HIGHER – ADDITIONAL QUESTION BANK
Please decide which Unit you would like to revise:
UNIT 1
Straight Line
Functions & Graphs
Trig Graphs & Equations
Basic Differentiation
Recurrence Relations
EXIT
UNIT 2
Polynomials
Quadratic Functions
Integration
Addition Formulae
The Circle
UNIT 3
Vectors
Further Calculus
Exponential /
Logarithmic Functions
The Wave Function
HIGHER – ADDITIONAL QUESTION BANK
UNIT 1 :
Straight Line
Trig Graphs
& Equations
Functions &
Graphs
Basic
Differentiation
Recurrence
Relations
EXIT
HIGHER – ADDITIONAL QUESTION BANK
You have chosen to study:
UNIT 1 :
Straight Line
Please choose a question to attempt from the following:
1
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2
3
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Unit 1 Menu
4
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
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EXIT
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 Marker’s Comments
Go to Straight Line Menu
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EXIT
y = -5/3x - 6
Question 1
3x – 5y = 4
Find the equation of the
3x - 4 = 5y
straight line which is
5y = 3x - 4
perpendicular to the line with
equation 3x – 5y = 4 and
which passes through
the point (-6,4).
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(5)
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)
y – 4 = -5/3x - 10
y = -5/3x - 6
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
•State the gradient clearly.
(5)
5y = 3x - 4
• State the condition for
perpendicular lines m1 m2 = -1.
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)
y – 4 = -5/3x - 10
y =
-5/ x
3
•When finding m2 simply invert
and change the sign on m1
m1 =
3
5
m2 =
-5
3
• Use the y - b = m(x - a) form
to obtain the equation of the
line.
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-6
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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
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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
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EXIT
y = -2x + 7
Question 2
Find the equation of the
8x + 4y – 7 = 0
4y = -8x + 7
(4)
straight line which is parallel
to the line with equation
y = -2x + 7/4
8x + 4y – 7 = 0 and which
Using y = mx + c , gradient of line is -2
passes through the point
So required gradient = -2 as parallel lines
have equal gradients.
(5,-3).
We now have (a,b) = (5,-3) & m = -2.
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Using
y – b = m(x – a)
We get
y – (-3) = -2(x – 5)
y + 3 = -2x + 10
y = -2x + 7
Markers Comments
8x + 4y – 7 = 0
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)
y + 3 = -2x + 10
y = -2x + 7
• An attempt must be made to
put the original equation into
the form y = mx + c to
read off the gradient.
• State the gradient clearly.
• State the condition for
parallel lines m1 = m2
• Use the y - b = m(x - a) form
to obtain the equation of
the line.
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STRAIGHT LINE : Question 3
Y
In triangle ABC, A is (2,0),
B is (8,0) and C is (7,3).
C
(a) Find the gradients of AC and BC.
(b) Hence find the size of ACB.
Reveal answer only
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A
B
X
STRAIGHT LINE : Question 3
Y
In triangle ABC, A is (2,0),
B is (8,0) and C is (7,3).
C
(a) Find the gradients of AC and BC.
(b) Hence find the size of ACB.
Reveal answer only
A
(a)
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mAC
= 3/5
mBC =
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X
B
(b)
= 77.4°
-3
Question 3
(a) Using the gradient formula: m 
In triangle ABC, A is (2,0),
B is (8,0) and C is (7,3).
(a) Find the gradients of AC
and BC.
mAC = 3 – 0
7-2
= 3/5
mBC = 3 – 0
7-8
=
y2  y1
x2  x1
-3
(b) Hence find the size of ACB. (b) Using tan = gradient
Y
If tan = 3/5 then CAB = 31.0°
C
A
B
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X
If tan = -3
then CBX = (180-71.6)°
= 108.4 o
so ABC = 71.6°
Hence :
ACB = 180° – 31.0° – 71.6°
= 77.4°
y2  y1
a) Using the gradient formula: m 
x2  x1
mAC = 3 – 0
7-2
=
mBC = 3 – 0
7-8
=
3/
5
• In calculating gradients state
the gradient formula.
• 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.
-3
b) Using tan = gradient
If tan = 3/5
Markers Comments
• If no diagram is given draw a
neat labelled diagram.
then CAB = 31.0°
B m = tanØ °
AB
If tan = -3 then CBX = (180-71.6)°
so ABC = 71.6°
= 108.4 o
Ø°
Ø ° = tan-1 mAB
A
Hence :
ACB = 180° – 31.0° – 71.6°
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= 77.4°
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STRAIGHT LINE : Question 4
In triangle PQR the vertices
are P(4,-5), Q(2,3) and R(10-1).
Find
(a) the equation of the line e, the
median from R of triangle PQR.
(b) the equation of the line f, the
perpendicular bisector of QR.
Y
Q(2,3)
X
R(10,-1)
P(4,-5)
Reveal answer only
(c) The coordinates of the point of
intersection of lines e & f.
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STRAIGHT LINE : Question 4
In triangle PQR the vertices
are P(4,-5), Q(2,3) and R(10-1).
Find
(a) the equation of the line e, the
median from R of triangle PQR.
(b) the equation of the line f, the
perpendicular bisector of QR.
Y
Q(2,3)
X
R(10,-1)
P(4,-5)
Reveal answer only
(c) The coordinates of the point of
intersection of lines e & f.
(a)
y = -1
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(b)
y = 2x – 11
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(c)
(5,-1)
Question 4 (a)
In triangle PQR the vertices
are P(4,-5), Q(2,3) and R(10-1).
Find
(a) the equation of the line e, the
median from R of triangle PQR.
Y
Q(2,3)
(a) Midpoint of PQ is (3,-1): let’s call this S
Using the gradient formula m = y2 – y1
x2 – x 1
mSR = -1 – (-1) = 0 (ie line is horizontal)
10 - 3
Since it passes through (3,-1)
equation of e is
y = -1
X
R(10,-1)
P(4,-5)
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Solution to 4 (b)
Question 4 (b)
(b) Midpoint of QR is (6,1)
In triangle PQR the vertices
are P(4,-5), Q(2,3) and R(10-1).
Find
(b) the equation of the line f,
the perpendicular bisector of QR.
Y
mQR = 3 – (-1) =
2 - 10
4/
-8
required gradient = 2
= - 1/ 2
(m1m2 = -1)
Using y – b = m(x – a) with (a,b) = (6,1)
&m=2
Q(2,3)
X
R(10,-1)
we get
y – 1 = 2(x – 6)
so f is
y = 2x – 11
P(4,-5)
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Solution to 4 (c)
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).
Find
(c) The coordinates of the point of
intersection of lines e & f.
Y
Q(2,3)
X
R(10,-1)
P(4,-5)
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so
2x – 11 = -1
ie
2x = 10
ie
x=5
Point of intersection is (5,-1)
Markers Comments
a) Midpoint of PQ is (3,-1): let’s call this S
Using the gradient formula m = y2 – y1
x2 – x1
mSR = -1 – (-1)
(ie line is horizontal)
10 - 3
• If no diagram is given
draw a neat labelled diagram.
•Sketch the median and the
perpendicular bisector
median
y
Perpendicular bisector
Since it passes through (3,-1)
equation of e is
y = -1
Q
R
x
P
Comments for 4 (b)
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Markers Comments
• To find midpoint of QR
(b) Midpoint of QR is (6,1)
mQR = 3 – (-1) =
2 - 10
4/
= - 1/2
-8
required gradient = 2
2 + 10 , 3 + (-1)
2
2
(m1m2 = -1)
Using y – b = m(x – a) with (a,b) = (6,1)
&m=2
we get
so f is
• Look for special cases:
Horizontal lines in the form y = k
Vertical lines in the form x = k
y
Q
y – 1 = 2(x – 6)
R
y = 2x – 11
x
P
Comments for 4 (c)
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Markers Comments
(c) e & f meet when y = -1 & y = 2x -11
so
2x – 11 = -1
ie
2x = 10
ie
x=5
Point of intersection is (5,-1)
• To find the point of
intersection of the two
lines solve the two
equations:
y = -1
y = 2x - 11
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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.
(c) The point of intersection of the
altitude and median.
Y
E(6,-3)
G(2,-5)
X
F(12,-5)
Reveal answer only
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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.
(c) The point of intersection of the
altitude and median.
Y
E(6,-3)
G(2,-5)
X
F(12,-5)
Reveal answer only
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(a)
x=6
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(b)
x + 8y + 28 = 0
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(c)
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(6,-4.25)
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Question 5(a)
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.
Y
X
E(6,-3)
G(2,-5)
F(12,-5)
Begin Solution
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y2  y1
(a) Using the gradient formula m 
x2  x1
mFG = -5 – (-5)
=0
12
-2
(ie line is horizontal so altitude is vertical)
Altitude is vertical line through (6,-3)
ie
x=6
Solution to 5 (b)
Question 5(b)
(b) Midpoint of EG is (4,-4)- let’s call this H
In triangle EFG the vertices are
E(6,-3), F(12,-5) and G(2,-5).
Find
(b) the equation of the median
from vertex F.
Y
X
E(6,-3)
G(2,-5)
F(12,-5)
mFH = -5 – (-4) =
12 - 4
-1/
8
Using y – b = m(x – a) with (a,b) = (4,-4)
& m = -1/8
we get
y – (-4) = -1/8(x – 4)
or
8y + 32 = -x + 4
Median is
x + 8y + 28 = 0
Begin Solution
Solution to 5 (c)
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(X8)
Question 5(c)
(c)
In triangle EFG the vertices are
E(6,-3), F(12,-5) and G(2,-5).
Find
(c) The point of intersection of the
altitude and median.
Y
X
E(6,-3)
G(2,-5)
F(12,-5)
Begin Solution
Continue Solution
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Lines meet when x = 6 & x + 8y + 28 = 0
put x =6 in 2nd equation
8y + 34 = 0
ie
8y = -34
ie
y = -4.25
Point of intersection is (6,-4.25)
Markers Comments
(a) Using the gradient formula m  y2  y1
x2  x1
mFG = -5 – (-5) = 0
12
-2
(ie line is horizontal so altitude is vertical)
• If no diagram is given draw a
neat labelled diagram.
• Sketch the altitude and
the median.
y
Altitude is vertical line through (6,-3)
ie
x
E
x=6
F
G
median
Comments for 5 (b)
altitude
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Markers Comments
• To find midpoint of EG
(b) Midpoint of EG is (4,-4)- call this H
mFH = -5 – (-4) =
12 - 4
-1/
8
Using y – b = m(x – a) with (a,b) = (4,-4)
& m = -1/8
we get
y – (-4) = -1/8(x – 4)
or
8y + 32 = -x + 4
Median is
x + 8y + 28 = 0
2 + 6 , -3 + (-5)
H
2
2
• Look for special cases:
Horizontal lines in the form y = k
Vertical lines in the form
x=k
y
(X8)
x
E
F
G
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Comments for 5 (c)
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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
put x =6 in 2nd equation
8y + 34 = 0
ie
8y = -34
ie
y = -4.25
x=6
x + 8y = -28
Point of intersection is (6,-4.25)
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HIGHER – ADDITIONAL QUESTION BANK
You have chosen to study:
UNIT 1 :
Basic
Differentiation
Please choose a question to attempt from the following:
1
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2
3
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Unit 1 Menu
4
5
BASIC DIFFERENTIATION : Question 1
Find the equation of the tangent to the curve
at the point where x = 4.
16
y x
x
Reveal answer only
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(x>0)
BASIC DIFFERENTIATION : Question 1
Find the equation of the tangent to the curve
at the point where x = 4.
16
y x
x
Reveal answer only
y = 5/4x – 7
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(x>0)
Question 1
Find the equation of the tangent to
NB: a tangent is a line so we need a
point of contact and a gradient.
the curve
Point
y = x – 16
x
(x>0)
at the point where x = 4.
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If x = 4 then y = 4 – 16
4
= 2 – 4 = -2
so (a,b) = (4,-2)
Gradient: y = x – 16 = x1/2 – 16x -1
x
dy/ = 1/ x-1/2 + 16x-2 = 1
+ 16
dx
2
2x
x2
If x = 4 then:
dy/
1 + 16
dx =
24
16
= ¼ + 1 = 5/4
Question 1
If x = 4 then:
Find the equation of the tangent to
dy/ =
dx
the curve
y = x – 16
x
(x>0)
1 + 16
24
16
Gradient of tangent = gradient of curve
at the point where x = 4.
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= ¼ + 1 = 5/4
so m = 5/4 .
We now use
y – b = m(x – a)
this gives us
y – (-2) = 5/4(x – 4)
or
y+2
or
y = 5/4x – 7
= 5/4x – 5
Markers Comments
• Prepare expression for differentiation.
NB: a tangent is a line so we need a
point of contact and a gradient.
1
2
16
y = x   x  16 x 1
x
• Find gradient of the tangent
Point
If x = 4 then y = 4 – 16
4
= 2 – 4 = -2
so (a,b) = (4,-2)
Gradient: y = x – 16 = x1/2 – 16x -1
x
dy/ = 1/ x-1/2 + 16x-2 = 1
+ 16
dx
2
2x
x2
If x = 4 then:
dy/
=
1 + 16
dx
24
16
= ¼ + 1 = 5/4
using rule:
dy
dx
“multiply by the power and reduce
the power by 1”
• Find gradient =
dy
dx
at x = 4.
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Markers Comments
• Find y coordinate at x = 4 using:
If x = 4 then:
dy/ =
dx
1 + 16
24
16
= ¼ + 1 = 5/4
Gradient of tangent = gradient of curve
so m = 5/4 .
16
16
y  x   4   2
x
4
• Use m = 5/4 and (4,-2) in
y - b = m(x - a)
We now use
y – b = m(x – a)
this gives us
y – (-2) = 5/4(x – 4)
or
y+2
or
y = 5/4x – 7
= 5/4x – 5
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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
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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
(2,4)
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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
Line
Using gradient = tan
we get gradient of line = tan135°
= -tan45°
positive direction of the X-axis.
= -1
Curve
Continue Solution
Gradient of curve = dy/dx = 2x - 5
It now follows that
2x – 5 = -1
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Or
Or
2x = 4
x = 2
Question 2
Find the coordinates of the point
Using
y = x2 – 5x + 10 with x = 2
on the curve y = x2 – 5x + 10
we get y = 22 – (5 X 2) + 10
where the tangent to the curve
ie
y=4
makes an angle of 135° with the
positive direction of the X-axis.
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So required point is
(2,4)
NB: gradient of line = gradient of curve
Line
Using gradient = tan
we get gradient of line = tan135°
= -tan45°
= -1
Curve
Markers Comments
dy
• Find gradient of the tangent
dx
using rule:
“multiply by the power and reduce
the power by 1”
• 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.
y
Gradient of curve = dy/dx = 2x - 5
m = tan135 ° = -1
It now follows that
2x – 5 = -1
Or
Or
2x = 4
x = 2
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135 °
x
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Markers Comments
dy
• Set m =
dx
i.e. 2x - 5 = -1
It now follows that
2x – 5 = -1
Or
Or
Using
2x = 4
x = 2
and solve for x.
y = x2 – 5x + 10 with x = 2
we get y = 22 – (5 X 2) + 10
ie
y=4
So required point is
(2,4)
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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
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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
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y
y = g(x)
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
r
(p,q)
either side of 0 and p:
x

0

p

g(x)
-
0
-
0
+
Make a sketch of the graph of
new y-values
y = g(x).
Begin Solution
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Click for graph
Question 3
y = g(x)
This now gives us the following graph
y = g(x)
y
r
0
p
(p,q)
Make a sketch of the graph of
y = g(x).
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dy
Stationary points occur at x = 0 and x = p.
To sketch the graph of f ' ( x ) 
dx
the gradient function:
1 Mark the stationary points on the
x axis i.e. f ' ( x)  dy  0
dx
y = g(x)
y
y
f ' ( x)
0
x
p
a
x
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dy
Stationary points occur at x = 0 and x = p.
To sketch the graph of f ' ( x ) 
dx
the gradient function:
1 Mark the stationary points on the
x axis i.e. f ' ( x)  dy  0
(We can ignore r.)
dx
We now consider the sign of the gradient
either side of 0 and p:
x

0

p

g(x)
-
0
-
0
+
new y-values
2 For each interval decide if the
value of f ' ( x)  dy is - or +
y
dx
'
f ( x)
+
a
-
x
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Markers Comments
dy
Stationary points occur at x = 0 and x = p.
y = g(x)
To sketch the graph of f ' ( x ) 
dx
the gradient function:
1 Mark the stationary points on the
x axis i.e. f ' ( x)  dy  0
dx
y
2 For each interval decide if the
value of f ' ( x)  dy is - or +
dx
0
x
p
3 Draw in ycurve' to fit information
f ( x)
+
a
• In any curve sketching
question use a ruler and
annotate the sketch
i.e. label all known coordinates.
x
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BASIC DIFFERENTIATION : Question 4
Here is part of the graph of
y = x3 - 3x2 - 9x + 2.
Find the coordinates of the
stationary points and determine
their nature algebraically.
Reveal answer only
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y = x3 - 3x2 - 9x + 2
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
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EXIT
BASIC DIFFERENTIATION : Question 4
Here is part of the graph of
y = x3 - 3x2 - 9x + 2.
Find the coordinates of the
stationary points and determine
their nature algebraically.
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EXIT
y = x3 - 3x2 - 9x + 2
Question 4
Here is part of the graph of
y = x3 - 3x2 - 9x + 2.
Find the coordinates of the
stationary points and determine
their nature algebraically.
SPs occur where
dy/
dx
= 0
ie
3x2 – 6x – 9 = 0
ie
ie
3(x2 – 2x – 3) = 0
3(x – 3)(x + 1) = 0
ie
x = -1 or x = 3
Using y = x3 - 3x2 - 9x + 2
when x = -1
Continue Solution
y = -1 – 3 + 9 + 2 = 7
& when x = 3
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y = 27 – 27 - 27 + 2 = -25
So stationary points are at
(-1,7) and (3,-25)
Question 4
Here is part of the graph of
y = x3 - 3x2 - 9x + 2.
Find the coordinates of the
stationary points and determine
their nature algebraically.
Back to graph
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We now consider the sign of the gradient
either side of -1 and 3.
x

-1

3

(x + 1)
-
0
+
+
+
(x - 3)
-
-
-
0
+
dy/
dx
+
0
-
0
+
Hence (-1,7) is a maximum TP
and (3,-25) is a minimum TP
Markers Comments
SPs occur where
dy/
dx
= 0
ie
3x2 – 6x – 9 = 0
ie
ie
3(x2 – 2x – 3) = 0
3(x – 3)(x + 1) = 0
ie
x = -1 or x = 3
Using y = x3 - 3x2 - 9x + 2
when x = -1
y = -1 – 3 + 9 + 2 = 7
& when x = 3
y = 27 – 27 - 27 + 2 = -25
So stationary points are at
(-1,7) and (3,-25)
Continue Comments
• Make the statement:
“At stationary points dy  0
“
dx
• Must attempt to find dy
dx
and set equal to zero
“multiply by the power and
reduce the power by 1”
• Find the value of y from
y = x3 -3x2-9x+2
not from dy
dx
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Markers Comments
We now consider the sign of the gradient
either side of -1 and 3.
x

-1

3

• Justify the nature of each
stationary point using a table
of “signs”
x
dy
dx
-1
+
0
-
(x + 1)
-
0
+
+
+
(x - 3)
-
-
-
0
+
Minimum requirement
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
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BASIC DIFFERENTIATION : Question 5
When a company launches a new product its share of the market after x
months is calculated by the formula
S(x) = 2 - 4
x
So after 5 months the share is
(x  2)
x2
S(5) = 2/5 – 4/25 = 6/25
Find the maximum share of the market that the company can achieve.
Reveal answer only
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BASIC DIFFERENTIATION : Question 5
When a company launches a new product its share of the market after x
months is calculated by the formula
S(x) = 2 - 4
x
So after 5 months the share is
(x  2)
x2
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
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EXIT
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Question 5
End points
When a company launches a new
product its share of the market
after x months is calculated as:
S(x) = 2 - 4
(x  2)
x x2
S(2) = 1 – 1 = 0
There is no upper limit but as x  
S(x)  0.
Stationary Points
S(x) = 2 - 4
Find the maximum share of the
market that the company can
achieve.
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= 2x-1 – 4x-2
x x2
So S (x) = -2x-2 + 8x-3
= -2 +8 = 8 - 2
x2 x3
x3 x2
Continue Solution
Question 5
When a company launches a new
SPs occur where S (x) = 0
product its share of the market
after x months is calculated as:
S(x) = 2 - 4
or
(x  2)
x x2
8 - 2 =0
x3 x2
8 = 2
( cross mult!)
3
2
x x
8x2 = 2x3
Find the maximum share of the
8x2 - 2x3 = 0
2x2(4 – x) = 0
market that the company can
x = 0 or x = 4
achieve.
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NB: x  2
In required interval
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Continue Solution
Question 5
When a company launches a new
product its share of the market
after x months is calculated as:
S(x) = 2 - 4
(x  2)
We now check the gradients either side of
X=4

4

S (3.9 ) = 0.00337…
S (x) +
0
-
S (4.1) = -0.0029…
x
x x2
Find the maximum share of the
Hence max TP at x = 4
market that the company can
achieve.
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So max share of market
= S(4) = 2/4 – 4/16
= 1/2 – 1/4
=
1/
4
Markers Comments
• Must look for key word to spot
the optimisation question i.e.
Maximum, minimum, greatest ,
least etc.
End points
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
= 2x-1 – 4x-2
x x2
• Prepare expression for
differentiation.
So S (x) = -2x-2 + 8x-3
= -2 +8 = 8 - 2
x2 x3
x3 x2
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Markers Comments
dy
 S(x)
• Must attempt to find
dx
SPs occur where S (x) = 0
8 - 2 =0
x3 x2
8 = 2
( cross mult!)
3
2
x x
8x2 = 2x3
or
8x2 - 2x3 = 0
2x2(4 – x) = 0
x = 0 or x = 4
NB: x  2
In required interval
( Note: No marks are allocated
for trial and error solution.)
dy
• Must attempt to find
dx
set equal to zero
and
“multiply by the power and
reduce the power by 1”
• Usually easier to solve resulting
equation using
cross-multiplication.
• Take care to reject “solutions”
outwith the domain.
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Markers Comments
We now check the gradients either side of
X=4

4

S (x) +
0
-
x
S (3.9 ) = 0.00337…
S (4.1) = -0.0029…
•Must show a maximum value
using a table of “signs”.
x
4
S(x)
+
0
-
Minimum requirement.
Hence max TP at x = 4
So max share of market
• State clearly:
Maximum T.P at x = 4
= S(4) = 2/4 – 4/16
=
=
1/
2
–
1/
4
1/
4
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HIGHER – ADDITIONAL QUESTION BANK
You have chosen to study:
UNIT 1 :
Recurrence
Relations
Please choose a question to attempt from the following:
1
EXIT
2
3
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RECURRENCE RELATIONS : Question 1
A recurrence relation is defined by the formula un+1 = aun + b,
where -1<a<1 and u0 = 20.
(a) If u1 = 10 and u2 = 4 then find the values of a and b.
(b) Find the limit of this recurrence relation as n  .
Reveal answer only
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RECURRENCE RELATIONS : Question 1
A recurrence relation is defined by the formula un+1 = aun + b,
where -1<a<1 and u0 = 20.
(a) If u1 = 10 and u2 = 4 then find the values of a and b.
(b) Find the limit of this recurrence relation as n  .
(a)
Reveal answer only
b = -2
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a = 0.6
(b)
L = -5
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Question 1
A recurrence relation is defined
by the formula un+1 = aun + b,
where -1<a<1 and u0 = 20.
(a) If u1 = 10 and u2 = 4 then
find the values of a and b.
(b) Find the limit of this recurrence
relation as n  .
Continue Solution
(a)
Using
un+1 = 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 
or
10a = 6
a = 0.6
Replacing a by 0.6 in 10a + b = 4
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gives
6+b=4
or
b = -2
Question 1
(b)
A recurrence relation is defined
un+1 = aun + b
is now un+1 = 0.6un - 2
by the formula un+1 = aun + b,
This has a limit since -1<0.6<1
where -1<a<1 and u0 = 20.
At this limit, L,
un+1 = un = L
So we now have
L = 0.6 L - 2
(a) If u1 = 10 and u2 = 4 then
find the values of a and b.
(b) Find the limit of this recurrence
relation as n  .
or
or
L = -2  0.4
or
L = -20  4
so
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0.4L = -2
L = -5
Markers Comments
(a)
Using
un+1 = 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 
or
10a = 6
a = 0.6
• 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.
Comments for 1(b)
Replacing a by 0.6 in 10a + b = 4
gives
6+b=4
or
b = -2
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Markers Comments
• Must state condition for limit
i.e. -1 < 0.6 < 1
(b)
un+1 = aun + b
is now un+1 = 0.6un - 2
This has a limit since -1<0.6<1
At this limit, L,
un+1 = un = L
So we now have
L = 0.6 L - 2
or
• Substitute L for un+1 and un
and solve for L.
0.4L = -2
or
L = -2  0.4
or
L = -20  4
so
• At limit L, state
un+1 = un = L
L = -5
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RECURRENCE RELATIONS : Question 2
Two different recurrence relations are known to have the same
limit as n  .
The first is defined by the formula
un+1 = -5kun + 3.
The second is defined by the formula
vn+1 = k2vn + 1.
Find the value of k and hence this limit.
Reveal answer only
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RECURRENCE RELATIONS : Question 2
Two different recurrence relations are known to have the same
limit as n  .
The first is defined by the formula
un+1 = -5kun + 3.
The second is defined by the formula
vn+1 = k2vn + 1.
Find the value of k and hence this limit.
Reveal answer only
k = 1/3
Go to full solution
L = 9/8
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Question 2
Two different recurrence relations
If the limit is L then as n   we have
un+1 = un = L and vn+1 = vn = L
are known to have the same limit
as n  .
The first is defined by the formula
First Sequence
Second Sequence
un+1 = -5kun + 3
vn+1 = k2vn + 1
un+1 = -5kun + 3.
becomes
becomes
L = k2 L + 1
The second is defined by
2
Vn+1 = k vn + 1.
Find the value of k and hence this
limit.
Begin Solution
L = -5kL + 3
L + 5kL = 3
L(1 + 5k) = 3
L = 3 .
.
(1 + 5k)
L - k2 L = 1
L(1 - k2) = 1
L = 1 .
.
(1 - k2)
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Continue Solution
Question 2
Two different recurrence relations
L = 3 .
.
(1 + 5k)
are known to have the same limit
as n  .
The first is defined by the formula
un+1 = -5kun + 3.
The second is defined by
Vn+1
= k2vn + 1.
Find the value of k and hence this
limit.
It follows that
.
L = 1 .
.
(1 - k2)
.
3 .
(1 + 5k)
Cross multiply to get
=
1 + 5k = 3 – 3k2
3k2 + 5k – 2 = 0
This becomes
Or
(3k – 1)(k + 2) = 0
So
k = 1/3 or k = -2
Since -1<k<1 then
k = 1/3
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. 1 .
(1 – k2)
Continue Solution
Question 1
Since -1<k<1 then
Two different recurrence relations
are known to have the same limit
as n  .
The first is defined by the formula
Using
gives us
L= . 1 . .
(1 – 1/9)
or
L = 1  8/9
ie
L = 9/8
un+1 = -5kun + 3.
The second is defined by
Vn+1
= k2vn + 1.
Find the value of k and hence this
limit.
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L = 1 .
.
(1 - k2)
k = 1/3
Markers Comments
If the limit is L then as n   we have
un+1 = un = L and vn+1 = vn = L
First Sequence
Second Sequence
un+1 = -5kun + 3
vn+1 = k2vn + 1
becomes
becomes
L = k2 L + 1
L = -5kL + 3
L + 5kL = 3
L(1 + 5k) = 3
L = 3 .
.
(1 + 5k)
L - k2L = 1
L(1 -
k2 )
• Since both recurrence relations
have the same limit, L, find the
limit for both and set equal.
Continue Comments
=1
L = 1 .
.
(1 - k2)
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Markers Comments
L = 3 .
.
(1 + 5k)
L = 1 .
.
(1 - k2)
It follows that
.
.
3 .
(1 + 5k)
Cross multiply to get
3k2
This becomes
=
. 1 .
(1 – k2)
1 + 5k = 3 – 3k2
+ 5k – 2 = 0
Or
(3k – 1)(k + 2) = 0
So
k = 1/3 or k = -2
Since -1<k<1 then
k = 1/3
• Since both recurrence relations
have the same limit, L, find the
limit for both and set equal.
• 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.
-1< k < 1.
• Take care to reject the
“solution” which is outwith the
range.
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Markers Comments
Find L from either formula.
Since -1<k<1 then
Using
k = 1/3
L = 1 .
.
(1 - k2)
gives us
L= . 1 . .
(1 – 1/9)
or
L = 1  8/9
ie
L = 9/8
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RECURRENCE RELATIONS : Question 3
A man plants a row of fast growing trees between his own house
and his neighbour’s. 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.
(a) Using the 30% pruning scheme what height should he expect
the trees to grow to in the long run?
(b) 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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RECURRENCE RELATIONS : Question 3
A man plants a row of fast growing trees between his own house
and his neighbour’s. 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.
(a) Using the 30% pruning scheme what height should he expect
the trees to grow to in the long run?
(b) 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
Height of trees in long run is
Go to full solution
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(b)
331/3%
31/3m.
Question 3
(a) Removing 30% leaves 70% or 0.7
The trees are known grow at a rate
of 1m per annum. He therefore
If Hn is the tree height in year n then
Hn+1 = 0.7Hn + 1
decides to prune 30% from their
height at the beginning of each year.
Since -1<0.7<1 this sequence has a limit, L.
At the limit Hn+1 = Hn = L
(a) Using the 30% pruning scheme
what height should he expect
So
L= 0.7L + 1
or
0.3 L = 1
the trees to grow to in the long
run?
Begin Solution
L = 1  0.3 = 10  3
= 31/3
Height of trees in long run is
31/3m.
ie
Continue Solution
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Continue Solution
Question 3
The trees are known grow at a rate
of 1m per annum. He therefore
(b) If fraction left after pruning is a and
we need the limit to be 3
then we have
3=aX3+1
or
3a = 2
decides to prune 30% from their
height at the beginning of each year.
(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?
Begin Solution
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a = 2/3
or
This means that the fraction pruned is
1/
3
or 331/3%
Markers Comments
(a) Removing 30% leaves 70% or 0.7
If Hn is the tree height in year n then
Hn+1 = 0.7Hn + 1
Since -1<0.7<1 this sequence has a limit, L.
At the limit Hn+1 = Hn = L
So
L= 0.7L + 1
or
0.3 L = 1
ie
• State the recurrence relation,
with the starting value.
Hn+1 = 0.7 Hn + 1,
H0 = 1
• State the condition for the limit
-1< 0.7< 1
• At limit L, state Hn+1 = Hn = L
L = 1  0.3 = 10  3
Height of trees in long run is
• Do some numerical work to
get the “feel” for the problem.
H0 = 1 (any value acceptable)
H1 = 0.7 x1 + 1 = 1.7
H2 = 0.7 x1.7 + 1 = 2.19 etc.
31/3m.
• Substitute L for Hn+1 and Hn
and solve for L.
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Markers Comments
(b) If fraction left after pruning is a and
we need the limit to be 3
then we have
• Since we know the limit we are
working backwards to %.
L = 0.7L + 1
3=aX3+1
New limit, L = 3 and multiplier a
or
3a = 2
or
a=
2/
3
This means that the fraction pruned is
1/
3
or 331/3%
3 = a x3 + 1
etc.
• Take care to subtract from 1
to get fraction pruned.
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HIGHER – ADDITIONAL QUESTION BANK
You have chosen to study:
UNIT 1 :
Trig Graphs
& Equations
Please choose a question to attempt from the following:
1
EXIT
2
3
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TRIG GRAPHS & EQUATIONS : Question 1
y = acosbx + c
This diagram shows the
graph of y = acosbx + c.
Determine the values of a, b
& c.
/
2

Reveal answer only
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EXIT
TRIG GRAPHS & EQUATIONS : Question 1
y = acosbx + c
This diagram shows the
graph of y = acosbx + c.
Determine the values of a, b
& c.
/
2

Reveal answer only
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a=3
b=2
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c = -1
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Question 1
a = ½(max – min)
This diagram shows the
graph of y = acosbx + c.
= ½(2 – (-4))
Determine the values of a, b
& c.
y = acosbx
+c
/
2

=½X6
= 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.
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ie 1 less
so c = -1
a = ½(max – min)
= ½(2 – (-4))
=½X6

= 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
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.
so c = -1
 


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TRIG GRAPHS & EQUATIONS : Question 2
Solve
3tan2 + 1 = 0
( where 0 <  <  ).
Reveal answer only
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EXIT
TRIG GRAPHS & EQUATIONS : Question 2
Solve
3tan2 + 1 = 0
( where 0 <  <  ).
Reveal answer only
 =
5/
12
Go to full solution
 =
11/
12
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EXIT
Question 2
Solve
3tan2 + 1 = 0
3tan2 + 1 = 0
3tan2 = -1
tan2 = -1/3
( where 0 <  <  ).
tan -1(1/3) = /6
Q2: angle =  - /6
so 2 = 5/6
ie  =
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-
sin

all
+
tan
5/
12
Q4: angle = 2 - /6
Begin Solution
Q2 or Q4
3
2 - 
cos
/
6
so 2 = 11/6
ie  =
11/
12
tan2 repeats every /2 radians but
repeat values are not in interval.
2
1
Markers Comments
Full marks
can be obtained
by.
•• Solve
the equation
for tan 2
working in degrees and
•changing
Use the positive
value back
when
final answers
tan-1.
tofinding
radians.
3tan2 + 1 = 0
3tan2 = -1
tan2 = -1/3
Q2 or Q4
-
sin
tan -1(1/3) = /6
Q2: angle =  - /6
+
tan
so 2 = 5/6
ie  =

all
2 - 
cos
5/
12
1   /180 radians
• Must learn special angles or
be able to calculate from
triangles.
• Take care
to reject60°
”solutions”
2
2
1
1
outwith domain.
45°
1
Q4: angle = 2 - /6
3
so 2 = 11/6
ie  =
 radians
 180
• Use the
quadrant
rule to find
the solutions.
/
6
11/
12
2
1
/
tan2 repeats every 2 radians but
repeat values are not in interval.
3
30°
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TRIG GRAPHS & EQUATIONS : Question 3
The diagram shows a the
graph of a sine function
from 0 to 2/3.
y=2
P
Q
(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
2/
3
TRIG GRAPHS & EQUATIONS : Question 3
The diagram shows a the
graph of a sine function
from 0 to 2/3.
y=2
P
Q
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
Reveal answer only
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P is (/18, 2)
and
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EXIT
y = 4sin3x
Q is (5/18, 2).
Question 3
(a) One complete wave from 0 to 2/3
The diagram shows a the
graph of a sine function
from 0 to 2/3.
so 3 waves from 0 to 2.
(a) State the equation of the graph.
Max/min = ±4
y=
2
Graph is
P
Q
4sin(…)
y = 4sin3x
2/
3
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Continue Solution
Question 3
(b) The line y = 2 meets the graph
Graph is
y = 4sin3x
(b) At P & Q y = 4sin3x and y = 2
at points P & Q.
Find the coordinates of these two
points.
y=
2
P
Q
2/
3
so
4sin3x = 2
or
sin3x = 1/2
-
sin
sin-1(1/2) = /6
Q1: angle = /6
so 3x = /6
so 3x =
Markers Comments
ie x =
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3
2 - 
cos
/
6
2
Q2: angle =  - /6
Continue Solution
Trig Graphs etc. Menu

all
+
tan
ie x = /18
Begin Solution
Q1 or Q2
1
5/
6
5/
18
P is (/18, 2)
and
Q is (5/18, 2).
Markers Comments
• Identify graph is of the form
y = asinbx.
(a) One complete wave from 0 to 2/3
so 3 waves from 0 to 2.
• Must justify choice of a and b.
• Possible justification of a
Max/min = ±4
Graph is
4sin(…)
y = 4sin3x
Max = 4, Min = -4
4sin(…)
y = sinx stretched by
a factor of 4
• Possible justification for b
Period = 2 / 3
3 waves from 0 to
2
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Markers Comments
Graph is
• At intersection y1 = y2
4sin3x = 2
y = 4sin3x
(b) At P & Q y = 4sin3x and y = 2
so
or
4sin3x = 2
sin3x =
Q1: angle = /6

all
+
tan
/
6
ie x = /18
Q2: angle =  - /6
3
2 - 
cos
/
6
2
5/
18
P is (/18, 2)
• Use the quadrant rule to find
the solutions.
• Must learn special angles or be
able to calculate from triangles.
1
2
45°
1
1
60°
2
30°
3
• Take care to state coordinates.
1
so 3x = 5/6
ie x =
Q1 or Q2
-
sin
sin-1(1/2) = /6
so 3x =
1/
2
• Solve for sin3x
Next Comment
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and
Q is (5/18, 2).
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HIGHER – ADDITIONAL QUESTION BANK
You have chosen to study:
UNIT 1 :
Functions
& Graphs
Please choose a question to attempt from the following:
1
EXIT
2
3
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Unit 1 Menu
4
FUNCTIONS & GRAPHS : Question 1
This graph shows the
the function y = g(x).
(-p,q)
Make a sketch of the
graph of the function
y = 4 – g(-x).
y = g(x)
-8
12
(u,-v)
Reveal answer only
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EXIT
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).
y = g(x)
(-p,q)
-8
12
(u,-v)
Reveal answer only
y = 4 – g(-x)
Go to full solution
(-u,v+4)
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EXIT
(0,4)
(p,-q+4)
(8,4)
Question 1
y = 4 – g(-x) = -g(-x) + 4
This graph shows the
the function y = g(x).
A
Reflect in
X-axis
Make a sketch of the
graph of the function
y = 4 – g(-x).
C
Reflect in Y-Slide 4 up
axis
y = g(x)
(-p,q)
Known Points
-8
12
(u,-v)
Begin Solution
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B
(-8,0), (-p,q), (0,0), (u,-v), (12,0)
A
(-8,0), (-p,-q), (0,0), (u,v), (12,0)
B
(8,0), (p,-q), (0,0), (-u,v), (-12,0)
C
(8,4), (p,-q+4), (0,4), (-u,v+4), (-12,4)
Question 1
(8,4), (p,-q+4), (0,4), (-u,v+4), (-12,4)
This graph shows the
the function y = g(x).
Now plot points and draw curve through
them.
Make a sketch of the
graph of the function
y = 4 – g(-x).
(-u,v+4)
y = g(x)
(-p,q)
-8
12
(u,-v)
Begin Solution
Continue Solution
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y = 4 – g(-x)
(-12,4)
(0,4)
(p,-q+4)
(8,4)
Markers Comments
y = 4 – g(-x) = -g(-x) + 4
A
Reflect in
X-axis
B
C
Reflect in Y-Slide 4 up
axis
Known Points
•Change order to give form:
y = k.g(x) + c
•When the function is being
changed by more than one
related function take each
change one at a time either
listing the coordinates or
sketching the steps to final
solution.
(-8,0), (-p,q), (0,0), (u,-v), (12,0)
A
(-8,0), (-p,-q), (0,0), (u,v), (12,0)
B
(8,0), (p,-q), (0,0), (-u,v), (-12,0)
C
(8,4), (p,-q+4), (0,4), (-u,v+4), (-12,4)
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Markers Comments
y = 4 – g(-x) = -g(-x) + 4
•Learn Rules: Not given on
formula sheet
A
Reflect in
X-axis
B
C
f(x) + k
Slide k units
parallel to y-axis
kf(x)
Stretch by a factor
=k
-f(x)
Reflect in the x-axis
(-8,0), (-p,q), (0,0), (u,-v), (12,0)
f(x-k)
A
(-8,0), (-p,-q), (0,0), (u,v), (12,0)
Slide k units
parallel to the x-axis
f(-x)
Reflect in y-axis
Reflect in Y-Slide 4 up
axis
Known Points
B
(8,0), (p,-q), (0,0), (-u,v), (-12,0)
C
(8,4), (p,-q+4), (0,4), (-u,v+4), (-12,4)
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Markers Comments
(8,4), (p,-q+4), (0,4), (-u,v+4), (-12,4)
Now plot points and draw curve through
them.
-12,4)
•In any curve sketching question
use a ruler and annotate the
sketch i.e. label all known
coordinates.
y = 4 – g(-x)
(-u,v+4)
(0,4)
(8,4)
(p,-q+4)
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FUNCTIONS & GRAPHS : Question 2
This graph shows the
the function y = ax.
Make sketches of the
graphs of the functions
(I) y = a(x+2)
(II) y = 2ax - 3
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EXIT
y = ax
(1,a)
FUNCTIONS & GRAPHS : Question 2
This graph shows the
the function y = ax.
ANSWER TO PART (I)
y = a(x+2)
y = ax
Make sketches of the
graphs of the functions
(-1,a)
(I) y = a(x+2)
(-2,1)
(II) y = 2ax - 3
Reveal answer only
ANSWER to PART (II)
Go to full solution
y = ax
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(1,2a-3)
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(0,-1)
y = 2ax - 3
Question 2
(I) y = a(x+2)
Make sketches of the
graphs of the functions
f(x) = ax
so a(x+2) = f(x+2)
(I) y = a(x+2)
y = ax
move f(x) 2 to left
(0,1)(-2,1) & (1,a) (-1,a)
(1,a)
y = a(x+2)
y = ax
(-1,a)
Begin Solution
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(-2,1)
(II) y = 2ax - 3
Question 2
Make sketches of the
graphs of the functions
f(x) = ax
so 2ax - 3 = 2f(x) - 3
(II) y = 2ax - 3
y = ax
double y-coords
(1,a)
slide 3 down
(0,1)(0,2)(0,-1)
(1,a) (1,2a) (1,2a-3)
y = ax
Begin Solution
(1,2a-3)
Continue Solution
(0,-1)
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y = 2ax - 3
Markers Comments
• When the problem is given in
terms of a specific function
rather in terms of the general
f(x), change back to f(x) eg.
y = 2x, y = log3x, y = x2 + 3x,
each becomes y = f(x)
(I) y = a(x+2)
f(x) = ax
so a(x+2) = f(x+2)
move f(x) 2 to left
(0,1)(-2,1) & (1,a) (-1,a)
y = a(x+2)
y = ax
• In any curve sketching
question use a ruler and
annotate the sketch
i.e. label all known coordinates.
(-1,a)
(-2,1)
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Markers Comments
(I) y = a(x+2)
•Learn Rules: Not given on
formula sheet
f(x) = ax
so
a(x+2)
= f(x+2)
f(x) + k
Slide k units
parallel to y-axis
kf(x)
Stretch by a factor
=k
-f(x)
Reflect in the x-axis
f(x-k)
Slide k units
parallel to the x-axis
f(-x)
Reflect in y-axis
move f(x) 2 to left
(0,1)(-2,1) & (1,a) (-1,a)
y = a(x+2)
y = ax
(-1,a)
(-2,1)
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Markers Comments
• When the problem is given in
terms of a specific function
rather in terms of the general
f(x), change back to f(x) eg.
y = 2x, y = log3x, y = x2 + 3x,
each becomes y = f(x)
(II) y = 2ax - 3
f(x) = ax
so 2ax - 3 = 2f(x) - 3
double y-coords
slide 3 down
(0,1)(0,2)(0,-1)
(1,a) (1,2a) (1,2a-3)
y = ax
(1,2a-3)
(0,-1)
y = 2ax - 3
• In any curve sketching
question use a ruler and
annotate the sketch
i.e. label all known
coordinates.
Next Comment
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Markers Comments
•Learn Rules: Not given on
formula sheet
(II) y = 2ax - 3
f(x) = ax
f(x) + k
Slide k units
parallel to y-axis
kf(x)
Stretch by a factor
=k
-f(x)
Reflect in the x-axis
f(x-k)
Slide k units
parallel to the x-axis
f(-x)
Reflect in y-axis
so 2ax - 3 = 2f(x) - 3
double y-coords
slide 3 down
(0,1)(0,2)(0,-1)
(1,a) (1,2a) (1,2a-3)
y = ax
(1,2a-3)
(0,-1)
y = 2ax - 3
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FUNCTIONS & GRAPHS : Question 3
Two functions f and g are defined on the set of real numbers by the
formulae
f(x) = 2x - 1 and g(x) = x2 .
(a)
Find formulae for (i) f(g(x)) (ii) g(f(x)) .
(b)
Hence show that the equation g(f(x)) = f(g(x))
has only one real solution.
Reveal answer only
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FUNCTIONS & GRAPHS : Question 3
Two functions f and g are defined on the set of real numbers by the
formulae
f(x) = 2x - 1 and g(x) = x2 .
(a)
Find formulae for (i) f(g(x)) (ii) g(f(x)) .
(b)
Hence show that the equation g(f(x)) = f(g(x))
has only one real solution.
Reveal answer only
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(a) (i) =
(ii) =
2x2 - 1
4x2 – 4x + 1
Question 3
(a)(i)
Two functions f and g are defined
on the set of real numbers by the
f(g(x))
= f(x2)
= 2x2 - 1
formulae
f(x) = 2x - 1 and g(x) = x2 .
(a) Find formulae for (i) f(g(x))
(ii) g(f(x)) .
(ii)
g(f(x))
= g(2x-1)
= (2x – 1)2
= 4x2 – 4x + 1
Begin Solution
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Continue Solution
Question 3
f(g(x)) = 2x2 - 1
Two functions f and g are defined
on the set of real numbers by the
formulae
f(x) = 2x - 1 and g(x) = x2 .
(b) Hence show that the equation
g(f(x)) = f(g(x)) has only one
real solution.
g(f(x)) = 4x2 – 4x + 1
(b)
g(f(x)) = f(g(x))
4x2 – 4x + 1 = 2x2 - 1
2x2 – 4x + 2 = 0
x2 – 2x + 1 = 0
(x – 1)(x – 1) = 0
x=1
Hence only one real solution!
Begin Solution
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Markers Comments
(a)(i)
f(g(x))
= f(x2)
= 2x2 - 1
(ii)
g(f(x))
= g(2x-1)
= (2x – 1)2
= 4x2 – 4x + 1
(a)
• In composite function problems
take at least 3 lines to
answer the problem:
State required composite function: f(g(x))
Replace g(x) without simplifying:
f(x2)
In f(x) replace each x by g(x):
2 x2 - 1
(II)
State required composite function: g(f(x))
Replace f(x) without simplifying:
g(2x-1)
In g(x) replace each x by f(x):
(2x – 1)2
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Markers Comments
f(g(x)) = 2x2 - 1
g(f(x)) = 4x2 – 4x + 1
(b)
g(f(x)) = f(g(x))
4x2 – 4x + 1 = 2x2 - 1
(b)
• Only one way to solve
resulting equation:
Terms to the left,
simplify and factorise.
2x2 – 4x + 2 = 0
x2 – 2x + 1 = 0
(x – 1)(x – 1) = 0
x=1
Hence only one real solution!
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FUNCTIONS & GRAPHS : Question 4
A function g is defined by the formula
g(x) = . 2
(x1)
(x – 1)
(a) Find a formula for
in its simplest form.
h(x) = g(g(x))
(b) State a suitable domain for h.
Reveal answer only
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FUNCTIONS & GRAPHS : Question 4
A function g is defined by the formula
g(x) = . 2
(x1)
(x – 1)
(a) Find a formula for
in its simplest form.
h(x) = g(g(x))
(b) State a suitable domain for h.
h(x)
= (2x - 2)
. . (3 – x)
Reveal answer only
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Domain = {x  R: x  3}
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Question 4
. 2 .
(a) g(g(x)) = g (x – 1)
(
)
A function g is defined by the
formula
g(x) = . 2
(x1)
(x – 1)
(a) Find a formula for
=
. 2 .
(x – 1)
=
- 1
2
2 - (x – 1)
.
(x – 1)
h(x) = g(g(x))
in its simplest form.
2
=
2
(3 - x)
.(x – 1)
= 2 (x - 1)
.(3 – x)
Begin Solution
Continue Solution
= (2x - 2)
. . (3 – x)
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Continue Solution
Question 4
h(x)
= (2x - 2)
. . (3 – x)
A function g is defined by the
formula
g(x) = . 2
(x1)
(x – 1)
(b)
(a) Find a formula for
h(x) = g(g(x))
in its simplest form.
(b) State a suitable domain for h.
Begin Solution
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For domain 3 - x  0
Domain = {x  R: x  3}
Markers Comments
. 2 .
(a) g(g(x)) = g (x – 1)
(
=
)
2
. 2 .
(x – 1)
=
- 1
2
2 - (x – 1)
.
(x – 1)
=
2
(a)
• In composite function problems
take at least 3 lines to
answer the problem:
State required composite function: g(g(x))
Replace g(x) without simplifying: g(2/(x-1))
In g(x) replace each x by g(x):
2
2
(x-1)
-1
(3 - x)
.(x – 1)
= 2 (x - 1)
.(3 – x)
= (2x - 2)
. . (3 – x)
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Markers Comments
h(x)
(b)
= (2x - 2)
. . (3 – x)
For domain 3 - x  0
Domain = {x  R: x  3}
(b)
• In finding a suitable domain it
is often necessary to restrict R
to prevent either
division by zero
or the root of a negative number:
In this case:
3-x=0
i.e. preventing division by zero.
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HIGHER – ADDITIONAL QUESTION BANK
UNIT 2 :
Polynomials
Quadratics
Integration
Addition
Formulae
The Circle
EXIT
HIGHER – ADDITIONAL QUESTION BANK
You have chosen to study:
UNIT 2 :
Polynomials
Please choose a question to attempt from the following:
1
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2
3
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4
POLYNOMIALS : Question 1
Show that
x=3
is a root of the equation
x3 + 3x2 – 10x – 24 = 0.
Hence find the other roots.
Reveal answer only
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EXIT
POLYNOMIALS : Question 1
Show that
x=3
is a root of the equation
x3 + 3x2 – 10x – 24 = 0.
Hence find the other roots.
Reveal answer only other roots are
Go to full solution
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EXIT
x = -4 & x = -2
Question 1
Show that x = 3 is a root of the
equation x3 + 3x2 – 10x – 24 = 0.
Hence find the other roots.
Using the nested method coefficients are 1, 3, -10, -24
f(3) = 3
1
3
3
-10
18
-24
24
1
6
8
0
f(3) = 0 so x = 3 is a root.
Also (x – 3) is a factor.
Other factor: x2 + 6x + 8 or (x + 4)(x + 2)
Begin Solution
If (x + 4)(x + 2) = 0 then x = -4 or x = -2
Continue Solution
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Hence other roots are
x = -4 & x = -2
Markers Comments
• State clearly in solution that
f(3) = 0
x = 3 is a root
Using the nested method coefficients are 1, 3, -10, -24
f(3) = 3
1
3
3
-10
18
-24
24
1
6
8
0
f(3) = 0 so x = 3 is a root.
Also (x – 3) is a factor.
Other factor: x2 + 6x + 8 or (x + 4)(x + 2)
If (x + 4)(x + 2) = 0 then x = -4 or x = -2
Hence other roots are
x = -4 & x = -2
• Show completed
factorisation of cubic i.e.
(x - 3)(x + 4)(x + 2) = 0
•Take care to set factorised
expression = 0
•List all the roots of the
polynomial
x = 3, x = -4, x = -2
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POLYNOMIALS : Question 2
Given that (x + 4) is a factor of the polynomial
f(x) = 3x3 + 8x2 + kx + 4
Hence solve the equation
of k.
find the value of k.
3x3 + 8x2 + kx + 4 = 0 for this value
Reveal answer only
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EXIT
POLYNOMIALS : Question 2
Given that (x + 4) is a factor of the polynomial
f(x) = 3x3 + 8x2 + kx + 4
Hence solve the equation
of k.
find the value of k.
3x3 + 8x2 + kx + 4 = 0 for this value
k = -15
Reveal answer only
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So full solution of equation is
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EXIT
x = -4 or x = 1/3 or x = 1
Question 2
Since (x + 4) a factor
then
f(-4) = 0 .
Given that (x + 4) is a factor of
the polynomial
f(x) = 3x3 + 8x2 + kx + 4
find the value of k.
Now using the nested method coefficients are 3, 8, k, 4
f(-4) = -4
3
8
-12
3
-4
Hence solve the equation
3x3 + 8x2 + kx + 4 = 0
for this value of k.
Since
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4
(-4k – 64)
(k + 16) (-4k – 60)
-4k – 60 = 0
then
-4k = 60
so
k = -15
Begin Solution
Continue Solution
k
16
Question 2
If k = -15 then we now have
Given that (x + 4) is a factor of
the polynomial
f(-4) = -4
3
8
-12
-15
16
4
-4
3
-4
1
0
f(x) = 3x3 + 8x2 + kx + 4
find the value of k.
Hence solve the equation
3x3 + 8x2 + kx + 4 = 0
for this value of k.
Other factor is
3x2 – 4x + 1
or (3x - 1)(x – 1)
If (3x - 1)(x – 1) = 0
then x = 1/3 or x = 1
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So full solution of equation is:
x = -4 or x = 1/3 or x = 1
Markers Comments
• The working in the nested
solution can sometimes be
Since (x + 4) a factor then f(-4) = 0 .
eased by working in both
directions toward the variable:
Now using the nested method -4
coefficients are 3, 8, k, 4
f(-4) = -4
3
3
Since
then
8
-12
-4
k
16
4
(-4k – 64)
(k + 16) (-4k – 60)
-4k – 60 = 0
3
8
-12
k
16
4
-4
3
-4
1
0
k + 16 = 1
k = -15
-4k = 60
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so
k = -15
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Markers Comments
•
Since (x + 4) a factor
then
f(-4) = 0 .
i.e.
3(-4)3 + 8(-4)2 + k(-4) + 4 = 0
k = -15
Now using the nested method coefficients are 3, 8, k, 4
f(-4) = -4
Since
then
3
8
-12
3
-4
k
16
Simply making f(-4) = 0
will also yield k
4
(-4k – 64)
(k + 16) (-4k – 60)
-4k – 60 = 0
-4k = 60
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so
k = -15
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Markers Comments
• Show completed factorisation
of the cubic:
If k = -15 then we now have
f(-4) = -4
3
3
Other factor is
8
-15
4
-12
16
-4
-4
1
0
(x + 4)(3x - 1)(x - 1) = 0
3x2 – 4x + 1
or (3x - 1)(x – 1)
If (3x - 1)(x – 1) = 0
So full solution of equation is:
x = -4 or x = 1/3 or x = 1
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POLYNOMIALS : Question 3
Given that
f(x).
f(x) = 6x3 + 13x2 - 4
show that (x + 2) is a factor of
Hence express f(x) in its fully factorised form.
Reveal answer only
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EXIT
POLYNOMIALS : Question 3
Given that
f(x).
f(x) = 6x3 + 13x2 - 4
show that (x + 2) is a factor of
Hence express f(x) in its fully factorised form.
6x3 + 13x2 - 4 = (3x + 2)(2x - 1)(x + 2)
Reveal answer only
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EXIT
Question 3
Given that f(x) = 6x3 + 13x2 - 4
Using the nested method coefficients are 6, 13, 0, -4
show that (x + 2) is a factor
of f(x).
f(-2) = -2
6
6
13
-12
1
0
-2
-2
Hence express f(x) in its fully
factorised form.
f(-2) = 0 so
(x + 2) is a factor
Begin Solution
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-4
4
0
Question 3
Given that f(x) = 6x3 + 13x2 - 4
Using the nested method coefficients are 6, 13, 0, -4
show that (x + 2) is a factor
of f(x).
f(-2) = -2
6
13
-12
1
6
0
-2
-2
-4
4
0
Hence express f(x) in its fully
factorised form.
6x2 + x – 2
Other factor is
or
Hence
Begin Solution
Continue Solution
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(3x + 2)(2x - 1)
6x3 + 13x2 - 4
= (3x + 2)(2x - 1)(x + 2)
Markers Comments
• State clearly in solution that
f(-2) = 0
x = -2 is a root
Using the nested method coefficients are 6, 13, 0, -4
f(-2) = -2
6
6
13
-12
1
0
-2
-2
-4
4
0
f(-2) = 0 so
(x + 2) is a factor
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Markers Comments
• Show completed factorisation
of cubic i.e.
Using the nested method coefficients are 6, 13, 0, -4
f(-2) = -2
6
13
-12
1
6
-4
4
0
6x2 + x – 2
Other factor is
or
Hence
0
-2
-2
(3x + 2)(2x - 1)(x +2).
•Can show (x + 2) is a factor
by showing f(-2) = 0 but still
need nested method for
quadratic factor.
(3x + 2)(2x - 1)
6x3 + 13x2 - 4
= (3x + 2)(2x - 1)(x + 2)
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POLYNOMIALS : Question 4
A busy road passes through several small villages so it is
decided to build a by-pass to reduce the volume of traffic.
Relative to a set of coordinate axes the road can be modelled by
the curve y = -x3 + 6x2 – 3x – 10. The by-pass is a tangent to this
curve at point P and rejoins the original road at Q as shown
below.
bypass
Q
P
4
y = -x3 + 6x2 – 3x – 10
(a) Find the coordinates of P and the equation of the bypass
PQ.
(b) Hence find the coordinates of Q – the point where the
bypass rejoins the original road.
EXIT
Reveal answer only
Go to full solution
POLYNOMIALS : Question 4
A busy road passes through several small villages so it is
decided to build a by-pass to reduce the volume of traffic.
Relative to a set of coordinate axes the road can be modelled by
the curve y = -x3 + 6x2 – 3x – 10. The by-pass is a tangent to this
curve at point P and rejoins the original road at Q as shown
below.
(a) Find the coordinates of P and the equation of the bypass
PQ.
P is (4,10)
PQ is y = -3x + 22
(b) Hence find the coordinates of Q – the point where the
bypass rejoins the original road.
Q is (-2,28)
EXIT
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Question 4
(a) At point P, x = 4 so using the
y = -x3 + 6x2 – 3x – 10
(a) Find the coordinates of P
and the equation of the
bypass PQ.
equation of the curve we get …..
y = -43 + (6 X 42) – (3 X 4) - 10
= -64 + 96 – 12 - 10
= 10
Q
P
y=
+
6x2
P is (4,10)
Gradient of tangent = gradient of curve
= dy/dx= -3x2 + 12x - 3
4
-x3
ie
– 3x – 10
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When x = 4 then
dy/
dx
= (-3 X 16) + (12 X 4) – 3
= -48 + 48 – 3 = -3
Question 4
y=
-x3
+
6x2
P is (4,10)
– 3x – 10
(a) Find the coordinates of P
Now using :
and the equation of the
dy/dx = -3
y – b = m(x – a)
where (a,b) = (4,10) & m = -3
bypass PQ.
Q
P
We get
y – 10 = -3(x – 4)
or
y – 10 = -3x + 12
So PQ is
y = -3x + 22
4
y = -x3 + 6x2 – 3x – 10
Begin Solution
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Question 4
(b) The tangent & curve meet whenever
y = -x3 + 6x2 – 3x – 10
(b)Hence find the coordinates of Q –
the point where the bypass rejoins
the original road.
PQ is y = -3x + 22
y = -3x + 22 and y = -x3 + 6x2 – 3x – 10
ie
or
-3x + 22 = -x3 + 6x2 – 3x – 10
x3 - 6x2 + 32 = 0
We already know that x = 4 is one
Q
P
solution to this so using the nested
method we get …..
4
y=
-x3
+
6x2
– 3x – 10
Begin Solution
f(4) = 4
1
-6
4
0
-8
32
-32
1
-2
-8
0
Continue Solution
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Other factor is x2 – 2x - 8
Question 4
(b) The other factor is x2 – 2x - 8
y = -x3 + 6x2 – 3x – 10
= (x – 4)(x + 2)
(b)Hence find the coordinates of Q –
the point where the bypass rejoins
Solving (x – 4)(x + 2) = 0
the original road.
PQ is y = -3x + 22
we get x = 4 or x = -2
Q
It now follows that Q has an x-coordinate
P
of -2
4
Using y = -3x + 22 if x = -2
y = -x3 + 6x2 – 3x – 10
Begin Solution
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then y = 6 + 22 = 28
Hence
Q is (-2,28)
Markers Comments
(a) At point P, x = 4 so using the
equation of the curve we get …..
y = -43 + (6 X 42) – (3 X 4) - 10
= -64 + 96 – 12 - 10
= 10
ie
P is (4,10)
(a)
• Must use differentiation to
find gradient.
Learn rule:
“Multiply by the power
then reduce the power by 1”
Gradient of tangent = gradient of curve
= dy/dx= -3x2 + 12x - 3
When x = 4 then
dy/
dx
= (-3 X 16) + (12 X 4) – 3
= -48 + 48 – 3 = -3
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Markers Comments
(a)
P is (4,10)
Now using :
dy/dx = -3
y – b = m(x – a)
where (a,b) = (4,10) & m = -3
We get
y – 10 = -3(x – 4)
or
y – 10 = -3x + 12
So PQ is
y = -3x + 22
•
Use :
1. the point of contact (4,10)&
2. Gradient of curve at this
point (m = -3) in equation
y - b = m(x - a)
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Markers Comments
(b)
(b) The tangent & curve meet whenever
• At intersection
y1 = y2
y = -3x + 22 and y = -x3 + 6x2 – 3x – 10
ie
or
-3x + 22 =
-x3
+
6x2
– 3x – 10
Terms to the left,
simplify and factorise
x3 - 6x2 + 32 = 0
We already know that x = 4 is one
solution to this so using the nested
method we get …..
f(4) = 4
1
-6
4
0
-8
32
-32
1
-2
-8
0
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Other factor is x2 – 2x - 8
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Markers Comments
(b)
other factor is x2 – 2x - 8
= (x – 4)(x + 2)
Solving (x – 4)(x + 2) = 0
• Note solution x = 4 appears
twice:
Repeated root
tangency
we get x = 4 or x = -2
It now follows that Q has an x-coordinate
of -2
Using y = -3x + 22 if x = -2
then y = 6 + 22 = 28
Hence
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Q is (-2,28)
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HIGHER – ADDITIONAL QUESTION BANK
You have chosen to study:
UNIT 2 :
Quadratics
Please choose a question to attempt from the following:
1
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2
3
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4
5
6
QUADRATICS : Question 1
(a) Express f(x) = x2 – 8x + 21 in the form (x – a)2 + b.
(b) Hence, or otherwise, sketch the graph of y = f(x).
Reveal answer only
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EXIT
QUADRATICS : Question 1
(a) Express f(x) = x2 – 8x + 21 in the form (x – a)2 + b.
(b) Hence, or otherwise, sketch the graph of y = f(x).
(a)
= (x – 4)2 + 5
y = x2 – 8x + 21
Reveal answer only
(b)
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(4,5)
EXIT
Question 1
(a) Express f(x) = x2 – 8x + 21
in the form (x – a)2 + b.
(a)
f(x) = x2 – 8x + 21
= (x2 – 8x + 16) + 21 - 16
(-82)2
(b) Hence, or otherwise,
sketch the graph of y = f(x).
Begin Solution
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= (x – 4)2 + 5
Question 1
(a) Express f(x) = x2 – 8x + 21
in the form (x – a)2 + b.
(b) f(x) = (x – 4)2 + 5 has a minimum
value of 5 when (x – 4)2 = 0 ie x = 4
So the graph has a minimum
(b) Hence, or otherwise,
turning point at (4,5).
sketch the graph of y = f(x).
When x = 0 , y = 21 (from original formula!)
so Y-intercept is (0,21).
Begin Solution
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Question 1
(b) Graph looks like….
(a) Express f(x) = x2 – 8x + 21
in the form (x – a)2 + b.
(b) Hence, or otherwise,
y = x2 – 8x + 21
(0,21)
sketch the graph of y = f(x).
(4,5)
Begin Solution
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Markers Comments
(a)
f(x) = x2 – 8x + 21
= (x2 – 8x + 16) + 21 - 16
(-82)2
= (x – 4)2 + 5
• Move towards desired form in
stages:
f(x) = x2 - 8x + 21
= (x2 - 8x) + 21
= (x2 - 8x +16) + 21 - 16
Find the number to complete the
perfect square and balance the
expression.
(a + b) 2 = a2 + 2ab + b2
= (x - 4)2 + 5
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Markers Comments
(b) f(x) = (x –
4)2
+ 5 has a minimum
value of 5 when (x – 4)2 = 0 ie x = 4
• The sketch can also be obtained
by calculus:
f(x) = x 2 -8 x  21
 f (x) = 2 x-8
So the graph has a minimum
turning point at (4,5).
When x = 0 , y = 21 (from original formula!)
so Y-intercept is (0,21).
dy
At stationary points
0
dx
2 x 8  0
x4
f(4)  42 - 8.4 + 21 = 5
(0,21)
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(4,5)
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Markers Comments
• Min. Turning Point (4,5),
(b) f(x) = (x – 4)2 + 5 has a minimum
value of 5 when (x – 4)2 = 0 ie x = 4
coefficient of x2 is positive.
f(0)  21
So the graph has a minimum
turning point at (4,5).
When x = 0 , y = 21 (from original formula!)
• Hence sketch.
21
so Y-intercept is (0,21).
(4,5)
(0,21)
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(4,5)
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QUADRATICS : Question 2
(a) Express f(x) = 7 + 8x - 4x2 in the form a - b(x - c)2.
(b) Hence find the maximum turning point on the graph of y = f(x).
Reveal answer only
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QUADRATICS : Question 2
(a) Express f(x) = 7 + 8x - 4x2 in the form a - b(x - c)2.
(b) Hence find the maximum turning point on the graph of y = f(x).
Reveal answer only
(a)
= 11 - 4(x – 1)2
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(b)
maximum t p is at (1,11) .
Question 2
(a) Express f(x) = 7 + 8x - 4x2 in
the form a - b(x - c)2.
(a) f(x) = 7 + 8x - 4x2
= - 4x2 + 8x + 7
= -4[x2 – 2x] + 7
= -4[(x2 – 2x + 1 ) - 1 ] + 7
(b) Hence find the maximum
turning point on the graph of
y = f(x).
(-22)2
= -4[(x – 1)2 - 1] + 7
= -4(x – 1)2 + 4 + 7
= 11 - 4(x – 1)2
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Question 2
= 11 - 4(x – 1)2
(a) Express f(x) = 7 + 8x - 4x2 in
the form a - b(x - c)2.
(b) Maximum value is 11 when
(x – 1)2 = 0 ie x = 1.
(b) Hence find the maximum
turning point on the graph of
y = f(x).
Begin Solution
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so maximum turning point is at (1,11) .
Markers Comments
(a) f(x) = 7 + 8x - 4x2
= - 4x2 + 8x + 7
= -4[x2 – 2x] + 7
= -4[(x2 – 2x + 1 ) - 1 ] + 7
(-22)2
= -4[(x – 1)2 - 1] + 7
= -4(x – 1)2 + 4 + 7
• Move towards desired form
in stages:
f(x) = 7 + 8x - 4x2
= - 4x2 + 8x + 7
= (- 4x2 + 8x) + 7
• Must reduce coefficient of
x2 to 1
= -4(x2 - 2x) + 7
= 11 - 4(x – 1)2
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Markers Comments
(a) f(x) = 7 + 8x - 4x2
= - 4x2 + 8x + 7
= -4[x2 – 2x] + 7
= -4[(x2 – 2x + 1 ) - 1 ] + 7
(-22)2
= -4[(x – 1)2 - 1] + 7
= -4(x – 1)2 + 4 + 7
= 11 - 4(x – 1)2
• Must reduce coefficient of
x2 to 1
= -4(x2 - 2x) + 7
Find the number to
complete the perfect
square and balance the
expression.
(a + b) 2 =
2 - 2x +1) -1] +7 )
=
-4[(x
2
a + 2ab + b2
= -4(x-1)2 + 7 + 4
= 11 - 4(x-1)2
Max. TP at (1,11)
(b) Maximum value is 11 when
(x – 1)2 = 0 ie x = 1.
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so maximum turning point is at (1,11) .
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QUADRATICS : Question 3
For what value(s) of k does the equation 4x2 – kx + (k + 5) = 0 have
equal roots.
Reveal answer only
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QUADRATICS : Question 3
For what value(s) of k does the equation 4x2 – kx + (k + 5) = 0 have
equal roots.
Reveal answer only
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EXIT
k = -4 or k = 20
Question 3
Let
4x2 – kx + (k + 5) = ax2 + bx + c
For what value(s) of k does the
equation 4x2 – kx + (k + 5) = 0
then
have equal roots.
For equal roots we need discriminant = 0
ie
a = 4, b = -k & c = (k + 5)
b2 – 4ac = 0
(-k)2 – (4 X 4 X (k + 5)) = 0
k2 – 16(k + 5) = 0
k2 – 16k - 80 = 0
(k + 4)(k – 20) = 0
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k = -4 or k = 20
Markers Comments
Let
then
4x2 – kx + (k + 5) = ax2 + bx + c
a = 4, b = -k & c = (k + 5)
For equal roots we need discriminant = 0
ie
b2 – 4ac = 0
(-k)2 – (4 X 4 X (k + 5)) = 0
k2 – 16(k + 5) = 0
k2 – 16k - 80 = 0
• Learn Rules relating to
discriminant b2- 4ac
b2- 4ac = 0
b2- 4ac > 0
 Equal roots
 2 Real,
distinct roots
b2- 4ac < 0  No real roots
b2- 4ac  0
 Real roots,
equal or distinct
(k + 4)(k – 20) = 0
k = -4 or k = 20
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Markers Comments
Let
then
4x2 – kx + (k + 5) = ax2 + bx + c
a = 4, b = -k & c = (k + 5)
For equal roots we need discriminant = 0
ie
• Must use factorisation to solve
resulting quadratic.
Trial and error receives no
credit.
b2 – 4ac = 0
(-k)2 – (4 X 4 X (k + 5)) = 0
k2 – 16(k + 5) = 0
k2 – 16k - 80 = 0
(k + 4)(k – 20) = 0
k = -4 or k = 20
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QUADRATICS : Question 4
The equation of a parabola is f(x) = px2 + 5x – 2p .
Prove that the equation f(x) = 0 always has two distinct
roots.
Reveal answer only
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EXIT
QUADRATICS : Question 4
The equation of a parabola is f(x) = px2 + 5x – 2p .
Prove that the equation f(x) = 0 always has two distinct
roots.
discriminant = b2 – 4ac
Reveal answer only
= 8p2 + 25
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Since p2  0 for all values of p
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then 8p2 + 25 > 0.
The discriminant is always positive
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EXIT
so there are always two distinct roots.
Question 4
Let
px2 + 5x – 2p = ax2 + bx + c
then
a = p, b = 5 & c = -2p.
The equation of a parabola is
f(x) =
px2
+ 5x – 2p .
Prove that the equation f(x) = 0
always has two distinct roots.
So discriminant = b2 – 4ac
= 52 – (4 X p X (-2p))
= 25 – (-8p2)
= 8p2 + 25
Since p2  0 for all values of p
then 8p2 + 25 > 0.
Begin Solution
Continue Solution
The discriminant is always positive
Markers Comments
so there are always two distinct roots.
Quadratics Menu
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Markers Comments
Let
px2 + 5x – 2p = ax2 + bx + c
then
a = p, b = 5 & c = -2p.
So discriminant = b2 – 4ac
= 52 – (4 X p X (-2p))
= 25 – (-8p2)
=
8p2
+ 25
• Learn Rules relating to
discriminant b2- 4ac
b2- 4ac = 0
b2- 4ac > 0
 Equal roots
 2 Real,
distinct roots
b2- 4ac < 0  No real roots
b2- 4ac  0
 Real roots,
equal or distinct
Since p2  0 for all values of p
then 8p2 + 25 > 0.
The discriminant is always positive
so there are always two distinct roots.
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Markers Comments
Let
px2 + 5x – 2p = ax2 + bx + c
then
a = p, b = 5 & c = -2p.
So discriminant = b2 – 4ac
• State condition you require
to show true explicitly:
For two distinct roots
b2- 4ac > 0
= 52 – (4 X p X (-2p))
= 25 – (-8p2)
= 8p2 + 25
Since p2  0 for all values of p
then 8p2 + 25 > 0.
The discriminant is always positive
so there are always two distinct roots.
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Markers Comments
Let
px2 + 5x – 2p = ax2 + bx + c
then
a = p, b = 5 & c = -2p.
•To show 8p2 + 25> 0 use
algebraic logic or show the
graph of 8p2 + 25 is always
above the “x-axis”.
So discriminant = b2 – 4ac
= 52 – (4 X p X (-2p))
= 25 – (-8p2)
25
= 8p2 + 25
Since p2  0 for all values of p
Min. T.P. at (0,25) hence
graph always above the “x - axis.”
then 8p2 + 25 > 0.
The discriminant is always positive
so there are always two distinct roots.
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QUADRATICS : Question 5
Given that the roots of 3x(x + p) = 4p(x – 1)
then show that p = 0 or p = 48.
Reveal answer only
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are equal
QUADRATICS : Question 5
Given that the roots of 3x(x + p) = 4p(x – 1)
are equal
then show that p = 0 or p = 48.
For equal roots we need discriminant = 0
p(p - 48) = 0
Reveal answer only
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ie
p = 0 or p = 48
Question 5
Rearranging
3x(x + p) = 4p(x – 1)
Given that the roots of
3x2 + 3px = 4px - 4p
3x(x + p) = 4p(x – 1) are equal
3x2 - px + 4p = 0
then show that p = 0 or p = 48.
Let 3x2 - px + 4p = ax2 + bx + c
then a = 3, b = -p & c = 4p
For equal roots we need discriminant = 0
ie
b2 – 4ac = 0
(-p)2 - (4 X 3 X 4p) = 0
p2 - 48p = 0
Begin Solution
p(p - 48) = 0
Continue Solution
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ie
p = 0 or p = 48
Markers Comments
Rearranging
3x(x + p) = 4p(x – 1)
3x2 + 3px = 4px - 4p
3x2 - px + 4p = 0
• Must put the equation into
standard quadratic form before
reading off a,b and c.
i.e.
ax2 + bx +c = 0
Let 3x2 - px + 4p = ax2 + bx + c
then a = 3, b = -p & c = 4p
For equal roots we need discriminant = 0
ie
b2 – 4ac = 0
(-p)2 - (4 X 3 X 4p) = 0
p2 - 48p = 0
p(p - 48) = 0
ie
p = 0 or p = 48
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Markers Comments
Rearranging
3x(x + p) = 4p(x – 1)
• Learn Rules relating to
discriminant b2- 4ac
3x2 + 3px = 4px - 4p
3x2
- px + 4p = 0
Let 3x2 - px + 4p = ax2 + bx + c
b2- 4ac = 0
b2- 4ac > 0
distinct roots
b2- 4ac < 0  No real roots
then a = 3, b = -p & c = 4p
For equal roots we need discriminant = 0
ie
b2 – 4ac = 0
 Equal roots
 2 Real,
b2- 4ac  0
 Real roots,
equal or distinct
(-p)2 - (4 X 3 X 4p) = 0
p2 - 48p = 0
p(p - 48) = 0
ie
p = 0 or p = 48
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Markers Comments
Rearranging
3x(x + p) = 4p(x – 1)
3x2 + 3px = 4px - 4p
3x2
- px + 4p = 0
•State condition you require
explicitly:
For two equal roots
b2- 4ac = 0
Let 3x2 - px + 4p = ax2 + bx + c
then a = 3, b = -p & c = 4p
For equal roots we need discriminant = 0
ie
• Must use factorisation to
solve resulting quadratic.
b2 – 4ac = 0
(-p)2 - (4 X 3 X 4p) = 0
p2 - 48p = 0
p(p - 48) = 0
ie
p = 0 or p = 48
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QUADRATICS : Question 6
The diagram below shows the parabola y = -2x2 + 3x + 2 and
the line x + y – 4 = 0.
Reveal answer only
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x+y–4=0
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y = -2x2 + 3x + 2
Prove that the line is a tangent to the curve.
EXIT
QUADRATICS : Question 6
The diagram below shows the parabola y = -2x2 + 3x + 2 and
the line x + y – 4 = 0.
Reveal answer only
Go to full solution
x+y–4=0
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2 + 3x + 2
y = -2xis
Since the discriminant = 0 then there
only one solution to the equation
so only one point of contact and it follows that the line is a tangent.
Prove that the line is a tangent to the curve.
EXIT
Question 6
Linear equation can be changed from
x + y – 4 = 0 to y = -x + 4.
The line and curve meet when
x+y–4=
0
y = -x + 4 and y = -2x2 + 3x + 2 .
So -x + 4 = -2x2 + 3x + 2
Or
y = -2x2 + 3x + 2
Prove that the line is a tangent.
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Let
2x2 - 4x + 2 = 0
2x2 - 4x + 2 = ax2 + bx + c
then a = 2, b = -4 & c = 2.
Question 6
then a = 2, b = -4 & c = 2.
So discriminant = b2 – 4ac
= (-4)2 – (4 X 2 X 2)
x+y–4=0
= 16 - 16
= 0
Since the discriminant = 0 then there is
only one solution to the equation so only
y = -2x2 + 3x + 2
Prove that the line is a tangent.
Begin Solution
Continue Solution
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one point of contact and it follows that
the line is a tangent.
Markers Comments
Linear equation can be changed from
x + y – 4 = 0 to y = -x + 4.
The line and curve meet when
y = -x + 4 and y = -2x2 + 3x + 2 .
• For intersection of line and
polynomial
y1 = y2
Terms to the left, simplify and
factorise.
So -x + 4 = -2x2 + 3x + 2
Or
Let
2x2 - 4x + 2 = 0
2x2 - 4x + 2 = ax2 + bx + c
then a = 2, b = -4 & c = 2.
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Markers Comments
Or
-x + 4 = -2x2 + 3x + 2
then a = 2, b = -4 & c = 2.
2x2 - 4x + 2 = 0
– 4ac
2(x2 - 2x + 1) = 0
So discriminant =
b2
= (-4)2 – (4 X 2 X 2)
2(x - 1)(x - 1) = 0
x = 1 (twice)
= 16 - 16
= 0
Since the discriminant = 0 then there is
only one solution to the equation so only
one point of contact and it follows that
the line is a tangent.
Equal roots
 tangency
• To prove tangency
“equal roots” may be used in
place of the discriminant.
The statement must be made
explicitly.
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HIGHER – ADDITIONAL QUESTION BANK
You have chosen to study:
UNIT 2 :
Integration
Please choose a question to attempt from the following:
1
EXIT
2
3
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Unit 2 Menu
4
5
INTEGRATION : Question 1
The diagram below shows the curve y = x2 - 8x + 18 and the lines
x = 3 and x = k.
y = x2 - 8x + 18
Reveal answer only
Go to full solution
x=3
x=k
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Show that the shaded area is given by
1 / k3
3
–
4k2
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+ 18k - 27
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EXIT
INTEGRATION : Question 1
The diagram below shows the curve y = x2 - 8x + 18 and the lines
x = 3 and x = k.
y = x2 - 8x + 18
Reveal answer only
Go to full solution
x=3
x=k
Go to Marker’s Comments
Go to Integration Menu
Show that the shaded area is given by
1 / k3
3
EXIT
– 4k2 + 18k - 27
Area =
Go to Main Menu

k
3
(x2 - 8x + 18) dx =
1 / k3
3
– 4k2 + 18k – 27 as required.
Question 1
The diagram shows the
curve y = x2 - 8x + 18 and the
Area =
k

(x2 - 8x + 18) dx
3
=
lines x = 3 and x = k.
[
Show that the shaded area is
[
given by
1 / k3
3
– 4k2 + 18k - 27
=
x3 - 8x2 + 18x
3 2
1/ x3
3
–
4x2
k
]
]
+ 18x
3
k
3
= (1/3k3 – 4k2 + 18k)
– ((1/3 X 27) – (4 X 9) + 54)
=
1 / k3
3
– 4k2 + 18k – 27
Begin Solution
Continue Solution
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as required.
Markers Comments
b
Area =

 f ( x)dx
• Learn result
k
(x2 - 8x + 18) dx
a
3
=
=
[
[
x3 - 8x2 + 18x
3 2
1/ x3
3
k
]
]
– 4x2 + 18x
3
can be used to find the
enclosed area shown:
k
3
f(x)
= (1/3k3 – 4k2 + 18k)
– ((1/3 X 27) – (4 X 9) + 54)
a
=
1 / k3
3
b
– 4k2 + 18k – 27
Next Comment
as required.
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Markers Comments
• Learn result for integration:
Area =
k

(x2 - 8x + 18) dx
“Add 1 to the power and
divide by the new power.”
3
=
=
[
[
x3 - 8x2 + 18x
3 2
1/ x3
3
k
]
]
– 4x2 + 18x
3
k
3
= (1/3k3 – 4k2 + 18k)
– ((1/3 X 27) – (4 X 9) + 54)
=
1 / k3
3
– 4k2 + 18k – 27
Next Comment
as required.
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INTEGRATION : Question 2
Given that
dy/
dx
= 12x2 – 6x and the curve y = f(x) passes
through the point (2,15) then find the equation of the curve
y = f(x).
Reveal answer only
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EXIT
INTEGRATION : Question 2
Given that
dy/
dx
= 12x2 – 6x and the curve y = f(x) passes
through the point (2,15) then find the equation of the curve
y = f(x).
Equation of curve is
Reveal answer only
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EXIT
y = 4x3 – 3x2 - 5
Question 2
Given that
dy/
dx
dy/
dx
= 12x2 – 6x and
So
the curve y = f(x) passes through
= 12x2 – 6x
y   (12 x 2  6 x)dx
the point (2,15) then find the
= 12x3 – 6x2 + C
3
2
equation of the curve y = f(x).
= 4x3 – 3x2 + C
Substituting (2,15) into y = 4x3 – 3x2 + C
We get 15 = (4 X 8) – (3 X 4) + C
So
C + 20 = 15
ie
C = -5
Begin Solution
Continue Solution
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Equation of curve is
y = 4x3 – 3x2 - 5
Markers Comments
dy/
dx
So
= 12x2 – 6x
y   (12 x 2  6 x)dx
= 12x3 – 6x2 + C
3
2
= 4x3 – 3x2 + C
Substituting (2,15) into y = 4x3 – 3x2 + C
We get 15 = (4 X 8) – (3 X 4) + C
So
C + 20 = 15
ie
C = -5
Equation of curve is
y = 4x3 – 3x2 - 5
• Learn the result that integration
undoes differentiation:
i.e.
given
dy
= f(x)  y =  f(x) dx
dx
• Learn result for integration:
“Add 1 to the power and
divide by the new power”.
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Markers Comments
dy/
dx
So
= 12x2 – 6x
• Do not forget the constant
of integration!!!
y   (12 x 2  6 x)dx
= 12x3 – 6x2 + C
3
2
= 4x3 – 3x2 + C
Substituting (2,15) into y = 4x3 – 3x2 + C
We get 15 = (4 X 8) – (3 X 4) + C
So
C + 20 = 15
ie
C = -5
Equation of curve is
y = 4x3 – 3x2 - 5
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INTEGRATION : Question 3
Find

x2 - 4 dx
2xx
Reveal answer only
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EXIT
INTEGRATION : Question 3
Find

x2 - 4 dx
2xx
Reveal answer only
=
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EXIT
xx + 4 + C
3
x
Question 3
Find

x2 - 4 dx
2xx
Begin Solution
Continue Solution
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
x2 - 4 dx
2xx
=

x2 - 4
dx
2x3/2
2x3/2
=

1 / x 1 /2
2
=
2/
3
X 1/2x3/2 - (-2) X 2x-1/2 + C
=
1 / x 3 /2
3
=
- 2x-3/2
dx
+ 4x-1/2 + C
xx + 4 + C
3
x
Markers Comments
• Prepare expression by:

x2 - 4 dx
2xx
=

x2 - 4
dx
2x3/2
2x3/2
=

1 / x 1 /2
2
dx
X 1/2x3/2 - (-2) X 2x-1/2 + C
=
2/
3
=
1 / x 3 /2
3
=
- 2x
-3/2
1 Dividing out the fraction.
2 Applying the laws of indices.
+ 4x-1/2 + C
xx + 4 + C
3
x
• Learn result for integration:
Add 1 to the power and divide
by the new power.
• Do not forget the constant
of integration.
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INTEGRATION : Question 4
2
Evaluate
(
1
x2 - 2
x
)
2
dx
Reveal answer only
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EXIT
INTEGRATION : Question 4
2
Evaluate
(
1
x2 - 2
x
)
2
dx
Reveal answer only
Go to full solution
Go to Marker’s Comments
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EXIT
= 21 / 5
Question 4
2
(
1
2
2
Evaluate
(
1
x2 - 2
x
)
2
dx
x4
(

=
1
x2 - 2
x
)
2
dx
- 4x + 4 )dx
x2
2
=
x4
(

)
- 4x + 4x-2 dx
1
= x5 - 4x2 + 4x-1
5 2
-1
[
= x5 - 2x2 - 4
5
x
[
]
]
2
1
2
1
= (32/5 - 8 - 2) - (1/5 - 2 - 4)
Begin Solution
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= 21/5
Markers Comments
• Prepare expression by:
2
(
1
2
x4
(

=
1
=
=
x2 - 2
x

)
2
dx
- 4x + 4 )dx
x2
2
(x4 - 4x + 4x-2 ) dx
1
x5
[5
-
4x2
2
+
4x-1
= x5 - 2x2 - 4
5
x
[
]
-1
]
2
1
2
1
= (32/5 - 8 - 2) - (1/5 - 2 - 4)
= 21/5
1 Expanding the bracket
2 Applying the laws of indices.
• Learn result for integration:
“Add 1 to the power and
divide by the new power”.
• When applying limits show
substitution clearly.
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INTEGRATION : Question 5
The diagram below shows the parabola y = -x2 + 8x - 10 and the
line y = x. They meet at the points A and B.
y=x
Reveal answer only
B
Go to full solution
A
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y=
-x2
+ 8x - 10
(a) Find the coordinates of A and B.
(b)Hence find the shaded area between the
EXIT
curves.
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INTEGRATION : Question 5
The diagram below shows the parabola y = -x2 + 8x - 10 and the
line y = x. They meet at the points A and B.
y=x
Reveal answer only
B
Go to full solution
A
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y=
-x2
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+ 8x - 10
(a) Find the coordinates of A and B.
A is (2,2) and B is (5,5) .
(b)Hence find the shaded area between the
EXIT
curves.
= 41/2units2
Question 5
The diagram shows the parabola
(a) Line & curve meet when
y = x and y = -x2 + 8x - 10 .
y = -x2 + 8x - 10 and the line y = x.
They meet at the points A and B.
(a) Find the coordinates of A and B.
(b) Hence find the shaded area
So
x = -x2 + 8x - 10
or
x2 - 7x + 10 = 0
ie
(x – 2)(x – 5) = 0
ie
x = 2 or x = 5
between the curves.
Since points lie on y = x then
Begin Solution
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A is (2,2) and B is (5,5) .
Question 5
A is (2,2) and B is (5,5) .
The diagram shows the parabola
y = -x2 + 8x - 10 and the line y = x. (b) Curve is above line between limits so
They meet at the points A and B.

5
Shaded area =
(-x2 + 8x – 10 - x) dx
2
5
(a) Find the coordinates of A and B.
 (-x + 7x – 10) dx
[ -x3 + 7x2 - 10x ]
2
=
2
(b) Hence find the shaded area
between the curves.
=
3
2
= (-125/3 + 175/2 – 50)
– (-8/3 +14 – 20)
Begin Solution
Continue Solution
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= 41/2units2
5
2
Markers Comments
(a) Line & curve meet when
y = x and y = -x2 + 8x - 10 .
-x2
So
x=
+ 8x - 10
or
x2 - 7x + 10 = 0
ie
(x – 2)(x – 5) = 0
ie
x = 2 or x = 5
• At intersection of line and
curve
y1 = y2
Terms to the left, simplify and
factorise.
Since points lie on y = x then
A is (2,2) and B is (5,5) .
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Markers Comments
• Learn result
Area =  (y 2  y1 )dx
(b) Curve is above line between limits so
5

Shaded area =
b
a
(-x2 + 8x – 10 - x) dx
can be used to find the
2
5

[ -x3
(-x2
=
enclosed area shown:
+ 7x – 10) dx
2
=
3
+ 7x2 - 10x
2
]
5
y1
area
2
lower curve
= (-125/3 + 175/2 – 50)
– (-8/3 +14 – 20)
= 41/2units2
y2
a
upper curve
b
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HIGHER – ADDITIONAL QUESTION BANK
You have chosen to study:
UNIT 2 :
Addition
Formulae
Please choose a question to attempt from the following:
1
EXIT
2
3
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Unit 2 Menu
4
ADDITION FORMULAE : Question 1
In triangle PQR show that the exact value of cos(a - b) is 4/5.
Q
2
P
b
a
1
4
R
Reveal answer only
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ADDITION FORMULAE : Question 1
In triangle PQR show that the exact value of cos(a - b) is 4/5.
Q
cos(a – b) = cosacosb + sinasinb
2
P
b
a
1
4
= (2/5 X 1/5 ) + (1/5 X 2/5 )
R
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= 4/5
Question 1
PQ2 = 12 + 22 = 5
In triangle PQR show that the
so PQ = 5
exact value of cos(a - b) is 4/5.
QR2 = 42 + 22 = 20
so QR = 20 = 45 = 25
Q
sina = 2/25 = 1/5 & sinb = 2/5
cosa = 4/25 = 2/5 & cosb = 1/5
2
P
b
a
1
4
R
cos(a – b) = cosacosb + sinasinb
= (2/5 X 1/5 ) + (1/5 X 2/5 )
Begin Solution
= 2/5 + 2/5
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= 4/5
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PQ2 = 12 + 22 = 5
so PQ = 5
• Use formula sheet to check
correct expansion and relate
to given variables:
QR2 = 42 + 22 = 20
so QR = 20 = 45 = 25
cos(a - b) = cosacosb + sina sinb
sina = 2/25 = 1/5 & sinb = 2/5
• Work only with exact values
when applying Pythagoras’:
cosa = 4/25 = 2/5 & cosb = 1/5
cos(a – b) = cosacosb + sinasinb
= (2/5 X 1/5 ) + (1/5 X 2/5 )
= 2/5 + 2/5
= 4/5
11
2
a= 7
a
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ADDITION FORMULAE : Question 2
Find the exact value of cos(p + q) in the diagram below
Y
F(4,4)
O
p
q
X
G(3,-1)
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ADDITION FORMULAE : Question 2
Find the exact value of cos(p + q) in the diagram below
Y
F(4,4)
O
p
q
X
G(3,-1)
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= 1/5
Question 2
OF2 = 42 + 42 = 32
Find the exact value of cos(p + q)
so OF = 32 = 162 = 42
in the diagram below
OG2 = 32 + 12 = 10
so OG = 10
Y
sinp = 4/42 = 1/2 & sinq = 1/10
F(4,4)
O
cosp = 4/42 = 1/2 & cosq = 3/10
cos(p + q) = cospcosq - sinpsinq
p
q
X
G(3,-1)
Begin Solution
= (1/2 X 3/10 ) - (1/2 X 1/10
) 3
= /20 - 1/20
= 2/20
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= 2/4 5
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= 2/2 5
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= 1/5
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OF2 = 42 + 42 = 32
so OF = 32 = 162 = 42
• Use formula sheet to check
correct expansion and relate
to given variables:
OG2 = 32 + 12 = 10
Formula Sheet:
so OG = 10
cos(a + b) = cosacosb - sina sinb
sinp = 4/42 = 1/2 & sinq = 1/10
cosp = 4/42 = 1/2 & cosq = 3/10
cos(p + q) = cospcosq - sinpsinq
becomes:
cos(p + q) = cospcosq - sinpsinq
= (1/2 X 3/10 ) - (1/2 X 1/10
) 3
= /20 - 1/20
= 2/20
= 2/4 5
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= 2/2 5
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= 1/5
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OF2 = 42 + 42 = 32
so OF = 32 = 162 = 42
• Work only with exact values
when applying Pythagoras’:
OG2 = 32 + 12 = 10
so OG = 10
sinp = 4/42 = 1/2 & sinq = 1/10
11
2
a= 7
a
cosp = 4/42 = 1/2 & cosq = 3/10
cos(p + q) = cospcosq - sinpsinq
= (1/2 X 3/10 ) - (1/2 X 1/10
) 3
= /20 - 1/20
= 2/20
= 2/4 5
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= 2/2 5
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= 1/5
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ADDITION FORMULAE : Question 3
Solve the equation
2 - sinx° = 3cos2x°
where 0<x<360.
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ADDITION FORMULAE : Question 3
Solve the equation
2 - sinx° = 3cos2x°
where 0<x<360.
Solution = {30, 150, 199.5, 340.5}
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Question 3
2 - sinx° = 3cos2x°
Solve the equation
2 - sinx° = 3(1 – 2sin2x°)
2 - sinx° = 3cos2x°
2 - sinx° = 3 – 6sin2x°
where 0<x<360.
Rearrange into quadratic form
6sin2x° - sinx° - 1 = 0
6s2 – s – 1 = (3s + 1)(2s – 1)
(3sinx° + 1)(2sinx° - 1) = 0
sinx° = -1/3 or sinx° = 1/2
Q3 or Q4
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Q1 or Q2
S
(180 - a)°
A
a°
(180 + a)° (360 - a)°
T
C
Question 3
sinx° = -1/3 or sinx° = 1/2
Solve the equation
2 - sinx° = 3cos2x°
where 0<x<360.
S
(180 - a)°
A
a°
(180 + a)° (360 - a)°
T
C
sin-1 (1/2) = 30°
Q1: x = 30
Q2: x = 180 – 30 = 150
sin-1 (1/3) = 19.5°
Begin Solution
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Q3: x = 180 + 19.5 = 199.5
Q4: x = 360 - 19.5 = 340.5
Solution = {30, 150, 199.5, 340.5}
Markers Comments
2 - sinx° = 3cos2x°
2 - sinx° = 3(1 – 2sin2x°)
•Use formula sheet to check
correct expansion and relate to
given variables:
2 - sinx° = 3 – 6sin2x°
Rearrange into quadratic form
6sin2x° - sinx° - 1 = 0
6s2 – s – 1 = (3s + 1)(2s – 1)
(3sinx° + 1)(2sinx° - 1) = 0
sinx° = -1/3 or sinx° = 1/2
Q3 or Q4
cos2x° =1 - 2sin2x°
•Formula must be consistent with
the rest of the equation.
2 - sinx°
i.e. choose formula with sinx°
Q1 or Q2
S
(180 - a)°
A
a°
(180 + a)° (360 - a)°
T
C
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2 - sinx° = 3cos2x°
2 - sinx° = 3(1 – 2sin2x°)
2 - sinx° = 3 – 6sin2x°
Rearrange into quadratic form
•Only one way to solve the
resulting quadratic:
Terms to the left, put in standard
quadratic form and factorise.
6sin2x° - sinx° - 1 = 0
6s2 – s – 1 = (3s + 1)(2s – 1)
(3sinx° + 1)(2sinx° - 1) = 0
sinx° = -1/3 or sinx° = 1/2
Q3 or Q4
Q1 or Q2
S
(180 - a)°
A
a°
(180 + a)° (360 - a)°
T
C
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sinx° = -1/3 or sinx° = 1/2
S
(180 - a)°
A
a°
•Take care to relate “solutions”
to the given domain.
Since 0  x  360 all 4 values
are included.
(180 + a)° (360 - a)°
T
C
sin-1 (1/2) = 30°
Q1: x = 30
Q2: x = 180 – 30 = 150
sin-1 (1/3) = 19.5°
Q3: x = 180 + 19.5 = 199.5
Q4: x = 360 - 19.5 = 340.5
Solution = {30, 150, 199.5, 340.5}
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ADDITION FORMULAE : Question 4
Solve
sin2 = cos 
where 0 <  < 2
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ADDITION FORMULAE : Question 4
Solve
sin2 = cos 
where 0 <  < 2
Soltn = {/6 , /2 , 5/6 , 3/2}
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Question 4
sin2 = cos
Solve sin2 = cos 
sin2 - cos = 0
where 0 <  < 2
2sin cos - cos = 0
(common factor cos)
cos(2sin - 1) = 0
cos = 0 or (2sin - 1) = 0
cos = 0 or sin = 1/2
(roller-coaster graph)
Begin Solution
=
/
2
or
S
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A
-

+
2 - 
Continue Solution
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Q1 or Q2
3/
2
T
C
Question 4
Solve sin2 = cos 
cos = 0 or sin = 1/2
 = /2 or
3/
2
Q1 or Q2
where 0 <  < 2
S
A
-

+
2 - 
T
C
sin-1(1/2) = /6
Q1:  = /6
Begin Solution
Q2:  =  - /6 = 5/6
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Soltn = {/6 , /2 , 5/6 , 3/2}
Markers Comments
• Use formula sheet to check
correct expansion and relate to
given variables:
sin2x = 2sinxcosx
sin2 = cos
sin2 - cos = 0
2sin cos - cos = 0
(common factor cos)
•Although equation is in radians
possible to work in degrees
and convert to radians using:
cos(2sin - 1) = 0
cos = 0 or (2sin - 1) = 0
1 
cos = 0 or sin = 1/2

180
radians
(roller-coaster graph)
 = /2 or
Q1 or Q2
3/
2
S
A
-
+
T

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2 - 
C
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Markers Comments
•Only one way to solve the
result:
sin2 = cos
sin2 - cos = 0
Terms to the left, put in standard
quadratic form and factorise.
2sin cos - cos = 0
(common factor cos)
cos(2sin - 1) = 0
cos = 0 or (2sin - 1) = 0
cos = 0 or sin = 1/2
(roller-coaster graph)
 = /2 or
Q1 or Q2
3/
2
S
A
-
+
T

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2 - 
C
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• For trig. equations
cos = 0 or sin = 1/2
 = /2 or
3/
2
sinx = 0 or 1 or cosx = 0 or 1
Q1 or Q2
S
A
-
solutions:
y

+
use sketch of graph to obtain
cosx = 0
y = cos x
2 - 
T
C

2
3
2
x
sin-1(1/2) = /6
Q1:  = /6
Q2:  =  -
x = 2
/ = 5/
6
6
3
,2
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Soltn =
{/6
,
/
2
,
5/
6
,
3/ }
2
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Markers Comments
cos = 0 or sin =
 = /2 or
3/
2
• Take care to relate “solutions”
to the given domain.
1/
2
Q1 or Q2
S
A
-

+
2 - 
T
Since 0    2 all 4 values
are included
C
sin-1(1/2) = /6
Q1:  = /6
Q2:  =  - /6 = 5/6
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Soltn =
{/6
,
/
2
,
5/
6
,
3/ }
2
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HIGHER – ADDITIONAL QUESTION BANK
You have chosen to study:
UNIT 2 :
The Circle
Please choose a question to attempt from the following:
1
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2
3
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4
THE CIRCLE : Question 1
The face of a stopwatch can be modelled by the
circle with equation x2 + y2 – 10x – 4y + 5 = 0.
The centre is at C and the winder is at W.
W
The dial for the second hand is 1/3 the size of
the face and is located half way between C and
W.
(a) Find the coordinates of C and W.
C
EXIT
(b) Hence find the equation of the dial
for the second hand.
Reveal answer only
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THE CIRCLE : Question 1
The face of a stopwatch can be modelled by the
circle with equation x2 + y2 – 10x – 4y + 5 = 0.
The centre is at C and the winder is at W.
W
The dial for the second hand is 1/3 the size of
the face and is located half way between C and
W.
C is (5,4)
W is (5,10).
(a) Find the coordinates of C and W.
C
(b) Hence find the equation of the dial
for the second hand.
(x – 5)2 + (y – 7)2 = 4
EXIT
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Question 1
(a) Find the coordinates of C and W.
The face of a stopwatch can be
Large circle is x2 + y2 – 10x – 4y + 5 = 0.
modelled by the circle with
Comparing
equation x2 + y2 – 10x – 4y + 5 = 0. Coefficients
The centre is at C and the winder
x2 + y2 + 2gx + 2fy + c = 0
2g = -10, 2f = -8 and
c=5
is at W.
The dial for the second hand is
1/
3
So g = -5,
f = -4 and c = 5
the size of the face and is located
Centre is (-g,-f) and r = (g2 + f2 – c)
half way between C and W.
C is (5,4)
r = (25 + 16 – 5)
r = 36 = 6
Begin Solution
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W is 6 units above C so W is (5,10).
Question 1
The face of a stopwatch can be
modelled by the circle with
equation x2 + y2 – 10x – 4y + 5 = 0.
(b) Hence find the equation of the dial
for the second hand.
Radius of small circle = 6  3 = 2.
Centre is midpoint of CW ie (5,7).
The centre is at C and the winder
is at W.
The dial for the second hand is 1/3
the size of the face and is located
half way between C and W.
Begin Solution
Continue Solution
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Using (x – a)2 + (y – b)2 = r2 we get
(x – 5)2 + (y – 7)2 = 4
Markers Comments
(a) Find the coordinates of C and W.
Large circle is x2 + y2 – 10x – 4y + 5 = 0.
Comparing
Coefficients
x2 + y2 + 2gx + 2fy + c = 0
2g = -10, 2f = -8 and
So g = -5,
• Use formula sheet to check
correct formulas and relate to
general equation of the circle.
c=5
f = -4 and c = 5
Centre is (-g,-f) and r = (g2 + f2 – c)
C is (5,4)
r = (25 + 16 – 5)
r = 36 = 6
W is 6 units above C so W is (5,10).
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Markers Comments
(b) Hence find the equation of the dial
for the second hand.
Radius of small circle = 6  3 = 2.
Centre is midpoint of CW ie (5,7).
Using (x – a)2 + (y – b)2 = r2 we get
(x – 5)2 + (y – 7)2 = 4
• Identify centre and radius
before putting values into
circle equation:
Centre (5,7), radius = 2
(x - a)2 + (y - b)2= r2
(x - 5)2 + (y - 7)2= 22
There is no need to expand
this form of the circle
equation.
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THE CIRCLE : Question 2
In a factory a conveyor belt passes through several rollers. Part of this
mechanism is shown below.
(a) The small roller has centre P and
equation x2 + y2 – 6x + 2y - 7 = 0.
belt
A(7,0)
(i)
P
Q
The belt is a common tangent
which meets the small roller at
A(7,0)and the large roller at B.
Find the equation of the tangent.
(ii) B has an x-coordinate of 10.
Find its y-coordinate.
B
(b) Find the equation of the larger roller
given that its diameter is twice that of
the smaller roller, and has centre Q.
EXIT
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THE CIRCLE : Question 2
In a factory a conveyor belt passes through several rollers. Part of this
mechanism is shown below.
(a) The small roller has centre P and
equation x2 + y2 – 6x + 2y - 7 = 0.
belt
A(7,0)
(i)
P
Q
(ii) B has an x-coordinate of 10.
Find its y-coordinate.
B
(x – 18)2 + (y + 10)2 = 68
EXIT
The belt is a common tangent
which meets the small roller at
A(7,0)and the large roller at B.
Find the equation of the tangent.
y = -4x + 28
B is (10,-12)
(b) Find the equation of the larger roller
given that its diameter is twice that of
the smaller roller, and has centre Q.
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Question 2
(a) The small roller has centre P
and equation
(a)
(i) Small circle is x2 + y2 – 6x + 2y - 7 = 0.
Comparing
coefficients
x2 + y2 + 2gx + 2fy + c = 0
x2 + y2 – 6x + 2y - 7 = 0.
(i) The belt is a common tangent
2g = -6, 2f = 2 and
c = -7
which meets the small roller at
A(7,0)and the large roller at B. So g = -3, f = 1 and
c = -7
Find the equation of the
tangent.
Centre is (-g,-f) and r = (g2 + f2 – c)
(ii) B has an x-coordinate of 10.
Find its y-coordinate.
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P is (3,-1)
r = (9 + 1 + 7)
r = 17
For equation of tangent:
Gradient of PA = 0 – (-1)
7-3
= 1/4
Question 2
(a) The small roller has centre P
and equation
x2 + y2 – 6x + 2y - 7 = 0.
(i) The belt is a common tangent
which meets the small roller at
A(7,0)and the large roller at B.
Find the equation of the
tangent.
(ii) B has an x-coordinate of 10.
Find its y-coordinate.
Begin Solution
Continue Solution
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P is (3,-1)
For equation of tangent:
Gradient of PA = 0 – (-1)
7-3
= 1/4
Gradient of tangent = -4 (as m1m2 = -1)
Using y – b = m(x – a)
with (a,b) = (7,0) & m = -4 we get ….
y – 0 = -4(x – 7) or
y = -4x + 28
Question 2
(a) The small roller has centre P
(a)(ii) At B x = 10
so y = (-4 X 10) + 28 = -12.
and equation
x2 + y2 – 6x + 2y - 7 = 0.
(i) The belt is a common tangent
which meets the small roller at
A(7,0)and the large roller at B.
Find the equation of the
tangent.
(ii) B has an x-coordinate of 10.
Find its y-coordinate.
Begin Solution
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ie B is (10,-12)
Question 2
(a) The small roller has centre P
and equation
x2
+
y2
– 6x + 2y - 7 = 0.
(b) Find the equation of the larger
Small Circle:
r = 17
P is (3,-1)
A(7,0)
(b) From P to A is 4 along and 1 up.
So from B to Q is 8 along and 2 up.
Q is at (10+8, -12+2) ie (18,-10)
Radius of larger circle is 217
roller given that its diameter is
twice that of the smaller roller,
so r2 = (217)2 = 4 X 17 = 68
and has centre Q.
Using (x – a)2 + (y – b)2 = r2 we get
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(x – 18)2 + (y + 10)2 = 68
Markers Comments
(a)
(i) Small circle is x2 + y2 – 6x + 2y - 7 = 0.
Comparing
coefficients
x2 + y2 + 2gx + 2fy + c = 0
2g = -6, 2f = 2 and
So
g = -3,
• Use formula sheet to check
correct formulas and relate to
general equation of the circle.
f = 1 and
c = -7
c = -7
Centre is (-g,-f) and r = (g2 + f2 – c)
P is (3,-1)
r = (9 + 1 + 7)
r = 17
For equation of tangent:
Gradient of PA = 0 – (-1)
7-3
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= 1/4
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Markers Comments
(b) From P to A is 4 along and 1 up.
• Identify centre and radius
before putting values into
circle equation:
So from B to Q is 8 along and 2 up.
Q is at (10+8, -12+2) ie (18,-10)
Radius of larger circle is 217
Centre (18,-10), radius = 2 17
(x - a)2 + (y - b)2= r2
(x - 18)2 + (y + 10)2= (2 17 ) 2
so r2 = (217)2 = 4 X 17 = 68
Using (x – a)2 + (y – b)2 = r2 we get
(x – 18)2 + (y + 10)2 = 68
There is no need to expand this
form of the circle equation.
If the radius is “squared out”
it must be left as an exact value.
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THE CIRCLE : Question 3
Part of the mechanism inside an alarm clock consists of
three different sized collinear cogwheels.
The wheels can be represented by circles with centres C, D and E as shown.
The small circle with centre C has equation x2 + (y –12)2 = 5.
The medium circle with centre E has equation (x - 18)2 + (y –3)2 = 20.
Find the equation of the large circle.
EXIT
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THE CIRCLE : Question 3
Part of the mechanism inside an alarm clock consists of
three different sized collinear cogwheels.
The wheels can be represented by circles with centres C, D and E as shown.
The small circle with centre C has equation x2 + (y –12)2 = 5.
The medium circle with centre E has equation (x - 18)2 + (y –3)2 = 20.
Find the equation of the large circle.
EXIT
(x – 8)2 + (y – 8)2 = 45
Reveal answer only
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Question 3
The small circle with centre C
has equation x2 + (y –12)2 = 5.
Small circle has centre (0,12) and radius 5
Med. circle has centre (18,3) and radius 20
= 4 X 5 = 25
The medium circle with centre E
has equation
(x - 18)2 + (y –3)2 = 20.
Find the equation of the
large circle.
CE2 = (18 – 0)2 + (3 – 12)2 (Distance form!!)
= 182 + (-9)2 = 324 + 81 = 405
CE = 405 = 81 X 5 = 95
Diameter of large circle = 95 - 25 - 5
= 65
Begin Solution
C
5
35
D
Continue Solution
95
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So radius of large circle = 35
25
E
Question 3
C
5
35
D
25
The small circle with centre C
95
has equation x2 + (y –12)2 = 5.
The medium circle with centre E
It follows that CD = 4/9CE
= 4/9[(
has equation
(x - 18)2 + (y –3)2 = 20.
) - (120 )]
18
= 4/9( -9 ) = ( -48
Find the equation of the
large circle.
18
3
)
So D is the point (0 + 8,12- 4) or (8,8)
Using (x – a)2 + (y – b)2 = r2 we get
(x – 8)2 + (y – 8)2 = (35)2
Begin Solution
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or
(x – 8)2 + (y – 8)2 = 45
E
Markers Comments
Small circle has centre (0,12) and radius 5
Med. circle has centre (18,3) and radius 20
= 4 X 5 = 25
• Use formula sheet to check
correct formulas and relate to
equation of the circle.
CE2 = (18 – 0)2 + (3 – 12)2 (Distance form!!)
= 182 + (-9)2 = 324 + 81 = 405
CE = 405 = 81 X 5 = 95
Diameter of large circle = 95 - 25 - 5
C
= 65
5
35
D
95
25
E
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Markers Comments
It follows that CD = 4/9CE
= 4/9[(
18
3
) - (120 )]
• The section formula is used to
find Centre D
C
45
18
= 4/9( -9 )
So D is the point (0 + 8,12- 4) or (8,8)
Using (x – a)2 + (y – b)2 = r2 we get
(x – 8)2 + (y – 8)2 = (35)2
or
(x – 8)2 + (y – 8)2 = 45
D
55
CD : DE = 4 : 5
5CD = 4DE
5(d - c) = 4(e - d)
9d = 5c + 4e
 8
d = 
 8
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E
THE CIRCLE : Question 4
The central driving wheels on a steam
locomotive are linked by a piston rod.
A
C
B
D
E
(a) The rear wheel has centre A & equation
the front wheel has centre C & equation
x2 + y2 – 4x – 10y + 4 = 0
x2 + y2 – 52x – 10y + 676 = 0.
If one unit is 5cm then find the size of the minimum gap between the
wheels given that the gaps are equal and the wheels are identical.
(b) The central wheel has equation
x2 + y2 – 28x – 10y + 196 = 0
and the midpoint of DE is M(13.5,1.5). Find the equation of DE and
hence find the coordinates of D & E.
EXIT
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Reveal answer only
THE CIRCLE : Question 4
The central driving wheels on a steam
locomotive are linked by a piston rod.
(a) The rear wheel has centre A & equation
the front wheel has centre C & equation
x2 + y2 – 4x – 10y + 4 = 0
x2 + y2 – 52x – 10y + 676 = 0.
If one unit is 5cm then find the size of the minimum gap between the
wheels given that the gaps are equal and the wheels are identical.
Each gap = 2 units = 10cm.
(b) The central wheel has equation
x2 + y2 – 28x – 10y + 196 = 0
and the midpoint of DE is M(13.5,1.5). Find the equation of DE and
hence find the coordinates of D & E.
Equation of DE
EXIT
x + 7y = 24
Go to full solution
Hence D is (10,2) and E is (17,1).
Reveal answer only
Question 4
(a)
The rear wheel has centre A &
(a) Rear wheel is x2 + y2 – 4x – 10y + 4 = 0 .
Comparing
coefficients
x2 + y2 + 2gx + 2fy + c = 0
equation x2 + y2 – 4x – 10y + 4 = 0
the front wheel has centre C &
equation
2g = -4, 2f = -10 and
So g = -2,
f = -5
c=4
and c = 4
x2 + y2 – 52x – 10y + 676 = 0
Centre is (-g,-f) and r = (g2 + f2 – c)
If one unit is 5cm then find the
A is (2,5)
size of the minimum gap between
the wheels.
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r = (4 + 25 – 4)
r = 25 = 5
Question 4
(a)
The rear wheel has centre A &
Front wheel is x2 + y2 – 52x – 10y + 676 = 0
Comparing
coefficients
equation x2 + y2 – 4x – 10y + 4 = 0
x2 + y2 + 2gx + 2fy + c = 0
the front wheel has centre C &
equation
x2 + y2 – 52x – 10y + 676 = 0
If one unit is 5cm then find the
size of the minimum gap between
the wheels.
Begin Solution
Continue Solution
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2g = -52, 2f = -10 and c = 676
So g = -26,
f = -5 and c = 676
Centre is (-g,-f)
C is (26,5)
AC = 24 units
r = 5 as wheels
are identical
{A is (2,5)}
Both gaps = AC – 2 radii - diameter
= 24 – 5 – 5 – 10 = 4 units
Each gap = 2 units = 10cm.
Question 4
(b)
The central wheel has equation
x2
+
y2
Middle wheel is x2 + y2 – 28x – 10y + 196 = 0
.
x2 + y2 + 2gx + 2fy + c = 0
– 28x – 10y + 196 = 0
and the midpoint of DE is
2g = -28, 2f = -10 and c = 196
M(13.5,1.5).
Find the equation of DE and hence
find the coordinates of D & E.
So g = -14,
f = -5
and c = 196
Centre is (-g,-f)
B is (14,5)
Equation of DE
Begin Solution
Gradient of BM =
5 – 1.5
14 – 13.5
= 3.5/0.5 = 7
Continue Solution
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Gradient of DE = -1/7 (m1m2 = -1)
Question 4
(b)
Equation of DE
y – b = m(x – a)
The central wheel has equation
Using
x2 + y2 – 28x – 10y + 196 = 0
we get y – 1.5 = -1/7 (x – 13.5)
and the midpoint of DE is
Or
M(13.5,1.5).
( X 7)
7y – 10.5 = -x + 13.5
So DE is
x + 7y = 24
Find the equation of DE and hence
find the coordinates of D & E.
x + 7y = 24 can be written as x = 24 – 7y
Line & circle meet when
x2 + y2 – 28x – 10y + 196 = 0 & x = 24 – 7y.
Begin Solution
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Substituting (24 – 7y) for x in the
circle equation we get ….
Question 4
(b)
The central wheel has equation
x2
+
y2
– 28x – 10y + 196 = 0
(24 – 7y)2 + y2 – 28(24 – 7y) – 10y + 196 = 0
576 – 336y + 49y2 + y2 – 672 + 196y – 10y + 196 = 0
50y2 – 150y + 100 = 0
and the midpoint of DE is
y2 – 3y + 2 = 0
M(13.5,1.5).
(y – 1)(y – 2) = 0
Find the equation of DE and
(50)
So y = 1 or y = 2
Hence find the coordinates
of D & E.
Using x = 24 – 7y
If y = 1 then x = 17
If y = 2 then x = 10
Begin Solution
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Hence D is (10,2) and E is (17,1).
Markers Comments
• Use formula sheet to check
correct formulas and relate to
(a) Rear wheel is x2 + y2 – 4x – 10y + 4 = 0 .
Comparing
general equation of the circle.
2
2
coefficients
x + y + 2gx + 2fy + c = 0
2g = -4, 2f = -10 and
So g = -2,
f = -5
c=4
and c = 4
Centre is (-g,-f) and r = (g2 + f2 – c)
A is (2,5)
r = (4 + 25 – 4)
r = 25 = 5
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Markers Comments
Equation of DE
Using
y – b = m(x – a)
we get y – 1.5 = -1/7 (x – 13.5)
Or
7y – 10.5 = -x + 13.5
So DE is
• In part b) avoid fractions when
substituting into the circle
equation:
Use
1
x = (24 - 7y) not y = (24 -x)
7
x + 7y = 24
x + 7y = 24 can be written as x = 24 – 7y
Line & circle meet when
x2 + y2 – 28x – 10y + 196 = 0 & x = 24 – 7y.
Substituting (24 – 7y) for x in the
circle equation we get ….
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Markers Comments
(24 – 7y)2 + y2 – 28(24 – 7y) – 10y + 196 = 0
576 – 336y + 49y2 + y2 – 672 + 196y – 10y + 196 = 0
50y2 – 150y + 100 = 0
(50)
y2 – 3y + 2 = 0
(y – 1)(y – 2) = 0
So y = 1 or y = 2
Using x = 24 – 7y
If y = 1 then x = 17
• Even if it appears an error
has occurred continue the
solution even if it means
applying the quadratic
formula:
x=
b  b 2  4ac
2a
If y = 2 then x = 10
Hence D is (10,2) and E is (17,1).
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HIGHER – ADDITIONAL QUESTION BANK
UNIT 3 :
EXIT
Wave Function
Vectors
Logs &
Exponential
Further Calculus
HIGHER – ADDITIONAL QUESTION BANK
You have chosen to study:
UNIT 3 :
Wave Function
Please choose a question to attempt from the following:
1
EXIT
2
3
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WAVE FUNCTION: Question 1
(a) Express cos(x°) + 7sin (x°) in the form kcos(x° - a°) where
k>0 and 0  a  90.
(b) Hence solve
cos(x°) + 7sin (x°) = 5
for 0  x  90.
Reveal answer only
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EXIT
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WAVE FUNCTION: Question 1
(a) Express cos(x°) + 7sin (x°) in the form kcos(x° - a°) where
k>0 and 0  a  90.
(b) Hence solve
cos(x°) + 7sin (x°) = 5
for 0  x  90.
Hence
Reveal answer only
(a)
cos(x°) + 7sin (x°) = 52cos(x° - 81.9°)
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EXIT
(b)
Solution is
{36.9}
Question 1
Let
(a) Express cos(x°) + 7sin (x°)
= k(cosx°cosa° + sinx°sina° )
in the form kcos(x° - a°)
where k>0 and 0  a  90.
(b) Hence solve
cos(x°) + 7sin (x°) = 5
for 0  x  90.
Begin Solution
Continue Solution
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cos(x°) + 7sin (x°) = kcos(x° - a°)
= (kcosa°)cosx° + (ksina°)sinx°
Comparing
coefficients
kcosa° = 1 & ksina° = 7
So
(kcosa°)2 + (ksina°)2 = 12 + 72
or
or
k2cos2a° + k2sin2a° = 1 + 49
k2(cos2a° + sin2a°) = 50
or
k2 = 50
so
k = 50 = 252 = 52
(cos2a° + sin2a°) = 1
Question 1
(a) Express cos(x°) + 7sin (x°)
in the form kcos(x° - a°)
where k>0 and 0  a  90.
(b) Hence solve
cos(x°) + 7sin (x°) = 5
for 0  x  90.
k = 50 = 252 = 52
so
ksina°
7
=
kcosa° 1
kcosa° > 0 so a in Q1 or Q4
ksina° > 0 so a in Q1 or Q2



so a in Q1!
tana° = 7
a° = tan-1(7) = 81.9°
Hence
cos(x°) + 7sin (x°) = 52cos(x° - 81.9°)
Begin Solution
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Question 1
(a) Express cos(x°) + 7sin (x°)
Hence
cos(x°) + 7sin (x°) = 52cos(x° - 81.9°)
in the form kcos(x° - a°)
where k>0 and 0  a  90.
(b) Hence solve
cos(x°) + 7sin (x°) = 5
for 0  x  90.
(b)
If
cos(x°) + 7sin (x°) = 5
then
52cos(x° - 81.9°) = 5
or
cos(x° - 81.9°) = 1/2
Q1 or Q4 & cos-1(1/2) = 45°
Q1: angle = 45° so x° - 81.9° = 45°
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so x° = 126.9°  not in range.
Q4: angle = 360° - 45° so x° - 81.9° = 315°
so x° = 396.9° (**)
Question 1
(a) Express cos(x°) + 7sin (x°)
in the form kcos(x° - a°)
where k>0 and 0  a  90.
(b) Hence solve
cos(x°) + 7sin (x°) = 5
for 0  x  90.
Begin Solution
Continue Solution
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Q4: angle = 360° - 45° so x° - 81.9° = 315°
so x° = 396.9° (**)
(**) function repeats every 360°
& 396.9° - 360° = 36.9°
Solution is
{36.9}
Markers Comments
Let
cos(x°) + 7sin (x°) = kcos(x° - a°)
= k(cosx°cosa° + sinx°sina° )
• Use the formula sheet for
the correct expansion:
kcos(x° - a°)
= (kcosa°)cosx° + (ksina°)sinx°
Comparing
coefficients
kcosa° = 1 & ksina° = 7
So
(kcosa°)2 + (ksina°)2 = 12 + 72
or
or
k2cos2a° + k2sin2a° = 1 + 49
k2(cos2a° + sin2a°) = 50
or
k2 = 50
so
k = 50 = 252 = 52
= kcosx°cosa° + ksinx°sina°
(Take care not to omit the “k”
term)
(cos2a° + sin2a°) = 1
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Markers Comments
Let
cos(x°) + 7sin (x°) = kcos(x° - a°)
= k(cosx°cosa° + sinx°sina° )
• When equating coefficients
“square” and “ring”
corresponding coefficients:
= (kcosa°)cosx° + (ksina°)sinx°
Comparing
coefficients
kcosa° = 1 & ksina° = 7
So
(kcosa°)2 + (ksina°)2 = 12 + 72
or
or
k2cos2a° + k2sin2a° = 1 + 49
k2(cos2a° + sin2a°) = 50
or
k2 = 50
so
k = 50 = 252 = 52
1cosx° + 7sinx° =
kcosx°cosa° + ksinx°sina°
(cos2a° + sin2a°) = 1
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Markers Comments
Let
cos(x°) + 7sin (x°) = kcos(x° - a°)
= k(cosx°cosa° + sinx°sina° )
• State the resulting equations
explicitly:
kcosa° = 1
ksina° = 7
= (kcosa°)cosx° + (ksina°)sinx°
Comparing
coefficients
kcosa° = 1 & ksina° = 7
So
(kcosa°)2 + (ksina°)2 = 12 + 72
or
or
k2cos2a° + k2sin2a° = 1 + 49
k2(cos2a° + sin2a°) = 50
or
k2 = 50
so
k = 50 = 252 = 52
(cos2a° + sin2a°) = 1
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Markers Comments
•
Let
k can be found directly:
cos(x°) + 7sin (x°) = kcos(x° - a°)
= k(cosx°cosa° + sinx°sina° )
k  12  72  50
= (kcosa°)cosx° + (ksina°)sinx°
Comparing
coefficients
kcosa° = 1 & ksina° = 7
So
(kcosa°)2 + (ksina°)2 = 12 + 72
or
or
k2cos2a° + k2sin2a° = 1 + 49
k2(cos2a° + sin2a°) = 50
or
k2 = 50
so
k = 50 = 252 = 52
Note: k is always positive
(cos2a° + sin2a°) = 1
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WAVE FUNCTION: Question 2
Express 4sin(x°)–2cos(x°) in the form ksin(x° + a°)
where k>0 and 0 < a < 360.
Reveal answer only
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WAVE FUNCTION: Question 2
Express 4sin(x°)–2cos(x°) in the form ksin(x° + a°)
where k>0 and 0 < a < 360.
Hence
4sin(x°) – 2cos(x°) = 25sin(x° + 333.4°)
Reveal answer only
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Question 2
Let
Express 4sin(x°)–2cos(x°)
4sin(x°)–2cos(x°) = ksin(x° + a°)
= k(sinx°cosa° + cosx°sina° )
in the form ksin(x° + a°)
= (kcosa°)sinx° + (ksina°)cosx°
where k>0 and 0 < a < 360.
Comparing coefficients kcosa° = 4 &
ksina° = -2
Begin Solution
So
(kcosa°)2 + (ksina°)2 = 42 + (-2)2
or
k2cos2a° + k2sin2a° = 16 + 4
or
k2(cos2a° + sin2a°) = 20
or
k2 = 20
so
k = 20 = 45 = 25
(cos2a° + sin2a°) = 1
Continue Solution
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Question 2
Express 4sin(x°)–2cos(x°)
in the form ksin(x° + a°)
so
k = 20 = 45 = 25
ksina°
-2
=
kcosa°
4



where k>0 and 0 < a < 360.
so a in Q4!
tana° = (-1/2) so Q2 or Q4
kcosa° > 0 so a in Q1 or Q4
ksina° < 0 so a in Q3 or Q4
tan-1(1/2) = 26.6°
Q4: a° = 360° – 26.6° = 333.4°
Begin Solution
Continue Solution
Hence
Markers Comments
4sin(x°) – 2cos(x°) = 25sin(x° + 333.4°)
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Markers Comments
Let
4sin(x°)–2cos(x°) = ksin(x° + a°)
= k(sinx°cosa° + cosx°sina° )
• Use the formula sheet for
the correct expansion:
ksin(x° + a°)
= (kcosa°)sinx° + (ksina°)cosx°
= ksinx°cosa° + kcosx°sina°
Comparing coefficients kcosa° = 4 &
ksina° = -2
So
(kcosa°)2 + (ksina°)2 = 42 + (-2)2
or
k2cos2a° + k2sin2a° = 16 + 4
or
k2(cos2a° + sin2a°) = 20
or
k2 = 20
so
k = 20 = 45 = 25
(cos2a° + sin2a°) = 1
(Take care not to omit the “k”
term)
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Markers Comments
Let
4sin(x°)–2cos(x°) = ksin(x° + a°)
= k(sinx°cosa° + cosx°sina° )
= (kcosa°)sinx° + (ksina°)cosx°
Comparing coefficients kcosa° = 4 &
• When equating coefficients
“square” and “ring”
corresponding coefficients:
4sinx° - 2cosx°
= ksinx°cosa° + kcosx°sina°
ksina° = -2
So
(kcosa°)2 + (ksina°)2 = 42 + (-2)2
or
k2cos2a° + k2sin2a° = 16 + 4
or
k2(cos2a° + sin2a°) = 20
or
k2 = 20
so
k = 20 = 45 = 25
(cos2a° + sin2a°) = 1
• State the resulting
equations explicitly:
kcosa° = 4
ksina° = -2
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Markers Comments
•
Let
k can be found directly:
4sin(x°)–2cos(x°) = ksin(x° + a°)
= k(sinx°cosa° + cosx°sina° )
k  42  (2)2  20
= (kcosa°)sinx° + (ksina°)cosx°
Note: k is always positive
Comparing coefficients kcosa° = 4 &
ksina° = -2
So
(kcosa°)2 + (ksina°)2 = 42 + (-2)2
or
k2cos2a° + k2sin2a° = 16 + 4
or
k2(cos2a° + sin2a°) = 20
or
k2 = 20
so
k = 20 = 45 = 25
(cos2a° + sin2a°) = 1
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Markers Comments
so
k = 20 = 45 = 25
ksina°
-2
=
kcosa°
4


• Use the sign of the equations
to determine the correct
quadrant:

kcosa° = 4
(cos +ve)
so a in Q4!
ksina° = -2
(sin -ve)
tana° = (-1/2) so Q2 or Q4
kcosa° > 0 so a in Q1 or Q4
ksina° < 0 so a in Q3 or Q4
tan-1(1/2) = 26.6°
Q4: a° = 360° – 26.6° = 333.4°

cos +ve
&sin -ve
Hence
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4sin(x°) – 2cos(x°) = 25sin(x° + 333.4°)
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WAVE FUNCTION: Question 3
The graph below is that of y = 5sin(x°) - 12cos(x°)
P
Reveal answer only
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y = 5sin(x°) - 12cos(x°)
(a) Express y = 5sin(x°) - 12cos(x°) in the form
k > 0 and 0  a < 360
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y = ksin(x° + a°) where
(b) Hence find the coordinates of the maximum turning point at P.
EXIT
WAVE FUNCTION: Question 3
The graph below is that of y = 5sin(x°) - 12cos(x°)
P
Reveal answer only
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y = 5sin(x°) - 12cos(x°)
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(a) Express y = 5sin(x°)Hence
- 12cos(x°) in the form y = ksin(x° + a°) where
k > 0 and 0  a < 360
5sin(x°) – 12cos(x°) = 13sin(x° + 292.6°)
(b) Hence find the coordinates of the maximum turning point at P.
EXIT
Max TP is at (157.4,13)
Question 3
(a) Let 5sin(x°) – 12cos(x°) = ksin(x° + a°)
(a) Express y = 5sin(x°) - 12cos(x°)
in the form
= k(sinx°cosa° + cosx°sina° )
y = ksin(x° + a°)
= (kcosa°)sinx° + (ksina°)cosx°
where k > 0 and 0  a < 360
Comparing
coefficients kcosa° = 5 & ksina° = -12
(b) Hence find the coordinates of
the maximum turning point at P.
Begin Solution
So
(kcosa°)2 + (ksina°)2 = 52 + (-12)2
or
k2cos2a° + k2sin2a° = 25 + 144
or
k2(cos2a° + sin2a°) = 169
or
k2 = 169
so
k = 13
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(cos2a° + sin2a°) = 1
Question 3
so
(a) Express y = 5sin(x°) - 12cos(x°)
in the form
y = ksin(x° + a°)
where k > 0 and 0  a < 360
(b) Hence find the coordinates of
the maximum turning point at P.
k = 13
ksina°
-12
=
kcosa°
5



so a in Q4!
tana° = (-12/5) so Q2 or Q4
kcosa° > 0 so a in Q1 or Q4
ksina° < 0 so a in Q3 or Q4
tan-1(12/5) = 67.4°
Q4: a° = 360° – 67.4° = 292.6°
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Hence
5sin(x°) – 12cos(x°) = 13sin(x° + 292.6°)
Question 3
Hence
(a) Express y = 5sin(x°) - 12cos(x°)
in the form
5sin(x°) – 12cos(x°) = 13sin(x° + 292.6°)
y = ksin(x° + a°)
where k > 0 and 0  a < 360
(b) The maximum value of
13sin(x° + 292.6°) is 13.
(b) Hence find the coordinates of
the maximum turning point at P.
Maximum on a sin graph occurs
when angle = 90°.
ie
x + 292.6 = 90
or
Begin Solution
x = -202.6 (**)
This is not in the desired range but the
Continue Solution
function repeats every 360°.
Markers Comments
Taking -202.6 + 360 = 157.4
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Max TP is at (157.4,13)
Markers Comments
• Use the formula sheet for
(a) Let 5sin(x°) – 12cos(x°) = ksin(x° + a°)
the correct expansion:
= k(sinx°cosa° + cosx°sina° )
= (kcosa°)sinx° + (ksina°)cosx°
Comparing
coefficients kcosa° = 5 & ksina° = -12
So
(kcosa°)2 + (ksina°)2 = 52 + (-12)2
or
k2cos2a° + k2sin2a° = 25 + 144
or
k2(cos2a° + sin2a°) = 169
or
k2 = 169
so
k = 13
(cos2a° + sin2a°) = 1
ksin(x° + a°)
= ksinx°cosa° + kcosx°sina°
(Take care not to omit the “k”
term)
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Markers Comments
• When equating coefficients
(a) Let 5sin(x°) – 12cos(x°) = ksin(x° + a°)
“square” and “ring”
corresponding coefficients:
= k(sinx°cosa° + cosx°sina° )
5sinx° - 12cosx°
= (kcosa°)sinx° + (ksina°)cosx°
Comparing
coefficients kcosa° = 5 & ksina° = -12
So
(kcosa°)2 + (ksina°)2 = 52 + (-12)2
or
k2cos2a° + k2sin2a° = 25 + 144
or
k2(cos2a° + sin2a°) = 169
or
k2 = 169
so
k = 13
(cos2a° + sin2a°) = 1
= ksinx°cosa° + kcosx°sina°
•
State the resulting
equations explicitly:
kcosa° = 5
ksina° = -12
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Markers Comments
•
k can be found directly:
(a) Let 5sin(x°) – 12cos(x°) = ksin(x° + a°)
= k(sinx°cosa° + cosx°sina° )
k  52  (12) 2
= (kcosa°)sinx° + (ksina°)cosx°
 169  13
Comparing
coefficients kcosa° = 5 & ksina° = -12
So
(kcosa°)2
or
k2cos2a° + k2sin2a° = 25 + 144
or
k2(cos2a° + sin2a°) = 169
or
k2 = 169
so
k = 13
+
(ksina°)2
=
52
+
(-12)2
(cos2a° + sin2a°) = 1
Note: k is always positive
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Markers Comments
so
k = 13
ksina°
-12
=
kcosa°
5


• Use the sign of the equations
to determine the correct
quadrant:

so a in Q4!
tana° = (-12/5) so Q2 or Q4
kcosa° = 5
(cos +ve)
ksina° = -12
(sin -ve)

cos +ve
&sin -ve
kcosa° > 0 so a in Q1 or Q4
ksina° < 0 so a in Q3 or Q4
tan-1(12/5) = 67.4°
Q4: a° = 360° – 67.4° = 292.6°
Hence
5sin(x°) – 12cos(x°) = 13sin(x° + 292.6°)
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Markers Comments
(b) The maximum value of
13sin(x° + 292.6°) is 13.
• The maximum value of
13sin(x°+292.6°) can also
be found by considering
rules for related functions:
Maximum on a sin graph occurs
when angle = 90°.
ie
x + 292.6 = 90
or
13sin(x° + 292.6°) slides
13sinx° graph 292.6° to the
left.
x = -202.6 (**)
This is not in the desired range but the
Maximum value of sinx°
occurs at 90°
function repeats every 360°.
Taking -202.6 + 360 = 157.4
Max TP is at (157.4,13)
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Markers Comments
(b) The maximum value of
• Maximum value of
sin(x°+292.6°) occurs at
(90°- 292.6°)
13sin(x° + 292.6°) is 13.
Maximum on a sin graph occurs
when angle = 90°.
ie
x + 292.6 = 90
or
i.e at x = - 202.6°
Add 360° to bring into
domain
x = 157.4°
x = -202.6 (**)
This is not in the desired range but the
function repeats every 360°.
Taking -202.6 + 360 = 157.4
Max TP is at (157.4,13)
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HIGHER – ADDITIONAL QUESTION BANK
You have chosen to study:
UNIT 3 :
Logs &
Exponentials
Please choose a question to attempt from the following:
1
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2
3
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4
5
LOGS & EXPONENTIALS: Question 1
As a radioactive substance decays the amount of radioactive material
remaining after t hours, At , is given by the formula
At = A0e-0.161t
where A0 is the original amount of material.
(a) If 400mg of material remain after 10 hours then determine how much
material there was at the start.
(b) The half life of the substance is the time required for exactly half of the
initial amount to decay. Find the half life to the nearest minute.
Reveal answer only
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LOGS & EXPONENTIALS: Question 1
As a radioactive substance decays the amount of radioactive material
remaining after t hours, At , is given by the formula
At = A0e-0.161t
where A0 is the original amount of material.
(a) If 400mg of material remain after 10 hours then determine how much
material there was at the start. Initial amount of material = 2000mg
(b) The half life of the substance is the time required for exactly half of the
initial amount to decay. Find the half life to the nearest minute.
Reveal answer only
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t = 4hrs 18mins
Question 1
At  A0 e
0.161t
(a) If 400mg of material remain after
10 hours then determine how
much material there was at the
start.
(a) When t = 10, A10 = 400 so At = A0e-0.161t
becomes
A10 = A0e-0.161X10
or
400 = A0e-1.61
and
A0 = 400  e-1.61
ie
A0 = 2001.12…
or 2000 to 3 sfs
Initial amount of material = 2000mg
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Question 1
At  A0 e
0.161t
(b) For half life At = 1/2A0
so
becomes
At = A0e-0.161t
1/ A
2 0
= A0e-0.161t
(b) The half life of the substance is
the time required for exactly
or
e-0.161t = 0.5
half of the initial amount to
or
ln(e-0.161t ) = ln0.5
ie
-0.161t = ln0.5
so
t = ln0.5  (-0.161)
ie
t = 4.3052…hrs
decay. Find the half life to the
nearest minute.
( 0.3052 X 60 = 18.3..)
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or
t = 4hrs 18mins
Markers Comments
(a) When t = 10, A10 = 400 so At = A0e-0.161t
becomes
A10 = A0e-0.161X10
or
400 = A0e-1.61
and
A0 = 400  e-1.61
ie
A0 = 2001.12…
• The ex is found on the
calculator:
2nd
ex
ln
or 2000 to 3 sfs
Initial amount of material = 2000mg
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Markers Comments
(b)
(b) For half life At = 1/2A0
so
becomes
At = A0e-0.161t
1/ A
2 0
= A0e-0.161t
or
e-0.161t = 0.5
or
ln(e-0.161t ) = ln0.5
ie
-0.161t = ln0.5
so
t = ln0.5  (-0.161)
ie
t = 4.3052…hrs
( 0.3052 X 60 = 18.3..)
or
t = 4hrs 18mins
• The half life can be found
using any real value:
e.g. A0 = 2000
At = 1000
• This results in the equation
At = A0e-0.161t
1000 = 2000e-0.161t
etc.
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Markers Comments
(b)
(b) For half life At = 1/2A0
so
becomes
At = A0e-0.161t
1/ A
2 0
= A0e-0.161t
or
e-0.161t = 0.5
or
ln(e-0.161t ) = ln0.5
ie
-0.161t = ln0.5
so
t = ln0.5  (-0.161)
ie
t = 4.3052…hrs
( 0.3052 X 60 = 18.3..)
or
t = 4hrs 18mins
• To solve an exponential
equation must use logs.
A trial and error solution will
only be given minimum
credit:
e.g.
12 = e2x
take logs. to both sides
ln12 = ln (e2x)
log and exponential are inverse
functions
ln 12 = 2x
etc.
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LOGS & EXPONENTIALS: Question 2
log10y
The graph illustrates the law y = k
xn
0.3
The line passes through (0,0.3) and
(1,0). Find the values of k and n.
1
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log10x
LOGS & EXPONENTIALS: Question 2
log10y
The graph illustrates the law y = k
xn
0.3
The line passes through (0,0.3) and
(1,0). Find the values of k and n.
1
log10x
Reveal answer only
Hence
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k = 2 and n = 0.3
Question 2
Intercept = (0,0.3)
The graph illustrates the law
y= k
xn
The line passes through (0,0.3) and
(1,0). Find the values of k and n.
log10y
gradient = 0 – 0.3 = -0.3
1-0
Straight line so in form Y = mX + c
with Y = log10y and X = log10x
This becomes
log10y = -0.3log10x + 0.3
0.3
1
Begin Solution
log10x
Or
log10y = -0.3log10x + log10100.3
or
log10y = -0.3log10x + log102
or
-0.3 + log 2
law3 log10y = log10x
10
or
-0.3
law1 log10y = log102x
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Question 2
or
-0.3
law1 log10y = log102x
The graph illustrates the law
y= k
xn
The line passes through (0,0.3) and
(1,0). Find the values of k and n.
log10y
0.3
1
Begin Solution
Continue Solution
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log10x
so
y = 2x-0.3 = 2
x0.3
Hence
k = 2 and n = 0.3
Markers Comments
Intercept = (0,0.3)
gradient = 0 – 0.3
1-0
Straight line so in form Y = mX + c
with Y = log10y and X = log10x
This becomes
log10y = -0.3log10x + 0.3
Or
log10y = -0.3log10x + log10100.3
or
log10y = -0.3log10x + log102
or
log10y = log10x-0.3 + log102
• It is also possible to find the
values of k and n by applying
the laws of logs to the given
equation and substituting two
coordinates from the graph:
e.g. y = kxn
Take logs to
both sides
log y = log kxn Apply Law 1:
log AB
= log A + logB
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or
log10y = log102x-0.3
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Markers Comments
Intercept = (0,0.3)
gradient = 0 – 0.3
1-0
Straight line so in form Y = mX + c
with Y = log10y and X = log10x
This becomes
log y = logk + logxn
Apply Law 3:
logxn = nlogx
log y = logk + nlogx
log y = nlogx + logk
log10y = -0.3log10x + 0.3
Or
log10y = -0.3log10x + log10100.3
or
log10y = -0.3log10x + log102
or
log10y = log10x-0.3 + log102
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or
log10y = log102x-0.3
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Markers Comments
or
so
Hence
log10y = log102x-0.3
y = 2x-0.3 = 2
x0.3
• Two coordinates from the graph:
log10y
0.3
k = 2 and n = 0.3
1
log10x
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Markers Comments
or
log10y = log102x-0.3
so
y = 2x-0.3 = 2
x0.3
Hence
k = 2 and n = 0.3
• Two coordinates from the graph:
log y = nlogx + logk
(0,0.3)
0.3 = n.0 + logk - 1
(1,0)
0
= n.1 + logk
- 2
Hence solve 1 and 2 to
find k and n
logk = 0.3
hence k = 2, and
n = - logk
hence n= -0.3.
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LOGS & EXPONENTIALS: Question 3
Solve the equation
log3(5) – log3(2x + 1) = -2
Reveal answer only
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LOGS & EXPONENTIALS: Question 3
Solve the equation
log3(5) – log3(2x + 1) = -2
x = 22
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Question 3
Solve the equation
log3(5) – log3(2x + 1) = -2
If
then
law2
log3(5) – log3(2x + 1) = -2
log3
(
5
2x + 1
) = -2
so
5
= 3-2
(2x + 1)
or
5
= 1
(2x + 1)
9
Cross mult
we get
or
2x + 1 = 45
2x = 44
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ie
x = 22
Markers Comments
log3(5) – log3(2x + 1) = -2
If
then
log3
so
(
5
2x + 1
) = -2
5
= 3-2
(2x + 1)
• To solve an equation involving
logs apply the laws so that the
equation is reduced to
log=log and the log can be
removed and the equation
solved:
log35 - log3(2x+1) = -2
or
Cross mult
we get
5
= 1
(2x + 1)
9
Law 2: log A
B
= logA - logB
2x + 1 = 45
Must know:
or
ie
2x = 44
x = 22
log33 = 1
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Markers Comments
log3(5) – log3(2x + 1) = -2
If
then
log3
so
or
ie
) = -2
5
= 3-2
(2x + 1)
Cross mult
we get
or
(
5
2x + 1
5
= 1
(2x + 1)
9
log3
5 = -2 log 3
3
2x+1
Law 3: log xn = nlogx
log3
5 = log 3-2
3
2x+1
The log terms can now be
dropped from both sides of the
equation and the equation
solved:
2x + 1 = 45
2x = 44
x = 22
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Markers Comments
log3(5) – log3(2x + 1) = -2
If
then
log3
so
(
5
2x + 1
) = -2
5
= 3-2
(2x + 1)
Drop log3 terms
5
= 3-2
2x+1
5
=
2x+1
1
9
etc.
or
Cross mult
we get
or
ie
5
= 1
(2x + 1)
9
2x + 1 = 45
2x = 44
x = 22
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LOGS & EXPONENTIALS: Question 4
The pressure in a leaky tyre drops according to the formula
Pt = P0e-kt
sssss
where P0 is the initial tyre pressure and Pt is the pressure after t hours.
(a) If the tyre is inflated to 35psi but this drops to 31 psi in 30 minutes
then find the value of k to 3 decimal places.
(b) By how many more psi will the pressure drop in the next 15 mins?
Reveal answer only
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LOGS & EXPONENTIALS: Question 4
The pressure in a leaky tyre drops according to the formula
Pt = P0e-kt
sssss
where P0 is the initial tyre pressure and Pt is the pressure after t hours.
(a) If the tyre is inflated to 35psi but this drops to 31 psi in 30 minutes
then find the value of k to 3 decimal places.
(b) By how many more psi will the pressure drop in the next 15 mins?
Reveal answer only
(a)
k = 0.243
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(b)
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1.8psi
to 3 dps
Question 4
Pt  P0 e
 kt
(a) We have P0 = 35, Pt = 31 and t = 0.5
(30mins = ½ hr)
(a) If the tyre is inflated to 35psi
So
Pt = P0e-kt
becomes
31 = 35e(-0.5k)
this becomes
e(-0.5k) =
but this drops to 31 psi in 30
minutes then find the value of
k to 3 decimal places.
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31/
35
or
lne(-0.5k) = ln(31/35)
or
-0.5k = ln(31/35)
or
k = ln(31/35)  (-0.5)
ie
k = 0.243
to 3 dps
Question 4
Pt  P0 e
 kt
ie
k = 0.243
to 3 dps
(b) We now have k = 0.243,
(b) By how many more psi will
the pressure drop in the next
15 mins?
P0 = 31 and t = 0.25 (15mins = 1/4hr)
Pt = P0e-0.243t
using
we get
P0.25 = 31e(-0.243X0.25)
so
P0.25 = 29.2
Pressure drop in next 15 mins is
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31 – 29.2 or 1.8psi
Markers Comments
(a) We have P0 = 35, Pt = 31 and t = 0.5
(30mins = ½ hr)
So
Pt = P0e-kt
becomes
31 = 35e(-0.5k)
this becomes
e(-0.5k) =
(a)
• The ex is found on the
calculator:
ex
ln
2nd
31/
35
or
lne(-0.5k) = ln(31/35)
or
-0.5k = ln(31/35)
or
k = ln(31/35)  (-0.5)
ie
k = 0.243
to 3 dps
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Markers Comments
(a) We have P0 = 35, Pt = 31 and t = 0.5
(30mins = ½ hr)
So
Pt = P0e-kt
becomes
31 = 35e(-0.5k)
this becomes
e(-0.5k) =
31/
35
or
lne(-0.5k) = ln(31/35)
or
-0.5k = ln(31/35)
or
k = ln(31/35)  (-0.5)
ie
k = 0.243
to 3 dps
• To solve an exponential
equation must use logs.
A trial and error solution will
only be given minimum credit:
e.g. e(-0.5k) = 31/35
take logs. to both sides
ln e(-0.5k) = ln 31/35
log and exponential are
inverse functions
-0.5k = ln31/35
etc.
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LOGS & EXPONENTIALS: Question 5
Given that
sin = au and cos = av
show that
u – v = logatan.
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LOGS & EXPONENTIALS: Question 5
Given that
sin = au and cos = av
show that
u – v = logatan.
u – v = logatan
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Question 5
Given that
sin = au
and cos = av
Since
sin = au and cos = av
then
u = logasin and v = logacos
so
u – v = logasin - logacos
or
u – v = loga(sin/cos)
show that
u – v = logatan.
logx – logy = log(x/y)
hence
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u – v = logatan
Markers Comments
Since
sin = au and cos = av
then
u = logasin and v = logacos
so
u – v = logasin - logacos
or
u – v = loga(sin/cos)
logx – logy = log(x/y)
hence
u – v = logatan
• There are different routes to the
same result but all involve
correct application of formulas
and laws of logs: Should be
known from
standard
grade
e.g.
tan =
sin 
cos 
au
tan = av
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Since
sin = au and cos = av
then
u = logasin and v = logacos
so
u – v = logasin - logacos
or
Take logs to base a to both sides
au
loga tan = loga av
A
Law2: log B =logA-logB
loga tan = loga au - loga av
u – v = loga(sin/cos)
logx – logy = log( /y)
x
hence
u – v = logatan
Law 3: log xn = nlogx
loga tan = uloga a - vloga a
logaa=1
loga tan = u - v
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HIGHER – ADDITIONAL QUESTION BANK
You have chosen to study:
UNIT 3 :
Vectors
Please choose a question to attempt from the following:
1
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2
3
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Unit 3 Menu
4
5
VECTORS: Question 1
(a) P, Q & R are the points (3,0,-5), (-1,1,-2) & (-13,4,7) respectively.
Prove that these points are collinear.


(b) Given that PS = 7 PQ then find the coordinates of S.
Reveal answer only
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VECTORS: Question 1
(a) P, Q & R are the points (3,0,-5), (-1,1,-2) & (-13,4,7) respectively.
Prove that these points are collinear.


(b) Given that PS = 7 PQ then find the coordinates of S.
(a)
Reveal answer only


Since PQ and PR are multiples of the same
vector and have P as a common point then
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it follows that P, Q & R are collinear.
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(b) S = (-25, 7, 16)
Question 1

(a) PQ = q - p =
-1
1
-2
[ ]- [ ]
(a) P, Q & R are the points (3,0,-5),
(-1,1,-2) & (-13,4,7) respectively.
3
0
-2
=
-4
1
3
[ ]
Prove that these points are
collinear.

PR = r - p =
(b) Given that


PS = 7 PQ then find
the coordinates of S.
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=
-13
4
7
3
0
-2
[ ]- [ ]
-16
4
9
-4
1
3
[ ]= 4[ ]


Since PQ and PR are multiples of the same
vector and have P as a common point then
it follows that P, Q & R are collinear.
Question 1


(b) PS = 7 PQ
=7
(a) P, Q & R are the points (3,0,-5),
(-1,1,-2) & (-13,4,7) respectively.
S is (3,0,-5) +
-4
1
3
[ ] =[ ]
-28
7
21
[ ]
Prove that these points are
collinear.
(b) Given that


PS = 7 PQ then find
the coordinates of S.
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= (3-28, 0+7, -5+21)
= (-25, 7, 16)
-28
7
21
Markers Comments
a)

(a) PQ = q - p =
-1
1
-2
3
0
-2
[ ]- [ ]
• Must know result:
Given coordinates of A and B
=

PR = r - p =
=
-4
1
3
[ ]
-13
4
7
AB = b  a
3
0
-2
[ ]- [ ]
-16
4
9
-4
1
3
[ ]= 4[ ]


Since PQ and PR are multiples of the same
vector and have P as a common point then
it follows that P, Q & R are collinear.
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Markers Comments

(a) PQ = q - p =
-1
1
-2
3
0
-2
[ ]- [ ]
=

PR = r - p =
=
Since PR = 4PQ with common
-4
1
3
[ ]
-13
4
7
point P  P,Q and R are collinear
3
0
-2
[ ]- [ ]
-16
4
9
• Must be able to state explicitly
the result for collinearity:
Beware common error 4PR = PQ
-4
1
3
[ ]= 4[ ]


Since PQ and PR are multiples of the same
vector and have P as a common point then
it follows that P, Q & R are collinear.
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b)


(b) PS = 7 PQ
=7
S is (3,0,-5) +
-4
1
3
-28
7
21
[ ] =[ ]
-28
7
21
[ ]
• An alternative approach is to
form a vector equation and
solve it:
= (3-28, 0+7, -5+21)
= (-25, 7, 16)
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b)


(b) PS = 7 PQ
=7
S is (3,0,-5) +
-4
1
3
[ ] =[ ]
-28
7
21
[ ]
= (3-28, 0+7, -5+21)
= (-25, 7, 16)
-28
7
21
PS = 7PQ
s - p = 7(q - p)
s = 7 q - 7p + p
 -1   3 
   
s = 7 q - 6p = 7  1   6  0 
 -2   5 
   
 25 


 7 
 16 


S(-25,7,16)
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VECTORS: Question 2
(a) E, F & G are the points (4,3,-1), (5,-1,2) & (8,-1,-1) respectively.


Find FE and FG in component form.
(b) Hence, or otherwise, find the size of angle EFG.
Reveal answer only
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VECTORS: Question 2
(a) E, F & G are the points (4,3,-1), (5,-1,2) & (8,-1,-1) respectively.


Find FE and FG in component form.
(b) Hence, or otherwise, find the size of angle EFG.

(a) FE =
Reveal answer only

FG
=
-1
4
-3
[ ]
3
0
-3
[ ]
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so 
= 73.9°
Question 2
(a) E, F & G are the points

(a) FE = e - f =
4
3
-1
5
-1
2
[ ]- [ ]
(4,3,-1), (5,-1,2) & (8,-1,-1)
respectively.

Find FE and

FG in
=
-1
4
-3
[ ]
component form.
(b) Hence, or otherwise, find the

FG = g - f =
size of angle EFG.
=
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8
-1
-1
5
-1
2
[ ]- [ ]
3
0
-3
[ ]
Question 2
(b) Let angle EFG = 
(a) E, F & G are the points
E
(4,3,-1), (5,-1,2) & (8,-1,-1)
respectively.

Find FE and

FG in
ie
F

G
component form.
(b) Hence, or otherwise, find the
size of angle EFG.
 
FE . FG =
-1
4
-3
3
0
-3
[ ].[ ]
= (-1 X 3) + (4 X 0) + (-3 X (-3))
= -3 + 0 + 9
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= 6

|FE| = ((-1)2 + 42 + (-3)2) = 26

= 18
|FG| = (32 + 02 + (-3)2)
Question 2
(a) E, F & G are the points
(4,3,-1), (5,-1,2) & (8,-1,-1)
respectively.

Find FE and

FG in
component form.
(b) Hence, or otherwise, find the
size of angle EFG.
Begin Solution
Continue Solution
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 
Given that FE.FG = |FE||FG|cos
 
6
FE.FG
=
then cos =
|FE||FG|
26 18
so  = cos-1(6  26  18) = 73.9°
Markers Comments
(b) Let angle EFG = 
• Ensure vectors are calculated
from the vertex:
E
ie
F
E

F
G
 
FE . FG =
-1
4
-3
3
0
-3
[ ].[ ]
G
Vectors FE and FG
= (-1 X 3) + (4 X 0) + (-3 X (-3))
= -3 + 0 + 9
= 6

|FE| = ((-1)2 + 42 + (-3)2) = 26

= 18
|FG| = (32 + 02 + (-3)2)
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 
Given that FE.FG = |FE||FG|cos
 
FE.FG
then cos =
|FE||FG|
so  =
cos-1(6
 26  18) = 73.9°
• Refer to formula sheet and
relate formula to given
variables:
a.b= a . b .cosθ
 FE.FG= FE . FG .cosθ
a.b=a1b1 +a 2 b 2 +a 3 b3
 FE.FG=a1b1 +a 2 b 2 +a 3 b3
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VECTORS: Question 3
PQRSTUVW is a cuboid in which
 

PQ , PS & PW are represented by the
vectors
4
2
0
-2
4
0
0
0
9
[ ], [ ]and [ ]resp.
U
T
A
B
V
W
R
A is 1/3 of the way up ST & B is the
S
P
midpoint of UV.
ie SA:AT = 1:2 & VB:BU = 1:1.


Find the components of PA & PB and hence the size of angle APB.
Reveal answer only
Go to full solution
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Q
VECTORS: Question 3
PQRSTUVW is a cuboid in which
 

PQ , PS & PW are represented by the
vectors
4
2
0
-2
4
0
0
0
9
[ ], [ ]and [ ]resp.
U
B
T
A
V
W
R
A is 1/3 of the way up ST & B is the
S
P
midpoint of UV.
ie SA:AT = 1:2 & VB:BU = 1:1.


Find the components of PA & PB and hence the size of angle APB.
Reveal answer only

|PA| = 29
Go to full solution

|PB| = 106
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APB = 48.1°
Q
Question 3
PQRSTUVW is a cuboid in which
 

PQ , PS & PW are represented by
vectors
4
2
0
-2
4
0
0
0
9
[ ], [ ]and [ ]resp.
A is 1/3 of the way up ST & B is the
midpoint of UV.
ie SA:AT = 1:2 & VB:BU = 1:1.


Find the components of PA & PB
and hence the size of angle APB.
Begin Solution
Continue Solution
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  


1
PA = PS + SA = PS + /3ST


1
= PS + /3PW
=
-2
4
0
0
0
3
-2
4
3
[ ]+ [ ] = [ ]


 
PB = PQ + QV + VB

 1 
= PQ + PW + /2PS
=
4
2
0
0
0
9
-1
2
0
3
4
9
[ ]+ [ ]+[ ]= [ ]
Question 3
PQRSTUVW is a cuboid in which
 

PQ , PS & PW are represented by
vectors
4
2
0
-2
4
0
0
0
9
[ ], [ ]and [ ]resp.
A is 1/3 of the way up ST & B is the
midpoint of UV.
ie SA:AT = 1:2 & VB:BU = 1:1.


Find the components of PA & PB
and hence the size of angle APB.
Begin Solution
Continue Solution
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
PA =

PB =
-2
4
3
[ ]
3
4
9
[ ]
(b) Let angle APB = 
A
ie
P

B
 
PA . PB =
-2
4
3
3
4
9
[ ].[ ]
= (-2 X 3) + (4 X 4) + (3 X 9)
= -6 + 16 + 27
= 37
Question 3
 
PA . PB = 37
PQRSTUVW is a cuboid in which
 

PQ , PS & PW are represented by
vectors

|PA| = ((-2)2 + 42 + 32) = 29
4
2
0
-2
4
0
0
0
9
[ ], [ ]and [ ]resp.
A is 1/3 of the way up ST & B is the
midpoint of UV.
ie SA:AT = 1:2 & VB:BU = 1:1.


Find the components of PA & PB

|PB| = (32 + 42 + 92)
= 106
 
Given that PA.PB = |PA||PB|cos
 
PA.PB
then cos =
|PA||PB|
=
37
29 106
and hence the size of angle APB.
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so  = cos-1(37  29  106)
= 48.1°
Markers Comments

PA =

PB =
-2
4
3
[ ]
• Ensure vectors are calculated
3
4
9
[ ]
from the vertex:
B
(b) Let angle APB = 
A
P
A
ie
P

Vectors PB and PA
B
 
PA . PB =
-2
4
3
3
4
9
[ ].[ ]
= (-2 X 3) + (4 X 4) + (3 X 9)
Next Comment
= -6 + 16 + 27
Vectors Menu
= 37
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Markers Comments
•
 
PA . PB =

|PA| = ((-2)2 + 42 + 32) = 29

|PB| = (32 + 42 + 92)
= 106
 
Given that PA.PB = |PA||PB|cos
 
PA.PB
then cos =
|PA||PB|
so  = cos-1(37  29  106)
= 48.1°
Refer to formula sheet and
relate formula to given
variables:
a.b= a . b .cosθ
 PB.PA= PB . PA .cosθ
a.b=a1b1 +a 2 b 2 +a 3 b3
 PB.PA=a1b1 +a 2 b 2 +a 3b3
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VECTORS: Question 4
An equilateral triangle has sides 4 units long
which are represented by the vectors a , b & c
as shown.
b.(a + c) & comment on your answer
(ii) b.(a – c)
Reveal answer only
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c
a
Find
(i)
b
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VECTORS: Question 4
b
An equilateral triangle has sides 4 units long
which are represented by the vectors a , b & c
as shown.
a
Find
(i)
c
b.(a + c) & comment on your answer
Dot product = 0 so
(ii) b.(a – c)
(ii)
b.(a – c) = b.b
b & (a + c) are perpendicular.
= 42 = 16
Reveal answer only
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Question 4
NB: each angle is 60° but vectors should
An equilateral triangle has
be “tail to tail”
b
sides 4 units long which are
b
represented by the vectors
a,b&c.
120°
60°
c
Find
(i)
(ii)
b.(a + c) & comment
b.(a – c)
(i) b.(a + c) = b.a + b.c
= |b||a|cos1 + |b||c|cos2
= 4X4Xcos60° + 4X4Xcos120°
= 16 X ½ + 16 X (-½ )
Begin Solution
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= 8 + (-8)
= 0
Markers Comments
Dot product = 0 so
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b & (a + c) are perpendicular.
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Question 4
NB:
a – c = a + (-c) or b .
(ii)
b.(a – c) = b.b
An equilateral triangle has
sides 4 units long which are
represented by the vectors
= |b||b|cos0
a,b&c.
= |b|2
Find
= 42 = 16
(i)
(ii)
b.(a + c) & comment
b.(a – c)
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Markers Comments
NB: each angle is 60° but vectors should
be “tail to tail”
b
b
• This geometric question is based
on the scalar product definition
and the distributive law:
120°
60°
a.b= a . b .cosθ
c
and
a(b+c) = ab + ac
(i) b.(a + c) = b.a + b.c
= |b||a|cos1 + |b||c|cos2
= 4X4Xcos60° + 4X4Xcos120°
= 16 X ½ + 16 X (-½ )
= 8 + (-8)
= 0
Dot product = 0 so
b & (a + c) are perpendicular.
No other results should
be applied.
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Markers Comments
NB: each angle is 60° but vectors should
be “tail to tail”
b
b
• Ensure all scalar products are
calculated from the vertex:
120°
60°
c
(i) b.(a + c) = b.a + b.c
= |b||a|cos1 + |b||c|cos2
a
θ
b
= 4X4Xcos60° + 4X4Xcos120°
= 16 X ½ + 16 X (-½ )
= 8 + (-8)
= 0
Dot product = 0 so
b & (a + c) are perpendicular.
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VECTORS: Question 5
a = 2i + j – 3k,
b = -i + 10k & c = -2i + j + k.
(a) Find (i) 2a + b in component form (ii) | 2a + b |
(b) Show that 2a + b and
c are perpendicular.
Reveal answer only
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VECTORS: Question 5
a = 2i + j – 3k,
b = -i + 10k & c = -2i + j + k.
(a) Find (i) 2a + b in component form (ii) | 2a + b |
(b) Show that 2a + b and
c are perpendicular.
(a)(i)
Reveal answer only
Go to full solution
2a + b
=
3
2
4
[]
(ii) | 2a + b | = 29
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(b)
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Since the dot product is zero it
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follows that the vectors are
perpendicular.
Question 5
a = 2i + j – 3k,
(a)(i)
b = -i + 10k
& c = -2i + j + k.
2a + b = 2
=
(a) Find
(i) 2a + b in component form
=
(ii) | 2a + b |
(b)Show that 2a + b and
are perpendicular.
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c
2
1
-3
-1
0
10
[ ]+ [ ]
4
2
-6
-1
0
10
[ ]+ [ ]
3
2
4
[]
Question 5
a = 2i + j – 3k,
(ii) | 2a + b | = (32 + 22 + 42)
b = -i + 10k
= 29
& c = -2i + j + k.
(a) Find
(i) 2a + b in component form
(ii) | 2a + b |
(b)Show that 2a + b and
are perpendicular.
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c
Question 5
a = 2i + j – 3k,
(b) (2a + b ).c =
b = -i + 10k
3
2
4
-2
1
1
[ ].[ ]
= (3 X (-2))+(2 X 1)+(4 X 1)
& c = -2i + j + k.
(a) Find
= -6 + 2 + 4
(i) 2a + b in component form
= 0
(ii) | 2a + b |
Since the dot product is zero it follows
(b)Show that 2a + b and
are perpendicular.
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c
that the vectors are perpendicular.
Markers Comments
(a)(i)
2a + b = 2
=
=
2
1
-3
-1
0
10
[ ]+ [ ]
4
2
-6
-1
0
10
[ ]+ [ ]
• When a vector is given in
i , j , k form change to
component form:
 5
 
e.g. 5i - 2j + k =  -2 
 1
 
3
2
4
[]
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(ii) | 2a + b | = (32 + 22 + 42)
= 29
• Must know formula for the
magnitude of a vector:
a 
 
n = b
c 
 

n  a 2 +b 2 +c 2
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Markers Comments
(b) (2a + b ).c =
3
2
4
-2
1
1
[ ].[ ]
• Must know condition for
perpendicular vectors:
= (3 X (-2))+(2 X 1)+(4 X 1)
a.b = 0  a is perpendicular to b
= -6 + 2 + 4
= 0
Since the dot product is zero it follows
that the vectors are perpendicular.
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HIGHER – ADDITIONAL QUESTION BANK
You have chosen to study:
UNIT 3 :
Further Calculus
Please choose a question to attempt from the following:
1
EXIT
2
3
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FURTHER CALCULUS : Question 1
Given that
y = 3sin(2x) – 1/2cos(4x) then find
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dy/
dx
.
FURTHER CALCULUS : Question 1
Given that
y = 3sin(2x) – 1/2cos(4x) then find
Reveal answer only
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EXIT
dx
.
= 6cos(2x) + 2sin(4x)
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dy/
Question 1
1/ cos(4x)
3sin(2x)
2
OUTER / INNER
Differentiate outer then inner
Given that
y = 3sin(2x) – 1/2cos(4x)
then find
dy/
dx
.
y = 3sin(2x) – 1/2cos(4x)
dy/
dx
= 3cos(2x) X 2 - (-1/2sin(4x)) X 4
= 6cos(2x) + 2sin(4x)
Begin Solution
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• Check formula sheet for
correct result:
1/ cos(4x)
3sin(2x)
2
OUTER / INNER
Differentiate outer then inner
y
sin(ax)
cos(ax)
dy
dx
acos(ax)
-asin(ax)
y = 3sin(2x) – 1/2cos(4x)
dy/
dx
= 3cos(2x) X 2 - (-1/2sin(4x)) X 4
• Relate formula to given
variables
= 6cos(2x) + 2sin(4x)
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• When applying the “chain rule”
“Peel an onion”
1/ cos(4x)
3sin(2x)
2
OUTER / INNER
Differentiate outer then inner
y = 3sin(2x) – 1/2cos(4x)
dy/
dx
= 3cos(2x) X 2 - (-1/2sin(4x)) X 4
= 6cos(2x) + 2sin(4x)
 
1/ cos(4x)
3sin(2x)
2
OUTER / INNER
Differentiate outer then inner
e.g. 3sin
2x
-
3cos
2
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FURTHER CALCULUS : Question 2
Given that
g(x) = (6x – 5)
then evaluate g´(9) .
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FURTHER CALCULUS : Question 2
Given that
g(x) = (6x – 5)
then evaluate g´(9) .
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=
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3/
7
Question 2
Given that
g(x) = (6x – 5) = (6x – 5)1/2
g(x) = (6x – 5)
then evaluate g´(9) .
(6x – 5)1/2
outer / inner
diff outer then inner
g´(x) = 1/2(6x – 5)-1/2 X 6
= 3(6x – 5)-1/2
=
3
(6x – 5)
Begin Solution
g´(9) =
3
(6X9 – 5)
Continue Solution
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=
49
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=
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Markers Comments
g(x) = (6x – 5) = (6x – 5)1/2
• Apply the laws of indices to
replace the
sign
xx
(6x – 5)1/2
outer / inner
1
2
diff outer then inner
• Apply the chain rule
g´(x) = 1/2(6x – 5)-1/2 X 6
= 3(6x – 5)-1/2
• Learn the rule for differentiation
=
Multiply by the power then
reduce the power by 1
3
(6x – 5)
g´(9) =
3
(6X9 – 5)
=
3
49
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=
3/
7
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g(x) = (6x – 5) = (6x – 5)1/2
(6x – 5)1/2
outer / inner
• Apply the laws of indices to
return power to a positive value
then the root
x
diff outer then inner

1
2

1
x
1
2

1
x
g´(x) = 1/2(6x – 5)-1/2 X 6
• Will usually work out to an
exact value without need to
use calculator.
= 3(6x – 5)-1/2
=
3
(6x – 5)
g´(9) =
3
(6X9 – 5)
=
3
49
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=
3/
7
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FURTHER CALCULUS : Question 3
A curve for which dy/dx = -12sin(3x) passes through the
point (/3,-2). Express y in terms of x.
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FURTHER CALCULUS : Question 3
A curve for which dy/dx = -12sin(3x) passes through the
point (/3,-2). Express y in terms of x.
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So
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y = 4cos(3x) + 2
Question 3
dy/
dx
if
A curve for which
then
dy/
dx
= -12sin(3x)
y =  -12sin(3x) dx
= -12sin(3x) passes
= 1/3 X 12cos(3x) + C
through the point (/3,-2).
= 4cos(3x) + C
Express y in terms of x.
At the point (/3,-2)
Begin Solution
y = 4cos(3x) + C
becomes
-2 = 4cos(3 X /3) + C
or
-2 = 4cos + C
or
ie
-2 = -4 + C
C = 2
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So
y = 4cos(3x) + 2
Markers Comments
dy/
dx
if
then
= -12sin(3x)
y =  -12sin(3x) dx
= 1/3 X 12cos(3x) + C
= 4cos(3x) + C
At the point (/3,-2)
y = 4cos(3x) + C
becomes
-2 = 4cos(3 X /3) + C
or
-2 = 4cos + C
or
ie
-2 = -4 + C
C = 2
• Learn the result that
integration undoes
differentiation:
i.e.
given
dy
= f(x)
dx

y =  f(x) dx
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So
y = 4cos(3x) + 2
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dy/
dx
if
then
= -12sin(3x)
y =  -12sin(3x) dx
= 1/3 X 12cos(3x) + C
= 4cos(3x) + C
At the point (/3,-2)
y = 4cos(3x) + C
becomes
-2 = 4cos(3 X /3) + C
or
-2 = 4cos + C
or
ie
-2 = -4 + C
C = 2
• Check formula sheet for
correct result:
y

sinax
-cos(ax)
+ c
a
cos(ax)
sin(ax)
+ c
a
• Do not forget the constant
of integration.
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So
y = 4cos(3x) + 2
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FURTHER CALCULUS : Question 4
(a) Find the derivative of y = (2x3 + 1)2/3 where x > 0.
(b) Hence find

x2
(2x3 + 1)1/3
dx
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FURTHER CALCULUS : Question 4
(a) Find the derivative of y =
(b) Hence find

(2x3
x2
(2x3 + 1)1/3
+ 1)
2 /3
=
where x > 0.
dx
=
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4x2
(2x3 + 1)1/3
1/ (2x3
4
+ 1)2/3 + C
Question 4
(a) if
y = (2x3 + 1)2/3
(2x3 + 1)2/3
outer inner
(a) Find the derivative of
y = (2x3 + 1)2/3 where x > 0.
diff outer then inner
then
dy/
dx
= 2/3 (2x3 + 1)-1/3 X 6x2
=
Begin Solution
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4x2
(2x3 + 1)1/3
Question 4
(b) From (a) it follows that

(b) Hence find

x2
(2x3 + 1)1/3
dx
4x2
dx
(2x3 + 1)1/3
now

x2
(2x3 + 1)1/3
= ¼ X
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= (2x3 + 1)2/3 + C
=

1/ (2x3
4
dx
4x2
(2x3 + 1)1/3
+ 1)2/3 + C
dx
Markers Comments
a)
(a) if
y=
(2x3
+ 1)
2 /3
•.
Apply the chain rule
“Peel an onion”
 
(2x3 + 1)2/3
outer inner
diff outer then inner
then
dy/
dx
= 2/3 (2x3 + 1)-1/3 X 6x2
(2x3 + 1)2/3
outer inner
diff outer then inner
=
4x2
(2x3 + 1)1/3
e.g.
(
2x 3 +1
)
2
3
2
(
3
)
-
1
3
6x 2
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b)
(b) From (a) it follows that

4x2
dx
(2x3 + 1)1/3
= (2x3 + 1)2/3 + C
• Learn the result that
integration undoes
differentiation:
i.e.
given
now

x2
(2x3 + 1)1/3
= ¼ X
=

1/ (2x3
4
dy
= f(x)
dx
dx
4x2
(2x3 + 1)1/3
+ 1)2/3 + C

y =  f(x) dx
dx
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