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2

*P44825A0232*

1. Given that

tan ° = p, where p is a constant, p 1

use standard trigonometric identities, to find in terms of p,

(a) tan2 °(2)

(b) cos °(2)

(c) cot( – 45)°(2)

Write each answer in its simplest form.

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C3 JUNE 2015C3 and C4 June 2015

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4

*P44825A0432*

2. Given that

f (x) = 2ex – 5, x

(a) sketch, on separate diagrams, the curve with equation

(i) y = f (x)

(ii) y = | f (x) |

On each diagram, show the coordinates of each point at which the curve meets or cuts the axes.

On each diagram state the equation of the asymptote.(6)

(b) Deduce the set of values of x for which f (x) = | f (x) |(1)

(c) Find the exact solutions of the equation | f (x) | = 2(3)

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*P44825A0832*

3. g( ) = 4cos2 + 2sin2

Given that g( ) = R cos(2 ), where > 0 and 0 < < 90°,

(a) find the exact value of and the value of to 2 decimal places.(3)

(b) Hence solve, for –90° < < 90°,

4cos2 + 2sin2 1

giving your answers to one decimal place.(5)

Given that k is a constant and the equation g( ) = k has no solutions,

(c) state the range of possible values of k.(2)

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*P44825A01232*

4. Water is being heated in an electric kettle. The temperature, °C, of the water t seconds after the kettle is switched on, is modelled by the equation

= 120 – 100e , 0 T

(a) State the value of when t = 0(1)

Given that the temperature of the water in the kettle is 70°C when t = 40,

(b) find the exact value of , giving your answer in the form ln ab

, where a and b are integers.

(4)

When t = T, the temperature of the water reaches 100°C and the kettle switches off.

(c) Calculate the value of T to the nearest whole number.(2)

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*P44825A01432*

5. The point P lies on the curve with equation

x = (4y – sin2y)2

Given that P has (x, y) coordinates ⎛ ⎞⎜ ⎟⎝ ⎠

,2πp , where is a constant,

(a) find the exact value of p.(1)

The tangent to the curve at P cuts the y-axis at the point A.

(b) Use calculus to find the coordinates of A.(6)

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*P44825A01832*

6.

O x

y

A

y = 17 – xy = 2x + 1 – 3

Figure 1

Figure 1 is a sketch showing part of the curve with equation y = 2x+1 – 3 and part of the line with equation y = 17 – x.

The curve and the line intersect at the point A.

(a) Show that the x coordinate of satisfies the equation

x x= − −lnln

( )202

1

(3)

(b) Use the iterative formula

x xn

n+ =

−−1

202

1ln

ln( )

, x0 = 3

to calculate the values of x1, x2 and x3, giving your answers to 3 decimal places.(3)

(c) Use your answer to part (b) to deduce the coordinates of the point A, giving your answers to one decimal place.

(2)

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*P44825A02232*

7.

y

O x

Figure 2

Figure 2 shows a sketch of part of the curve with equation

g(x) = x2(1 – x)e–2x, x 0

(a) Show that g' ( )x = f (x)e–2x, where f(x) is a cubic function to be found.(3)

(b) Hence find the range of g.(6)

(c) State a reason why the function g–1(x) does not exist.(1)

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*P44825A02632*

8. (a) Prove that

sec tan cos sincos sin

2 2A A A AA A

+ ≡ +−

, (2 1)4n πA +≠ , n

(5)

(b) Hence solve, for 0 < 2 ,1sec 2 tan 22

θ θ+ =

Give your answers to 3 decimal places.

(4)

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30

*P44825A03032*

9. Given that k is a negative constant and that the function f(x) is defined by

f ( ) ( )( )x x k x kx kx k

= − − −− +

2 53 22 2

, x 0

(a) show that f ( )x x kx k

= +− 2

(3)

(b) Hence find f ' ( )x , giving your answer in its simplest form.

(3)

(c) State, with a reason, whether f (x) is an increasing or a decreasing function.

Justify your answer.

(2)

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2

*P44827A0232*

1. (a) Find the binomial expansion of

( ) ,4 5 45

12+ <x x

in ascending powers of x, up to and including the term in x2. Give each coefficient in its simplest form.

(5)

(b) Find the exact value of ( )4 512+ x when x = 1

10 Give your answer in the form k 2 , where k is a constant to be determined.

(1)

(c) Substitute x = 110

into your binomial expansion from part (a) and hence find an

approximate value for 2

Give your answer in the form pq where p and q are integers.(2)

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C4 JUNE 2015

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*P44827A0432*

2. The curve C has equation

x2 – 3xy – 4y2 + 64 = 0

(a) Find ddyx

in terms of x and y.(5)

(b) Find the coordinates of the points on C where ddyx

= 0

(Solutions based entirely on graphical or numerical methods are not acceptable.)(6)

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*P44827A0832*

3.

Figure 1

Figure 1 shows a sketch of part of the curve with equation y x x x= −412e , x 0

The curve meets the x-axis at the origin O and cuts the x-axis at the point A.

(a) Find, in terms of ln2, the x coordinate of the point A. (2)

(b) Find

x xxe d12∫

(3)

The finite region R, shown shaded in Figure 1, is bounded by the x-axis and the curve with equation

y x x x= −412e , x 0

(c) Find, by integration, the exact value for the area of R. Give your answer in terms of ln2

(3)

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y

O

R

xA

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*P44827A01232*

4. With respect to a fixed origin O, the lines l1 and l2 are given by the equations

l1: r = 53−

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

p +

013−

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ , l2: r =

852−

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ +

345−

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

where and are scalar parameters and p is a constant.

The lines l1 and l2 intersect at the point A.

(a) Find the coordinates of A.(2)

(b) Find the value of the constant p.(3)

(c) Find the acute angle between l1 and l2, giving your answer in degrees to 2 decimal places.

(3)

The point B lies on l2 where = 1

(d) Find the shortest distance from the point B to the line l1, giving your answer to 3 significant figures.

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*P44827A01632*

5. A curve C has parametric equations

x = 4t + 3, y = 4t + 8 + 52t

, t 0

(a) Find the value of ddyx

at the point on C where t = 2, giving your answer as a fraction

in its simplest form.(3)

(b) Show that the cartesian equation of the curve C can be written in the form

y x ax bx

= + +−

2

3, x 3

where a and b are integers to be determined.(3)

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20

*P44827A02032*

6.y

O

R

3 x

Diagram not to scale

Figure 2

Figure 2 shows a sketch of the curve with equation y x x= − +( )( )3 1 , 0 x 3

The finite region R, shown shaded in Figure 2, is bounded by the curve, the x-axis, and the y-axis.

(a) Use the substitution x = 1 + 2sin to show that

∫ ∫3 2

06

π

π−

2(3 )( 1) d cos dx x x k θ θ− + =

where k is a constant to be determined.(5)

(b) Hence find, by integration, the exact area of R. (3)

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24

*P44827A02432*

7. (a) Express 22P P( )−

in partial fractions.

(3)

A team of biologists is studying a population of a particular species of animal.

The population is modelled by the differential equation

ddPt

P P t= −12

2 2( )cos , t 0

where P is the population in thousands, and t is the time measured in years since the start of the study.

Given that P = 3 when t = 0,

(b) solve this differential equation to show that

Pt

=−

6

312 2e sin

(7)

(c) find the time taken for the population to reach 4000 for the first time. Give your answer in years to 3 significant figures.

(3)

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28

*P44827A02832*

8.

y

x

l

P

Q

R

O

C

Diagram not to scale

Figure 3

Figure 3 shows a sketch of part of the curve C with equation

y = 3x

The point P lies on C and has coordinates (2, 9).

The line l is a tangent to C at P. The line l cuts the x-axis at the point Q.

(a) Find the exact value of the x coordinate of Q.(4)

The finite region R, shown shaded in Figure 3, is bounded by the curve C, the x-axis, the y-axis and the line l. This region R is rotated through 360° about the x-axis.

(b) Use integration to find the exact value of the volume of the solid generated.

Give your answer in the form pq where p and q are exact constants.

[You may assume the formula V = 13

r 2h for the volume of a cone.](6)

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(a) M1 Attempt to use the double angle formula for tangent followed by the substitution tan p .

For example accept tan

tantan

θ pθ

θ p

= =

2 2

2 22

1 1

Condone unconventional notation such as tantan

tan

θθ

θ

=

22

21

followed by an attempt to

substitute tan p for the M mark. Recovery from this notation is allowed for the A1.

Alternatively use tan tan

tan( )1 tan tan

A BA B

A B

with an attempt at substituting

tan tanA B p . The unsimplified answer 1

p pp p

is evidence

It is possible to use sin sin costan

cos cos

p

p pθ θ θθ

θ θp

´ ´

= = = - ´ -

2 2

2

2

12

1 12 22

12 2 1 2 11

but it is

unlikely to succeed.

A1 Correct simplified answer of tan or( )( )

p pθ

p p p=

- - +2

2 22

1 1 1.

Do not allow if they ''simplify'' to p-2

1

Allow the correct answer for both marks as long as no incorrect working is seen.

Question Number Scheme Marks

1.(a) tan

tantan

θ pθ

θ p

= =

- -2 2

2 22

1 1 Final answer M1A1

(2)

(b) cossec tan

θθ θ p

= = = + +2 2

1 1 1

1 1 Final answer M1A1

(2)

(c) ( )( )

tan tancot

tan tan tan

θ pθ

θ θ p

+ +- = = =

- - -1 1 45 1

4545 45 1

Final answer M1A1

(2) (6 marks)

C3 JUNE 2015 MARK SCHEME

Question Number Scheme Marks

2.(ai)

Shape

5 andln , 0 0, 32

5y

B1 B1 B1

(3)

(aii)

Shape inc cusp

5ln , 0 0,32

and

5y

B1ft B1ft B1ft

(3)

(b) 5

ln2

xæ ö÷ç ÷ç ÷çè ø

B1 ft (1)

(c) ( ) 32e 5 2 ln

2x x

æ ö÷ç- =- = ÷ç ÷çè ø M1A1

( ) 7ln

2x

æ ö÷ç= ÷ç ÷çè ø B1

(3)

(10 marks)

Question Number Scheme Marks

3(a) 4 cos 2 2 sin 2 cos(2 )R

2 24 2 20 2 5R B1

1

arctan 26.565 ... awrt 26.572

M1A1

(3)

(b) ( ) ( )cos 26.6 cos 26.57θ θ- = - =

120 2 1 2

20 M1

( )2 26.57 77.1.. ...θ θ - =+ = dM1

awrt 51.8θ = A1

2 26.57 ' 77.1... ' awrt 25.3θ θ- = - = - ddM1A1

(5)

(c) ,k k<- >20 20 B1ft either

B1ft both

(2)

(10 marks)

You can marks parts (a) and (b) together as one. (a) B1 For 2 0 2 5R . Condone 20R

M1 For 1arctan2

or a rc tan 2 leading to a solution of

Condone any solutions coming from cos 4, sin 2

Condone for this mark 12 arctan ..2

If R has been used to find award for only 4arcos' 'R

2arcsin' 'R

A1 awrt 26.57

Question Number Scheme Marks

4(a) ( )θ= 20 B1 (1)

(b) Sub 4 04 0 , 7 0 7 0 1 2 0 1 0 0 e λt θ -= = = -

4 0 0 .5λe- = M1A1

ln 2

40λ = M1A1

(4)

(c) ln 0.2100

their ' 'θ T

λ= =

- M1

awrt 93T = A1

(2)

(7 marks)

Alt (b) Sub 4 04 0 , 7 0 1 0 0 e 5 0λt θ -= = =

ln100 40 ln 50λ - = M1A1

ln100 ln 50 ln 2

40 40λ

- = = M1A1

(4)

Question Number Scheme Marks

5.(a) p π= 24 or ( )π 22 B1

(1)

(b) 2 d4 sin2 2 4 sin2 4 2cos2

d

xx y y y y y

y M1A1

Sub πy=2

into d2 4 sin2 4 2cos2

d

xy y y

y

d

d

xπ

y =24 (= 75.4) / ( )d 1

0.013d 24

y

x π= = M1

Equation of tangent ( )πy x π

π- = - 21

42 24

M1

Using ( )πy x π

π- = - 21

42 24

with πx y= =0

3 cso M1, A1

(6)

(7 marks)

Alt (b) I 2 0.54 sin 2 4 sin 2x y y x y y

0.5 d

0.5 4 2cos2d

xx y

y

M1A1

Alt (b) II 2 216 8 sin 2 sin 2x y y y y

d d d d

1 32 8 sin 2 16 cos 2 4 sin 2 cos 2d d d d

y y y yy y y y y y

x x x x M1A1

Or 1d 32 d 8 sin 2 d 16 cos 2 d 4 sin 2 cos 2 dx y y y y y y y y y y

(a)

B1 p π= 24 or exact equivalent ( )π 22

Also allow x π= 24 (b)

Question Number Scheme Marks

6.(a) 1 12 3 17 2 20x xx x+ +- = - = - M1

( ) ln ln( ) ...x x x+ = - =1 2 20 dM1

ln(20 )1

ln 2

xx

A1*

(3)

(b) Sub 0 3x into 1 1ln(20 )

1, 3.087ln 2

nn

xx x

(awrt) M1A1

2 33.080, 3.081x x (awrt) A1

(3)

(c) ( )3.1,13.9A= cao M1,A1

(2)(8 marks)

6.(a)Alt 1 202 3 17 2

2x x x

x+ -- = - = M1

ln ln ...x

x x-

= =20

22

dM1

ln(20 )1

ln 2

xx

A1*

(3)

6.(a) ln(20 )1

ln 2

xx

( ) ln ln( )x x+ = -1 2 20 M1

backwards 12 20x x+ = - dM1

Hence 12 3xy += - meets 17y x= - A1*

(3)

Question Number Scheme Marks

7.(a) Applies ' 'vu uv to 2 3 2e xx x

2 3 2 2 2g '( ) -2e 2 3 ex xx x x x x M1 A1

3 2 2g '( ) 2 5 2 e xx x x x A1

(3)

(b) Sets 3 2 2 3 22 5 2 e 0 2 5 2 0xx x x x x x M1

2 12 5 2 0 (0), , 2

2x x x x M1,A1

Sub 1, 2

2x into 2 3 2

4

1 1 4( ) e , (2)

2 8e exg x x x g g

dM1,A1

Range ( )e e

g x-4

4 1

8 A1

(6)

(c) Accept g( )x is NOT a ONE to ONE function

Accept g( )x is a MANY to ONE function B1

Accept g ( )x- 1 would be ONE to MANY (1)

(10 marks)

Note that parts (a) and (b) can be scored together. Eg accept work in part (b) for part (a) (a) M1 Uses the product rule ' 'vu uv with 2 3u x x and 2 xv e or vice versa. If the rule is

quoted it must be correct. It may be implied by their .. .. ' .. ' ..u v u v followed by their ' 'vu uv . If the rule is not quoted nor implied only accept expressions of the form

2 3 2 2 2e ex xx x A Bx Cx condoning bracketing issues

Method 2: multiplies out and uses the product rule on each term of 2 2 3 2e ex xx x Condone issues in the signs of the last two terms for the method mark

Uses the product rule for ' ' 'uvw u vw uv w uvw= + + applied as in method 1

Method 3:Uses the quotient rule with 2 3u x x and 2 xv e . If the rule is quoted it must

be correct. It may be implied by their .. .. ' .. ' ..u v u v followed by their 2

' 'vu uv

v

If the

rule is not quoted nor implied accept expressions of the form

2 2 2 3 2

22

e e

e

x x

x

Ax Bx x x C

condoning missing brackets on the numerator and 22e x on the denominator.

Method 4: Apply implicit differentiation to 2 2 3 2 2 2de e 2e 2 3

dx x xy

y x x y x xx

Condone errors on coefficients and signs

Question Number Scheme Marks

8(a) sinsec tan

cos cos

AA A

A A+ = +

1 22 2

2 2 B1

sin

cos

A

A

+=1 2

2 M1

sin cos

cos sin

A A

A A

+=

-2 2

1 2 M1

cos sin sin cos

cos sin

A A A A

A A

+ +=

-

2 2

2 2

2

(cos sin )(cos sin )

(cos sin )(cos sin )

A A A A

A A A A

+ +=

+ - M1

cos sin

cos sin

A A

A A

+=

- A1*

(5)

(b) sec tanθ θ+ = 1

2 22

cos sin

cos sin

θ θ

θ θ

+=

-1

2

cos sin cos sinθ θ θ θ + = -2 2

tan θ =-1

3 M1 A1

2.820, 5.961θ awrt = dM1A1 (4)

(9 marks)

(a)

B1 A correct identity for seccos

AA

=1

22

OR sintan

cos

AA

A=

22

2.

It need not be in the proof and it could be implied by the sight of seccos sin

AA A

=-2 2

12

M1 For setting their expression as a single fraction. The denominator must be correct for their fractions and at least two terms on the numerator.

This is usually scored for cos tan

cos

A A

A

+1 2 2

2 or sin

cos

A

A

+1 2

2

M1 For getting an expression in just sin A and cos A by using the double angle identities

sin2 2sin cosA A A= and 2 2cos2 cos sinA A A= - , 22 cos 1A- or 21 2sin A- . Alternatively for getting an expression in just sin A and cos A by using the double angle

identities sin 2 2sin cosA A A= and tantan

tan

AA

A=

- 2

22

1with sin

tancos

AA

A= .

For example sin

cossincos sin

cos

AA

AA AA

= +- -

22 2

2

21

1 is B1M0M1 so far

M1 In the main scheme it is for replacing 1 by cos sinA A+2 2 and factorising both numerator and denominator

Question Number Scheme Marks

9.(a) 2 23 2 ( 2 )( )x kx k x k x k B1

( 5 )( ) ( 5 ) 2( 2 ) ( 5 )

2 2( 2 )( ) ( 2 ) ( 2 )

x k x k x k x k x k

x k x k x k x k

M1

( 2 )

x k

x k

A1*

(3)

(b) Applies 2

' 'vu uv

v

to 2

x ky

x k

with u x k and 2v x k

2

( 2 ) 1 ( ) 1f '( )

( 2 )

x k x kx

x k

M1, A1

2

3f '( )

( 2 )

kx

x k

A1

(3)

(c) If 2f '( )

( 2 )

Ckx

x k

f(x) is an increasing function as f '( ) 0x ,

M1

2

3f '( ) 0

( 2 )

kx

x k

for all values of x as negative×negative

= positivepositive

A1

(2)

(8 marks)

(a)

B1 For seeing 2 23 2 ( 2 )( )x kx k x k x k anywhere in the solution M1 For writing as a single term or two terms with the same denominator

Score for ( 5 ) 2( 2 ) ( 5 )

2( 2 ) ( 2 )

x k x k x k

x k x k

or

2 2

2 2

( 5 )( ) 2( 2 )( ) ( 5 )( )2

( 2 )( ) ( 2 )( ) 3 2

x k x k x k x k x k x k x k

x k x k x k x k x kx k

A1* Proceeds without any errors (including bracketing) to ( 2 )

x k

x k

9

June 2015

6666/01 Core Mathematics 4 Mark Scheme

Question Number

Scheme Marks

1. (a) 1 1

1 1 2 22 2

5 5(4 5 ) 4 1 2 1

4 4

x xx

1

2(4) or 2 B1

2 1

1

2

(kx)( 1

2 )( 12 )

2!(kx)2 ...

see notes M1 A1ft

2 1

1

2

5x

4

( 12 )( 1

2 )

2!

5x

4

2

...

25 25

2 1 ...8 128

x x

See notes below!

25 252 ; ...

4 64x x isw A1; A1

[5]

(b) x 1

10 (4 5(0.1))

12 4.5

9

2

3

2

3

2

2

2

3

22

32

2 or k

3

2or 1.5 o.e. B1

[1]

(c) 3

22 or 1.5 2 or

3

2 2

5

4

1

10

25

64

1

10

2

... 2.121... See notes M1

So, 3

22

543

256 or

3

2

543

256

yields, 181

2128

or 2 256

181

181

128 or

362

256 or

543

384 or

256

181 etc. A1 oe

[2] 8 Question 1 Notes

1. (a) B1 1

2(4) or 2 outside brackets or 2 as candidate’s constant term in their binomial expansion.

M1 Expands ... k x 1

2 to give any 2 terms out of 3 terms simplified or un-simplified,

Eg: 11

2

(kx) or 1

2

(kx)( 1

2 )( 12 )

2!(kx)2 or 1 ...

( 12 )( 1

2 )

2!(kx)2

where k is a numerical value and where k 1.

A1 A correct simplified or un-simplified 1 1

2

(kx)( 1

2 )( 12 )

2!(kx)2 expansion with consistent (kx).

Note (kx) , k 1, must be consistent (on the RHS, not necessarily on the LHS) in a candidate’s expansion.

12

Question Number

Scheme Marks

2. 2 23 4 64 0x xy y

(a) dy

dx

2x 3y 3xdy

dx

8y

dy

dx 0

M1A1

M1

dM1

dy

dx

2x 3y

3x 8yor

3y 2x

3x 8y o.e. A1 cso

[5]

(b) dy

dx 0

2x 3y 0 M1

2

3y x

3

2x y A1ft

22 2 2

3 4 64 03 3

x x x x

2

23 33 4 64 0

2 2y y y y

dM1

2 2 2 216 252 64 0 64 0

9 9x x x x 2 2 2 29 9 25

4 64 0 64 04 2 4

y y y y

x2

576

25

x 24

5or

24

5 y2

256

25

y 16

5or

16

5 A1 cso

When

24

5x , y

2

3

24

5

and 2

3

24

5

When 16

5y , x

3

2

16

5

and 3

2

16

5

24

5,16

5

and 24

5,

16

5

or x 24

5, y

16

5 and x

24

5, y

16

5

ddM1

cso A1

[6] 11 Alternative method for part (a)

(a) dx

dy

2xdx

dy 3y

dx

dy 3x

8y 0

M1A1

M1

(2x 3y)

dx

dy 3x 8y 0 dM1

dy

dx

2x 3y

3x 8yor

3y 2x

3x 8y o.e. A1 cso

[5] Question 2 Notes

2. (a) General Note Writing down

dy

dx

2x 3y

3x 8yor

3y 2x

3x 8y from no working is full marks

Note Writing down

dy

dx

2x 3y

3x 8yor

3y 2x

3x 8y from no working is M1A0B1M1A0

Note Few candidates will write 2xdx 3ydx 3xdy 8ydy 0 leading to

dy

dx

2x 3y

3x 8y, o.e.

This should get full marks.

d2 3 ( 3 8 ) 0

d

yx y x y

x

15

Question Number

Scheme Marks

3.

(a) y 0 4x x e12

x 0 x(4 e12

x) 0

e12

x 4 xA 4ln2

Attempts to solve e12

x 4 giving x ...

in terms of ln where 0 M1

4ln2 cao (Ignore x 0 ) A1 [2]

(b) x e12xdx 2xe

1

2x 2e

1

2x

dx xe

1

2x e

1

2x

dx , 0, 0 M1

2xe1

2x 2e

1

2x

dx , with or without dx A1 (M1 on ePEN)

2xe

1

2x 4e

1

2x

c 1 1

2 22 e 4ex x

x o.e. with or without c A1

[3]

(c) 4xdx 2x2 4x 2x2 or4x2

2o.e. B1

(4x x e12x) dx

0

4ln2

2x2 2xe

1

2x 4e

1

2x

0

4ln2 or ln16 or their limits

2(4ln2)2 2(4ln2)e

1

2(4ln2)

4e1

2(4ln 2)

2(0)2 2(0)e

1

2(0) 4e

1

2(0)

See notes M1

232(ln 2) 32(ln 2) 16 4

232(ln 2) 32(ln 2) 12 32(ln2)2 32(ln2) 12 , see notes A1

[3] 8

Question 3 Notes

3. (a) M1 Attempts to solve e12

x 4 giving x ... in terms of ln where 0

A1 4ln2 cao stated in part (a) only (Ignore x 0 )

(b) NOT

E Part (b) appears as M1M1A1 on ePEN, but is now marked as M1A1A1.

M1 Integration by parts is applied in the form xe1

2x e

1

2x

dx , where 0, 0 .

(must be in this form) with or without dx

A1 2xe1

2x 2e

1

2x

dx or equivalent, with or without dx . Can be un-simplified.

A1 2xe1

2x 4e

1

2x or equivalent with or without + c. Can be un-simplified.

Note You can also allow 2e1

2x(x 2) or e

1

2x(2x 4) for the final A1.

isw You can ignore subsequent working following on from a correct solution.

SC SPECIAL CASE: A candidate who uses u x ,dv

dx e

1

2x, writes down the correct “by parts”

formula, but makes only one error when applying it can be awarded Special Case M1. (Applying their v counts for one consistent error.)

17

Question Number

Scheme Marks

4. 1 2

5 0 8 3

: 3 1 , : 5 4 .

3 2 5

l l

p

r r Let acute angle between 1l and 2 .l

Note: You can mark parts (a) and (b) together. (a) 1 2 : 5 8 3 1l l i

So,

Finds and substitutes their into 2l M1

5 3 i j k or

5

1

3

or (5, 1, 3) A1

[2]

(b) : 3 5 4 j 3 5 4( 1) 4 Equates j components, substitutes their and solves to give ... M1

k : p 3 2 5

p 3(4) 2 5(1) p 15

or k : p 3 3

p 3(4) 3 p 15

Equates k components, substitutes their and their and solves to give ...p or

equates k components to give their " p 3 the k value of A found in part (a)”,

substitutes their and solves to give ...p

M1

15p A1 [3]

(c)

0

1

3

1d , 2

3

4

5

d

0 3

1 4

3 5

Realisation that the dot product is

required between and .

M1

cos K

0(3) (1)(4) (3)(5)

(0)2 (1)2 (3)2 . (3)2 (4)2 (5)2

An attempt to apply the dot product formula between Ad

1

and Bd2.

dM1 (A1 on ePEN)

cos 19

10 . 50 31.8203116... 31.82 (2 dp) anything that rounds to 31.82 A1

[3]

(d) See

notesM1

sin10 2

d Writes down a correct trigonometric equation involving

the shortest distance, d. Eg: sintheir

d

AB , oe. dM1

d 10 2 sin31.82... d 7.456540753... 7.46 (3sf ) anything that rounds to 7.46 A1

[3] 11

A 1d 2B d

20

Question Number

Scheme Marks

5. Note: You can mark parts (a) and (b) together.

(a) 5

4 3, 4 82

x t y tt

dx

dt 4 , 2d 5

4d 2

yt

t Both

dx

dt 4 or

dt

dx

1

4 and 2d 5

4d 2

yt

t B1

So,

2

22

54d 5 52 1 1

d 4 8 8

tyt

x t

Candidate’s

d

d

y

t divided by a candidate’s

d

d

x

tM1 o.e.

d 27When 2,

d 32

yt

x

27

32 or 0.84375 cao A1

[3] Way 2: Cartesian Method

dy

dx 1

10

(x 3)2

dy

dx 1

10

(x 3)2 , simplified or un-simplifed. B1

dy

dx

(x 3)2, 0, 0 M1

When t 2, x 11 dy

dx

27

32

27

32 or 0.84375 cao A1

[3] Way 3: Cartesian Method

dy

dx

(2x 2)(x 3) (x2 2x 5)

(x 3)2

x2 6x 1

(x 3)2

Correct expression fordy

dx, simplified or un-simplified. B1

dy

dx

f (x)(x 3)1f (x)

(x 3)2 ,

where f (x) their "x2 ax b", g(x) x 3

M1

When t 2, x 11 dy

dx

27

32

27

32 or 0.84375 cao A1

[3]

(b) t

x 3

4

y 4x 3

4

8

5

2x 3

4

Eliminates t to achieve an equation in only x and y

M1

y x 3 8

10

x 3

y

(x 3)(x 3) 8(x 3) 10

x 3 or y(x 3) (x 3)(x 3) 8(x 3) 10

or y (x 5)(x 3) 10

x 3 or y

(x 5)(x 3)

x 3

10

x 3

See notes dM1

y

x2 2x 5

x 3 , { a 2 and b 5}

Correct algebra leading to

or

A1 cso

[3] 6

2 2 5

3

x xy

x

a 2 and b 5

23

Question Number

Scheme Marks

6. (a) A (3 x)(x 1)0

3

dx , x 1 2sin

dx

d 2cos

dx

d 2cos or used correctly

in their working. Can be implied.B1

(3 x)(x 1) dx or (3 2x x2 ) dx

3 (1 2sin ) (1 2sin ) 1 2cos d Substitutes for both x and dx ,

where dx d . Ignore d M1

2 2sin 2 2sin 2cos d

4 4sin2 2cos d

4 4(1 cos2 2cos d or 4cos2 2cos d Applies 2 2cos 1 sin

see notes M1

4 cos2 d , 4k

4 cos2 d or

4cos2 d

Note: d is required here. A1

0 1 2sin or 1 2sin or sin

1

2

6

and 3 1 2sin or 2 2sin or sin 1 2

See notes B1

[5]

(b) k cos2 d

k 1 cos2

2

d Applies cos2 2cos2 1

to their integral M1

k 1

2

1

4sin2

Integrates to give sin2 , 0, 0

or k( sin2 ) M1 (A1 on ePEN)

So 4 cos2

6

2

d 2 sin2 6

2

2 22 sin 2 sin

2 2 6 6

3 4 3

3 2 3 2

or

1

68 3 3

A1 cao cso

[3] 8

2cos

4 3

3 2

25

Question Number

Scheme Marks

7. (a) 2

( 2) ( 2)

A B

P P P P

2 ( 2)A P B P Can be implied. M1

1, 1A B Either one. A1

giving 1 1

( 2)P P

See notes. cao, aef A1

[3]

(b) d 1

( 2)cos 2d 2

PP P t

t

2

d cos 2 d( 2)

P t tP P

can be implied by later working B1 oe

1ln ( 2) ln sin 2

2P P t c

ln(P 2) ln P ,

0, 0M1

1ln ( 2) ln sin 2

2P P t A1

0, 3t P ln1 ln 3 0 c c ln3 or ln( 13) See notes M1

1ln ( 2) ln sin 2 ln 3

2P P t

3( 2) 1ln sin 2

2

Pt

P

12 sin 23( 2)

e tP

P

Starting from an equation of the form ln(P ) ln P K sin t c ,

, applies a fully correct method to eliminate their logarithms.

Must have a constant of integration that need not be evaluated (see note)

M1

1 12 2sin 2 sin 23( 2) e 3 6 et tP P P P

gives 3P Pe12

sin2t 6 P(3 e12

sin2t) 6

12 sin 2

6

(3 e )tP

*

A complete method of rearranging to make P the subject.

Must have a constant of integration that need not be evaluated (see note)

dM1

Correct proof. A1 * cso

[7]

(c) population = 4000 4P States P 4 or applies 4P M1

1 3(4 2) 3

sin 2 ln ln2 4 2

t

Obtains sin2t ln k or

0, k 0where and k are numerical

values and can be 1

M1

0.4728700467...t

anything that rounds to 0.473 Do not apply isw here

A1

[3] 13

,, , K , 0

sin t ln k ,

28

Question Number

Scheme Marks

8. (a) y 3x dy

dx 3x ln3

dy

dx 3x ln3 or ln3 ex ln3 or y ln3 B1

Either T: 29 3 ln3( 2)y x

or T: y (32 ln3)x 918ln3, where 9 (32 ln3)(2) c See notes M1

{Cuts x-axis y 0 }

9 9ln3(x 2) or 0 (32 ln3)x 918ln3, Sets 0y in their tangent equation

and progresses to M1

So, 1

2ln 3

x 2 1

ln3or

2ln3 1

ln3 o.e. A1 cso

[4]

(b) V 3x 2

dx or 32x dx or 9x dx

V 3x 2

with or without dx ,

which can be implied B1 o.e.

32x

2ln3

or 9x

ln9

Eg: either 2

2 233 or (ln 3)3

(ln 3)

xx x

or 9x

9x

(ln9)or (ln9)9x ,

M1

22 3

32ln 3

xx or

99

ln 9

xx or e2x ln3

1

2ln3e2x ln3 A1 o.e.

V 32x dx0

2

32x

2ln3

0

2

34

2ln3

1

2ln3

40ln3

Dependent on the previous method mark. Substitutes

and and subtracts

the correct way round.

dM1

V

cone

1

3 (9)2 1

ln3

27ln3

2cone

1(9) 2 their ( )

3V a . See notes. B1ft

Vol(S)

40ln3

27ln3

13ln3

or or 262ln3

etc., isw A1 o.e.

Eg: p 13 , q ln3 [6]

10 (b) Alternative Method 1: Use of a substitution

V 3x 2

dx B1 o.e.

u 3x

du

dx 3x ln3 u ln3

V u2

u ln3du u

ln3du

u2

2ln3

3x 2

u2

(ln3)or (ln3)u2, where u 3x M1

3x 2

u2

2(ln3), where u 3x A1

V 3x 2dx

0

2

u2

2ln3

1

9

92

2ln3

1

2ln3

40ln3

Substitutes limits of 9 and 1 in u (or 2 and 0 in x)

and subtracts the correct way round.

dM1

then apply the main scheme.

x ...

2x 0x

13

ln 3

26

ln 9