎤ ⎦ ⎥

2

We know that r 3

r =1

n

∑ = 13 + 22 + 33 + ....+ (n – 1)3 + n3

Proof :- for n = 1, r 3

r =1

n

∑ = r 3

r =1

1

∑ = 13 = 1

and 14

n2 n + 1( )2 =14×12 × 22 = 1 => True for n = 1

Assume true for n = k, k ≥ 1 i.e. r 3

r =1

k

∑ =14

k2 k + 1( ) 2

Hence prove true for n = k + 1 i.e. r 3

r =1

k+ 1

∑ =14

(k + 1)2 k + 2( )2

=> r 3

r =1

k+1

∑ = r 3

r =1

k

∑ + k + 1( )3

= 14

k2 k + 1( ) 2 + k + 1( )3

= 14

k + 1( )2 k2 + 4 k + 1( )[ ] by factorisation

= 14

k + 1( )2 k2 + 4k + 4( ) = 1

4k + 1( )2 k + 2( )2

Hence if true for k, it is also true for n = k + 1True for n = 1 => True for n = 2 since k ≥ 1True for n = 2 => True for n = 3 etc. for all n.and so true for all n, by induction.

3. For all integers n ≥ 10, n2 ≥ 10n.Proof :- for n = 10, n2 =100 and 10n = 100

=> True for n = 10.

Assume true for n = k, k ≥ 10 i.e. k2 ≥ 10k.Prove true for n = k + 1 also i.e. (k + 1)2 ≥ 10(k + 1).=> (k + 1)2 = k2 +2k + 1 ≥ 10k + 2k + 1

≥ 10(k + 1) + (2k – 9) ≥ 10(k + 1)

since 2k – 9 ≥ 11Hence if true for k, it is true for n = k + 1

True for n = 10 => True for n = 11 since k ≥ 10True for n = 11 => True for n = 12and so on true for all values of n, by induction.


Page 84

4. Show that 4n + 6n – 1 is divisible by 9 for all n ≥ 1Proof :- for n = 1,

4n + 6n – 1 = 4 + 6 – 1 = 9 , divisible by 9.True for n = 1Assume true for n = k, k ≥ 1 (We want to show that 4k + 6k –1 is divisible by 9=> 4k+1 + 6(k + 1) – 1 is divisible by 9) (= 9p), p ∈ W4k+1 + 6(k + 1) – 1 = 4.4k + 6k + 5

= 4(4k + 6k – 1) – 18k + 9But since 4k + 6k – 1 is divisible by 9 (= 9p), p ∈ W

=> 4k+1 + 6(k + 1) – 1 = 4.9p – 18k + 9 = 9(4p – 2k + 1) (= 9t) t ∈ W

=> 4k+1 + 6(k + 1) – 1 is divisible by 9. (= 9t) t ∈ WHence if true for k, it is true for n = k + 1.True for n = 1 => True for n = 2 since k ≥ 1True for n = 2 => True for n = 3and so on for all values of n.

5. Prove that Sn of the series 11 × 2

+1

2 × 3+

13 × 4

+ ...... =n

n + 1Proof :- for n = 1, S1 = 1

1 × 2 = 1

2, n

n + 1 = 1

2 (true)

for n = 2, S2 = S1 + 12 × 3

= 12

+ 16

= 23

, nn + 1

= 23

True for n = 1 and n = 2

Assume it is true for n = k, i.e. Sk = kk + 1

Sk+1 = Sk + 1k + 1( ) k + 2( )

= kk + 1

+ 1k + 1( ) k + 2( )

= k k + 2( ) + 1k + 1( ) k + 2( )

= k2 + 2k + 1k + 1( ) k + 2( )

= k + 1( )2

k + 1( ) k + 2( ) = k + 1

k + 2

= k + 1k + 1( ) + 1

Hence if true for k, it is true for n = k + 1True for n = 1 => True for n = 2 since k ≥ 1True for n = 2 => True for n = 3and so on for all values of n.


Page 85

Exercise 5 Prove the following by induction :-

1. r r + 1( )r =1

n

∑ =13

n n + 1( ) n + 2( ).

2. r r + 1( )r =1

n

∑ r + 2( ) =14

n n + 1( ) n + 2( ) n + 3( ).

3. Sn of the series 11 × 3

+1

3 × 5+

15 × 7

+ ...... =n

2n + 1

4. Sn of the series 11 × 2 × 3

+1

2 × 3 × 4+

13 × 4 × 5

+ ...... =14−

12 n + 1( ) n + 2( )

5. Use the results of rr =1

n

∑ , r 2

r =1

n

∑ and r 3

r =1

n

∑ to prove by direct method

(a) r r + 1( )r =1

n

∑ =13

n n + 1( ) n + 2( )

(b) r r + 1( )r =1

n

∑ r + 2( ) =14

n n + 1( ) n + 2( ) n + 3( )

6. n3 + 3n2 – 10n is divisible by 3.

7. 7n + 4n + 1n is divisible by 6.

8. For all integers n > 2, 2n > 2n.

9. For all integers n ≥ 4, 3n > n3.


Page 86contd...

Number Theory

The Division Algorithm

If a is a non-negative integer and b a positive integer, then there exists uniquenon-negative integers q and r such that

a = bq + r and 0 ≤ r < b

Proof :- On the real number line, the integers a, a – b, a – 2b, a – 3b, .....form an decreasing sequence of integers.

Since only finitely many of these are ≥ 0, there is a unique integerq ≥ 0 for which

a – (q + 1)b < 0 ≤ a – bqand so 0 ≤ a – bq < b

If we write r = a – bq, then a = bq + r with 0 ≤ r < b.

Thus we have found non-negative integers q and r for whicha = bq + r and 0 ≤ r < b both hold.

To show that q and r are unique, suppose that a = bq1 + r1 and 0 ≤ r1 < b.

Then r1 = a – bq1 and 0 ≤ a – bq1 < b.It follows that a – (q1 + 1)b < 0 ≤ a – bq1 so that q1 is the integerdetermined above and r1 = a – bq = r.Thus the theorem is proved.

Examples 1. a = 193 and b = 17

193 = 11.17 + 6 q = 11, r = 6

2. a = 581 and b = 23

581 = 25.23 + 6 q = 25, r = 6

Exercise 6 Use the division identity for the following

1. a = 75 and b = 12

2. a = 327 and b = 13

3. a = 392 and b = 19

If r = 0 then we say that b is a divisor of a.

The notation used is b | a which means “b is the divisor of a”.


Page 87

Euclidean Algorithm

The Euclidean Algorithm is used to find the greatest common divisor (G.C.D) of 2 or more positive integers where this cannot be done simply.

For integers a and b, a = bq1 + r1 and 0 ≤ r1 < b b = r1q2 + r2 and 0 ≤ r2 < r1 r1 = r2q3 + r3 and 0 ≤ r3 < r2

and so on until rn- 2 = rn- 1qn + rn rn- 1 = rnqn+1 + 0. (i.e. r eventually becomes 0)

To find the G.C.D. for small numbers, you use factorisation as follows :-

Example Find the G.C.D. of 15 and 24

15 = 3 x 5 and 24 = 23 x 3

so the G.C.D. of 15 and 24 is 3

Notation (15, 24) = 3 (i.e. G.C.D. of 15 and 24 is 3)

To find the G.C.D. for large numbers, use the Euclidean Algorithm.

ExamplesUse the Euclidean Algorithm to find the G.C.D. of (1147, 851).

Use repeated application of the division identity until r = 0.=> The last non-zero remainder is the G.C.D.

1147 = 1 x 851 + 296

851 = 2 x 296 + 259

296 = 1 x 259 + 37

259 = 7 x 37 + 0

Hence (1147, 851) = 37

Exercise 7

1. Find the G.C.D. of

(i) (15, 27) (ii) (16, 42) (iii) (72, 108)

2. Use the Euclidean Algorithm to find the G.C.D. of

(i) (1219, 901) (ii) (4277, 2821) (iii) (5213, 2867)


Page 88

Expressing the G.C.D. of two Positive Integers as a Linear Combination of the two Integers.

Having found the G.C.D. of two positive integers a and b, it is possible, by working backwards, to express the divisor (d) in terms of the two integers in the form of a linear combination.

i.e. d = xa + yb where x and y are integers.

Example

1. (a) Use the Euclidean Algorithm to find the G.C.D.of (1147, 851).

(b) Hence find the integers x and y to write this G.C.D. in the form x.1147 + y.851.

(a) 1147 = 1 x 851 + 296 (1) 851 = 2 x 296 + 259 (2) 296 = 1 x 259 + 37 (3) 259 = 7 x 37 + 0 (4)

Hence (1147, 851) = 37

(b) From (3) 37 = 296 – 1 x 259From (2) = 296 – 1 x (851 – 2 x 296)

= 296 – 1 x 851 + 2 x 296= 3 x 296 – 1 x 851

From (1) = 3 x (1147 – 1 x 851) – 1 x 851= 3 x 1147 – 3 x 851 – 1 x 851= 3 x 1147 – 4 x 851

x = 3 and y = – 4

Exercise 8

1. Express the G.C.D.’s found in Exercise 7 Question 2 as a linearcombination of the original numbers.

2. (a) Use the Euclidean Algorithm to find the G.C.D.of (7293, 798).

(b) Hence find the integers x and y to write this G.C.D. inthe form x.7293 + y.798.


Page 89

Expressing Base 10 integers in other Bases

Examples1. Express 235ten in the base 6.

235 = 6 x 39 + 1 39 = 6 x 6 + 3 6 = 6 x 1 + 0 1 = 6 x 0 + 1

Reading the remainders in reverse gives 1031six or 235 = 6 x 39 + 1

= 6 x (6 x 6 + 3) + 1= 6 x 6 x 6 + 3 x 6 + 1

= 1 x 63 + 0 x 6 + 3 x 6 + 1= 1031six

2. Express 423ten in the base 8.

423 = 8 x 52 + 7 52 = 8 x 6 + 4 6 = 8 x 0 + 6

Reading the remainders in reverse gives 647eight

or 423 = 8 x 52 + 7= 8 x (8 x 6 + 4) + 7= 6 x 8 x 8 + 4 x 8 + 7= 6 x 82 + 4 x 8 + 7= 647eight

Exercise 9

1. Express 81 in the base 2.





Page 90

AnswersExercise 1

1. (c) 2. (d)3. (a) For some real x, x2 is not positive.

(b) No pupils find mathematics difficult.(c) Some dogs like cats.(d) There is no positive integer x such that x + 3 > 0.(e) Some parallelograms do not have half turn symmetry.(f) Some school boys lie.(g) Some numbers withh a zero in the units place are not divisible by 5.(h) Some numbers of the form 2n – 1, (n an integer), are not prime.

Exercise 2

1. (a) If a number is divisible by 5, it ends in zero. (e.g. 15)(b) If a number is odd, it is aprime number greater than 2. (e.g. 21)(c) If the diagonals of a quadrilateral intersect at right angles, the

quadrilateral is a square. (e.g. a rhombus)(d) If x2 = 9, x = 3 (e.g. x = 3)(e) If the sum of two numbers is even, => the numbers are odd. (6 & 4)(f) If the product of two numbers is even, the numbers are even. (6 & 5)

Exercise 3

1. necessary 2. neither

3. (a) necessary (b) none of these (c) sufficient(d) necessary and sufficient (e) sufficient(f) necessary and sufficient (g) neither (h) sufficient

Exercise 4 All proofs

Exercise 5 All Proofs

Exercise 6

1. 75 = 6.12 + 3 2. 327 = 25.13 + 2 3. 392 = 20.19 + 12

Exercise 7

1. (i) 3 (ii) 2 (iii) 36 2. (i) 53 (ii) 91 (iii) 1

Exercise 8

1. (a) 53 = 3 x 1219 – 4 x 901 (b) 91 = 2 x 4277 – 3 x 2821(c) 1 = –952 x 5213 + 1731 x 2867

2. (a) 3 (b) x = –115, y = 1051

Exercise 9

1. 1010001two 2. 4304five 3. 3050seven 4. 22865nine

maths.lanark.s-lanark.sch.ukmaths.lanark.s-lanark.sch.uk/.../2018/02/advanced-higher-all.pdf ·...

Documents