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Maths4Scotland Higher

Select topic – from which the questions are taken.

The Straight Line Functions

Differentiation

Sequences

Integration

Polynomials

Quadratic Theory

Circle

Compound Angles

Vectors Exponential & Log Function

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The Straight Line

Maths4Scotland The Straight Line Higher

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The Straight Line

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Non-calculator questions will be indicated

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You will need a pencil, paper, ruler and rubber.


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Hint

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Find the equation of the line which passes through the point (-1, 3)

and is perpendicular to the line with equation 4 1 0x y

Find gradient of given line: 4 1 0 4 1 4x y y x m

Find gradient of perpendicular:1

4m

Find equation:1 3

1 4( 3) 1 4 124 ( 1)

yx y x y

x

4 13 0y x


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Find the equation of the straight line which is parallel to the line with

equation and which passes through the point (2, –1).

Find gradient of given line:

Gradient of parallel line is same:2

3m

Find equation: 2 ( 1)2 4 3 3

3 2

yx y

x

3 2 1y x

2 3 5x y

2 2

3 33 2 5 5y x y x m


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Hint

Previous NextTable of exact values

Find gradient of the line:

1tan

3

Use table of exact values1 1

tan 303

2 ( 1) 3 1

3 3 0 3 3 3m

Use tanm

Find the size of the angle a° that the line joining the

points A(0, -1) and B(33, 2) makes with the

positive direction of the x-axis.


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A and B are the points (–3, –1) and (5, 5).Find the equation ofa) the line AB.

b) the perpendicular bisector of AB

Find gradient of the AB: 4 3 5y x

Find mid-point of AB 1, 3

3

4m Find equation of AB

Gradient of AB (perp):4

3m

Use gradient and mid-point to obtain perpendicular bisector AB

3 4 13y x


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Hint


The line AB makes an angle of radians with

the

y-axis, as shown in the diagram.

Find the exact value of the gradient of AB.

Find angle between AB and x-axis:2 3 6

Use table of exact values

3

Use tanm tan6

m

1

3m

(x and y axes are perpendicular)


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A triangle ABC has vertices A(4, 3), B(6, 1)

and C(–2, –3) as shown in the diagram.

Find the equation of AM, the median from A.

Find mid-point of BC: (2, 1)

Find equation of median AM

Find gradient of median AM 2m

2 5y x


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P(–4, 5), Q(–2, –2) and R(4, 1) are the vertices

of triangle PQR as shown in the diagram.

Find the equation of PS, the altitude from P.

Find gradient of QR:1

2m

Find equation of altitude PS

Find gradient of PS (perpendicular to QR) 2m

2 3 0y x


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The lines and makeangles of a and b with the positivedirection of the x-axis, as shown in the diagram.a) Find the values of a and bb) Hence find the acute angle

between the two given lines.

2m

Find supplement of b 180 135 45

2 4y x 13x y

Find gradient of 2 4y x

Find gradient of 13x y 1m

Find a° tan 2 63a a

Find b° tan 1 135b b

Angle between two lines

Use angle sum triangle = 180°

72°


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Triangle ABC has vertices A(–1, 6), B(–3, –2) and C(5, 2)Find: a) the equation of the line p, the median from C of triangle ABC. b) the equation of the line q, the perpendicular bisector of BC. c) the co-ordinates of the point of intersection of the lines p and q.

Find mid-point of AB

Find equation of p 2y

Find gradient of p(-2, 2)

Find mid-point of BC (1, 0) Find gradient of BC1

2m

0m

Find gradient of q 2m Find equation of q 2 2y x

Solve p and q simultaneously for intersection (0, 2)


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Triangle ABC has vertices A(2, 2), B(12, 2) and C(8, 6). a) Write down the equation of l1, the perpendicular bisector of AB

b) Find the equation of l2, the perpendicular bisector of AC.

c) Find the point of intersection of lines l1 and l2

d) Hence find the equation of the circle passing through A, B and C.

7, 2Mid-point AB

Find mid-point AC (5, 4) Find gradient of AC2

3m

Equ. of perp. bisector AC

26r

Gradient AC perp.3

2m 2 3 23y x

Point of intersection (7, 1) This is the centre of circle

Find radius (intersection to A)

Equation of circle: 2 27 1 26x y

7x Perpendicular bisector AB


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A triangle ABC has vertices A(–4, 1), B(12,3) and C(7, –7). a) Find the equation of the median CM. b) Find the equation of the altitude AD. c) Find the co-ordinates of the point of intersection of CM and AD

4, 2Mid-point AB

Equation of median CM

Gradient of perpendicular ADGradient BC 2m 1

2m

Equation of AD

3m Gradient CM (median)

3 14y x

Solve simultaneously for point of intersection (6, -4)

2 2 0y x


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A triangle ABC has vertices A(–3, –3), B(–1, 1) and C(7,–3). a) Show that the triangle ABC is right angled at B.

b) The medians AD and BE intersect at M. i) Find the equations of AD and BE.

ii) Hence find the co-ordinates of M. 2m Gradient AB

Product of gradients

Gradient of median ADMid-point BC 3, 11

3m Equation AD

1

2m Gradient BC

12 1

2

Solve simultaneously for M, point of intersection

3 6 0y x

Hence AB is perpendicular to BC, so B = 90°

Gradient of median BEMid-point AC 2, 34

3m Equation AD 3 4 1 0y x

51,

3


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Graphs & Functions

Strategies

Higher Maths

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Graphs & Functons



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Maths4Scotland Graphs & Functions Higher

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The diagram shows the graph of a function f. f has a minimum turning point at (0, -3) and a point of inflexion at (-4, 2).

a) sketch the graph of y = f(-x).

b) On the same diagram, sketch the graph of y = 2f(-x)

a) Reflect across the y axis

b) Now scale by 2 in the y direction-1 3 4

2

y = f(-x)

-3

y

x

4

y = 2f(-x)

-6


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The diagram shows a sketch of part ofthe graph of a trigonometric function

whose equation is of the form

Determine the values of a, b and c

sin( )y a bx c

a is the amplitude: a = 4

b is the number of waves in 2 b = 2

c is where the wave is centred vertically c = 1

2a

1 in 2 in 2

1


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Functions and are defined on suitable domains.

a) Find an expression for h(x) where h(x) = f(g(x)).

b) Write down any restrictions on the domain of h.

1( )

4f x

x

( ) 2 3g x x

( ( )) (2 3)f g x f x a)1

2 3 4x

1

( )2 1

h xx

b) 2 1 0x 1

2x


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a) Express in the form

b) On the same diagram sketch

i) the graph of

ii) the graph of

c) Find the range of values of x for

which is positive

2( ) 4 5f x x x 2( )x a b

( )y f x10 ( )y f x

10 ( )f x

a) 2( 2) 4 5x 2( 2) 4 5x 2( 2) 1x

c) Solve:210 ( 2) 1 0x

2( 2) 9x ( 2) 3x 1 or 5x

10 - f(x) is positive for -1 < x < 5

b)

(2, 1)

(2, -1)

(2, 9)

5

y=f(x)

y= -f(x)

y= 10 - f(x)

-5


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The graph of a function f intersects the x-axis at (–a, 0)and (e, 0) as shown.

There is a point of inflexion at (0, b) and a maximum turningpoint at (c, d).

Sketch the graph of the derived function f

m is + m is + m is -

f(x)


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Functions f and g are defined on suitable domains by

and

a) Find expressions for:

i)

ii)

b) Solve

( ) sin( )f x x ( ) 2g x x

( ( ))f g x

( ( ))g f x

2 ( ( )) ( ( )) 0 360f g x g f x for x

( ( )) (2 )f g x f xa) sin 2x ( ( )) (sin )g f x g x 2sin x

b) 2sin 2 2sinx x sin 2 sin 0x x

2sin cos sin 0x x x sin (2cos 1) 0x x 1

or2

sin 0 cosx x 0 , 180 , 360x 60 , 300x


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The diagram shows the graphs of two quadraticfunctions

Both graphs have a minimum turning point at (3, 2).

Sketch the graph of

and on the same diagram

sketch the graph of

and( ) ( )y f x y g x

( )y f x

( )y g x

y=g(x)

y=f(x)


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Functions are defined on a suitable set of real numbers.a) Find expressions for

i) ii)

b) i) Show that

ii) Find a similar expression for

and hence solve the equation

4and( ) sin , ( ) cos ( )f x x g x x h x x

( ( ))f h x ( ( ))g h x1 1

2 2( ( )) sin cosf h x x x

( ( ))g h x

for( ( )) ( ( )) 1 0 2f h x g h x x

4( ( )) ( )f h x f x a) 4

sin( )x 4

( ( )) cos( )g h x x

sin cos4 4 4

sin( ) sin cos xx x b) Now use exact values

Repeat for ii)

equation reduces to2

sin 12

x 2 1sin

2 2x

3,

4 4x


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A sketch of the graph of y = f(x) where is shown.The graph has a maximum at A and a minimum at B(3, 0)

a) Find the co-ordinates of the turning point at A.

b) Hence, sketch the graph of Indicate the co-ordinates of the turning points. There is no need to calculate the co-ordinates of the points of intersection with the axes.

c) Write down the range of values of k for which g(x) = k has 3 real roots.

3 2( ) 6 9f x x x x

( ) ( 2) 4g x f x

a) Differentiate2( ) 3 12 9f x x x for SP, f(x) = 0 1 3x or x

when x = 1 4y t.p. at A is: (1, 4)

b) Graph is moved 2 units to the left, and 4 units up(3, 0) (1, 4)

(1, 4) ( 1, 8)

t.p.’s are:

c) For 3 real roots, line y = k has to cut graph at 3 points

from the graph, k 4


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3( ) 3 ( ) , 0x

f x x and g x x

a) Find

b) If find in its simplest form.

( ) where ( ) ( ( ))p x p x f g x3

3( ) , 3

xq x x

( ( ))p q x

3( ) ( ( ))x

p x f g x f

a)3

3x

3 3x

x

3( 1)x

x

b)

33 1

3333

3

( ( )) x

xx

p q x p

9 3

33 3x x

9 3(3 ) 3

3 3

x x

x

3 3

3 3

x x

x

x


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Part of the graph of is shown in the diagram.

On separate diagrams sketch the graph of

a) b)

Indicate on each graph the images of O, A, B, C, and D.

( )y f x

( 1)y f x 2 ( )y f x

a)

b)

graph moves to the left 1 unit

graph is reflected in the x axis

graph is then scaled 2 units in the y direction


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Functions f and g are defined on the set of real numbersby

a) Find formulae for

i) ii)

b) The function h is defined by

Show that and sketch the graph of h. c) Find the area enclosed between this graph and the x-axis.

2( ) 1 and ( )f x x g x x

( ( ))f g x ( ( ))g f x

( ) ( ( )) ( ( ))h x f g x g f x 2( ) 2 2h x x x

a)

b)

2 2( ( )) ( ) 1f g x f x x 2( ( )) ( 1) 1g f x g x x

22( ) 1 1h x x x 2 2( ) 1 2 1h x x x x 22 2x x

c) Graph cuts x axis at 0 and 1 Now evaluate1 2

02 2x x dx

2unit

1

3Area


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The functions f and g are defined on a suitable domain by

a) Find an expression for

b) Factorise

2 2( ) 1 and ( ) 2f x x g x x ( ( ))f g x

( ( ))f g x

a) 22 2( ( )) ( 2) 2 1f g x f x x

2 22 1 2 1x x Difference of 2 squares

Simplify 2 23 1x x

b)


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Higher Maths

Strategies


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Sequences

Maths4Scotland Sequences Higher

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Sequences



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Hint

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Put u1 into recurrence relation

Solve simultaneously:1

31 3p p

A recurrence relation is defined by where -1 < p < -1 and u0 = 12

a) If u1 = 15 and u2 = 16 find the values of p and q

b) Find the limit of this recurrence relation as n

1n nu pu q

..... (1)15 12 p q

Put u2 into recurrence relation ..... (2)16 15 p q

(2) – (1) substitute into (1) 11q

Hence1

and 113

qp

State limit condition -1 < p < 1, so a limit L exists

Use formula1

cL

m

1

3

11

1L

Limit = 16½


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Construct a recurrence relation 1 0.8 0.5n nu u

State limit condition -1 < 0.8 < 1, so a limit L exists

Use formula1

cL

m

0.5

1 0.8L

Limit = 2.5 metres

A man decides to plant a number of fast-growing trees as a boundary between his property and the property ofhis neighbour. He has been warned however by the local garden centre, that during any year, the treesare expected to increase in height by 0.5 metres. In response to this warning, he decides to trim 20% off the height of the trees at the start of any year.

(a) If he adopts the “20% pruning policy”, to what height will he expect the trees to grow in the long run.

(b) His neighbour is concerned that the trees are growing at an alarming rate and wants assurancethat the trees will grow no taller than 2 metres. What is the minimum percentage that the treeswill need to be trimmed each year so as to meet this condition.

un = height at the start of year

Use formula again 1

cL

m

0.5

21 m

Minimum prune = 25%m = 0.75


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Construct a recurrence relation 1 1.015 300n nu u

On the first day of March, a bank loans a man £2500 at a fixed rate of interest of 1.5% per month. This interest is added on the last day of each month and is calculated on the amount due on the first day of the month. He agrees to make repayments on the first day of each subsequent month. Each repayment is £300except for the smaller final amount which will pay off the loan. a) The amount that he owes at the start of each month is taken to be the amount still owing just after the monthly repayment has been made. Let un and un+1 and represent the amounts that he owes at the starts of two successive months.

Write down a recurrence relation involving un and un+1

b) Find the date and amount of the final payment.

u0 = 2500

Calculate each term in the recurrence relation

1 Mar u0 = 2500.001 Apr u1 = 2237.501 May u2 = 1971.061 Jun u3 = 1700.621 Jul u4 = 1426.14

1 Aug u5 = 1147.531 Sept u6 = 864.741 Oct u7 = 577.711 Nov u8 = 286.381 Dec Final payment £290.68


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Equate the two limits Cross multiply

Sequence 1

Since limit exists a 1, so

Use formula for each sequence1

cL

m

Limit = 25

Two sequences are generated by the recurrence relations and

The two sequences approach the same limit as n .

Determine the value of a and evaluate the limit.

1 10n nu au 21 16n nv a v

10

1L

a

Sequence 2 2

16

1L

a

2

10 16

1 1a a

210 1 16(1 )a a

Simplify 210 16 6 0a a 25 8 3 0a a

Solve 5 3 1 0a a 3hence or

51a a

Deduction3

5a


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Equate the two limits Cross multiply

Sequence 1

Use formula for each sequence1

cL

m

1 0.2

pL

Sequence 2

1 0.6

qL

0.8 0.4

p q 0.4 0.6p q

Rearrange0.6

0.4

qp

Two sequences are defined by the recurrence relations

If both sequences have the same limit, express p in terms of q.

1 0

1 0

0.2 , 1 and

0.6 , 1n n

n n

u u p u

v v q v

3

2

qp


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Sequence 2

Requirement for a limit

4

1 0.3L

List terms of 1st sequence

Two sequences are defined by these recurrence relations

a) Explain why only one of these sequences approaches a limit as n b) Find algebraically the exact value of the limit.c) For the other sequence find

i) the smallest value of n for which the nth term exceeds 1000, andii) the value of that term.

1 0with3 0.4 1n nu u u 1 0with0.3 4 1n nv v v

First sequence has no limit since 3 is not between –1 and 1

2nd sequence has a limit since –1 < 0.3 < 1

4

0.7L 5

7

405

7Limit

u0 = 1u1 = 2.6 u2 = 7.4

u3 = 21.8u4 = 65 u5 = 194.6

u6 = 583.4u7 = 1749.8

Smallest value of n is 8; value of 8th term = 1749.8


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Differentiation

Higher Mathematics


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Maths4Scotland Differentiation Higher

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Differentiate 2

2( )f x x

x

Differentiate

Straight line form

122( ) 2f x x x

1321

2( ) 4f x x x

Hint


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Differentiate 3 22 7 4 4y x x x

26 14 4y x x

Hint


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Differentiate 62siny x

Chain Rule

Simplify

62 cos 1

dyx

dx

62 cos

dyx

dx

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Differentiate3

(8 )4

A a a

multiply out

Differentiate

236

4A a a

36

2A a

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Differentiate

13 2( ) (8 )f x x

Chain Rule

Simplify

13 221

2( ) (8 ) ( 2 )f x x x

12 3 2( ) (8 )f x x x

Hint


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Differentiate16

, 0y x xx

Differentiate

Straight line form1

216y x x

3

21 8dy

xdx

Hint


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Differentiate 23 3 16( )

2A x x

x

Straight line form

Multiply out23 3 3 3 16

( )2 2

A x xx

Differentiate

2 13 3( ) 24 3

2A x x x

2( ) 3 3 24 3A x x x

Hint


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Differentiate 1

2( ) 5 4f x x

Chain Rule

Simplify

1

21

2( ) 5 4 5f x x

1

25

2( ) 5 4f x x

Hint


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Differentiate 216( ) 240

3A x x x

32( ) 240

3A x x

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Differentiate 2( ) 3 (2 1)f x x x

Multiply out

Differentiate

3 2( ) 6 3f x x x

2( ) 18 6f x x x

Hint


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Differentiate2 2( ) cos sinf x x x

( ) 2 cos sin 2sin cosf x x x x x Chain Rule

Simplify ( ) 4 cos sinf x x x

Hint


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Quadratic Theory

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Maths4Scotland Quadratic Theory Higher

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Show that the line with equation

does not intersect the parabola

with equation

2 1y x

2 3 4y x x

Put two equations equal

Use discriminant

Show discriminant < 0

No real roots


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a) Write in the form

b) Hence or otherwise sketch the graph of

2( ) 6 11f x x x 2x a b

( )y f x

a) 2( ) ( 3) 2f x x

b) This is graph of2y x moved 3 places to left and 2 units up.

minimum t.p. at (-3, 2) y-intercept at (0, 11)


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2(1 2 ) 5 2 0k x kx k Show that the equation

has real roots for all integer values of k

Use discriminant (1 2 ) 5 2a k b k c k

2 24 25 4 (1 2 ) 2b ac k k k

2 225 8 16k k k 29 8k k

Consider when this is greater than or equal to zero

Sketch graph cuts x axis at8

and9

0k k

Hence equation has real roots for all integer k


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The diagram shows a sketch of a parabolapassing through (–1, 0), (0, p) and (p, 0).

a) Show that the equation of the parabola is

b) For what value of p will the line

be a tangent to this curve?

2( 1)y p p x x

y x p

a) ( 1)( )y k x x p Use point (0, p) to find k (0 1)(0 )p k p

p pk 1k ( 1)( )y x x p 2y x px x p 21y p p x x

b) Simultaneous equations 21x p p p x x

20 2p x x Discriminant = 0 for tangency 2p


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Given , express in the form

Hint

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2( ) 2 8f x x x ( )f x 2x a b

2( ) ( 1) 10f x x


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For what value of k does the equation have equal roots? 2 5 ( 6) 0x x k

1 5 6a b c k Discriminant

2 4 25 4( 6)b ac k

0 25 4 24k

4 1k

1

4k

For equal rootsdiscriminant = 0


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Integration

Higher Mathematics


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Maths4Scotland Integration Higher

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Integrate 2 4 3x x dx 3 24

33 2

x xx c

3 212 3

3x x x c

Integrate term by term

simplify

Hint


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Find 3cos x dx3sin x c

Hint


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Integrate (3 1)( 5)x x dx 23 14 5x x dx

3 23 145

3 2

x xx c

3 27 5x x x c

Multiply out brackets


simplify

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Find 2sin d 2cos c

Hint


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Integrate2(5 3 )x dx

3(5 3 )

3 3

xc

31

9(5 3 )x c

Standard Integral(from Chain Rule)

Hint


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Find p, given

1

42p

x dx 1

2

1

42p

x dx 3

2

1

2

342

p

x

3 3

2 22 2

3 3(1) 42p

2 233 3

42p 32 2 126p

3 64p 3 8p 2p Hint


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Evaluate

2

21

dx

x2

2

1

x dx 21

1x 1 12 1

11

2

1

2

Straight line form

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Find

1

6

0

(2 3)x dx17

0

(2 3)

7 2

x

7 7(2 3) (0 3)

14 14

7 75 3

14 14

5580.36 156.21 (4sf)5424

Use standard Integral(from chain rule)

Hint


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Find1

3sin cos2

x x dx1

3cos sin2

x x c Integrate term by term

Hint


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Integrate3

2x dx

x

132 2x x dx

3 222

3

2

2

xx c

3222

3x x c

Straight line form

Hint


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Integrate3 1

x dxx

1

3 2x x dx

1

4 2

1

24

x xc

14 21

42x x c

Straight line form

Hint


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Integrate

3 5x xdx

x

1 1

2 2

3 5x xdx

x x

5 1

2 25x x dx 7 3

2 22

7

10

3cx x

Straight line form

Hint


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Integrate

324

2

x xdx

x

3

2

1 1

2 2

4

2 2

x xdx

x x

1

2 1

22x x dx

3

2 24 1

3 4x x c

Split into separate fractions

Hint


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Find 2

3

1

2 1x dx

24

1

2 1

4 2

x

4 44 1 2 1

4 2 4 2

4 45 3

8 8

68


Hint


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Find sin 2 cos 34

x x dx

1 1

2 3cos 2 sin 3

4x x c

Hint


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Find2

sin7

t dt2

7cos t c

Hint


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Integrate

1

2

1

0 3 1

dx

x

1

2

1

0

3 1x dx

Straight line form

1

2

1

0

13

2

3 1x

1

0

23 1

3x

2 23 1 0 1

3 3

2 24 1

3 3

4 2

3 3

2

3

Hint


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Given the acceleration a is: 1

22(4 ) , 0 4a t t

If it starts at rest, find an expression for the velocity v where dv

adt

1

22(4 )dv

tdt

3

2

31

2

2(4 )tv c

3

24

(4 )3

v t c

344

3v t c Starts at rest, so

v = 0, when t = 0 340 4

3c

340 4

3c

320

3c

32

3c

3

24 32

3 3(4 )v t

Hint


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A curve for which 3sin(2 )dy

xdx

5

12, 3passes through the point

Find y in terms of x. 3

2cos(2 )y x c

3 5

2 123 cos(2 ) c Use the point 5

12, 3

3 5

2 63 cos( ) c 3 3

2 23 c

3 33

4c

4 3 3 3

4 4c

3

4c

3 3

2 4cos(2 )y x

Hint


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Integrate 2 2

2

2 2, 0

x xdx x

x

4

2

4xdx

x

4

2 2

4xdx

x x 2 24x x dx

3 14

3 1

x xc

31

3

4x c

x

Split intoseparate fractions

Multiply out brackets

Hint


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If

( ) sin(3 )f x x passes through the point 9, 1

( )y f x express y in terms of x.

1

3( ) cos(3 )f x x c Use the point 9

, 1

1

31 cos 3

9c

1

31 cos

3c

1 1

3 21 c

7

6c 1 7

3 6cos(3 )y x

Hint


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Integrate2

1

(7 3 )dx

x2(7 3 )x dx

1(7 3 )

1 3

xc

11

3(7 3 )x c

Straight line form

Hint


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The graph of

32

1 1

4

dyx

dx x

( )y g x passes through the point (1, 2).

express y in terms of x. If

3 2 1

4

dyx x

dx

4 1 1

4 1 4

x xy x c

simplify

4 1 1

4 4

xy x c

x Use the point

41 1 1

2 14 1 4

c

3c Evaluate c41 1

4 4

13y x x

x Hint


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Integrate2 5x

dxx x

3

2

2 5xdx

x

3 3

2 2

2 5xdx

x x

1 3

2 25x x dx

3 1

2 2

3 1

2 2

5x xc

3 1

2 22

310x x c

Straight line form

Hint


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A curve for which 26 2

dyx x

dx passes through the point (–1, 2).

Express y in terms of x.

3 26 2

3 2

x xy c 3 22y x x c

Use the point3 22 2( 1) ( 1) c 5c

3 22 5y x x Hint


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Evaluate

222

1

1x dxx

1

222

1x x dx

Cannot use standard integralSo multiply out

4 22

12x x x dx

25 2 1

1

1

5x x x 5 2 5 21 1 1 1

5 2 5 12 2 1 1

32 1 1

5 2 54 64 40 20 2

10 10 10 10 82

10 1

58

Hint


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Evaluate4

1

x dx1

2

4

1

x dx 43

2

1

2

3x

4

1

32

3x

3 32 24 1

3 3

16 2

3 3

14

3

2

34

Straight line form

Hint


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Evaluate0

2

3

(2 3)x dx


03

3

(2 3)

3 2

x

3 3(2(0) 3) (2( 3) 3)

6 6

27 27

6 6

27 27

6 6

54

6 9

Hint


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The curve ( )y f x ,112

passes through the point

( ) cos 2f x x Find f(x)

1

2( ) sin 2f x x c

1

2 121 sin 2 c

use the given point ,112

1

2 61 sin c

1 1

2 21 c 3

4c 1 3

2 4( ) sin 2f x x

1

6 2sin

Hint


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2(6 cos )x x x dx Integrate

3 26sin

3 2

x xx c Integrate term by term

Hint


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Integrate 33 4x x dx4 23 4

4 2

x xc

4 23

42x x c


Hint


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Evaluate1

01 3x dx

1

21

01 3x dx

13

2

0

3

2

1 3

3

x

13

2

0

21 3

9x

13

0

21 3

9x

3 32 21 3(1) 1 3(0)

9 9

3 32 24 1

9 9

16 2

9 9

14

9 5

91

Hint


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Higher Maths

Strategies


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Compound Angles

Maths4Scotland Compound Angles Higher

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Compound Angles



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This presentation is split into two parts

Using Compound angle formula for

Exact values

Solving equations

Choose by clicking on the appropriate button


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A is the point (8, 4). The line OA is inclined at an angle p radians to the x-axis a) Find the exact values of: i) sin (2p) ii) cos (2p)

The line OB is inclined at an angle 2p radians to the x-axis. b) Write down the exact value of the gradient of OB.

Draw triangle Pythagoras80

Write down values for cos p and sin p8 4

cos sin80 80

p p

Expand sin (2p) sin 2 2sin cosp p p 4 8 64 42

80 580 80

Expand cos (2p) 2 2cos 2 cos sinp p p 2 28 4

80 80

64 16 3

80 5

Use m = tan (2p)sin 2

tan 2cos 2

pp

p 4 3 4

5 5 3

8

4p


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In triangle ABC show that the exact value of

Use Pythagoras

Write down values forsin a, cos a, sin b, cos b

1 1 1 3sin cos sin cos

2 2 10 10a a b b

Expand sin (a + b) sin( ) sin cos cos sina b a b a b

is2

sin( )5

a b

2 10AC CB

2 10

Substitute values1 3 1 1

2 10 2 10sin( )a b

Simplify3 1

20 20sin( )a b 4

20

4 4 2

4 5 2 5 5


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Using triangle PQR, as shown, find theexact value of cos 2x

Use Pythagoras

Write down values forcos x and sin x

2 7cos sin

11 11x x

Expand cos 2x2 2cos 2 cos sinx x x

11PR

11

Substitute values 222 7

11 11cos 2x

Simplify4 7

cos 211 11

x 3

11


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On the co-ordinate diagram shown, A is the point (6, 8) andB is the point (12, -5). Angle AOC = p and angle COB = q Find the exact value of sin (p + q).

Use Pythagoras

Write down values forsin p, cos p, sin q, cos q

8 6 5 12

10 10 13 13sin , cos , sin , cosp p q q

Expand sin (p + q) sin ( ) sin cos cos sinp q p q p q

10 13OA OB

Substitute values

Simplify 126 63

130 65

6

8

512

10

13

Mark up triangles

8 12 6 5

10 13 10 13sin ( )p q

96 30

130 130sin ( )p q


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Draw triangles Use Pythagoras

Expand sin 2A sin 2 2sin cosA A A

A and B are acute angles such that and .

Find the exact value of

a) b) c)

3

4tan A 5

12tan B

sin 2A cos 2A sin(2 )A B4

3A

12

5B

Hypotenuses are 5 and 13 respectively

5 13

Write down sin A, cos A, sin B, cos B 3 4 5 12

, , ,5 5 13 13

sin cos sin cosA A B B

3 4 24

5 5 25sin 2 2A

Expand cos 2A 2 2cos 2 cos sinA A A 2 2 16 9 74 3

25 25 255 5cos 2A

Expand sin (2A + B) sin 2 sin 2 cos cos 2 sinA B A B A B

Substitute 24 12 7 5 323sin 2

25 13 25 13 325A B


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Draw triangle Use Pythagoras

Expand sin (x + 30) sin( 30) sin cos30 cos sin 30x x x

If x° is an acute angle such that

show that the exact value of

4

3tan x

4 3 3sin( 30) is

10x

3

4

x

Hypotenuse is 5

5

Write down sin x and cos x4 3

,5 5

sin cosx x

Substitute

Simplify

Table of exact values

4 3 3 1sin( 30)

5 2 5 2x

4 3 3sin( 30)

10 10x 4 3 3

10


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Use Pythagoras

Expand cos (x + y) cos( ) cos cos sin sinx y x y x y

Write downsin x, cos x, sin y, cos y.

3 4 24 5, , ,

5 5 7 7sin cos sin cosx x y y

Substitute

Simplify20 3 4 6

35

The diagram shows two right angled trianglesABD and BCD with AB = 7 cm, BC = 4 cm and CD = 3 cm. Angle DBC = x° and angle ABD is y°.

Show that the exact value of 20 6 6

cos( )35

x y is

5, 24BD AD

24

5

4 5 3 24cos( )

5 7 5 7x y

20 3 24cos( )

35 35x y

20 6 6

35


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2 5

3 32 2sin , cosx x

The framework of a child’s swing has dimensionsas shown in the diagram. Find the exact value of sin x°

Write down sin ½ x and cos ½ x

5h

Substitute

Simplify


3 3

4

xDraw in perpendicular

2

2

x

h5Use fact that sin x = sin ( ½ x + ½ x)

Expand sin ( ½ x + ½ x) 2 2 2 2 2 22 2sin sin cos sin cos 2sin cosx x x x x xx x

2 5

3 32 2sin 2x x

4 5sin

9x


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Given that

find the exact value of

Write down values forcos a and sin a

3 11cos sin

20 20a a

Expand sin 2a sin 2 2 sin cosa a a

20

Substitute values11 3

sin 2 220 20

a

Simplify

11tan , 0

3 2

3a

11sin 2

Draw triangle Use Pythagoras hypotenuse 20

6 11sin 2

20a

3 11

10


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Find algebraically the exact value of

1 3cos 120 cos 60 cos 150 cos30

2 2

3 1sin 120 sin 60 sin 150 sin 30

2 2

Expand sin (+120)

sin 120 sin cos120 cos sin120

Use table of exact values

1 3 3 1

2 2 2 2sin sin . cos . cos . sin . Combine and substitute

sin sin 120 cos( 150)


Expand cos (+150)

cos 150 cos cos150 sin sin150

Simplify 1 3 3 1

2 2 2 2sin sin cos cos sin

0


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If find the exact value of

a) b)

Write down values forcos and sin

4 3cos sin

5 5

Expand sin 2 sin 2 2 sin cos


4cos , 0

5 2

5

4

3

Opposite side = 3

3 4 242

5 5 25

Expand sin 4 (4 = 2 + 2) sin 4 2 sin 2 cos 2

Expand cos 2 2 2cos 2 cos sin 16 9 7

25 25 25

Find sin 424 7

sin 4 225 25

336

625

sin 2 sin 4


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Draw triangles Use Pythagoras

Expand sin (P + Q) sin sin cos cos sinP Q P Q P Q

For acute angles P and Q

Show that the exact value of12

13

P

53

Q

Adjacent sides are 5 and 4 respectively

5 4

Write down sin P, cos P, sin Q, cos Q 12 5 3 4

, , ,13 13 5 5

sin cos sin cosP P Q Q

Substitute

12 3and

13 5sin sinP Q

63

65sin ( )P Q

12 4 5 3sin

13 5 13 5P Q

Simplify 48 15sin

65 65P Q 63

65


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Solving Equations


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Solve the equation for 0 ≤ x ≤ correct to 2 decimal places 3cos(2 ) 10cos( ) 1 0x x

Replace cos 2x with 2cos 2 2cos 1x x

Substitute 23 2cos 1 10cos 1 0x x

Simplify 26cos 10cos 4 0x x 23cos 5cos 2 0x x

Factorise 3cos 1 cos 2 0x x

Hence 1

3cos

cos 2

x

x

Discard

Find acute x 1.23acute radx

Determine quadrants

AS

CT

1.23 2 1.23or radsx

1.23

5.05

rads

rads

x

x


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Hint


Solve simultaneously 2cos 2 3x

Rearrange 3

2cos 2x

0 0 2 2x x

Find acute 2x 62acute x

Determine quadrants

AS

CT

6 6

6 6 6 62 or radsx

5 7

12 12orx

The diagram shows the graph of a cosine function from 0 to .

a) State the equation of the graph.

b) The line with equation y = -3 intersects this graphat points A and B. Find the co-ordinates of B.

Equation 2cos 2y x

Check range

7

12, 3isB B Deduce 2x


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Functions f and g are defined on suitable domains by f(x) = sin (x) and g(x) = 2x a) Find expressions for:

i) f(g(x)) ii) g(f(x)) b) Solve 2 f(g(x)) = g(f(x)) for 0 x 360°

Hint


2nd expression

Form equation 2sin 2 2sinx x

Rearrange

Determinequadrants

AS

CT60 , 300x

1st expression ( ( )) (2 ) sin 2f g x f x x

Common factor

( ( )) (sin ) 2sing f x g x x

Replace sin 2x 2sin cos sinx x x

sin 2 sinx x

2sin cos sin 0x x x

sin 2cos 1 0x x

Hence1

or2

sin 0 2cos 1 0 cosx x x

Determine x

sin 0 0 , 360x x

1

2cos 60acutex x

0 , 60 , 300 , 360x


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Functions are defined on a suitable set of real numbers

a) Find expressions for i) f(h(x)) ii) g(h(x))

b) i) Show that ii) Find a similar expression for g(h(x))

iii) Hence solve the equation

Hint


2nd expression

Simplify 1st expr.

Similarly for 2nd expr.

Determinequadrants

AS

CT3,

4 4x

1st expression 4 4( ( )) sinf h x f x x

Use exact values

and4

( ) sin , ( ) cos ( )f x x g x x h x x

1 1( ( )) sin cos

2 2f h x x x

for( ( )) ( ( )) 1 0 2f h x g h x x

4 4( ( )) cosg h x g x x

4 4( ( )) sin cos cos sinf h x x x

1 1

2 2( ( )) sin cosf h x x x

4 4( ( )) cos cos sin sing h x x x

1 1

2 2( ( )) cos sing h x x x

Form Eqn. ( ( )) ( ( )) 1f h x g h x

2

2sin 1x Simplifies to

2 2 1

2 2 2 2sin x Rearrange:

acute x 4acute x


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a) Solve the equation sin 2x - cos x = 0 in the interval 0 x 180°b) The diagram shows parts of two trigonometric graphs,

y = sin 2x and y = cos x. Use your solutions in (a) towrite down the co-ordinates of the point P.

Hint


Determine quadrantsfor sin x

AS

CT

30 , 150x

Common factor

Replace sin 2x 2sin cos cos 0x x x

cos 2sin 1 0x x

Hence1

or2

cos 0 2sin 1 0 sinx x x

Determine x cos 0 90 , ( 270 )out of rangex x 1

2sin 30acutex x

30 , 90 , 150x

Solutions for where graphs cross

150x By inspection (P)

cos150y Find y value3

2y

Coords, P

3

2150 ,P


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Solve the equation for 0 ≤ x ≤ 360°3cos(2 ) cos( ) 1x x


Substitute 23 2cos 1 cos 1x x

Simplify 26cos cos 2 0x x

Factorise 3cos 2 2cos 1 0x x

Hence2

3cos x

Find acute x 48acute x

Determine quadrants

AS

CT1

2cos x

60acute x


2

3cos x

AS

CT

1

2cos x

132

228

x

x

60

300

x

x

Solutions are: x= 60°, 132°, 228° and 300°

48acute x 60acute x


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Solve the equation for 0 ≤ x ≤ 2 62sin 2 1x

Rearrange

Find acute x 62

6acute x

Determine quadrantsAS

CT


Solutions are:

6

1sin 2

2x

62

6x

6

52

6x

Note range 0 2 0 2 4x x

and for range 2 2 4x

6

132

6x

6

172

6x

7 3, , ,

6 2 6 2x

for range 0 2 2x


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a) Write the equation cos 2 + 8 cos + 9 = 0 in terms of cos and show that for cos it has equal roots.

b) Show that there are no real roots for

Rearrange

Divide by 2

Deduction

Factorise cos 2 cos 2 0

Replace cos 2 with 2cos 2 2cos 1

22cos 8cos 8 0

2cos 4cos 4 0

Equal roots for cos

Try to solve:

cos 2 0

cos 2

Hence there are no real solutions for

No solution


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Solve algebraically, the equation sin 2x + sin x = 0, 0 x 360

Hint


Determine quadrantsfor cos x

AS

CT

120 , 240x

Common factor

Replace sin 2x 2sin cos sin 0x x x

sin 2cos 1 0x x

Hence1

or2

sin 0

2cos 1 0 cos

x

x x

Determine x sin 0 0 , 360x x

1

2cos 60acutex x

x = 0°, 120°, 240°, 360°


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Find the exact solutions of 4sin2 x = 1, 0 x 2

Hint


Determine quadrants for sin x

AS

CT

Take square roots

Rearrange 2 1

4sin x

1

2sin x

Find acute x6

acute x

+ and – from the square root requires all 4 quadrants

5 7 11, , ,

6 6 6 6x


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Solve the equation for 0 ≤ x ≤ 360°cos 2 cos 0x x


Substitute 22cos 1 cos 0x x

Simplify


Hence1

2cos x


Determine quadrants

AS

CTcos 1x

180x


1

2cos x

60

300

x

x

Solutions are: x= 60°, 180° and 300°

60acute x 22cos cos 1 0x x


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Solve algebraically, the equation for 0 ≤ x ≤ 360°cos 2 5cos 2 0x x


Substitute 22cos 1 5cos 2 0x x

Simplify 22cos 5cos 3 0x x


Hence1

2cos x


Determine quadrants

cos 3x


AS

CT

1

2cos x

60

300

x

x

Solutions are: x= 60° and 300°

60acute x

Discard above


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Higher Maths

Strategies


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The Circle

Maths4Scotland The Circle Higher

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The Circle



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Find the equation of the circle with centre

(–3, 4) and passing through the origin.

Find radius (distance formula): 5r

You know the centre: ( 3, 4)

Write down equation: 2 2( 3) ( 4) 25x y


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Explain why the equation

does not represent a circle.

Consider the 2 conditions

Calculate g and f:

2 2. . 0i e g f c

2 2 2 3 5 0x y x y

1. Coefficients of x2 and y2 must be the same.

31,

2g f

22 3 1

2 4( 1) 5 1 2 5 0

2. Radius must be > 0

Evaluate2 2g f c

Deduction: 2 2 2 20g f c so g f c not real

Equation does not represent a circle


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Calculate mid-point for centre:

Calculate radius CQ:

(1, 2)

2 21 2 18x y Write down equation;

Find the equation of the circle which has P(–2, –1) and Q(4, 5)

as the end points of a diameter.

18r

Make a sketch

P(-2, -1)

Q(4, 5)

C


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Calculate centre of circle:

Calculate gradient of OP (radius to tangent)

( 1, 2)

Gradient of tangent:

Find the equation of the tangent at the point (3, 4) on the circle

1

2m

2 2 2 4 15 0x y x y

2m

Equation of tangent: 2 10y x

Make a sketch O(-1, 2)

P(3, 4)


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Find centre of circle:

Calculate gradient of radius to tangent

( 1, 1)


The point P(2, 3) lies on the circle

Find the equation of the tangent at P.

2

3m

3

2m

Equation of tangent: 2 3 12y x

Make a sketch

2 2( 1) ( 1) 13x y

O(-1, 1)

P(2, 3)


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A is centre of small circle

O, A and B are the centres of the three circles shown inthe diagram. The two outer circles are congruent, eachtouches the smallest circle. Circle centre A has equation

The three centres lie on a parabola whose axis of symmetryis shown the by broken line through A. a) i) State coordinates of A and find length of line OA. ii) Hence find the equation of the circle with centre B. b) The equation of the parabola can be written in the form

2 212 5 25x y

( )y px x q

A(12, 5) Find OA (Distance formula) 13

Find radius of circle A from eqn.Use symmetry, find B B(24, 0) 5

Find radius of circle B 13 5 8

Find p and q.

Eqn. of B 2 2( 24) 64x y

Points O, A, B lie on parabola – subst. A and B in turn

0 24 (24 )

5 12 (12 )

p q

p q

Solve: 5

144, 24p q


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Find centre of circle P:

Gradient of radius of Q to tangent:

(4, 5)

Equation of tangent: 5y x

Solve eqns. simultaneously

Circle P has equation Circle Q has centre (–2, –1) and radius 22. a) i) Show that the radius of circle P is 42 ii) Hence show that circles P and Q touch. b) Find the equation of the tangent to circle Q at the point (–4, 1) c) The tangent in (b) intersects circle P in two points. Find the x co-ordinates of the points of intersection, expressing your answers in the form

2 2 8 10 9 0x y x y

3a b

Find radius of circle :P: 2 24 5 9 32 4 2

Find distance between centres 72 6 2 Deduction: = sum of radii, so circles touch

1m Gradient tangent at Q: 1m

2 2 8 10 9 0

5

x y x y

y x

Soln: 2 2 3


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2 2 4 2 2 0x y kx ky k For what range of values of k does the equation represent a circle ?

Determine g, f and c: 2 , , 2g k f k c k

State condition2 2 0g f c Put in values

2 2( 2 ) ( 2) 0k k k

Simplify 25 2 0k k

Complete the square

2

2

2

1

5

1 1

10 100

1 195

10 100

5 2

5 2

5

k k

k

k

So equation is a circle for all values of k.

Need to see the positionof the parabola

Minimum value is195 1

100 10when k

This is positive, so graph is:

Expression is positive for all k:


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2 2 6 4 0x y x y c For what range of values of c does the equation represent a circle ?

Determine g, f and c: 3, 2, ?g f c

State condition2 2 0g f c Put in values

2 23 ( 2) 0c

Simplify 9 4 0c

Re-arrange: 13c


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The circle shown has equation Find the equation of the tangent at the point (6, 2).

2 2( 3) ( 2) 25x y

Calculate centre of circle:

Calculate gradient of radius (to tangent)

(3, 2)


4

3m

3

4m

Equation of tangent: 4 3 26y x

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Hint

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When newspapers were printed by lithograph, the newsprint hadto run over three rollers, illustrated in the diagram by 3 circles. The centres A, B and C of the three circles are collinear.The equations of the circumferences of the outer circles are

Find the equation of the central circle.

2 2 2 2( 12) ( 15) 25 and ( 24) ( 12) 100x y x y

Find centre and radius of Circle A ( 12, 15) 5r

Find centre and radius of Circle C (24, 12) 10r

Find distance AB (distance formula) 2 236 27 45

Find diameter of circle B so radius of B = 45 (5 10) 30 15

Use proportion to find B25 25

relative to C45 45

27 15, 36 20

Centre of B (4, 3) Equation of B 2 24 3 225x y

(24, 12)

(-12, -15)

27

36

25

20B


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Vectors

Strategies

Higher Maths

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Maths4Scotland Vectors Higher

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Vectors



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The questions are in groups

Angles between vectors (5)

Points dividing lines in ratiosCollinear points (8)

General vector questions (15)


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General Vector Questions

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Vectors u and v are defined by andDetermine whether or not u and v are perpendicular to each other.

3 2 u i j 2 3 4 v i j k

Is Scalar product = 0

3 2

2 3

0 4

u.v

3 2 2 3 0 4 u.v 6 6 0 u.v

0u.v Hence vectors are perpendicular


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Hint

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For what value of t are the vectors and perpendicular ?2

3

t

u2

10

t

v

Put Scalar product = 0

2

2 10

3

t

t

u.v

2 2 10 3t t u.v 5 20t u.v

Perpendicular u.v = 0 0 5 20t

4t


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Hint

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VABCD is a pyramid with rectangular base ABCD.

The vectors are given by

Express in component form.

, andAB AD AV��

8 2 2AB ��

i j k

2 10 2AD ��

i j k

7 7AV ��

i j k

CV��

AC CV AV ��

CV AV AC ��

BC AD��

AB BC AC ��

Ttriangle rule ACV Re-arrange

Triangle rule ABC also

CV AV AB AD �� 1 8 2

7 2 10

7 2 2

CV

��9

5

7

CV

��

9 5 7CV ��

i j k


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Hint

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The diagram shows two vectors a and b, with | a | = 3 and | b | = 22.These vectors are inclined at an angle of 45° to each other.a) Evaluate

i) a.aii) b.biii) a.b

b) Another vector p is defined by Evaluate p.p and hence write down | p |.

2 3 p a b

cos0 a a a a 3 3 1 9 2 2 2 2 b b 8

cos 45 a b a b1

3 2 2 62

i) ii)

iii)

b) 2 3 2 3 p p a b a b 4 . 12 9 a a a.b b.b

36 72 72 180 Since p.p = p2

180 6 5 p


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Hint

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Vectors p, q and r are defined by

a) Express in component form

b) Calculate p.r

c) Find |r|

- , 4 , and 4 3 p i j k q i k r i j

2 p q r

2 p q r - 4 2 4 3 i j k i k i j 8 5 -5 i j ka)

b) . - . 4 3 p r i j k i j . 1 4 1 ( 3) ( 1) 0 p r . 1 p r

c)2 24 ( 3) r 16 9 5 r r


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Hint

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The diagram shows a point P with co-ordinates(4, 2, 6) and two points S and T which lie on the x-axis.

If P is 7 units from S and 7 units from T, find the co-ordinates of S and T.

Use distance formula ( , 0, 0)S a ( , 0, 0)T b

2 2 2 249 (4 ) 2 6PS a 249 (4 ) 40a 29 (4 )a

4 3a 7 1a or a

hence there are 2 points on the x axis that are 7 units from P

(1, 0, 0)S (7, 0, 0)T

i.e. S and T

and


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The position vectors of the points P and Q are

p = –i +3j+4k and q = 7 i – j + 5 k respectively.

a) Express in component form.

b) Find the length of PQ.

Hint

Previous Next

PQ��

PQ ��

q - p7 1

1 3

5 4

PQ

��

-a)8

4

1

PQ

��

8 4 i j k

2 2 28 ( 4) 1PQ ��

b) 64 16 1 81 9


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Hint

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PQR is an equilateral triangle of side 2 units.

Evaluate a.(b + c) and hence identifytwo vectors which are perpendicular.

, , andPQ PR QR ��

a b c

( ) a. b c a.b a.c

cos60 a.b a b1

22 2 a.b 2 a.b

Diagram

P

RQ60° 60°

60°a b

c

NB for a.c vectors must point OUT of the vertex ( so angle is 120° )

cos120 a.c a c1

22 2

a.c 2 a.c

Hence ( ) 0 a. b c so, a is perpendicular to b + c

Table ofExact Values


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Calculate the length of the vector 2i – 3j + 3k

22 22 ( 3) 3 Length 4 9 3

16

4


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Hint

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Find the value of k for which the vectors and are perpendicular 1

2

1

4

3

1k

Put Scalar product = 0

1 4

2 3

1 1

0k

0 4 6 ( 1)k

3k

0 2 1k


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A is the point (2, –1, 4), B is (7, 1, 3) and C is (–6, 4, 2).

If ABCD is a parallelogram, find the co-ordinates of D.

AD BC ��

c b6 7

4 1

2 3

BC

�� 13

3

1

BC

��

D is the displacement AD��

from A

hence2 13

1 3

4 1

d11

2

3

d 11, 2, 3D


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Hint

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If and write down the components of u + v and u – v

Hence show that u + v and u – v are perpendicular.

3

3

3

u1

5

1

v

2

8

2

u v4

2

4

u v

2 4

8 2

2 4

.

u v u v

look at scalar product

.

( 2) ( 4) 8 ( 2) 2 4

u v u v

8 16 8 0

Hence vectors are perpendicular


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The vectors a, b and c are defined as follows:

a = 2i – k, b = i + 2j + k, c = –j + k

a) Evaluate a.b + a.c

b) From your answer to part (a), make a deduction about the vector b + c

2 1

0 2

1 1

a.ba) 2 0 1 a.b 1a.b

2 0

0 1

1 1

a.c

b)

0 0 1 a.c 1a.c 0 a.b a.c

b + c is perpendicular to a


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A is the point ( –3, 2, 4 ) and B is ( –1, 3, 2 )Find:

a) the components of

b) the length of AB

AB��

AB ��

b aa)

1 3

3 2

2 4

AB

�� 2

1

2

AB

��

2 2 22 1 ( 2)AB b) 4 1 4AB

9AB 3AB


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In the square based pyramid,all the eight edges are of length 3 units.

Evaluate p.(q + r)

, , ,AV AD AB ��

p q r

Triangular faces are all equilateral

( ) p. q r p.q p.r

cos60 p.q p q1

23 3 p.q

1

24p.q

cos60 p.r p r1

23 3 p.r

1

24p.q

1 1

2 2( ) 4 4 p. q r ( ) 9 p. q r

Table ofExact Values


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Points dividing lines in ratios

Collinear Points

Continue


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A and B are the points (-1, -3, 2)and (2, -1, 1) respectively.

B and C are the points of trisection of AD.That is, AB = BC = CD.

Find the coordinates of D

1

3

AB

AD

��

�� 3AB AD �� 3 b a d a

3 3 b a d a 3 2 d b a

2 1

1 3

1 2

3 2

d8

3

1

d (8, 3, 1)D


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The point Q divides the line joining P(–1, –1, 0) to R(5, 2 –3) in the ratio 2:1.Find the co-ordinates of Q.

2

1

PQ

QR

��

�� 2PQ QR ��

2 2 q p r q

3 2 q r p

5 1

2 1

3 0

3 2

q9

3

6

1

3

q (3, 1, 2)Q

Diagram P

QR

21


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a) Roadmakers look along the tops of a set of T-rods to ensurethat straight sections of road are being created. Relative to suitable axes the top left corners of the T-rods are the points A(–8, –10, –2), B(–2, –1, 1) and C(6, 11, 5).Determine whether or not the section of road ABC has beenbuilt in a straight line.

b) A further T-rod is placed such that D has co-ordinates (1, –4, 4).Show that DB is perpendicular to AB.

AB ��

b aa)6 2

9 3 3

3 1

AB

�� 14 2

21 7 3

7 1

AC

��

andAB AC��

are scalar multiples, so are parallel. A is common. A, B, C are collinear

b) Use scalar product6 3

9 3

3 3

. .AB BD

��

. 18 27 9 0AB BD ��

Hence, DB is perpendicular to AB


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VABCD is a pyramid with rectangular base ABCD.Relative to some appropriate axis,

represents – 7i – 13j – 11k

represents 6i + 6j – 6k

represents 8i – 4j – 4k

K divides BC in the ratio 1:3

Find in component form.

VA��

AB��

AD��

VK��

VA AB VB ��

VK KB VB �� 1 1 1

4 4 4KB CB DA AD ��

VK VB KB �� 1

4VK VA AB AD ��

7 6 81

13 6 44

11 6 4

VK

�� 1

8

18

VK

��


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The line AB is divided into 3 equal parts bythe points C and D, as shown. A and B have co-ordinates (3, –1, 2) and (9, 2, –4).

a) Find the components of and

b) Find the co-ordinates of C and D.

AB��

AC��

AB ��

b a6

3

6

AB

�� 2

1

2

1

3AC AB

��

a)

b) C is a displacement of from AAC�� 3 2

1 1

2 2

c (5, 0, 0)C

similarly

5 2

0 1

0 2

d (7, 1, 2)D


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Relative to a suitable set of axes, the tops of three chimneys haveco-ordinates given by A(1, 3, 2), B(2, –1, 4) and C(4, –9, 8).Show that A, B and C are collinear

AB ��

b a1

4

2

AB

�� 3 1

12 3 4

6 2

AC

��

andAB AC��

are scalar multiples, so are parallel. A is common. A, B, C are collinear


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A is the point (2, –5, 6), B is (6, –3, 4) and C is (12, 0, 1).

Show that A, B and C are collinearand determine the ratio in which B divides AC

AB ��

b a4 2

2 2 1

2 1

AB

�� 6 2

3 3 1

3 1

BC

��

andAB BC��

are scalar multiples, so are parallel. B is common. A, B, C are collinear

2

3

AB

BC

��

��A

BC

23

B divides AB in ratio 2 : 3


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Relative to the top of a hill, three glidershave positions given by

R(–1, –8, –2), S(2, –5, 4) and T(3, –4, 6).

Prove that R, S and T are collinear

RS ��

s r3 1

3 3 1

6 2

RS

�� 4 1

4 4 1

8 2

RT

��

andRS RT��

are scalar multiples, so are parallel. R is common. R, S, T are collinear


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Angle between two vectors

Continue


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The diagram shows vectors a and b. If |a| = 5, |b| = 4 and a.(a + b) = 36Find the size of the acute anglebetween a and b.

cos a.b

a b( ) 36 36 a. a b a.a a.b

25 a.a a a 25 36 a.b 11 a.b

11cos

5 4

1 11

cos20

56.6


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The diagram shows a square based pyramid of height 8 units.Square OABC has a side length of 6 units.The co-ordinates of A and D are (6, 0, 0) and (3, 3, 8).C lies on the y-axis.a) Write down the co-ordinates of B

b) Determine the components of

c) Calculate the size of angle ADB.

andDA DB��

a) B(6, 6, 0) b)3

3

8

DA

�� 3

3

8

DB

��

c).

cosAD DB

AD DB

��

��3

3

8

AD DA

�� 3 3

3 . 3 64

8 8

.AD DB

��

64cos

82 82

141.3


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A box in the shape of a cuboid designed with circles of differentsizes on each face.

The diagram shows three of the circles, where the origin representsone of the corners of the cuboid.

The centres of the circles are A(6, 0, 7), B(0, 5, 6) and C(4, 5, 0)Find the size of angle ABC

6

5

1

BA

��4

0

6

BC

��Vectors to pointaway from vertex

. 24 0 6 18BA BC ��

36 25 1 62BA ��

16 36 52BC ��

18cos

62 52 71.5


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A cuboid measuring 11cm by 5 cm by 7 cm is placed centrally on top of another cuboid measuring 17 cm by 9 cm by 8 cm.Co-ordinate axes are taken as shown.

a) The point A has co-ordinates (0, 9, 8) and C has co-ordinates (17, 0, 8).

Write down the co-ordinates of B

b) Calculate the size of angle ABC.

(3, 2, 15)Ba) b)3

7

7

BA

��15

2

7

BC

��

. 45 14 49 10BA BC ��

225 4 49 278BC ��

9 49 49 107BA �� 10

cos278 107

93.3


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A triangle ABC has vertices A(2, –1, 3), B(3, 6, 5) and C(6, 6, –2).

a) Find and

b) Calculate the size of angle BAC.

c) Hence find the area of the triangle.

AB��

AC��

1

7

2

AB

��b aa)

4

7

5

AC

��c a

b) 2 2 21 7 2 54AB ��

90AC ��

. 4 49 10 43AB AC ��

43cos 0.6168

54 90 1cos 0.6168 51.9 51.9 BAC =

c) Area of ABC = 1

2sinab C

190 54

2sin 51.9 2

unit27.43


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Higher Maths

Strategies


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The Wave Function

Maths4Scotland The Wave Function Higher

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The Wave Function



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Part of the graph of y = 2 sin x + 5 cos x is shownin the diagram.a) Express y = 2 sin x + 5 cos x in the form k sin (x + a) where k > 0 and 0 a 360b) Find the coordinates of the minimum turning point P.

Hint

Expand ksin(x + a): sin( ) sin cos cos sink x a k x a k x a

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Equate coefficients: cos 2 sin 5k a k a

Square and add2 2 22 5 29k k

Dividing:

Put together: 2sin 5cos 29 sin( 68 )x x x

Minimum when: ( 68 ) 270 202x x

P has coords. (202 , 29)

5

2tan a acute 68a a is in 1st quadrant

(sin and cos are +) 68a


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2

2Hint

Expand ksin(x - a): sin( ) sin cos cos sink x a k x a k x a

Previous Next



Dividing:

Put together: 4 4sin cos 2 sin( ) 2x x x k a

Sketch Graph

a) Write sin x - cos x in the form k sin (x - a) stating the values of k and a where k > 0 and 0 a 2

b) Sketch the graph of sin x - cos x for 0 a 2 showing clearly the graph’s maximum and minimum values and where it cuts the x-axis and the y-axis.

max min2 2

3 7max at min at

4 4x x


tan 1a acute4

a a is in 1st quadrant(sin and cos are +) 4

a

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Hint

Expand kcos(x + a): cos( ) cos cos sin sink x a k x a k x a

Previous Next



Dividing:

Put together: 8cos 6sin 10cos( 37 )x x x

Express in the form where andcos( ) 0 0 360k x a k a 8cos 6sinx x

6


(sin and cos are +) 37a


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Hint

Express as Rcos(x - a): cos( ) cos cos sin sinR x a R x a R x a

Previous Next

Equate coefficients: cos 1 sin 1R a R a

Square and add 2 2 21 1 2R R

Dividing:

Put together: 7

4cos sin 2 cosx x x

Find the maximum value of and the value of x for which it occurs in the interval 0 x 2.

cos sinx x

tan 1a acute4

a a is in 4th quadrant(sin is - and cos is +)

7

4a

Max value: 2 when 7 7

4 40,x x



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Hint

Expand ksin(x - a): sin( ) sin cos cos sink x a k x a k x a

Previous Next



Dividing:

Put together: 2cos 5sin 29 sin 68x x x

5


(sin and cos are both +) 68a

Express in the form2sin 5cosx x sin( ) , 0 360 and 0k x k


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Hint

Max for sine occurs ,2

(...)

Previous Next

Max value of sine function:

Max value of function:

The diagram shows an incomplete graph of

3sin , for 0 23

y x x

Find the coordinates of the maximum stationary point.

5

6x

Sine takes values between 1 and -1

3

Coordinates of max s.p. 5,

63


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Hint

Expand kcos(x - a): cos( ) cos cos sin sink x a k x a k x a

Previous Next



Dividing:

Put together: 2cos 3sin 13 cos 56x x x

3


(sin and cos are both + )56a

( ) 2 cos 3sinf x x x a) Express f (x) in the form where andcos( ) 0 0 360k x k

for( ) 0.5 0 360f x x b) Hence solve algebraically

Solve equation. 13 cos 56 0.5x 0.5

13cos 56x

56 82acute x Cosine +, so 1st & 4th quadrants138 334x or x


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Hint

Use tan A = sin A / cos A

5

2tan x

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Divide

acute 68x

Sine and cosine are both + in original equations

68x

Solve the simultaneous equations

where k > 0 and 0 x 360

sin 5

cos 2

k x

k x

Find acute angle

Determine quadrant(s)

Solution must be in 1st quadrant

State solution


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Hint

Use Rcos(x - a): cos( ) cos cos sin sinR x a R x a R x a

Previous Next

Equate coefficients: cos 3 sin 2R a R a

Square and add 22 22 3 13R R

Dividing:

Put together: 2sin 3cos 13 cos 146x x x

2

3tan a acute 34a a is in 2nd quadrant

(sin + and cos - ) 146a

Solve equation. 13 cos 146 2.5x 2.5

13cos 146x

146 46acute x Cosine +, so 1st & 4th quadrants

or (out of range, so subtract 360°)192 460x x

Solve the equation in the interval 0 x 360. 2sin 3cos 2.5x x

or100 192x x


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Higher Maths

Logarithms & Exponential


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Maths4Scotland Logarithms & Exponential Higher

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Reminder

All the questions on this topic will depend

upon you knowing and being able to use,

some very basic rules and facts.

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When you see this buttonclick for more information


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Three Rules of logs

log log loga a ax y xy

log log loga a a

xx y

y

log logpa ax p x


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Two special logarithms

log 1a a

log 1 0a


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Relationship between log and exponential

log ya x y a x


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Graph of the exponential function


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Graph of the logarithmic function


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Related functions of

( )

( )

( )

( )

( )

( )

y f x a

y f x a

y f x

y f x

y f x a

y f x a

Move graph left a units

Move graph right a units

Reflect in x axis

Reflect in y axis

Move graph up a units

Move graph down a units

( )y f x

Click to show


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Calculator keys

ln = loge

log = 10log


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Calculator keys

ln=log 2.5e 2 . 5 = = 0.916…

log=10log 7.6 7 . 6 = = 0.8808…

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Solving exponential equations

2.4 3.1 xe

Show

log 2.4 log 3.1 xe e eTake loge both sides

log 2.4 log 3.1 log xe e e e

log 2.4 log 3.1 loge e ex e Use log ab = log a + log b

log 2.4 log 3.1e ex

Use log ax = x log a

Use loga a = 1 log 2.4 log 3.1e e x

0.25593... (2dp)0.26


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60 80 keSolving exponential equations

log 60 log 80 ke e eTake loge both sides

log 60 log 80 log ke e e e

log 60 log 80 loge e ek e Use log ab = log a + log b

log 80 log 60e ek

Use log ax = x log a

Use loga a = 1 log 60 log 80e e k

0.2876... (2dp)0.29

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Solving logarithmic equations

3log 0.5y 0.53y Change to exponential form

Change to exponential form

1

21

2

1 13

33

y

(2dp)0.577.... 0.58y

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3log (2 ) 2 log (3 )e ee elog loge eA B C

Simplify

expressing your answer in the form

where A, B and C are whole numbers.

3 2log (2 ) log (3 )e ee e 3 2log 8 log 9e ee e

3

2

8log

9e

e

e

8log

9e

e

log 8 log log 9e e ee 1 log 8 log 9e e

Show


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5 5 5log 2 log 50 log 4 Simplify

5

2 50log

4

5log 25

25log 5 52 log 5 2

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Find x if 4 log 6 2 log 4 1x x

4 2log 6 log 4 1x x 4

2

6log 1

4x

36log x

936

9

41

41 1 log 81 1x

1 81x 81x

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5 5log 3 log 4x Given find algebraically the value of x.

5log 3 4x 5log 12x

5 12x

10 10log 5 log 12x 10 10log 5 log 12x

10

10

log 12

log 5x 1.5439..x (2dp)1.54x

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Find the x co-ordinate of the point where the graph of the curve

with equation 3log ( 2) 1y x intersects the x-axis.

When y = 0 30 log ( 2) 1x

31 log ( 2)x

Exponential form

Re-arrange

13 2x

12 3x Re-arrange1

32x

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The graph illustrates the law

ny kxIf the straight line passes through A(0.5, 0)and B(0, 1).

Find the values of k and n.

5 5log log ny kx

5 5 5log log logy k n x

5logY k nX

Gradient1

0.5 y-intercept 1

5log 1k 5k

(gradient)2n

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Before a forest fire was brought under control, the spread of fire was described by a law of the form

0ktA A e

where

0A is the area covered by the fire when it was first detectedand A is the area covered by the fire t hours later.

If it takes one and a half hours for the area of the forest fire to double, find the value of the constant k.

1.50 02 kA A e 1.52 ke log 2 1.5 loge ek e

log 2 1.5e k log 2

1.5ek (2 dp)0.462.. 0.46k

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The results of an experimentgive rise to the graph shown.

a) Write down the equation of the line in terms of P and Q.

It is given that

logeP p and logeQ q

bp aq stating the values of a and b.

b) Show that p and q satisfy a relationship of the form

log log be ep aq

log log loge e ep a b q

logeP a bQ Gradient 0.6 y-intercept 1.8

0.6 1.8P Q 1.8 6.0o .8 5l g 1e a ea a

0.6b

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The diagram shows part of the graph of

log ( )by x a

.Determine the values of a and b.

...(1)1 log (7 )b a Use (7, 1)

...(2)0 log (3 )b a Use (3, 0)

Hence, from (2) 3 1a 2a

and from (1) 5b 1 log 5b

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The diagram shows a sketch ofpart of the graph of

2log ( )y x

a) State the values of a and b.

b) Sketch the graph of 2log ( 1) 3y x

1a 3b

Graph moves1 unit to the left and 3 units down

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a) i) Sketch the graph of 1, 2xy a a

ii) On the same diagram, sketch the graph of 1, 2xy a a

b) Prove that the graphs intersect at a point

where the x-coordinate is 1

log1a a

11x xa a 11 x xa a 1 1xa a

log 1 log log 1xa a aa a 0 log 1ax a

log 1ax a 1log 1ax a

1

log1ax

a

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Part of the graph of

105 log (2 10)y x is shown in the diagram.

This graph crosses thex-axis at the point A and

the straight line 8y at the point B. Find algebraically the x co-ordinates of A and B.

108 5 log (2 10)x 10

8log (2 10)

5x 101.6 log (2 10)x

1.610 2 10x 1.610 10 2x 14.9x B (14.9, 8)

100 5 log (2 10)x 2 10 1x 4.5x A ( 4.5, 0)

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The diagram is a sketch of part of

the graph of , 1xy a a a) If (1, t) and (u, 1) lie on this curve, write down the values of t and u.

b) Make a copy of this diagram and on it

sketch the graph of 2xy a

c) Find the co-ordinates of the point of intersection of 2xy a with the line 1x

a) t a 0u b)

c) 2 1y a 2y a 21, a

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The diagram shows part of the graph with equation

3xy and the straight line with equation 42y

These graphs intersect at P.

Solve algebraically the equation 3 42x

and hence write down, correct to 3 decimal places, the co-ordinates of P.

10 10log 3 log 42x 10 10log 3 log 42x

10

10

log 42

log 3x 3.40217...x

P (3.402, 42)

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