HIGHER

1
HIGHER ADDITIONAL QUESTION BANK
Please decide which Unit you would like to revise
UNIT 1
UNIT 2
UNIT 3
Straight Line Functions Graphs Trig Graphs
Equations Basic Differentiation Recurrence
Relations
Polynomials Quadratic Functions
Integration Addition Formulae The Circle
Vectors Further Calculus Exponential /
Logarithmic Functions The Wave Function
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2
HIGHER ADDITIONAL QUESTION BANK
Straight Line
Trig Graphs Equations
UNIT 1
Functions Graphs
Basic Differentiation
Recurrence Relations
EXIT
3
HIGHER ADDITIONAL QUESTION BANK
You have chosen to study
Straight Line
UNIT 1
Please choose a question to attempt from the
following
1
2
3
4
5
Back to Unit 1 Menu
EXIT
4
STRAIGHT LINE Question 1
Find the equation of the straight line which is
perpendicular to the line with equation 3x 5y
4 and which passes through the point (-6,4).
Reveal answer only
Go to full solution
Go to Markers Comments
Go to Straight Line Menu
Go to Main Menu
EXIT
5
STRAIGHT LINE Question 1
Find the equation of the straight line which is
perpendicular to the line with equation 3x 5y
4 and which passes through the point (-6,4).
Reveal answer only
Go to full solution
Go to Markers Comments
Go to Straight Line Menu
Go to Main Menu
EXIT
6
Question 1
3x 5y 4
Find the equation of the straight line which is
perpendicular to the line with equation 3x
5y 4 and which passes through the point
(-6,4).
3x - 4 5y
(?5)
5y 3x - 4
y 3/5x - 4/5
Using y mx c , gradient of line is 3/5
So required gradient -5/3 , ( m1m2 -1)
We now have (a,b) (-6,4) m -5/3 .
Using y b m(x a)
We get y 4 -5/3 (x (-6))
y 4 -5/3 (x 6)
Begin Solution
y 4 -5/3x - 10
Continue Solution
Markers Comments
Straight Line Menu
7
Markers Comments

• An attempt must be made to put the original
  equation into the form
• y mx c to read off the gradient.

3x 5y 4
3x - 4 5y
(?5)
5y 3x - 4

• State the gradient clearly.

• State the condition for perpendicular lines
  m1 m2 -1.

y 3/5x - 4/5
Using y mx c , gradient of line is 3/5

• When finding m2 simply invert and change the sign
  on m1

So required gradient -5/3 , ( m1m2 -1)
We now have (a,b) (-6,4) m -5/3 .
Using y b m(x a)

• Use the y - b m(x - a) form to obtain the
  equation of the line.

We get y 4 -5/3 (x (-6))
y 4 -5/3 (x 6)
Next Comment
y 4 -5/3x - 10
Straight Line Menu
8
STRAIGHT LINE Question 2
Find the equation of the straight line which is
parallel to the line with equation 8x 4y 7
0 and which passes through the point (5,-3).
Reveal answer only
Go to full solution
Go to Markers Comments
Go to Straight Line Menu
Go to Main Menu
EXIT
9
STRAIGHT LINE Question 2
Find the equation of the straight line which is
parallel to the line with equation 8x 4y 7
0 and which passes through the point (5,-3).
y -2x 7
Reveal answer only
Go to full solution
Go to Markers Comments
Go to Straight Line Menu
Go to Main Menu
EXIT
10
Question 2
8x 4y 7 0
Find the equation of the straight line which is
parallel to the line with equation 8x 4y 7
0 and which passes through the point (5,-3).
4y -8x 7
(?4)
y -2x 7/4
Using y mx c , gradient of line is -2
So required gradient -2 as parallel lines have
equal gradients.
We now have (a,b) (5,-3) m -2.
Using y b m(x a)
We get y (-3) -2(x 5)
Begin Solution
y 3 -2x 10
Continue Solution
Markers Comments
y -2x 7
Straight Line Menu
11
Markers Comments

• An attempt must be made to
• put the original equation into
• the form y mx c to
• read off the gradient.

8x 4y 7 0
4y -8x 7
(?4)

• State the gradient clearly.

y -2x 7/4
Using y mx c , gradient of line is -2

• State the condition for
• parallel lines m1 m2

So required gradient -2 as parallel lines have
equal gradients.

• Use the y - b m(x - a) form
• to obtain the equation of
• the line.

We now have (a,b) (5,-3) m -2.
Using y b m(x a)
We get y (-3) -2(x 5)
y 3 -2x 10
Next Comment
Straight Line Menu
y -2x 7
12
STRAIGHT LINE Question 3
In triangle ABC, A is (2,0), B is (8,0) and C
is (7,3).
(a) Find the gradients of AC and BC.
(b) Hence find the size of ACB.
Reveal answer only
Go to full solution
Go to Markers Comments
Go to Straight Line Menu
Go to Main Menu
EXIT
13
STRAIGHT LINE Question 3
In triangle ABC, A is (2,0), B is (8,0) and C
is (7,3).
(a) Find the gradients of AC and BC.
(b) Hence find the size of ACB.
mAC
3/5
(a)
Reveal answer only
mBC
- 3
Go to full solution
Go to Markers Comments
77.4
(b)
Go to Straight Line Menu
Go to Main Menu
EXIT
14
Question 3

1. Using the gradient formula

In triangle ABC, A is (2,0), B is (8,0) and C
is (7,3).
mAC 3 0 7 - 2
3/5

• Find the gradients of AC
• and BC.

mBC 3 0 7 - 8

- 3
(b) Hence find the size of ACB.
(b) Using tan? gradient
If tan? 3/5
then CAB 31.0
If tan? -3
then CBX (180-71.6)
108.4 o
so ABC 71.6
Begin Solution
Hence ACB 180 31.0 71.6
Continue Solution
Markers Comments
77.4
Straight Line Menu
15
Markers Comments

• If no diagram is given draw a
• neat labelled diagram.
• In calculating gradients state
• the gradient formula.

1. Using the gradient formula

mAC 3 0 7 - 2
3/5

• Must use the result that the
• gradient of the line is equal
• to the tangent of the angle
• the line makes with the
• positive direction of the x-axis.
• Not given on the formula sheet.

mBC 3 0 7 - 8

- 3
(b) Using tan? gradient
If tan? 3/5
then CAB 31.0
mAB tanØ Ø tan-1 mAB
then CBX (180-71.6)
108.4 o
If tan? -3
so ABC 71.6
Hence ACB 180 31.0 71.6
Next Comment
Straight Line Menu
77.4
16
STRAIGHT LINE Question 4
Y
In triangle PQR the vertices are P(4,-5), Q(2,3)
and R(10-1).
Q(2,3)
Find (a) the equation of the line e, the median
from R of triangle PQR.
X
R(10,-1)
P(4,-5)
(b) the equation of the line f, the perpendicular
bisector of QR.
Reveal answer only
(c) The coordinates of the point of intersection
of lines e f.
Go to full solution
Go to Markers Comments
Go to Straight Line Menu
Go to Main Menu
EXIT
17
STRAIGHT LINE Question 4
Y
In triangle PQR the vertices are P(4,-5), Q(2,3)
and R(10-1).
Q(2,3)
Find (a) the equation of the line e, the median
from R of triangle PQR.
X
R(10,-1)
P(4,-5)
(b) the equation of the line f, the perpendicular
bisector of QR.
Reveal answer only
(c) The coordinates of the point of intersection
of lines e f.
Go to full solution
Go to Markers Comments
y -1
(a)
Go to Straight Line Menu
(b)
y 2x 11
Go to Main Menu
EXIT
(5,-1)
(c)
18
Question 4 (a)

1. Midpoint of PQ is (3,-1) lets call this S

In triangle PQR the vertices are P(4,-5), Q(2,3)
and R(10-1).
Using the gradient formula m y2 y1
x2 x1
Find (a) the equation of the line e, the median
from R of triangle PQR.
mSR -1 (-1) 10 - 3
0
(ie line is horizontal)
Since it passes through (3,-1) equation of e is
y -1
Solution to 4 (b)
Begin Solution
Continue Solution
Markers Comments
Straight Line Menu
19
Question 4 (b)
In triangle PQR the vertices are P(4,-5), Q(2,3)
and R(10-1).
- 1/2
Find
required gradient 2 (m1m2 -1)
(b) the equation of the line f, the
perpendicular bisector of QR.
Using y b m(x a) with (a,b) (6,1)

m 2
we get y 1 2(x 6)
Solution to 4 (c)
Begin Solution
Continue Solution
Markers Comments
Straight Line Menu
20
Question 4 (c)
(c) e f meet when y -1 y 2x -11
In triangle PQR the vertices are P(4,-5), Q(2,3)
and R(10-1).
so 2x 11 -1
Find
ie 2x 10
(c) The coordinates of the point of intersection
of lines e f.
ie x 5
Point of intersection is (5,-1)
Begin Solution
Continue Solution
Markers Comments
Straight Line Menu
21
Markers Comments

• If no diagram is given draw a neat labelled
  diagram.

1. Midpoint of PQ is (3,-1) lets call this S

Using the gradient formula m y2 y1
x2 x1

• Sketch the median and the
• perpendicular bisector

mSR -1 (-1) 10 - 3
(ie line is horizontal)
median
Perpendicular bisector
Since it passes through (3,-1) equation of e is
y -1
Next Comment
Comments for 4 (b)
Straight Line Menu
22
Markers Comments

• To find midpoint of QR

• Look for special cases

- 1/2
Horizontal lines in the form y k Vertical
lines in the form x k
Next Comment
Comments for 4 (c)
Straight Line Menu
23
Markers Comments

• To find the point of intersection of the two
  lines solve the two equations

(c) e f meet when y -1 y 2x -11
so 2x 11 -1
ie 2x 10
y -1 y 2x - 11
ie x 5
Point of intersection is (5,-1)
Next Comment
Straight Line Menu
24
STRAIGHT LINE Question 5
In triangle EFG the vertices are E(6,-3),
F(12,-5) and G(2,-5).
Find (a) the equation of the altitude from vertex
E.
(b) the equation of the median from vertex F.
Reveal answer only
(c) The point of intersection of the altitude
and median.
Go to full solution
Go to Markers Comments
Go to Straight Line Menu
Go to Main Menu
EXIT
25
STRAIGHT LINE Question 5
In triangle EFG the vertices are E(6,-3),
F(12,-5) and G(2,-5).
Find (a) the equation of the altitude from vertex
E.
(b) the equation of the median from vertex F.
Reveal answer only
(c) The point of intersection of the altitude
and median.
Go to full solution
(a)
x 6
Go to Markers Comments
(b)
Go to Straight Line Menu
x 8y 28 0
Go to Main Menu
(c)
(6,-4.25)
EXIT
26
Question 5(a)

1. Using the gradient formula

In triangle EFG the vertices are E(6,-3),
F(12,-5) and G(2,-5).
mFG -5 (-5) 12 - 2
0
Find (a) the equation of the altitude from vertex
E.
(ie line is horizontal so altitude is vertical)
Y
Altitude is vertical line through (6,-3)
ie x 6
X
E(6,-3)
F(12,-5)
Solution to 5 (b)
G(2,-5)
Begin Solution
Continue Solution
Markers Comments
Straight Line Menu
27
Question 5(b)

• Midpoint of EG is (4,-4)- lets call this H

In triangle EFG the vertices are E(6,-3),
F(12,-5) and G(2,-5).
mFH -5 (-4) 12 - 4
-1/8
Find
Using y b m(x a) with (a,b) (4,-4)

m -1/8
(b) the equation of the median from vertex F.
Y
we get y (-4) -1/8(x 4)
(X8)
X
E(6,-3)
or 8y 32 -x 4
F(12,-5)
G(2,-5)
Median is x 8y 28 0
Begin Solution
Solution to 5 (c)
Continue Solution
Markers Comments
Straight Line Menu
28
Question 5(c)
(c) Lines meet when x 6 x 8y 28 0
In triangle EFG the vertices are E(6,-3),
F(12,-5) and G(2,-5).
put x 6 in 2nd equation 8y 34 0
Find
ie 8y -34
(c) The point of intersection of the altitude
and median.
ie y -4.25
Y
Point of intersection is (6,-4.25)
X
E(6,-3)
F(12,-5)
G(2,-5)
Begin Solution
Continue Solution
Markers Comments
Straight Line Menu
29
Markers Comments

• If no diagram is given draw a
• neat labelled diagram.
• Sketch the altitude and
• the median.

1. Using the gradient formula

mFG -5 (-5) 12 - 2
0
(ie line is horizontal so altitude is vertical)
Altitude is vertical line through (6,-3)
ie x 6
median
Comments for 5 (b)
altitude
Next Comment
Straight Line Menu
30
Markers Comments

• Midpoint of EG is (4,-4)- call this H

• Look for special cases

mFH -5 (-4) 12 - 4
-1/8
Horizontal lines in the form y k Vertical
lines in the form x k
Using y b m(x a) with (a,b) (4,-4)

m -1/8
we get y (-4) -1/8(x 4)
(X8)
or 8y 32 -x 4
Median is x 8y 28 0
Next Comment
Straight Line Menu
Comments for 5 (c)
31
Markers Comments

• To find the point of
• intersection of the two lines
• solve the two equations

(c) Lines meet when x 6 x 8y 28 0
x 6 x 8y -28
put x 6 in 2nd equation 8y 34 0
ie 8y -34
ie y -4.25
Point of intersection is (6,-4.25)
Next Comment
Straight Line Menu
32
HIGHER ADDITIONAL QUESTION BANK
You have chosen to study
Basic Differentiation
UNIT 1
Please choose a question to attempt from the
following
1
2
3
4
5
Back to Unit 1 Menu
EXIT
33
BASIC DIFFERENTIATION Question 1
Find the equation of the tangent to the curve
(xgt0) at the
point where x 4.
Reveal answer only
Go to full solution
Go to Markers Comments
Go to Basic Differentiation Menu
Go to Main Menu
EXIT
34
BASIC DIFFERENTIATION Question 1
Find the equation of the tangent to the curve
(xgt0)
at the point where x 4.
Reveal answer only
y 5/4x 7
Go to full solution
Go to Markers Comments
Go to Basic Differentiation Menu
Go to Main Menu
EXIT
35
Question 1
NB a tangent is a line so we need a point of
contact and a gradient.
Find the equation of the tangent to the curve
y ?x 16 x (xgt0)
at the point where x 4.
Point
If x 4 then
y ?4 16 4
2 4 -2
so (a,b) (4,-2)
Gradient
y ?x 16 x
x1/2 16x -1
Continue Solution
dy/dx 1/2x-1/2 16x-2
1 16 2?x x2
If x 4 then
Begin Solution
dy/dx
1 16 2?4 16
Continue Solution
¼ 1 5/4
Markers Comments
Basic Differentiation Menu
36
Question 1
If x 4 then
Find the equation of the tangent to the curve
y ?x 16 x (xgt0)
at the point where x 4.
dy/dx
1 16 2?4 16
¼ 1 5/4
Gradient of tangent gradient of curve
so m 5/4 .
We now use y b m(x a)
this gives us y (-2) 5/4(x 4)
Back to Previous
or y 2 5/4x 5
or y 5/4x 7
Begin Solution
Continue Solution
Markers Comments
Basic Differentiation Menu
37
Markers Comments

• Prepare expression for differentiation.

NB a tangent is a line so we need a point of
contact and a gradient.
Point

• Find gradient of the tangent
• using rule

If x 4 then
y ?4 16 4
2 4 -2
multiply by the power and reduce the power
by 1
so (a,b) (4,-2)
Gradient
y ?x 16 x
x1/2 16x -1

• Find gradient at x 4.

dy/dx 1/2x-1/2 16x-2
1 16 2?x x2
Continue Comments
If x 4 then
1 16 2?4 16
dy/dx
Next Comment
Differentiation Menu
¼ 1 5/4
38
Markers Comments

• Find y coordinate at x 4 using

If x 4 then
1 16 2?4 16
dy/dx
¼ 1 5/4

• Use m 5/4 and (4,-2) in
• y - b m(x - a)

Gradient of tangent gradient of curve
so m 5/4 .
We now use y b m(x a)
this gives us y (-2) 5/4(x 4)
or y 2 5/4x 5
or y 5/4x 7
Next Comment
Differentiation Menu
39
BASIC DIFFERENTIATION Question 2
Find the coordinates of the point on the curve
y x2 5x 10 where the tangent to the curve
makes an angle of 135 with the positive
direction of the X-axis.
Reveal answer only
Go to full solution
Go to Markers Comments
Go to Basic Differentiation Menu
Go to Main Menu
EXIT
40
BASIC DIFFERENTIATION Question 2
Find the coordinates of the point on the curve
y x2 5x 10 where the tangent to the curve
makes an angle of 135 with the positive
direction of the X-axis.
(2,4)
Reveal answer only
Go to full solution
Go to Markers Comments
Go to Basic Differentiation Menu
Go to Main Menu
EXIT
41
Question 2
NB gradient of line gradient of curve
Find the coordinates of the point on the curve
y x2 5x 10 where the tangent to the
curve makes an angle of 135 with the positive
direction of the X-axis.
Line
Using gradient tan?
we get gradient of line tan135
-tan45
-1
Curve
Gradient of curve dy/dx
2x - 5
Continue Solution
It now follows that 2x 5 -1
Begin Solution
Or 2x 4
Continue Solution
Or x 2
Markers Comments
Basic Differentiation Menu
42
Question 2
Using y x2 5x 10 with x 2
Find the coordinates of the point on the curve
y x2 5x 10 where the tangent to the
curve makes an angle of 135 with the positive
direction of the X-axis.
we get y 22 (5 X 2) 10
ie y 4
So required point is
(2,4)
Back to Previous
Begin Solution
Continue Solution
Markers Comments
Basic Differentiation Menu
43
Markers Comments

• Find gradient of the tangent
• using rule

NB gradient of line gradient of curve
multiply by the power and reduce the power by 1
Line
Using gradient tan?

• Must use the result that the gradient
• of the line is also equal to the tangent
• of the angle the line makes with the
• positive direction of the x- axis.
• Not given on the formula sheet.

we get gradient of line tan135
-tan45
-1
Curve
Gradient of curve dy/dx
2x - 5
m tan135 -1
It now follows that 2x 5 -1
Or 2x 4
Next Comment
Or x 2
Differentiation Menu
Continue Comments
44
Markers Comments
It now follows that 2x 5 -1
Or 2x 4
Or x 2
Using y x2 5x 10 with x 2
we get y 22 (5 X 2) 10
ie y 4
So required point is
(2,4)
Next Comment
Differentiation Menu
45
BASIC DIFFERENTIATION Question 3
y g(x)
The graph of y g(x) is shown here.
There is a point of inflection at the origin, a
minimum turning point at (p,q) and the graph also
cuts the X-axis at r.
r
Make a sketch of the graph of y g?(x).
(p,q)
Reveal answer only
Go to full solution
Go to Markers Comments
Go to Basic Differentiation Menu
Go to Main Menu
EXIT
46
BASIC DIFFERENTIATION Question 3
y g(x)
The graph of y g(x) is shown here.
There is a point of inflection at the origin, a
minimum turning point at (p,q) and the graph also
cuts the X-axis at r.
r
Make a sketch of the graph of y g?(x).
(p,q)
Reveal answer only
y g?(x)
Go to full solution
Go to Markers Comments
Go to Basic Differentiation Menu
Go to Main Menu
EXIT
47
Question 3
y g(x)
Stationary points occur at x 0 and x p.
(We can ignore r.)
We now consider the sign of the gradient either
side of 0 and p
r
(p,q)
Make a sketch of the graph of y g?(x).
new y-values
Begin Solution
Click for graph
Continue Solution
Markers Comments
Basic Differentiation Menu
48
Question 3
y g(x)
This now gives us the following graph
y g?(x)
r
0
p
(p,q)
Make a sketch of the graph of y g?(x).
Begin Solution
Return to Nature Table
Continue Solution
Markers Comments
Basic Differentiation Menu
49
Markers Comments
To sketch the graph of the gradient function
Stationary points occur at x 0 and x p.
1 Mark the stationary points on the x axis i.e.
Continue Comments
Next Comment
Differentiation Menu
50
Markers Comments
To sketch the graph of the gradient function
Stationary points occur at x 0 and x p.
1 Mark the stationary points on the x axis i.e.
(We can ignore r.)
We now consider the sign of the gradient either
side of 0 and p
2 For each interval decide if the value of

x
new y-values
-
-
Next Comment
Differentiation Menu
Continue Comments
51
Markers Comments
To sketch the graph of the gradient function
Stationary points occur at x 0 and x p.
1 Mark the stationary points on the x axis i.e.
2 For each interval decide if the value of
3 Draw in curve to fit information
Next Comment

• In any curve sketching question use a
  ruler and annotate the sketch
• i.e. label all known coordinates.

Differentiation Menu
52
BASIC DIFFERENTIATION Question 4
Here is part of the graph of y x3 - 3x2 - 9x
2.
y x3 - 3x2 - 9x 2
Find the coordinates of the stationary points and
determine their nature algebraically.
Reveal answer only
Go to full solution
Go to Markers Comments
Go to Basic Differentiation Menu
Go to Main Menu
EXIT
53
BASIC DIFFERENTIATION Question 4
Here is part of the graph of y x3 - 3x2 - 9x
2.
y x3 - 3x2 - 9x 2
Find the coordinates of the stationary points and
determine their nature algebraically.
Reveal answer only
(-1,7) is a maximum TP and (3,-25) is a
minimum TP
Go to full solution
Go to Markers Comments
Go to Basic Differentiation Menu
Go to Main Menu
EXIT
54
BASIC DIFFERENTIATION Question 4
Here is part of the graph of y x3 - 3x2 - 9x
2.
y x3 - 3x2 - 9x 2
Find the coordinates of the stationary points and
determine their nature algebraically.
Return to solution
EXIT
55
Question 4
SPs occur where dy/dx 0
ie 3x2 6x 9 0
Here is part of the graph of y x3 - 3x2 - 9x
2.
ie 3(x2 2x 3) 0
ie 3(x 3)(x 1) 0
Find the coordinates of the stationary points and
determine their nature algebraically.
ie x -1 or x 3
Using y x3 - 3x2 - 9x 2
when x -1
y -1 3 9 2 7
Continue Solution
when x 3
y 27 27 - 27 2 -25
Begin Solution
So stationary points are at
(-1,7) and (3,-25)
Continue Solution
Markers Comments
Basic Differentiation Menu
56
Question 4
We now consider the sign of the gradient either
side of -1 and 3.
Here is part of the graph of y x3 - 3x2 - 9x
2.
x ? -1 ? 3 ?
Find the coordinates of the stationary points and
determine their nature algebraically.
(x 1) - 0
(x - 3) - - - 0
dy/dx 0 - 0
Back to graph
Begin Solution
Hence (-1,7) is a maximum TP and (3,-25) is a
minimum TP
Continue Solution
Markers Comments
Basic Differentiation Menu
57
Markers Comments

• Make the statement
• At stationary points

SPs occur where dy/dx 0
ie 3x2 6x 9 0

• Must attempt to find
• and set equal to zero

ie 3(x2 2x 3) 0
ie 3(x 3)(x 1) 0
ie x -1 or x 3
multiply by the power and reduce the power by 1
Using y x3 - 3x2 - 9x 2
when x -1

• Find the value of y from
• y x3 -3x2-9x2
• not from

y -1 3 9 2 7
when x 3
y 27 27 - 27 2 -25
Next Comment
So stationary points are at
(-1,7) and (3,-25)
Differentiation Menu
Continue Comments
58
Markers Comments

• Justify the nature of each
• stationary point using a table
• of signs

We now consider the sign of the gradient either
side of -1 and 3.
x ? -1 ? 3 ?
(x 1) - 0
Minimum requirement
(x - 3) - - - 0
dy/dx 0 - 0

• State the nature of the
• stationary point
• i.e. Maximum T.P.

Hence (-1,7) is a maximum TP and (3,-25) is a
minimum TP
Next Comment
Differentiation Menu
59
BASIC DIFFERENTIATION Question 5
When a company launches a new product its share
of the market after x months is calculated by the
formula
So after 5 months the share is
S(5) 2/5 4/25
6/25
Find the maximum share of the market that the
company can achieve.
Reveal answer only
Go to full solution
Go to Markers Comments
Go to Basic Differentiation Menu
Go to Main Menu
EXIT
60
BASIC DIFFERENTIATION Question 5
When a company launches a new product its share
of the market after x months is calculated by the
formula
So after 5 months the share is
S(5) 2/5 4/25
6/25
Find the maximum share of the market that the
company can achieve.
Reveal answer only
1/4
Go to full solution
Go to Markers Comments
Go to Basic Differentiation Menu
Go to Main Menu
EXIT
61
Question 5
End points
When a company launches a new product its share
of the market after x months is calculated as
S(2) 1 1 0
There is no upper limit but as x ? ?
S(x) ? 0.
Stationary Points
S(x) 2 - 4 x x2
2x-1 4x-2
Find the maximum share of the market that the
company can achieve.
So S ?(x) -2x-2 8x-3
8 - 2 x3 x2
- 2 8 x2 x3
Begin Solution
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Basic Differentiation Menu
62
Question 5
SPs occur where S ?(x) 0
When a company launches a new product its share
of the market after x months is calculated as
8 - 2 x3 x2
0
8 2 x3 x2
( cross mult!)
or
8x2 2x3
8x2 - 2x3 0
2x2(4 x) 0
Find the maximum share of the market that the
company can achieve.
x 0 or x 4
In required interval
NB x ? 2
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Continue Solution
Continue Solution
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63
Question 5
We now check the gradients either side of X 4
When a company launches a new product its share
of the market after x months is calculated as
x ? 4 ?
S ?(3.9 ) 0.00337
S ?(x) 0 -
S ?(4.1) -0.0029
Find the maximum share of the market that the
company can achieve.
Hence max TP at x 4
So max share of market S(4)
2/4 4/16
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1/2 1/4
Continue Solution
1/4
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Basic Differentiation Menu
64
Markers Comments

• Must look for key word to spot
• the optimisation question i.e.

End points
Maximum, minimum, greatest , least etc.
S(2) 1 1 0
There is no upper limit but as x ? ?
S(x) ? 0.

• Must consider end points
• and stationary points.

Stationary Points
S(x) 2 - 4 x x2
2x-1 4x-2

• Prepare expression for
• differentiation.

So S ?(x) -2x-2 8x-3
8 - 2 x3 x2
- 2 8 x2 x3
Next Comment
Differentiation Menu
Continue Comments
65
Markers Comments

• Must attempt to find
• ( Note No marks are allocated
• for trial and error solution.)

SPs occur where S ?(x) 0
8 - 2 x3 x2
0

• Must attempt to find and
• set equal to zero

8 2 x3 x2
( cross mult!)
or
8x2 2x3
multiply by the power and reduce the power by 1
8x2 - 2x3 0

• Usually easier to solve resulting
• equation using
• cross-multiplication.
• Take care to reject solutions
• outwith the domain.

2x2(4 x) 0
x 0 or x 4
In required interval
NB x ? 2
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Differentiation Menu
Continue Comments
66
Markers Comments

• Must show a maximum value
• using a table of signs.

We now check the gradients either side of X 4
x ? 4 ?
S ?(3.9 ) 0.00337
S ?(x) 0 -
S ?(4.1) -0.0029
Minimum requirement.
Hence max TP at x 4

• State clearly
• Maximum T.P at x 4

So max share of market S(4)
2/4 4/16
Next Comment
1/2 1/4
Differentiation Menu
1/4
67
HIGHER ADDITIONAL QUESTION BANK
You have chosen to study
Recurrence Relations
UNIT 1
Please choose a question to attempt from the
following
1
2
3
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EXIT
68
RECURRENCE RELATIONS Question 1

• A recurrence relation is defined by the formula
  un1 aun b, where -1ltalt1 and u0 20.
• If u1 10 and u2 4 then find the values
  of a and b.
• Find the limit of this recurrence relation as n
  ? ?.

Reveal answer only
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EXIT
69
RECURRENCE RELATIONS Question 1

• A recurrence relation is defined by the formula
  un1 aun b, where -1ltalt1 and u0 20.
• If u1 10 and u2 4 then find the values
  of a and b.
• Find the limit of this recurrence relation as n
  ? ?.

(a)
a 0.6
Reveal answer only
b -2
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(b)
Go to Markers Comments
L -5
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EXIT
70
Question 1
(a)
Using un1 aun b

• A recurrence relation is defined
• by the formula un1 aun b,
• where -1ltalt1 and u0 20.
• If u1 10 and u2 4 then
• find the values of a and b.
• (b) Find the limit of this recurrence
• relation as n ? ?.

we get u1 au0 b
and u2 au1 b
Replacing u0 by 20, u1 by 10 u2 by 4
gives us 20a b 10
and 10a b 4
subtract ? 10a 6
a 0.6
or
Continue Solution
Replacing a by 0.6 in 10a b 4
Begin Solution
gives 6 b 4
Continue Solution
or
b -2
Markers Comments
Recurrence Relations Menu
71
Question 1
(b)
un1 aun b is now un1 0.6un - 2

• A recurrence relation is defined
• by the formula un1 aun b,
• where -1ltalt1 and u0 20.
• If u1 10 and u2 4 then
• find the values of a and b.
• (b) Find the limit of this recurrence
• relation as n ? ?.

This has a limit since -1lt0.6lt1
At this limit, L, un1 un L
So we now have L 0.6 L - 2
or 0.4L -2
or L -2 ? 0.4
or L -20 ? 4
L -5
so
Begin Solution
Continue Solution
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Recurrence Relations Menu
72
Markers Comments

• Must form the two
• simultaneous equations
• and solve.
• u1 is obtained from u0 and
• u2 is obtained from u1 .
• A trial and error solution
• would only score 1 mark.

(a)
Using un1 aun b
we get u1 au0 b
and u2 au1 b
Replacing u0 by 20, u1 by 10 u2 by 4
gives us 20a b 10
and 10a b 4
subtract ? 10a 6
Comments for 1(b)
a 0.6
or
Replacing a by 0.6 in 10a b 4
Next Comment
gives 6 b 4
Recurrence Menu
or
b -2
73
Markers Comments

• Must state condition for limit
• i.e. -1 lt 0.6 lt 1
• At limit L, state
• un1 un L
• Substitute L for un1 and un
• and solve for L.

(b)
un1 aun b is now un1 0.6un - 2
This has a limit since -1lt0.6lt1
At this limit, L, un1 un L
So we now have L 0.6 L - 2
or 0.4L -2
or L -2 ? 0.4
or L -20 ? 4
so
L -5
Next Comment
Recurrence Menu
74
RECURRENCE RELATIONS Question 2
Two different recurrence relations are known to
have the same limit as n ? ?.
The first is defined by the formula un1
-5kun 3. The second is defined by the
formula vn1 k2vn 1. Find the value of
k and hence this limit.
Reveal answer only
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EXIT
75
RECURRENCE RELATIONS Question 2
Two different recurrence relations are known to
have the same limit as n ? ?.
The first is defined by the formula un1
-5kun 3. The second is defined by the
formula vn1 k2vn 1. Find the value of
k and hence this limit.
k 1/3
Reveal answer only
L 9/8
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EXIT
76
Question 2
If the limit is L then as n ? ? we have un1
un L and vn1 vn L
Two different recurrence relations are known to
have the same limit as n ? ?.
The first is defined by the formula un1 -5kun
3. The second is defined by Vn1
k2vn 1. Find the value of k and hence this
limit.
First Sequence
Second Sequence
un1 -5kun 3 becomes
vn1 k2vn 1 becomes
L k2L 1
L -5kL 3
L - k2L 1
L 5kL 3
L(1 - k2) 1
L(1 5k) 3
L 1 . . (1 - k2)
L 3 . . (1 5k)
Begin Solution
Continue Solution
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Recurrence Relations Menu
77
Question 2
L 1 . . (1 - k2)
L 3 . . (1 5k)
Two different recurrence relations are known to
have the same limit as n ? ?.
The first is defined by the formula un1 -5kun
3. The second is defined by Vn1
k2vn 1. Find the value of k and hence this
limit.
. 3 . . 1 . .
(1 5k) (1 k2)
It follows that
Cross multiply to get 1 5k 3 3k2
This becomes 3k2 5k 2 0
Or (3k 1)(k 2) 0
So k 1/3 or k -2
Since -1ltklt1 then k 1/3
Begin Solution
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Recurrence Relations Menu
78
Question 1
Since -1ltklt1 then k 1/3
Two different recurrence relations are known to
have the same limit as n ? ?.
The first is defined by the formula un1 -5kun
3. The second is defined by Vn1
k2vn 1. Find the value of k and hence this
limit.
Using
L 1 . . (1 - k2)
gives us L . 1 . . (1
1/9)
or L 1 ? 8/9
ie L 9/8
Begin Solution
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79
Markers Comments

• Since both recurrence relations
• have the same limit, L, find the
• limit for both and set equal.

If the limit is L then as n ? ? we have un1
un L and vn1 vn L
First Sequence
Second Sequence
un1 -5kun 3 becomes
vn1 k2vn 1 becomes
L k2L 1
L -5kL 3
L - k2L 1
Continue Comments
L 5kL 3
L(1 - k2) 1
L(1 5k) 3
L 3 . . (1 5k)
L 1 . . (1 - k2)
Next Comment
Recurrence Menu
80
Markers Comments

• Since both recurrence relations
• have the same limit, L, find the
• limit for both and set equal.

L 1 . . (1 - k2)
L 3 . . (1 5k)

• Only one way to solve resulting
• equation i.e. terms to the left,
• form the quadratic and factorise.
• State clearly the condition for
• the recurrence relation to
• approach a limit. -1lt k lt 1.
• Take care to reject the
• solution which is outwith the
• range.

. 3 . . 1 . .
(1 5k) (1 k2)
It follows that
Cross multiply to get 1 5k 3 3k2
This becomes 3k2 5k 2 0
Or (3k 1)(k 2) 0
So k 1/3 or k -2
Since -1ltklt1 then k 1/3
Next Comment
Recurrence Menu
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81
Markers Comments
Find L from either formula.
Since -1ltklt1 then k 1/3
Using
L 1 . . (1 - k2)
gives us L . 1 . . (1
1/9)
or L 1 ? 8/9
ie L 9/8
Next Comment
Recurrence Menu
82
RECURRENCE RELATIONS Question 3
A man plants a row of fast growing trees between
his own house and his neighbours. These trees
are known grow at a rate of 1m per annum so
cannot be allowed to become too high. He
therefore decides to prune 30 from their height
at the beginning of each year.

• Using the 30 pruning scheme what height should
  he expect the trees to grow to in the long run?
• The neighbour is concerned at the growth rate and
  asks that the trees be kept to a maximum height
  of 3m. What percentage should be pruned to
  ensure that this happens?

Reveal answer only
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EXIT
83
RECURRENCE RELATIONS Question 3
A man plants a row of fast growing trees between
his own house and his neighbours. These trees
are known grow at a rate of 1m per annum so
cannot be allowed to become too high. He
therefore decides to prune 30 from their height
at the beginning of each year.

• Using the 30 pruning scheme what height should
  he expect the trees to grow to in the long run?
• The neighbour is concerned at the growth rate and
  asks that the trees be kept to a maximum height
  of 3m. What percentage should be pruned to
  ensure that this happens?

Reveal answer only
Go to full solution
331/3
(b)
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EXIT
84
Question 3
(a) Removing 30 leaves 70 or 0.7
The trees are known grow at a rate of 1m
per annum. He therefore decides to prune
30 from their height at the beginning of
each year.
If Hn is the tree height in year n then Hn1
0.7Hn 1
Since -1lt0.7lt1 this sequence has a limit, L.
At the limit Hn1 Hn L

• Using the 30 pruning scheme
• what height should he expect
• the trees to grow to in the long
• run?

So L 0.7L 1
or 0.3 L 1
ie L 1 ? 0.3
10 ? 3
31/3
Begin Solution
Continue Solution
Continue Solution
Markers Comments
Recurrence Relation Menu
85
Question 3

• If fraction left after pruning is a and
• we need the limit to be 3

The trees are known grow at a rate of 1m
per annum. He therefore decides to prune
30 from their height at the beginning of
each year.
then we have 3 a X 3 1
or 3a 2
or a 2/3
(b) The neighbour asks that the trees be
kept to a maximum height of 3m. What
percentage should be pruned to ensure that this
happens?
This means that the fraction pruned is
1/3 or 331/3
Begin Solution
Continue Solution
Markers Comments
Recurrence Relation Menu
86
Markers Comments

• Do some numerical work to
• get the feel for the problem.

(a) Removing 30 leaves 70 or 0.7
H0 1 (any value acceptable) H1 0.7 x1 1
1.7 H2 0.7 x1.7 1 2.19 etc.
If Hn is the tree height in year n then Hn1
0.7Hn 1

• State the recurrence relation,
• with the starting value.
• Hn1 0.7 Hn 1, H0 1
• State the condition for the limit
• -1lt 0.7lt 1
• At limit L, state Hn1 Hn L
• Substitute L for Hn1 and Hn
• and solve for L.

Since -1lt0.7lt1 this sequence has a limit, L.
At the limit Hn1 Hn L
So L 0.7L 1
or 0.3 L 1
ie L 1 ? 0.3
10 ? 3
Next Comment
Recurrence Menu
Continue Solution
87
Markers Comments

• Since we know the limit we are
• working backwards to .
• L 0.7L 1
• New limit, L 3 and multiplier a
• 3 a x3 1
• etc.
• Take care to subtract from 1
• to get fraction pruned.

• If fraction left after pruning is a and
• we need the limit to be 3

then we have 3 a X 3 1
or 3a 2
or a 2/3
This means that the fraction pruned is
1/3 or 331/3
Next Comment
Recurrence Menu
88
HIGHER ADDITIONAL QUESTION BANK
You have chosen to study
Trig Graphs Equations
UNIT 1
Please choose a question to attempt from the
following
1
2
3
Back to Unit 1 Menu
EXIT
89
TRIG GRAPHS EQUATIONS Question 1
This diagram shows the graph of y acosbx
c. Determine the values of a, b c.
Reveal answer only
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EXIT
90
TRIG GRAPHS EQUATIONS Question 1
This diagram shows the graph of y acosbx
c. Determine the values of a, b c.
Reveal answer only
a 3
Go to full solution
b 2
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c -1
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EXIT
91
Question 1
a ½(max min)
This diagram shows the graph of y acosbx
c. Determine the values of a, b c.
½(2 (-4))
½ X 6
3
Period of graph ? so two complete sections
between 0 2?
ie b 2
For 3cos() max 3 min -3.
This graph max 2 min -4.
ie 1 less
Begin Solution
so c -1
Continue Solution
Markers Comments
Trig Graphs etc. Menu
92
Markers Comments

• The values chosen for a,b and c
• must be justified.
• Possible justification of a 3
• a 1/2(max - min)
• y cosx graph stretched by a
• factor of 3 etc.
• Possible justification of b 2
• Period of graph
• 2 complete cycles in 2
• 2 2 2 complete
• cycles in 2 etc.
• Possible justification for c -1
• 3cos max 3, min -3
• This graph max 2, min -4
• i.e. -1 c -1
• y 3cosx graph slide down
• 1 unit etc.

a ½(max min)
½(2 (-4))
½ X 6
3
Period of graph ? so two complete sections
between 0 2?
ie b 2
For 3cos() max 3 min -3.
This graph max 2 min -4.
ie 1 less
so c -1
Next Comment
Trig Graphs Menu
93
TRIG GRAPHS EQUATIONS Question 2
Solve ?3tan2? 1 0 ( where 0 lt ? lt ? ).
Reveal answer only
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EXIT
94
TRIG GRAPHS EQUATIONS Question 2
Solve ?3tan2? 1 0 ( where 0 lt ? lt ? ).
? 5?/12
Reveal answer only
? 11?/12
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EXIT
95
Question 2
?3tan2? 1 0
?3tan2? -1
Solve ?3tan2? 1 0 ( where 0 lt ? lt ? ).
tan2? -1/?3
Q2 or Q4
tan -1(1/?3) ?/6
Q2 angle ? - ?/6
so 2? 5?/6
ie ? 5?/12
Q4 angle 2? - ?/6
so 2? 11?/6
Begin Solution
ie ? 11?/12
Continue Solution
Markers Comments
tan2? repeats every ?/2 radians but repeat values
are not in interval.
Trig Graphs etc. Menu
96
Markers Comments

• Full marks can be obtained by
• working in degrees and
• changing final answers back
• to radians.

?3tan2? 1 0

• Use the positive value when
• finding tan-1.

?3tan2? -1
tan2? -1/?3
Q2 or Q4

• Use the quadrant rule to find
• the solutions.

tan -1(1/?3) ?/6

• Must learn special angles or
• be able to calculate from
• triangles.

Q2 angle ? - ?/6
so 2? 5?/6

• Take care to reject solutions
• outwith domain.

ie ? 5?/12
Q4 angle 2? - ?/6
so 2? 11?/6
Next Comment
ie ? 11?/12
Trig Graphs Menu
tan2? repeats every ?/2 radians but repeat values
are not in interval.
97
TRIG GRAPHS EQUATIONS Question 3
The diagram shows a the graph of a sine function
from 0 to 2?/3.
(a) State the equation of the graph.
(b) The line y 2 meets the graph at
points P Q. Find the coordinates of these two
points.
Reveal answer only
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EXIT
98
TRIG GRAPHS EQUATIONS Question 3
The diagram shows a the graph of a sine function
from 0 to 2?/3.
y 2
2?/3
(a) State the equation of the graph.
(b) The line y 2 meets the graph at
points P Q. Find the coordinates of these two
points.
Graph is y 4sin3x
Reveal answer only
P is (?/18, 2) and Q is (5?/18, 2).
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EXIT
99
Question 3

• One complete wave from 0 to 2?/3
• so 3 waves from 0 to 2?.

The diagram shows a the graph of a sine function
from 0 to 2?/3.
(a) State the equation of the graph.
Max/min 4
4sin()
Graph is y 4sin3x
Begin Solution
Continue Solution
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Markers Comments
Trig Graphs etc. Menu
100
Question 3
Graph is y 4sin3x
(b) The line y 2 meets the graph at
points P Q. Find the coordinates of these two
points.
(b) At P Q y 4sin3x and y 2
so 4sin3x 2
or sin3x 1/2
Q1 or Q2
sin-1(1/2) ?/6
Q1 angle ?/6
so 3x ?/6
ie x ?/18
Q2 angle ? - ?/6
Begin Solution
so 3x 5?/6
Continue Solution
ie x 5?/18
Markers Comments
Trig Graphs etc. Menu
101
Markers Comments

• Identify graph is of the form
• y asinbx.
• Must justify choice of a and b.
• Possible justification of a

• One complete wave from 0 to 2?/3
• so 3 waves from 0 to 2?.

Max/min 4
4sin()
Max 4, Min -4 4sin() y sinx
stretched by a
factor of 4
Graph is y 4sin3x

• Possible justification for b

Period 3 waves from 0 to
Next Comment
Trig Graphs Menu
102
Markers Comments

• At intersection y1 y2
• 4sin3x 2
• Solve for sin3x

Graph is y 4sin3x
(b) At P Q y 4sin3x and y 2
so 4sin3x 2

• Use the quadrant rule to find
• the solutions.

or sin3x 1/2
Q1 or Q2
sin-1(1/2) ?/6

• Must learn special angles or be
• able to calculate from triangles.

Q1 angle ?/6
so 3x ?/6
ie x ?/18
Q2 angle ? - ?/6

• Take care to state coordinates.

so 3x 5?/6
Next Comment
ie x 5?/18
Trig Graphs Menu
103
HIGHER ADDITIONAL QUESTION BANK
You have chosen to study
Functions Graphs
UNIT 1
Please choose a question to attempt from the
following
1
2
3
4
Back to Unit 1 Menu
EXIT
104
FUNCTIONS GRAPHS Question 1
This graph shows the the function y g(x).
Make a sketch of the graph of the function y
4 g(-x).
Reveal answer only
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