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The Straight Line
Higher
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Higher Maths
Strategies
The Straight Line
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The Straight Line
Higher
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The Straight Line
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The Straight Line
Higher
Find the equation of the line which passes through the point (-1, 3)
and is perpendicular to the line with equation 4 x  y  1  0
Find gradient of given line:
4 x  y  1  0  y  4 x  1  m  4
m
Find gradient of perpendicular:
Find equation:
1
4
1
y 3

 x  1  4( y  3)  x  1  4 y  12
4 x ( 1)
4 y  x  13  0
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The Straight Line
Higher
Find the equation of the straight line which is parallel to the line with
equation
2 x  3 y  5 and which passes through the point (2, –1).
Find gradient of given line:
2
3
3 y  2 x  5  y   x  5  m  
Gradient of parallel line is same:
Find equation:

2
y ( 1)

3
x2
m

2
3
2
3
 2x  4  3y  3
3y  2x  1
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The Straight Line
Higher
Find the size of the angle a° that the line joining the
points A(0, -1) and B(33, 2) makes with the
positive direction of the x-axis.
Find gradient of the line:
Use m  tan 
m 
2  (1)
3


3 3 0
3 3
1
3
1
tan  
3
Use table of exact values
  tan
1
1
  30
3
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The Straight Line
Higher
A and B are the points (–3, –1) and (5, 5).
Find the equation of
a) the line AB.
b) the perpendicular bisector of AB
Find gradient of the AB:
Find mid-point of AB
m
3
4
Find equation of AB
Gradient of AB (perp):
(1,2)
4 y  3x  5
4
m
3
Use gradient and mid-point to obtain perpendicular bisector AB
3 y  4 x  13
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The Straight Line
Higher
 radians with the
3
y-axis, as shown in the diagram.
The line AB makes an angle of
Find the exact value of the gradient of AB.
Find angle between AB and x-axis:
Use m  tan 
m  tan
  
 
2 3 6
(x and y axes are perpendicular)

6
Use table of exact values
1
m
3
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The Straight Line
Higher
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 gradient of median AM
m2
Find equation of median AM
y  2x  5
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The Straight Line
Higher
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
m
2
Find gradient of PS (perpendicular to QR)
m  2
y  2x  3  0
Find equation of altitude PS
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The Straight Line
Higher
The lines y  2 x  4 and x  y  13 make
angles of a and b with the positive
direction of the x-axis, as shown in the diagram.
a) Find the values of a and b
b) Hence find the acute angle
between the two given lines.
Find gradient of
y  2x  4
m2
Find a°
tan a  2  a  63
Find gradient of
x  y  13
m  1
Find b°
tan b  1  b  135
Find supplement of b
Angle between two lines
 180  135  45
Use angle sum triangle = 180°
72°
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The Straight Line
Higher
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
Find mid-point of BC
Find gradient of q
(-2, 2)
Find gradient of p
m0
y2
(1, 0)
Find gradient of BC
m  2
Find equation of q
Solve p and q simultaneously for intersection
1
2
y  2x  2
m
(0, 2)
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The Straight Line
Higher
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.
Mid-point AB
 7, 2 
Find mid-point AC
(5, 4)
Gradient AC perp.
3
m
2
Point of intersection
(7, 1)
Find radius (intersection to A)
Equation of circle:
Previous
x7
2
m
Find gradient of AC
3
2 y  3x  23
Equ. of perp. bisector AC
Perpendicular bisector AB
This is the centre of circle
r  26
 x  7    y  1  26
2
2
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The Straight Line
Higher
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
Mid-point AB
 4, 2 
Gradient CM (median)
y  3 x  14
Equation of median CM
Gradient BC
Equation of AD
m  3
m2
Gradient of perpendicular AD
m
2y  x  2  0
Solve simultaneously for point of intersection
(6, -4)
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1
2
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The Straight Line
Higher
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.
m2
Gradient AB
Product of gradients
2  
Gradient BC
1
 1
2
1
m
2
Hence AB is perpendicular to BC, so B = 90°
Mid-point BC
 3, 1
Gradient of median AD
Mid-point AC
 2, 3
Gradient of median BE
1
3
4
m
3
m
Solve simultaneously for M, point of intersection
Equation AD
3y  x  6  0
Equation AD
3y  4x 1  0
5

1,



3

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The Straight Line
Higher
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Higher Maths
Graphs & Functions
Strategies
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Graphs & Functions
Higher
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Graphs & Functons
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Graphs & Functions
Higher
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).
y = 2f(-x)
b) On the same diagram, sketch the graph of y = 2f(-x)
a)
y
Reflect across the y axis
y = f(-x)
4
2
-1
b)
3
Now scale by 2 in the y direction
4
x
-3
-6
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Graphs & Functions
Higher
The diagram shows a sketch of part of
the graph of a trigonometric function
whose equation is of the form y  a sin(bx )  c
Determine the values of a, b and c
a is the amplitude:
1
2a
a=4
b is the number of waves in 2
1 in 
2 in 2 
b=2
c is where the wave is centred vertically
c=1
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Functions f ( x) 
Graphs & Functions
Higher
1
and g ( x)  2 x  3 are defined on suitable domains.
x4
a)
Find an expression for h(x) where h(x) = f(g(x)).
b)
Write down any restrictions on the domain of h.
a)
f ( g ( x))  f (2 x  3)
b)
2 x 1  0
 x

1
2x  3  4
 h( x ) 
1
2x 1
1
2
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Graphs & Functions
2
in the form ( x  a)  b
a) Express f ( x)  x  4 x  5
b) On the same diagram sketch
2
c)
i)
the graph of
ii)
the graph of
(2, 9)
y  f ( x)
y=f(x)
y  10  f ( x)
b)
5
Find the range of values of x for
which
10  f ( x)
Higher
is positive
y= 10 - f(x)
(2, 1)
a)
( x  2)2  4  5  ( x  2)2  4  5
(2, -1)
 ( x  2)2  1
-5
c)
Solve:
10  ( x  2)2  1  0
 ( x  2)2  9
 ( x  2)  3
y= -f(x)
 x  1 or 5
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10 - f(x) is positive for -1 < x < 5
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Graphs & Functions
Higher
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 turning
point at (c, d).
Sketch the graph of the derived function f 
m is +
m is +
m is -
f(x)
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Graphs & Functions
Higher
Functions f and g are defined on suitable domains by
f ( x)  sin( x) and g ( x )  2 x
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
a)
f ( g ( x))  f (2 x)  sin 2x
b)
2sin 2x  2sin x
 sin 2 x  sin x  0
 2sin x cos x  sin x  0
 sin x  0
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or
g ( f ( x))  g (sin x)  2sin x
cos x 
1
2
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 sin x(2 cos x  1)  0
x  0, 180, 360
x  60, 300
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Graphs & Functions
Higher
The diagram shows the graphs of two quadratic
functions y  f ( x) and y  g ( x)
Both graphs have a minimum turning point at (3, 2).
Sketch the graph of y  f ( x)
and on the same diagram
sketch the graph of y  g ( x )
y=f(x)
y=g(x)
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Graphs & Functions
Higher

Functions f ( x)  sin x, g ( x)  cos x and h( x)  x  4
are defined on a suitable set of real numbers.
a) Find expressions for
g (h( x))
i)
ii)
f ( h( x ))
b)
f (h( x)) 
1
2
sin x 
1
2
cos x
i)
Show that
ii)
Find a similar expression for g (h( x))
and hence solve the equation f (h( x))  g (h( x))  1 for 0  x  2

a)
f (h( x))  f ( x  )
b)
sin( x  )  sin x cos
4


 sin( x  )
g (h( x))  cos( x  )
4


4
4
 sin

4
cos x
4
Now use exact values
Repeat for ii)
equation reduces to
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2
sin x  1
2
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sin x 
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2
1

2
2
x
 3
4
,
4
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Graphs & Functions
Higher
A sketch of the graph of y = f(x) where f ( x)  x3  6 x 2  9 x 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 g ( x)  f ( x  2)  4
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.
a)
Differentiate
f ( x)  3x 2  12 x  9
when x = 1
y4
t.p. at A is:
for SP, f(x) = 0
(1, 4)
moved 2 units to the left, and 4 units up
b)
Graph is
c)
For 3 real roots, line y = k has to cut graph at 3 points
from the graph, k  4
Previous
x  1 or x  3
t.p.’s are:
(3, 0)  (1, 4)
(1, 4)  (1, 8)
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f ( x)  3  x
a) Find
3
x
g ( x)  ,
and
3
3 x
, x3
p(q( x)) 

x0
find
p( x)  f ( g ( x))  f
b)
Higher
p( x) where p( x)  f ( g ( x))
b) If q ( x) 
a)
Graphs & Functions
p  3 
 3 x 

 3
 
 x

 3

3
1
 3 x 
3
3 x
 9  3(3  x)  3  x


 3 x  3
in its simplest form.
p ( q ( x ))

3
3
x

3 x 3
x
3( x 1)
x

 9
 3

 3 
 3 x
 3 x
3x 3  x

3 x
3

x
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Graphs & Functions
Higher
Part of the graph of y  f ( x) is shown in the diagram.
On separate diagrams sketch the graph of
a) y  f ( x  1)
b) y  2 f ( x)
Indicate on each graph the images of O, A, B, C, and D.
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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Graphs & Functions
Higher
Functions f and g are defined on the set of real numbers
by f ( x)  x  1 and g ( x)  x 2
a) Find formulae for
i)
ii) g ( f ( x))
f ( g ( x))
b) The function h is defined by h( x)  f ( g ( x))  g ( f ( x))
Show that h( x)  2 x 2  2 x and sketch the graph of h.
c) Find the area enclosed between this graph and the x-axis.
a)
f ( g ( x))  f ( x 2 )  x 2  1
b)
h( x)  x 2  1   x  1
c)
2
g ( f ( x))  g ( x  1)   x  1
h( x)  x 2  1  x 2  2 x  1
Graph cuts x axis at 0 and 1
Area
Previous

1
3
unit
Now evaluate

1
0
2
 2 x2  2 x
2 x2  2 x dx
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Graphs & Functions
Higher
The functions f and g are defined on a suitable domain by
f ( x)  x 2  1 and g ( x)  x 2  2
a) Find an expression for f ( g ( x))
b) Factorise f ( g ( x))
a)
f ( g ( x))  f ( x  2)   x  2   1
b)
Difference of 2 squares
2
Simplify
2
2

 x
2
  x
 2  1
2

 2 1
  x 2  3 x 2  1
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Graphs & Functions
Higher
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Sequences
Higher
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Strategies
Sequences
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Sequences
Higher
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Sequences
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Sequences
Higher
A recurrence relation is defined by un 1  pun  q 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  
Put u1 into recurrence relation
15  12 p  q
..... (1)
Put u2 into recurrence relation
16  15 p  q
..... (2)
Solve simultaneously:
Hence
p
Previous
1  3p

p
1
3
substitute into (1)
q  11
1
and q  11
3
State limit condition
Use formula
(2) – (1)
c
L
1 m
-1 < p < 1, so a limit L exists
L 
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11
1
1
3
Limit = 16½
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Sequences
Higher
A man decides to plant a number of fast-growing trees as a boundary between his property and the property of
his neighbour. He has been warned however by the local garden centre, that during any year, the trees
are 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 assurance
that the trees will grow no taller than 2 metres. What is the minimum percentage that the trees
will need to be trimmed each year so as to meet this condition.
un1  0.8un  0.5
Construct a recurrence relation
-1 < 0.8 < 1, so a limit L exists
State limit condition
Use formula
L
c
1 m
L 
0.5
1  0.8
Use formula
again
L
c
1 m
2 
0.5
1 m
Previous
un = height at the start of year
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Limit = 2.5 metres
m = 0.75 Minimum prune = 25%
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Sequences
Higher
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 £300
except 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.
Construct a recurrence relation
un1  1.015un  300
u0 = 2500
Calculate each term in the recurrence relation
1 Mar
1 Apr
1 May
1 Jun
1 Jul
Previous
u0 = 2500.00
u1 = 2237.50
u2 = 1971.06
u3 = 1700.62
u4 = 1426.14
1 Aug
1 Sept
1 Oct
1 Nov
1 Dec
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u5 = 1147.53
u6 = 864.74
u7 = 577.71
u8 = 286.38
Final payment £290.68
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Hint
Maths4Scotland
Sequences
Higher
Two sequences are generated by the recurrence relations un1  aun  10 and vn 1  a 2vn  16
The two sequences approach the same limit as n  .
Determine the value of a and evaluate the limit.
L
Use formula for each sequence
L
Sequence 1
Equate the two limits
Simplify
Solve
10
1 a
10
16

1 a
1  a2
5a  3 a 1  0
Previous
L
Sequence 2
10a 2  16a  6  0
Deduction
c
1 m
16
1  a2
Cross multiply

5a 2  8a  3  0
hence
a 1
3
Since limit exists a  1, so a  5
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or
a
3
5
Limit = 25
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10 1  a 2   16(1  a)
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Hint
Maths4Scotland
Sequences
Higher
Two sequences are defined by the recurrence relations
un 1  0.2un  p, u0  1 and
vn 1  0.6vn  q, v0  1
If both sequences have the same limit, express p in terms of q.
L
Use formula for each sequence
Sequence 1
L
p
1  0.2

p
L
Sequence 2
p
q

0.8
0.4
Equate the two limits
Rearrange
c
1 m
0.6q
0.4

q
1  0.6
Cross multiply
p
0.4 p  0.6q
3q
2
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Hint
Maths4Scotland
Sequences
Higher
Two sequences are defined by these recurrence relations
un1  3un  0.4
with
u0  1
vn1  0.3vn  4
with
v0  1
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, and
ii) the value of that term.
Requirement for a limit
First sequence has no limit since 3 is not between –1 and 1
2nd sequence has a limit since –1 < 0.3 < 1
L
Sequence 2
List terms of
1st sequence
4
1  0.3
u0 = 1
u1 = 2.6
u2 = 7.4
L
4
0.7
u3 = 21.8
u4 = 65
u5 = 194.6
Limit 
40
5
 5
7
7
u6 = 583.4
u7 = 1749.8
Smallest value of n is 8; value of 8th term = 1749.8
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Sequences
Higher
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Differentiation
Higher
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Differentiation
Higher Mathematics
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Maths4Scotland
Differentiate
Differentiation
f ( x)  x 
Straight line form
Differentiate
Higher
2
x2
1
2
f ( x)  x  2 x 2
f ( x)  x
1
2

1
2
 4 x 3
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Hint
Maths4Scotland
Differentiate
Differentiation
Higher
y  2 x3  7 x 2  4 x  4
y  6 x 2  14 x  4
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Hint
Maths4Scotland
Differentiate
Differentiation

y  2 sin x 

6
Higher





Chain Rule
dy

 2 cos x  1
6
dx
Simplify
dy

 2 cos x 
6
dx
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Hint
Maths4Scotland
Differentiate
multiply out
Differentiate
Differentiation
Higher
3
A  a (8  a)
4
3
A  6a  a 2
4
A 6
3
a
2
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Hint
Maths4Scotland
Differentiate
Chain Rule
Simplify
Differentiation
Higher
1
3 2
f ( x)  (8  x )

1
2
f ( x)  (8  x3 )  (2 x 2 )
1
2
2
3

f ( x)   x (8  x )

1
2
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Hint
Maths4Scotland
Differentiate
Differentiation
Higher
16
y  x
, x0
x

1
2
Straight line form
y  x  16 x
Differentiate
3

dy
 1  8x 2
dx
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Hint
Maths4Scotland
Differentiate
Differentiation
Higher
3 3  2 16 
A( x) 
x  
2 
x
Multiply out
A( x) 
3 3 2 3 3 16
x 
2
2 x
3 3 2
x  24 3x 1
2
Straight line form
A( x) 
Differentiate
A( x)  3 3x  24 3x 2
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Hint
Maths4Scotland
Differentiate
Differentiation
f ( x)   5 x  4 
Chain Rule
f ( x) 
Simplify
f ( x) 
1
2
1
2
5x  4
5
2
Higher

5x  4
1
2

5
1
2
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Hint
Maths4Scotland
Differentiate
Differentiation
Higher
16 2
A( x)  240 x  x
3
32
A( x)  240 
x
3
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Hint
Maths4Scotland
Differentiate
Differentiation
Higher
f ( x)  3x 2 (2 x  1)
Multiply out
f ( x)  6 x 3  3 x 2
Differentiate
f ( x)  18 x 2  6 x
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Hint
Maths4Scotland
Differentiate
Chain Rule
Simplify
Differentiation
Higher
f ( x)  cos2 x  sin 2 x
f ( x)  2 cos x sin x  2 sin x cos x
f ( x)  4 cos x sin x
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Maths4Scotland
Quadratic Theory
Higher
Higher Maths
Quadratic Theory
Strategies
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Quadratic Theory
Higher
The following questions are on
Quadratic Theory
Non-calculator questions will be indicated
You will need a pencil, paper, ruler and rubber.
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Maths4Scotland
Quadratic Theory
Higher
Show that the line with equation y  2 x  1
does not intersect the parabola
with equation
y  x 2  3x  4
Put two equations equal
Use discriminant
Show discriminant < 0
No real roots
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Hint
Maths4Scotland
a) Write
Quadratic Theory
 x  a
f ( x)  x  6 x  11 in the form
2
b) Hence or otherwise sketch the graph of
a)
f ( x)  ( x  3)2  2
b)
This is graph of
y  x2
minimum t.p. at (-3, 2)
Higher
2
b
y  f ( x)
moved 3 places to left and 2 units up.
y-intercept at (0, 11)
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Maths4Scotland
Quadratic Theory
Show that the equation
Higher
(1  2k ) x 2  5kx  2k  0
has real roots for all integer values of k
Use discriminant
a  (1  2k )
b  5k
c  2k
b2  4ac  25k 2  4  (1  2k )   2k 
 9k 2  8k
 25k 2  8k  16k 2
Consider when this is greater than or equal to zero
Sketch graph
cuts x axis at
k 0
and
k 
Hence equation has real roots for all integer k
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8
9
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Maths4Scotland
Quadratic Theory
Higher
The diagram shows a sketch of a parabola
passing through (–1, 0), (0, p) and (p, 0).
a) Show that the equation of the parabola is
y  p  ( p  1) x  x 2
b) For what value of p will the line y  x  p
be a tangent to this curve?
a)
y  k ( x  1)( x  p) Use point (0, p) to find k
p   pk
k  1
y  ( x  1)( x  p)
 y  p   p 1 x  x2
 y   x 2  px  x  p
b)
Simultaneous equations
0   p  2 x  x2
p  k (0  1)(0  p)
x  p  p   p  1 x  x 2
Discriminant = 0 for tangency
p  2 Back to
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Maths4Scotland
Given
Quadratic Theory
f ( x)  x  2 x  8 , express f ( x ) in the form
2
Higher
 x  a
2
b
f ( x)  ( x  1)2  10
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Maths4Scotland
Quadratic Theory
Higher
For what value of k does the equation x 2  5x  (k  6)  0 have equal roots?
Discriminant
a  1 b  5 c  k  6
b2  4ac  25  4(k  6)
For equal roots
discriminant = 0
0  25  4k  24
4k  1
1
k
4
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Quadratic Theory
Higher
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Integration
Higher
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Integration
Higher Mathematics
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Integrate
x
Integration
2
Higher
 4 x  3 dx
3
2
x 4x

 3x  c
3
2
Integrate term by term
1 3
2
x  2 x  3x  c
3
simplify
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Hint
Maths4Scotland
Find

Integration
Higher
3cos x dx
3sin x  c
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Hint
Maths4Scotland
Integrate
Integration
Higher
 (3x  1)( x  5) dx
2
3
x
  14 x  5 dx
Multiply out brackets
3x3 14 x 2

 5x  c
3
2
Integrate term by term
x  7 x  5x  c
3
simplify
2
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Hint
Maths4Scotland
Find

Integration
Higher
2sin  d
2cos  c
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Maths4Scotland
Integrate

Integration
Higher
(5  3 x) 2 dx
(5  3x)
c
3 3
3
Standard Integral
(from Chain Rule)
1
3
(5  3 x)
9
c
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Maths4Scotland
Integration

Find p, given
Higher
p
x dx  42
1

p
x
1
2
dx  42
1


2
3
p3
p3

2
3
 2
 42
 64
p
2 
  x   42 
3

1
3
2
 p3  8
p3
2
3
3
2
p 
2
(1)
3
3
2
 2  126

p 2
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 42
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Maths4Scotland
Evaluate
Integration

2
1
dx
2
x
Straight line form
Higher

x 2 dx
1
   x 
1
  21    11 
 1
    1
 2
1

2
1 2

2
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Maths4Scotland
Find

Integration
1
Higher
Use standard Integral
(from chain rule)
(2 x  3) dx
6
0
1
 (2  3)7   (0  3)7 
 


 14   14 
 (2 x  3) 
 

7

2

0
7
 57   37 
   
 14   14 
 5580.36  156.21
 5424
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(4sf)
Hint
Maths4Scotland
Find

Integration
Higher
1
3sin x  cos x dx
2
Integrate term by term
1
3cos x  sin x  c
2
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
Integrate
Integration
2
x  3 dx
x
Straight line form

2
3
3
2
Higher


1
2
x  2 x dx
2
2x
x 
c
2
3

2
3
3
2
x  x 2  c
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Integrate

Integration
1
x 
dx
x
3

Straight line form
4
1
2
x
x

 1 c
4
2
Back
Higher


x x
3
1
4

1
2
dx
1
2
x4  2x  c
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
Integrate

Integration
x
x
3
1
2

Higher
x  5x
dx
x
3
5x
x
1
2
Straight line form

dx

2
7
x
7
2

x
5
2
10

x
3
1
2
 5x dx
3
2
c
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
Integrate


1
2
Integration
Higher
3
2
4x  x
dx
2 x
1
2
2x  x dx
Split into separate fractions


 2x
4x
4
3
x
3
2
1
2

1

4
x
3
2
2x
1
2
x c
2
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Find

2
Integration
 2 x  1
3
Higher
Use standard Integral
(from chain rule)
dx
1
  2 x  1
 
 4  2
4
 54   34 
   
8 8
  4  14    2  14 



 4 2   4 2 

 

2


1
 68
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Find

Integration
Higher


sin 2 x  cos  3 x   dx
4



1
1
 cos 2 x  sin  3 x 
2
3
4


c

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Find

Integration
Higher
2
sin t dt
7
2
 cos t  c
7
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Integration

Integrate
1
0
dx
 3x  1

Straight line form
1
2
1
 3x  1

1
2
dx
0
2
 
3
1
 3x  1 
0
2
 2 
 
4    1
3
 3 
Back
Higher
2
 
3
  3 x  1
  1

3
 2
 2
 3  1   
 3
4 2
   
3 3
1
2
1



0
 0  1 

2

3
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Integration
Higher
1
2
a  2(4  t ) , 0  t  4
Given the acceleration a is:
If it starts at rest, find an expression for the velocity v where
dv
 2(4  t )
dt
1
2
4
 v
3

4
 0
3
 4
Back
 v
4t
3
 c
3
c
2(4  t )
3
2
3
  1
2
c
Starts at rest, so
v = 0, when t = 0
32
 0  c
3
32
 c
3
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a
dv
dt
3
4
 v   (4  t ) 2  c
3
4
 0
3
 4
3
2
4
3
3
 v   (4  t ) 
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c
Hint
32
3
Maths4Scotland
Integration
dy
 3sin(2 x)
dx
A curve for which

3
3
2

5
,
12
3

3
2
3   cos(2 

3
5
)c
12
3
3

c
2 2 
3
 c
4
3
2

3
y   cos(2 x) 
3 3
c
4
3
4
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3
3
2
5
cos(  )  c
6
4 3 3 3


c
4
4

passes through the point
5
,
12
y   cos(2 x)  c
Find y in terms of x.
Use the point
Higher
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
Maths4Scotland
Integrate

Integration
2
2
x

2
x

  2 
x
2
dx, x  0

Multiply out brackets
Split into
separate fractions


Higher

x4 4
 2 dx
2
x x
x3 4 x 1


c
3
1
x4  4
dx
2
x



1
3
x 2  4 x 2 dx
4
x  c
x
3
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Hint
Maths4Scotland
Integration
f ( x)  sin(3x)
y  f ( x)
If
f ( x)  
1


7
6
c
passes through the point


9

, 1
express y in terms of x.
1
cos(3x)  c
3
 
1
 cos  3 
3
9
Higher

c

Use the point
 1 

1
cos 
3
3

1
3
y   cos(3 x) 



c

9

, 1
 1 
1 1

3 2
7
6
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c
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Hint
Maths4Scotland
Integrate


Integration
1
dx
2
(7  3 x)
Higher
Straight line form
(7  3 x) 1

c
 1   3 
2
(7  3 x) dx

1
(7  3x) 1  c
3
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Hint
Maths4Scotland
The graph of
If
y  g ( x)
dy
1 1
3
x  2
dx
x 4
dy
1
3
2
x x 
dx
4
Integration
passes through the point (1, 2).
express y in terms of x.
x 4 x 1 1
 y 
 xc
4 1 4
1

2
4
x 1 1
y    xc
4 x 4
Evaluate c
Higher
simplify
4
1 1

 1  c
4
1 4
Use the point
1 1
Hint
 y  x   x3
x Back4 to
c3
1
4
4
start
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Maths4Scotland

Integrate


x2
x
x
3
2
3
2
3
2

Integration
x2  5
dx
x x

5
x
5x

1

2
3
2
1
2
dx
Higher

Straight line form



c
1
2
x  5x
2
3
3
2

3
2
x  10 x

x2  5
x
3
2
dx

1
2
c
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dx
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Hint
Maths4Scotland
A curve for which
Integration
dy
 6 x2  2 x
dx
Higher
passes through the point (–1, 2).
Express y in terms of x.
6 x3 2 x 2
y

c
3
2
Use the point
 y  2 x3  x 2  c
 2  2(1)3  (1)2  c
 c5
Hint
 y  2 x3  x 2  5
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Maths4Scotland
Evaluate

2
1
Integration
2
 2 1
 x   dx
x



2

4
1
2
 

1
5


2

x2  x 1

2
dx
x 4  2 x  x 2 dx
1
1
  x5  x 2  x 1 
5
1
32
5

2
1
Cannot use standard integral
So multiply out

Higher
64
10



40
10

1
5
25  2 2 
20
10

2
10
1
2
 


1 5
1
5
82
10
 12 
1
1

 8 15
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Hint
Maths4Scotland
Evaluate

Integration
Higher
x dx
Straight line form

4
2 
  x 
3

1

2

3
1
2
dx
1
1
3
2
x
4
4
3  2
4 
 3
 

4
 2  x 3 
3
1
3
1 

 

 16   2 
  
 3   3
14

3
Hint

4
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2
3
Maths4Scotland
Evaluate

Integration
0
(2 x  3)
2
Higher
Use standard Integral
(from chain rule)
dx
3
0
 (2 x  3) 
 

 3  2  3
3
 27   27 
  

 6   6 
 (2(0)  3)3   (2(3)  3)3 
 


6
6

 

27 27


6
6
54

6

9
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Hint
Maths4Scotland
Integration
y  f ( x)
The curve
f ( x)  cos 2 x
 f ( x) 
 1
 1
passes through the point
1
sin 2 x
2
c
 
 ,1 
 12 
Find f(x)
1

sin 2 
2
12
1 1

2 2
Higher
c
use the given point
 1
 c
 c
3
4
1

sin
2
6
 
 ,1 
 12 
 c sin   1
6
2
1
2
 f ( x)  sin 2 x 
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3
4
Hint
Maths4Scotland
Integrate

Integration
Higher
(6 x 2  x  cos x) dx
6 x3 x 2

  sin x  c
3
2
Integrate term by term
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Hint
Maths4Scotland
Integrate

Integration
Higher
3x  4 x dx
3
3x 4 4 x 2


c
4
2
Integrate term by term

3
4
x  2x  c
4
2
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Hint
Maths4Scotland
Integration
1
0
Evaluate

1  3x 



 3 3
 2
2
 
9


1  3x
3
2

Back
1
0 1  3x 

dx
1
2
dx
1



 0
1
3
2

  1  3x  2 
9
0
 2
1  3(1)   
 9
16 2

9
9
Higher
3



2
 
9

1  3(0) 

14
9
2
 
9
3

1
 


 2
4  
 9
3
 
Quit

1 

3
5
9
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1

1  3x 
0
3
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Hint
Maths4Scotland
Compound Angles
Higher
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Compound Angles
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Compound Angles
Higher
The following questions are on
Compound Angles
Non-calculator questions will be indicated
You will need a pencil, paper, ruler and rubber.
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Compound Angles
Higher
This presentation is split into two parts
Using Compound angle formula for
Exact values
Solving equations
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Compound Angles
Higher
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
80
Pythagoras
4
p
8
8
sin p 
80
4
8
 2


80
80
cos p 
Write down values for cos p and sin p
Expand sin (2p)
sin 2 p  2sin p cos p
Expand cos (2p)
cos 2 p  cos p  sin p 
Use m = tan (2p)
tan 2 p 
Previous
2
2
sin 2 p
cos 2 p
Quit

   
8
80
2

4
80
4 3
4


5 5
3
Quit
2

4
80
64
4

80
5
64  16
3

80
5
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Hint
Maths4Scotland
Compound Angles
Higher
In triangle ABC show that the exact value of
sin(a  b) is
Use Pythagoras
Substitute values
Simplify
10
2
AC  2 CB  10
Write down values for
sin a, cos a, sin b, cos b
Expand sin (a + b)
2
5
sin a 
1
2
cos a 
1
2
sin b 
1
10
cos b 
3
10
sin(a  b)  sin a cos b  cos a sin b
sin(a  b) 
sin(a  b) 
3
20

1
20
1
3

2
10


4
20
1
1

2
10

4
4
2


45
2 5
5
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Hint
Maths4Scotland
Compound Angles
Using triangle PQR, as shown, find the
exact value of cos 2x
11
PR  11
Use Pythagoras
Write down values for
cos x and sin x
2
cos x 
11
7
sin x 
11
Expand cos 2x
cos 2 x  cos 2 x  sin 2 x
Substitute values
cos 2x 
Simplify
Previous
Higher
cos 2 x 
  
2
11
4
7

11
11
Quit
2
7
11
 
3
11
Quit
2
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Hint
Maths4Scotland
Compound Angles
Higher
On the co-ordinate diagram shown, A is the point (6, 8) and
B is the point (12, -5). Angle AOC = p and angle COB = q
Find the exact value of sin (p + q).
Mark up triangles
Use Pythagoras
OA  10
Write down values for
sin p, cos p, sin q, cos q
10
6
12
OB  13
sin p 
8
,
10
cos p 
6
,
10
sin q 
5
,
13
sin ( p  q)  sin p cos q  cos p sin q
Substitute values
sin ( p  q) 
Previous
sin ( p  q) 
96
130

30
130
Quit
5
13
Expand sin (p + q)
Simplify
8
8 12

10 13


12
13
6
5

10 13
126
130
Quit
cos q 

63
65
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Hint
Maths4Scotland
Compound Angles
A and B are acute angles such that tan A 
Find the exact value of
a) sin 2A
b) cos 2A
Draw triangles
c)
and tan B 
sin A 
sin 2 A  2sin A cos A
Expand cos 2A
cos 2 A  cos A  sin A
Substitute
Previous
.
5
13
3
A
5
B
4
12
Hypotenuses are 5 and 13 respectively
Expand sin 2A
2
Expand sin (2A + B)
5
12
sin(2 A  B )
Use Pythagoras
Write down sin A, cos A, sin B, cos B
3
4
Higher
2
3
,
5
cos A 
4
,
5
sin B 
3
5
sin 2 A  2  
cos 2A 
2
4
5
 4  3
   
 5 5
5
,
13
cos B 

24
25
2
16 9

25 25

12
13

7
25
sin  2 A  B   sin 2 A cos B  cos 2 A sin B
sin  2 A  B  
24 12
7
5
323




25 13
25 13 325
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Hint
Maths4Scotland
Compound Angles
If x° is an acute angle such that tan x 
4
3
Higher
5
4 3 3
sin(
x

30)

is
show that the exact value of
10
4
x
3
Draw triangle
Use Pythagoras
Write down sin x and cos x
Expand sin (x + 30)
Hypotenuse is 5
sin x 
4
,
5
cos x 
3
5
sin( x  30)  sin x cos 30  cos x sin 30
Substitute
sin( x  30) 
4
3
3 1

 
5 2
5 2
Simplify
sin( x  30) 
4 3
3

10
10

4 3 3
10
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Table of exact values
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Hint
Maths4Scotland
Compound Angles
Higher
The diagram shows two right angled triangles
ABD and BCD with AB = 7 cm, BC = 4 cm and CD = 3 cm.
Angle DBC = x° and angle ABD is y°.
20  6 6
Show that the exact value of cos( x  y ) is
35
24
5
BD  5, AD  24
Use Pythagoras
Write down
sin x, cos x, sin y, cos y.
Expand cos (x + y)
sin x 
3
,
5
cos x 
4
,
5
sin y 
24
,
7
cos y 
5
7
cos( x  y )  cos x cos y  sin x sin y
4 5
3
24
  
5 7
5
7
Substitute
cos( x  y ) 
Simplify
20  3 4  6
20  6 6
20 3 24


cos( x  y ) 

35
35 Back to
35
35
start
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Hint
Maths4Scotland
Compound Angles
Higher
The framework of a child’s swing has dimensions
as shown in the diagram. Find the exact value of sin x°
Draw triangle
h 5
Use Pythagoras
Draw in perpendicular
3
Use fact that sin x = sin ( ½ x + ½ x)

sin 
  sin
Write down sin ½ x and cos ½ x
Substitute
Simplify
sin

x x

2 2
sin x 
sin
x
2
x x

2 2
Expand sin ( ½ x + ½ x)

2
3
2 

2
3
, cos

x
x
cos
2
2
Table of exact values
x
2

2
5
3
x
2
 sin cos
3
4
x

2
x
2
2sin cos
x
2
5
3
4 5
9
Previous
x
x
2 h5
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Hint
Maths4Scotland
Given that
tan  
Compound Angles
11

, 0  
3
2
find the exact value of
sin 2
Draw triangle
Use Pythagoras
Write down values for
cos a and sin a
20
11
a
hypotenuse
3
cos a 
20

20
3
11
sin a 
20
Expand sin 2a
sin 2a  2 sin a cos a
Substitute values
sin 2a  2 
Simplify
6 11
sin 2a 
20
Previous
Higher
11
3

20
20
Quit

Quit
3 11
10
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Hint
Maths4Scotland
Compound Angles
Higher
Find algebraically the exact value of
sin    sin   120   cos(  150)
Expand sin ( +120)
sin   120  sin  cos120  cos  sin120
Expand cos ( +150)
cos   150  cos cos150  sin  sin150
Use table of exact values
Simplify
cos 150   cos 30  
sin 120 
sin 150 
sin 60 
sin   sin  .
Combine and substitute
1
2
3
2
cos 120   cos 60  
sin   sin  
1
2
3
cos 
2
   cos .   cos .   sin  . 


1
2
3
2
3
cos 
2
Table of exact values
Quit
Quit

3
2
1
2
 sin 
1
2
0
Previous
sin 30 
3
2
1
2
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Hint
Maths4Scotland
If cos  
a)
sin 2
Compound Angles
4

, 0  
5
2
b)
Draw triangle
Find sin 4
Previous
5
3

Use Pythagoras
4
Opposite side = 3
4
cos  
5
3
sin  
5
3 4
24
 2  
5 5
25
sin 2  2 sin  cos 
Expand sin 4 (4 = 2 + 2)
Expand cos 2
find the exact value of
sin 4
Write down values for
cos  and sin 
Expand sin 2
Higher
sin 4  2 sin 2 cos 2
cos 2  cos   sin 
2
24 7
sin 4  2  
25 25
Quit
2

336

625
Quit
16 9
7


25 25
25
Back to
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Hint
Maths4Scotland
Compound Angles
For acute angles P and Q
sin P 
12
and
13
sin Q 
Show that the exact value of sin ( P Q) 
Draw triangles
Use Pythagoras
Write down sin P, cos P, sin Q, cos Q
Expand sin (P + Q)
3
5
13
63
65
5
12
P
3
Q
5
4
Adjacent sides are 5 and 4 respectively
sin P 
12
,
13
cos P 
5
,
13
sin Q 
3
,
5
cos Q 
4
5
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Hint
sin  P  Q   sin P cos Q  cos P sin Q
Substitute
sin  P  Q  
12 4
5 3
 

13 5
13 5
Simplify
sin  P  Q  
48
15

65
65
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Higher
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
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63
65
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Maths4Scotland
Compound Angles
Higher
You have completed all 12 questions in this section
Using Compound angle formula for
Solving Equations
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Compound Angles
Higher
Using Compound angle formula for
Solving Equations
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Compound Angles
Higher
Solve the equation 3cos(2 x)  10cos( x)  1  0 for 0 ≤ x ≤  correct to 2 decimal places
Replace cos 2x with
Substitute
Simplify
cos 2 x  2 cos 2 x  1
3  2 cos x  1  10 cos x  1  0
2
Determine quadrants
6 cos x  10 cos x  4  0
2
S
A
T
C
3cos 2 x  5cos x  2  0
Factorise
Hence
3cos x 1 cos x  2  0
cos x 
1
3
cos x  2 Discard
Find acute x
Previous
acute
x  1.23 rad
Quit
x  1.23
or
x  1.23
rads
x  5.05
rads
2  1.23
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rads
Hint
Maths4Scotland
Compound Angles
Higher
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 graph
at points A and B. Find the co-ordinates of B.
Equation
y  2 cos 2 x
Determine quadrants
2cos 2 x   3
Solve simultaneously
cos 2x  
Rearrange
Check range
0 x 
Find acute 2x
Deduce 2x
acute
2x 
2x 

Table of exact values
T
C
x 
5
7
or
12
12
B

6
Previous
A
3
2
 0  2 x  2
6 

or
6
6
S
6 

rads
6
6
Quit
Quit
is
B
7
12
, 3

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Hint
Maths4Scotland
Compound Angles
Higher
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))
Determine x
b) Solve 2 f(g(x)) = g(f(x)) for 0  x  360°
1st
expression
2nd expression
Form equation
Replace sin 2x
f ( g ( x))  f (2 x)  sin 2 x
cos x 
g ( f ( x))  g (sin x)  2sin x
2sin 2x  2sin x  sin 2 x  sin x
2sin x cos x  sin x
Common factor
1
2

acute
x  60
S
A
T
C
Determine
quadrants
x  60, 300
2sin x cos x  sin x  0
Rearrange
Hence
sin x  0  x  0, 360
x  0, 60, 300, 360
sin x  2cos x 1  0
sin x  0
or
2 cos x  1  0  cos x 
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1
2
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Hint
Maths4Scotland
Compound Angles
Functions f ( x)  sin x, g ( x)  cos x
a)
Find expressions for
b)
i)
2nd
expression
expression
Simplify
h( x )  x 

4
are defined on a suitable set of real numbers
i) f(h(x)) ii) g(h(x))
1
1
sin x 
cos x ii) Find a similar expression for g(h(x))
2
2
Hence solve the equation f (h( x))  g (h( x))  1 for 0  x  2
Show that
iii)
1st
and
Higher
1st
f (h( x)) 
   
g (h( x))  g  x    cos  x  

f (h( x))  f x 
 sin x 
4

4


4
4
Rearrange:


4
4
f (h( x))  sin x cos  cos x sin
expr.
1
1
sin x 
2
2
Use exact values
f (h( x)) 
Similarly for 2nd expr.
g (h( x))  cos x cos  sin x sin
g (h( x)) 
Form Eqn.
1
2
acute x
cos x


4
4
cos x 
1
sin x
2
f (h( x))  g (h( x))  1
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2
2
Simplifies to
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acute
Determine
quadrants
x
 3
4
,
sin x 
4
sin x  1
2

2
x
2
1

2 2
2

4
S
A
T
C
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Hint
Maths4Scotland
a)
b)
Compound Angles
Higher
Solve the equation sin 2x - cos x = 0 in the interval 0  x  180°
The diagram shows parts of two trigonometric graphs,
y = sin 2x and y = cos x. Use your solutions in (a) to
write down the co-ordinates of the point P.
Replace sin 2x
2sin x cos x  cos x  0
Common factor
cos x  2sin x 1  0
Hence
cos x  0
Determine x
or
Solutions for where graphs cross
x  30, 90, 150
2sin x  1  0  sin x 
1
2
cos x  0  x  90, ( 270 out of range)
sin x 
1
2

acute
x  30
S
A
Determine quadrants
for sin x
Previous
Coords, P
Table of exact values
C
Quit
y  cos150
Find y value
y
x  30, 150
T
x  150
By inspection (P)
Quit

P 150, 
3
2

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3
2
Hint
Maths4Scotland
Solve the equation
Compound Angles
3cos(2 x)  cos( x)  1
for 0 ≤ x ≤ 360°
cos 2 x  2 cos 2 x  1
Replace cos 2x with
Determine quadrants
3  2 cos x  1  cos x  1
2
Substitute
Simplify
6 cos 2 x  cos x  2  0
Factorise
3cos x  2 2cos x 1  0
cos x  
Hence
Find acute x
acute
2
3
x  48
cos x 
acute
Higher
1
2
x  60
cos x  
Table of exact values
Quit
Quit
cos x 
1
2
acute
x  48
acute
x  60
S
A
S
A
T
C
x  132
x  228
Solutions are: x= 60°, 132°, 228° and 300°
Previous
2
3
T
C
x  60
x  300
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Hint
Maths4Scotland
Compound Angles



Solve the equation 2sin 2 x  6  1
Rearrange
sin
Find acute x
Note range


2x 
6
acute

2x 


6
for 0 ≤ x ≤ 2



Determine quadrants
2x 

6


and for range
6

0  x  2  0  2 x  4
S
0  2 x  2
for range
1

2

Higher
2x 

6




6
2x 

6


5
6


17
6
2  2 x  4

13
6
2x 

6
A
Solutions are:
T
x
C

6
,

2
,
7 3
,
6
2
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Table of exact values
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Hint
Maths4Scotland
Compound Angles
Higher
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 
Replace cos 2 with
cos 2  2 cos 2   1
Rearrange
2 cos 2   8cos   8  0
Divide by 2
cos 2   4 cos   4  0
Factorise
Deduction
Try to solve:
 cos  2  0
cos  2
No solution
Hence there are no real solutions for 
 cos  2 cos  2  0
Equal roots for cos 
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Hint
Maths4Scotland
Compound Angles
Higher
Solve algebraically, the equation sin 2x + sin x = 0, 0  x  360
Replace sin 2x
2sin x cos x  sin x  0
Common factor
sin x  2cos x  1  0
Hence
S
A
T
C
sin x  0
or
Determine x
Determine quadrants
for cos x
2cos x  1  0

cos x  
1
2
x  120, 240
sin x  0  x  0, 360
1
2
cos x   
acute
x  60
x = 0°,
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Table of exact values
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120°, 240°, 360°
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Hint
Maths4Scotland
Compound Angles
Higher
Find the exact solutions of 4sin2 x = 1, 0  x  2
Rearrange
sin 2 x 
1
4
Take square roots
sin x  
1
2
Find acute x
acute
x 

Determine quadrants for sin x
S
6
+ and – from the square root requires all 4 quadrants
A

T
C
5
7 11
x  ,
,
,
6
6
6
6
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start
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Table of exact values
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Hint
Maths4Scotland
Compound Angles
Solve the equation
cos 2x  cos x  0
Replace cos 2x with
cos 2 x  2 cos 2 x  1
for 0 ≤ x ≤ 360°
Determine quadrants
cos x 
2 cos 2 x  1  cos x  0
Substitute
Simplify
2 cos 2 x  cos x  1  0
Factorise
 2cos x 1 cos x  1  0
cos x 
Hence
Find acute x
acute
1
2
x  60
cos x  1
x  180
Table of exact values
Quit
1
2
acute
x  60
S
A
T
C
x  60
x  300
Solutions are: x= 60°, 180° and 300°
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Hint
Maths4Scotland
Compound Angles
cos2x  5cos x  2  0
Solve algebraically, the equation
Replace cos 2x with
Substitute
cos 2 x  2 cos 2 x  1
Higher
for 0 ≤ x ≤ 360°
Determine quadrants
2 cos x  1  5cos x  2  0
2
cos x 
acute
Simplify
2 cos 2 x  5cos x  3  0
Factorise
 2cos x 1 cos x  3  0
Hence
Find acute x
cos x 
acute
1
2
x  60
S
cos x  3
Discard above
Table of exact values
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Quit
x  60
A
T
C
x  60
x  300
Solutions are: x= 60° and 300°
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1
2
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Hint
Maths4Scotland
Compound Angles
Higher
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Maths4Scotland
The Circle
Higher
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Higher Maths
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The Circle
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The Circle
Higher
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The Circle
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Maths4Scotland
The Circle
Higher
Find the equation of the circle with centre
(–3, 4) and passing through the origin.
Find radius (distance formula):
You know the centre:
Write down equation:
r 5
(3, 4)
( x  3)2  ( y  4)2  25
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Hint
Maths4Scotland
The Circle
Explain why the equation x 2
does not represent a circle.
Consider the 2 conditions
Higher
 y2  2x  3 y  5  0
1. Coefficients of x2 and y2 must be the same.
2. Radius must be > 0
g  1,
Calculate g and f:
Evaluate
Deduction:
g  f c
2
2
f 
(1) 
2
g 2  f 2  c  0 so
i.e. g 2  f 2  c  0
3
2
  5
3

2
2
1
4
 1  2 5  0
g 2  f 2  c not real
Equation does not represent a circle
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Hint
Maths4Scotland
The Circle
Higher
Find the equation of the circle which has P(–2, –1) and Q(4, 5)
as the end points of a diameter.
Q(4, 5)
C
Make a sketch
P(-2, -1)
(1, 2)
Calculate mid-point for centre:
Calculate radius CQ:
Write down equation;
r  18
 x  1   y  2   18
2
2
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Hint
Maths4Scotland
The Circle
Higher
Find the equation of the tangent at the point (3, 4) on the circle
x 2  y 2  2 x  4 y  15  0
Calculate centre of circle:
P(3, 4)
(1, 2)
Make a sketch
O(-1, 2)
Calculate gradient of OP (radius to tangent)
Gradient of tangent:
m  2
Equation of tangent:
y  2 x  10
m
1
2
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Hint
Maths4Scotland
The Circle
Higher
The point P(2, 3) lies on the circle ( x  1)2  ( y  1)2  13
Find the equation of the tangent at P.
Find centre of circle:
P(2, 3)
(1, 1)
Make a sketch
O(-1, 1)
Calculate gradient of radius to tangent
Gradient of tangent:
3
m
2
Equation of tangent:
2 y  3 x  12
m
2
3
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Hint
Maths4Scotland
The Circle
Higher
O, A and B are the centres of the three circles shown in
the diagram. The two outer circles are congruent, each
touches the smallest circle. Circle centre A has equation
 x  12 
2
  y  5   25
2
The three centres lie on a parabola whose axis of symmetry
is 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 y  px( x  q )
A is centre of small circle
A(12,  5)
Find OA (Distance formula)
Use symmetry, find B
B(24, 0)
Find radius of circle A from eqn.
Find radius of circle B
13  5  8
Eqn. of B
Points O, A, B lie on parabola
– subst. A and B in turn
0  24 p (24  q)
5  12 p (12  q)
Solve:
Find p and q.
13
5
( x  24) 2  y 2  64
p
5
,
144
q  24
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Hint
Maths4Scotland
The Circle
Higher
Circle P has equation x 2  y 2  8x  10 y  9  0 Circle Q has centre (–2, –1) and radius 22.
a) i) Show that the radius of circle P is 42
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 a  b 3
Find centre of circle P:
Find radius of circle :P:
(4, 5)
Find distance between centres
72  6 2
Gradient of radius of Q to tangent:
Equation of tangent:
m  1
Deduction:
= sum of radii, so circles touch
Gradient tangent at Q:
m 1
y  x5
2
2
Solve eqns. simultaneously x  y  8 x  10 y  9  0
y  x5
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42  52  9  32  4 2
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Soln:
22 3
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Hint
Maths4Scotland
The Circle
For what range of values of k does the equation
represent a circle ?
g  2k ,
Determine g, f and c:
State condition
g  f c  0
2
2
5k 2  k  2  0
Simplify
f  k,
Higher
x 2  y 2  4kx  2ky  k  2  0
c  k  2
Put in values
(2k )2  k 2  (k  2)  0
Need to see the position
of the parabola
Complete the square

1
5



5 k2  k  2

5k 

1
5 k 
10

1
10
2
2
Previous
1 
2
100 

195
100
Minimum value is
195
1
when k  
100
10
This is positive, so graph is:
Expression is positive for all k:
So equation is a circle for all values of k.
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Hint
Maths4Scotland
The Circle
For what range of values of c does the equation
represent a circle ?
Determine g, f and c:
g  3,
State condition
g2  f 2  c  0
Simplify
94c  0
Re-arrange:
Higher
x2  y 2  6 x  4 y  c  0
f  2,
c?
Put in values
32  (2)2  c  0
c  13
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Hint
Maths4Scotland
The Circle
Higher
The circle shown has equation ( x  3)2  ( y  2)2  25
Find the equation of the tangent at the point (6, 2).
Calculate centre of circle:
(3,  2)
Calculate gradient of radius (to tangent)
4
m
3
3
4
Gradient of tangent:
m
Equation of tangent:
4 y  3 x  26
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Hint
Maths4Scotland
The Circle
Higher
When newspapers were printed by lithograph, the newsprint had
to 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
( x  12)2  ( y  15)2  25 and ( x  24)2  ( y  12)2  100
Find the equation of the central circle.
Find centre and radius of Circle A
(12,  15)
Find centre and radius of Circle C
(24, 12)
Find diameter of circle B
45  (5  10)  30
Use proportion to find B
25
 27
45
Previous
(4,  3)
r 5
25
r  10
 15,
Equation of B
Quit
27
B
20
362  272  45
Find distance AB (distance formula)
Centre of B
(24, 12)
(-12, -15)
36
so radius of B = 15
25
 36 
45
20
relative to C
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 x  4    y  3  225
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Hint
Maths4Scotland
The Circle
Higher
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Vectors
Higher
Higher Maths
Vectors
Strategies
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Vectors
Higher
The following questions are on
Vectors
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Vectors
Higher
The questions are in groups
General vector questions (15)
Points dividing lines in ratios
Collinear points (8)
Angles between vectors (5)
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Maths4Scotland
Vectors
Higher
General Vector Questions
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Maths4Scotland
Vectors
Higher
Vectors u and v are defined by u  3i  2 j and v  2i  3 j  4k
Determine whether or not u and v are perpendicular to each other.
Is Scalar product = 0
 3
u.v   2 
 0
 
u.v  3  2  2  3  0  4
u.v  0
2
 
 3 
4
 
 u.v  6   6   0
Hence vectors are perpendicular
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Hint
Maths4Scotland
Vectors
For what value of t are the vectors u
Put Scalar product = 0
 t 
  2 
 3
 
Perpendicular  u.v = 0
and v
 t 
u.v   2 
3
 
u.v  2t   2 10  3t
Higher
2
 
 10 
 t 
 
perpendicular ?
2
 
10 
t 
 
 u.v  5t  20
 0  5t  20
 t4
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Hint
Maths4Scotland
Vectors
Higher
VABCD is a pyramid with rectangular base ABCD.
The vectors AB, AD and AV are given by
AB  8i  2 j  2k
AD  2i  10 j  2k
AV  i  7 j  7k
Express CV
in component form.
Ttriangle rule  ACV
AC  CV  AV
Triangle rule  ABC
AB  BC  AC also

 CV  AV  AB  AD

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
Re-arrange
BC  AD
1 8  2 
     
 CV   7    2    10 
 7   2   2 
     
CV  9i  5 j  7k
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 CV  AV  AC
 9 
 
 CV   5 
7
 
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Maths4Scotland
Vectors
Higher
The diagram shows two vectors a and b, with | a | = 3 and | b | = 22.
These vectors are inclined at an angle of 45° to each other.
a) Evaluate
i) a.a
ii) b.b
iii) a.b
b) Another vector p is defined by p  2a  3b
Evaluate p.p and hence write down | p |.
i)
a  a  a a cos 0  3  3 1  9
iii) a  b  a b cos 45  3  2 2 
b)
p  p   2a  3b    2a  3b 
ii)
bb  2 2 2 2
1
6
2
 4a.a  12a.b  9b.b
 36  72  72  180 Since p.p = p2
p  180  6 5
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Maths4Scotland
Vectors
Vectors p, q and r are defined by p  i  j - k ,
a) Express p  q  2r in component form
b)
Calculate p.r
c)
Find |r|
Higher
q  i  4k ,
  i  j - k    i  4k   2  4i  3 j 
and
r  4i  3 j
 8i  5 j - 5 k
a)
p  q  2r
b)
p.r   i  j - k  .  4i  3 j   p.r  1 4  1 (3)  (1)  0  p.r  1
c)
r 
42  (3)2
 r  16  9 
r 5
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Maths4Scotland
Vectors
Higher
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
S ( a, 0, 0)
PS 2  49  (4  a)2  22  62
 a  43
T (b, 0, 0)
 49  (4  a)2  40
 9  (4  a)2
 a  7 or a  1
hence there are 2 points on the x axis that are 7 units from P
S (1, 0, 0)
and
i.e. S and T
T (7, 0, 0)
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Maths4Scotland
Vectors
Higher
The position vectors of the points P and Q are
p = –i +3j+4k and q = 7 i – j + 5 k respectively.
a) Express PQ in component form.
b) Find the length of PQ.
a)
b)
PQ  q - p
 PQ
 7   1
   
  1 -  3 
5 4
   
PQ  82  (4) 2  12
 PQ
 64  16  1
8
 
  4 
1
 
 81
 8i  4 j  k
9
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Maths4Scotland
Vectors
Higher
P
PQR is an equilateral triangle of side 2 units.
PQ  a, PR  b, and QR  c
Evaluate a.(b + c) and hence identify
two vectors which are perpendicular.
a
b
Diagram
Q
a.(b  c )  a.b  a.c
a.b  a b cos 60
60°
60°
60°
R
c
 a.b  2  2 
1
2
 a.b  2
NB for a.c vectors must point OUT of the vertex ( so angle is 120° )
 1

a.c

2

2

a.c  a c cos120
 
 2
Hence
a.(b  c )  0
 a.c   2
so, a is perpendicular to b + c
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Maths4Scotland
Vectors
Higher
Calculate the length of the vector 2i – 3j + 3k
Length  2  (3) 
2
2
 3
2
 493
 16
4
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Maths4Scotland
Vectors
Find the value of k for which the vectors
1
 
 2
 1 
 
1
0   2 
 1
 
 4 


3


 k 1


Put Scalar product = 0
0  4  6  (k  1)
Higher
and
 4 


 3 
 k 1 


are perpendicular
 0  2  k 1
 k 3
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Maths4Scotland
Vectors
Higher
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  b
D is the displacement
hence
d
BC
AD
 2   13 
  1   3 
 4   1 
  

Previous
 6   7 
   
  4   1
 2   3
   
BC
 13 


 3 
 1 


from A
d
 11


 2 
 3 


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 D  11, 2, 3
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Hint
Maths4Scotland
If
 3 
 
u 3
3
 
and v
1
  5 
 1
 
Vectors
Higher
write down the components of u + v and u – v
Hence show that u + v and u – v are perpendicular.
uv
 2 
 
 8 
2
 
 u  v  . u  v 
uv
 2   4 
  8    2 
2 4
   
 4 
 
  2 
 4
 
look at scalar product
 u  v  . u  v 
 (2)  (4)  8  (2)  2  4
 8 16  8
Hence vectors are perpendicular
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Maths4Scotland
Vectors
Higher
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
a)
 2  1
a.b   0    2 
 1  1 
   
a.c
b)
2 0
  0    1
 1  1 
   
a.b  2  0  1
a.b  1
a.c  0  0  1
a.c  1
a.b  a.c  0
b + c is perpendicular to a
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Maths4Scotland
Vectors
Higher
A is the point ( –3, 2, 4 ) and B is ( –1, 3, 2 )
Find:
a) the components of AB
b) the length of AB
a)
b)
AB  b  a
AB
 1  3 
   
  3  2 
2 4
   
AB  22  12  (2)2
AB
 2
 
 1 
 2 
 
AB  4  1  4
AB  3
AB  9
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Maths4Scotland
Vectors
Higher
In the square based pyramid,
all the eight edges are of length 3 units.
AV  p, AD  q, AB  r ,
Evaluate p.(q + r)
Triangular faces are all equilateral
p.(q  r )  p.q  p.r
p.q  p q cos 60
p.r  p r cos60
1
2
p.(q  r )  4  4
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Table of
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p.q
1
 3 3
2
p.r
1
 3 3
2
p.(q  r )  9
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p.q  4
p.q 
1
2
1
4
2
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Maths4Scotland
Vectors
Higher
You have completed all 15 questions in this section
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Maths4Scotland
Vectors
Higher
Points dividing lines in ratios
Collinear Points
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Maths4Scotland
Vectors
Higher
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
AB 1

AD 3
 3AB  AD
 3b  3a  d  a
 d
2
 1
 3  1  2  3 
1
2
 
 
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 3 b  a   d  a
 d  3b  2a
 d
8
  3 
 1
 
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 D(8, 3, 1)
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Maths4Scotland
Vectors
Higher
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.
R
Diagram
PQ 2

QR 1
 3q
P
1
 q  p  2r  2q
 PQ  2QR
 5   1
 2  2    1
 3   0 
   
2
Q
 3q  2r  p




1 9 
 q 3
3  6 
 Q(3, 1,  2)
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Maths4Scotland
Vectors
Higher
a) Roadmakers look along the tops of a set of T-rods to ensure
that 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 been
built 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.
a)
AB  b  a
AB
b)
and
AB
 6
 2
 
 
 9  3  3
 3
1
 
 
AC
 14 
 2
 
  21  7  3 
7
1
 
 
AC are scalar multiples, so are parallel. A is common. A, B, C are collinear
Use scalar product
AB.BD
 6  3 
  9  .  3 
 3  3 
   
AB.BD  18  27  9  0
Hence, DB is perpendicular to AB
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Hint
Maths4Scotland
Vectors
Higher
VABCD is a pyramid with rectangular base ABCD.
Relative to some appropriate axis,
VA represents – 7i – 13j – 11k
6i + 6j – 6k
AB represents
8i – 4j – 4k
AD represents
K divides BC in the ratio 1:3
Find VK in component form.
VA  AB  VB
VK  VB  KB
 VK
VK  KB  VB
1
4
1
4
1
4
 VK  VA  AB  AD
 7   6 
8
1
  13    6    4 
 11  6  4  4 

  
 
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1
4
KB  CB  DA   AD
 VK
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 1 
  8 
 18 


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Hint
Maths4Scotland
Vectors
Higher
The line AB is divided into 3 equal parts by
the points C and D, as shown.
A and B have co-ordinates (3, –1, 2) and (9, 2, –4).
a) Find the components of AB and AC
b) Find the co-ordinates of C and D.
a)
b)
AB  b  a
AB
6
 
 3
 6 
 
C is a displacement of AC from A
similarly
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d
 5  2 
   
  0   1 
 0   2 
   




2
1
 
AC  AB   1 
3
 2 
c
3 2
  1   1 
 2   2 
   
 C (5, 0, 0)
 D(7, 1,  2)
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Maths4Scotland
Vectors
Higher
Relative to a suitable set of axes, the tops of three chimneys have
co-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  a
AB
and
AB
1
 
  4 
 2
 
AC
 3 
1
  12   3  4 
 6 
2


 
AC are scalar multiples, so are parallel. A is common. A, B, C are collinear
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Maths4Scotland
Vectors
Higher
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 collinear
and determine the ratio in which B divides AC
AB  b  a
AB
and
BC
AB 2

BC 3
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AB
4
2
  2   2  1 
 2 
 1
 
 
BC
6
2
  3   3  1 
 3 
 1
 
 
are scalar multiples, so are parallel. B is common. A, B, C are collinear
A
2
B
3
C
B divides AB in ratio 2 : 3
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Maths4Scotland
Vectors
Higher
Relative to the top of a hill, three gliders
have 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  r
RS
and
RT
RS
 3
1
  3   3  1 
 6
 2
 
 
RT
 4
1
 
 
  4  4  1
8
 2
 
 
are scalar multiples, so are parallel. R is common. R, S, T are collinear
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Vectors
Higher
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Vectors
Higher
Angle between two vectors
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Vectors
Higher
The diagram shows vectors a and b.
If |a| = 5, |b| = 4 and a.(a + b) = 36
Find the size of the acute angle
between a and b.
cos  
a.b
a b
a.a  a a  25
11
cos  
5 4
Previous
a.(a  b)  36  a.a  a.b  36
 25  a.b  36
 11 
  cos  
 20 
  56.6
1
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 a.b  11
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Hint
Maths4Scotland
Vectors
Higher
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 DA and DB
c) Calculate the size of angle ADB.
a)
c)
B(6, 6, 0)
cos  
AD.DB
AD DB
cos  
Previous
64
82 82
b)
3
 
DA   3 
 8 
 
AD
DB
 3 
 
  DA   3 
8
 
  141.3
Quit
3
  3 
 8 
 
AD.DB
 3   3 
  3  .  3   64
 8   8 
   
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Vectors
Higher
A box in the shape of a cuboid designed with circles of different
sizes on each face.
The diagram shows three of the circles, where the origin represents
one 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
Vectors to point
away from vertex
6
 
BA   5 
1
 
 4
 
BC   0 
 6 
 
BA  36  25  1  62
cos  
18
62 52
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BA.BC  24  0  6  18
BC  16  36  52
  71.5
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Vectors
Higher
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.
a)
B(3, 2, 15)
b)
 15 
 
BC   2 
 7 
 
BA.BC  45  14  49   10
BA  9  49  49  107
Previous
 3 
 
BA   7 
 7 
 
BC  225  4  49  278
10
cos  
278 107
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  93.3
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Vectors
Higher
A triangle ABC has vertices A(2, –1, 3), B(3, 6, 5) and C(6, 6, –2).
a) Find
and
AB
AC
b) Calculate the size of angle BAC.
c) Hence find the area of the triangle.
a)
b)
1
 
AB  b  a   7 
 2
 
AB  12  7 2  22  54
cos  
c)
4
 
AC  c  a   7 
 5 
 
43
 0.6168
54 90
Area of  ABC =
Previous
AC  90
AB. AC  4  49  10  43
  cos1 0.6168  51.9
1
ab sin C
2
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
1
90 54 sin 51.9
2
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 BAC = 51.9

27.43
unit
2
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Vectors
Higher
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The Wave Function
Higher
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The Wave Function
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The Wave Function
Higher
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The Wave Function
Non-calculator questions will be indicated
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The Wave Function
Higher
Part of the graph of y = 2 sin x + 5 cos x is shown
in the diagram.
a) Express y = 2 sin x + 5 cos x in the form k sin (x + a)
where k > 0 and 0  a  360
b) Find the coordinates of the minimum turning point P.
Expand ksin(x + a):
k sin( x  a)  k sin x cos a  k cos x sin a
Equate coefficients:
k cos a  2
Square and add
Dividing:
k 2  2 2  52
tan a 
5
2
k sin a  5
k  29
acute
a  68
Put together:
2sin x  5cos x  29 sin( x  68)
Minimum when:
( x  68)  270  x  202
P has coords.
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(sin and cos are +)
a  68
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(202,  29)
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a is in 1st quadrant
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Hint
a)
b)
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
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.
Expand ksin(x - a):
k sin( x  a)  k sin x cos a  k cos x sin a
Equate coefficients:
k cos a  1
k 2  12  12
Square and add
tan a  1
Dividing:
k 2
acute
a


a is in 1st quadrant
4
(sin and cos are +)

2 a
4
sin x  cos x  2 sin( x  )
Put together:
4
Sketch Graph
max
max at
Previous
k sin a  1
 2
x
3
4
min
min at
Table of exact values
2
 2
x
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7
4
k
2
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a

4
Hint
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The Wave Function
Higher
Express 8cos x  6sin x in the form k cos( x  a) where k  0 and 0  a  360
Expand kcos(x + a):
k cos( x  a)  k cos x cos a  k sin x sin a
Equate coefficients:
k cos a  8
Square and add
k 2  82  62
Dividing:
Put together:
tan a 
6
8
k sin a  6
k  10
acute
a  37
a is in 1st quadrant
a  37
(sin and cos are +)
8cos x  6sin x  10 cos( x  37)
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The Wave Function
Higher
Find the maximum value of cos x  sin x and the value of x for which it occurs in
the interval 0  x  2.
Express as Rcos(x - a):
R cos( x  a)  R cos x cos a  R sin x sin a
Equate coefficients:
R cos a  1
Square and add
R 2  12  12
Dividing:
Put together:
Max value:
tan a  1
acute
R sin a  1
R 2
a

a is in 4th quadrant
4
(sin is - and cos is +)
 
  0, x 
cos x  sin x  2 cos x 
2
when
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Table of exact values

x
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7
4
a
7
4
7
4
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7
4
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Hint
Maths4Scotland
The Wave Function
Higher
Express 2sin x  5cos x in the form k sin( x   ), 0    360 and k  0
Expand ksin(x - a):
k sin( x  a)  k sin x cos a  k cos x sin a
Equate coefficients:
k cos a  2
Square and add
k 2  2 2  52
5
2
k sin a  5
k  29
a  68
a is in 1st quadrant
Dividing:
tan a 
Put together:
2cos x  5sin x  29 sin  x  68 
acute
(sin and cos are both +)
a  68
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The Wave Function
Higher
The diagram shows an incomplete graph of


y  3sin  x   , for 0  x  2
3

Find the coordinates of the maximum stationary point.
Max for sine occurs
(...) 
Coordinates of max s.p.
2
x
,
5
6
Sine takes values between 1 and -1
Max value of sine function:
Max value of function:

3

5
,3
6

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The Wave Function
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f ( x)  2 cos x  3sin x
a)
Express f (x) in the form k cos( x   )
b)
Hence solve algebraically f ( x)  0.5
Expand kcos(x - a):
k cos a  2
Square and add
k 2  22  32
Put together:
tan a 
 x  56  82
Previous
and
0    360
0  x  360
3
2
acute
k sin a  3
k  13
a  56
a is in 1st quadrant
(sin and cos are both + )
a  56
2cos x  3sin x  13 cos  x  56 
Solve equation.
acute
for
k 0
k cos( x  a)  k cos x cos a  k sin x sin a
Equate coefficients:
Dividing:
where
cos  x  56 
13 cos  x  56   0.5
Cosine +, so 1st & 4th quadrants
x  138 or x  334
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0.5
13
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The Wave Function
Solve the simultaneous equations
k sin x  5
where k > 0 and 0  x  360
k cos x  2
Higher
Use tan A = sin A / cos A
Divide
tan x 
Find acute angle
5
2
acute
Determine quadrant(s)
x  68
Sine and cosine are both + in original equations
Solution must be in 1st quadrant
State solution
x  68
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The Wave Function
Higher
Solve the equation 2sin x  3cos x  2.5 in the interval 0  x  360.
R cos( x  a)  R cos x cos a  R sin x sin a
Use Rcos(x - a):
R cos a  3
Equate coefficients:
R 2  22   3
Square and add
tan a  
Dividing:
Put together:
 x 146  46
x  192
or
Previous
acute
R  13
a  34
a is in 2nd quadrant
a  146
(sin + and cos - )
2sin x  3cos x  13 cos  x  146
Solve equation.
acute
2
3
2
R sin a  2
x  460
cos  x  146 
13 cos  x  146   2.5
2.5
13
Cosine +, so 1st & 4th quadrants
(out of range, so subtract 360°)
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x  100
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x  192start
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The Wave Function
Higher
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Logarithms & Exponential
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Logarithms & Exponential
Higher
Reminder
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upon you knowing and being able to use,
some very basic rules and facts.
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Logarithms & Exponential
Higher
Three Rules of logs
log a x  log a y  log a xy
x
log a x  log a y  log a
y
log a x  p log a x
p
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Logarithms & Exponential
Higher
Two special logarithms
log a a  1
log a 1  0
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Logarithms & Exponential
Higher
Relationship between log and exponential
log a x  y  a  x
y
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Logarithms & Exponential
Higher
Graph of the exponential function
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Logarithms & Exponential
Higher
Graph of the logarithmic function
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Logarithms & Exponential
Higher
Related functions of y  f ( x)
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
Move graph right
units
a
units
Reflect in x axis
Reflect in y axis
Move graph up a units
Move graph down a units
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Logarithms & Exponential
Higher
Calculator keys
ln
=
log e
log
=
log10
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Logarithms & Exponential
Higher
Calculator keys
loge 2.5
=
log10 7.6
=
ln
2
.
5
=
= 0.916…
log 7
.
6
=
= 0.8808…
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Logarithms & Exponential
Higher
Solving exponential equations
2.4  3.1e
x
Use log ab = log a + log b
log e 2.4  log e 3.1e x
x
log e 2.4  log e 3.1  log e e
Use log ax = x log a
loge 2.4  loge 3.1  x loge e
Use loga a = 1
log e 2.4  log e 3.1  x
Take loge both sides
x  log e 2.4  log e 3.1  0.25593...  0.26 (2dp)
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Logarithms & Exponential
Higher
Solving exponential equations
60  80e
k
log e 60  log e 80e k
Take loge both sides
log e 60  log e 80  log e e
Use log ab = log a + log b
k
Use log ax = x log a
loge 60  loge 80  k loge e
Use loga a = 1
loge 60  loge 80  k
k  loge 80  loge 60
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 0.2876...
 0.29 (2dp)
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Logarithms & Exponential
Higher
Solving logarithmic equations
log3 y  0.5
Change to exponential form
0.5
y 3

Change to exponential form
y 3
1
2

1
3
1
2
1

3
y  0.577....  0.58 (2dp)
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Logarithms & Exponential
Higher
3loge (2e)  2log e (3e)
A  log e B  log e C
expressing your answer in the form
Simplify
where A, B and C are whole numbers.
log e (2e)  log e (3e)
3
2
8e3
log e 2
9e
 log e 8e3  log e 9e2
8e
 log e
9
 loge 8  loge e  loge 9
 1  loge 8  log e 9
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Simplify
Logarithms & Exponential
Higher
log5 2  log5 50  log5 4
2  50
 log 5
4
 log 5 52
 log5 25
 2
 2 log5 5
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Find x if
Logarithms & Exponential
4 log x 6  2 log x 4  1
64
 log x 2  1
4
 log x 64  log x 42  1
9
 log x
Higher
36  36
1
4 4
1
9
 log x 81  1
1
 x  81
 x1  81
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Given
Logarithms & Exponential
x  log5 3  log5 4
Higher
find algebraically the value of x.
 x  log5 3  4
 x  log5 12
 5x  12
 log10 5x  log10 12
log10 12
 x
log10 5
 x log10 5  log10 12
 x  1.5439..
 x  1.54 (2 dp)
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Logarithms & Exponential
Higher
Find the x co-ordinate of the point where the graph of the curve
with equation
y  log3 ( x  2)  1
intersects the x-axis.
When y = 0
0  log3 ( x  2)  1
Re-arrange
1  log3 ( x  2)
Exponential form
Re-arrange
31  x  2
1
x  23
x2
1
3
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Logarithms & Exponential
Higher
The graph illustrates the law
y  kx n
If the straight line passes through A(0.5, 0)
and B(0, 1).
Find the values of k and n.
log 5 y  log 5 kx n
Gradient
log5 y  log5 k  n log5 x
Y  log5 k  nX

1
0.5
y-intercept
log5 k  1
 k 5
n  2 (gradient)
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Logarithms & Exponential
Higher
Before a forest fire was brought under control, the spread of fire was described
by a law of the form
A  A0e kt
where
A0
is the area covered by the fire when it was first detected
and 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.
2 A0  A0 e k 1.5
 loge 2  1.5k
 2  ek 1.5
 loge 2  1.5k loge e
log e 2
 k
 k  0.462..  0.46
1.5
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(2 dp)
Maths4Scotland
Logarithms & Exponential
The results of an experiment
Higher
give rise to the graph shown.
a) Write down the equation of the line
in terms of P and Q.
It is given that
P  log e p
and
Q  log e q
b) Show that p and q satisfy a relationship of the form
p  aqb
stating the values of a and b.
0.6
Gradient
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loge p  loge a  b loge q
y-intercept 1.8
P  0.6Q  1.8
log e p  log e aq b
P  log e a  bQ
log e a  1.8  a  e1.8
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b  0.6
 a  6.05
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Logarithms & Exponential
Higher
The diagram shows part of the graph of
y  logb ( x  a)
Determine the values of a and b.
.
Use (7, 1)
1  logb (7  a)
...(1)
Use (3, 0)
0  logb (3  a)
...(2)
Hence, from (2)
and from (1)
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3 a 1
1  log b 5
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a  2
b5
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Logarithms & Exponential
Higher
The diagram shows a sketch of
part of the graph of
y  log 2 ( x)
a) State the values of a and b.
b) Sketch the graph of
a 1
y  log 2 ( x  1)  3
b3
Graph moves
1 unit to the left and 3 units down
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Logarithms & Exponential
a) i) Sketch the graph of
Higher
y  a x  1, a  2
ii) On the same diagram, sketch the graph of
y  a x 1 , a  2
b) Prove that the graphs intersect at a point
where the x-coordinate is
a x  1  a x 1
 1 
log a 

a

1


 1  a x 1  a x
 loga 1  log a a x  log a  a 1
 x   loga  a  1
 1  a x  a 1
 0  x  log a  a  1
 x  log a  a  1
1
 1 
 x  log a 

a  1 


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Logarithms & Exponential
Higher
Part of the graph of
y  5 log10 (2 x  10)
is shown in the diagram.
This graph crosses the
x-axis at the point A and
y 8
the straight line
at the point B. Find algebraically the x co-ordinates of A and B.
8  5 log10 (2 x  10) 
 101.6  2 x  10
0  5 log10 (2 x  10)
8
 1.6  log10 (2 x  10)
 log10 (2 x  10)
5
 101.6  10  2x  x  14.9 B (14.9, 8)
 2 x  10  1
 x  4.5
A (4.5, 0)
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Logarithms & Exponential
Higher
The diagram is a sketch of part of
y  ax , a  1
the graph of
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
y  a 2x
c) Find the co-ordinates of the point of intersection of
y  a 2x
with the line
a)
ta
c)
y  a 21
x 1
u 0
y  a2
b)
2
1,
a
 
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Logarithms & Exponential
Higher
The diagram shows part of the graph with equation
y  3x
y  42
and the straight line with equation
These graphs intersect at P.
Solve algebraically the equation
3x  42
and hence write down, correct to 3 decimal places, the co-ordinates of P.
log10 3x  log10 42
log10 42
 x
log10 3
 x log10 3  log10 42
 x  3.40217...
 P (3.402, 42)
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Higher
Table of exact values
sin
cos
tan
30°
45°
60°

6
1
2

4

3
1
2
1
2
3
2
3
2
1
3
1
1
2
3
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Higher
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