higher – additional question bank exit unit 1unit 2unit 3 please decide which unit you would like...
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![Page 1: HIGHER – ADDITIONAL QUESTION BANK EXIT UNIT 1UNIT 2UNIT 3 Please decide which Unit you would like to revise: Straight Line Functions & Graphs Trig Graphs](https://reader038.vdocument.in/reader038/viewer/2022102906/56649d045503460f949d762a/html5/thumbnails/1.jpg)
HIGHER – ADDITIONAL QUESTION BANK
EXIT
UNIT 1 UNIT 2 UNIT 3
Please decide which Unit you would like to revise:
Straight LineFunctions & GraphsTrig Graphs & EquationsBasic DifferentiationRecurrence Relations
PolynomialsQuadratic Functions IntegrationAddition FormulaeThe Circle
VectorsFurther CalculusExponential / Logarithmic FunctionsThe Wave Function
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HIGHER – ADDITIONAL QUESTION BANK
UNIT 1 :
Functions & Graphs
Straight Line
Recurrence Relations
BasicDifferentiation
Trig Graphs & Equations
EXIT
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HIGHER – ADDITIONAL QUESTION BANK
UNIT 1 :Straight Line
You have chosen to study:
Please choose a question to attempt from the following:
1 2 3 4 5
EXITBack to
Unit 1 Menu
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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).
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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).
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Reveal answer only y = -5/3x - 6
EXIT
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3x – 5y = 4
3x - 4 = 5y
5y = 3x - 4 (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
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).
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3x – 5y = 4
3x - 4 = 5y
5y = 3x - 4 (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
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•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 perpendicular lines m1 m2 = -1.
•When finding m2 simply invert and change the sign on m1
m1 = 35 m2 =
-5 3
• Use the y - b = m(x - a) form to obtain the equation of the line.
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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).
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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).
y = -2x + 7
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Question 2
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8x + 4y – 7 = 0
4y = -8x + 7 (4)
y = -2x + 7/4
y = -2x + 7
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
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).
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• 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.
8x + 4y – 7 = 0
4y = -8x + 7 (4)
y = -2x + 7/4
y = -2x + 7
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
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STRAIGHT LINE : Question 3
In triangle ABC, A is (2,0),
B is (8,0) and C is (7,3).
(a) Find the gradients of AC and BC.
(b) Hence find the size of ACB. X
Y
A B
C
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STRAIGHT LINE : Question 3
In triangle ABC, A is (2,0),
B is (8,0) and C is (7,3).
(a) Find the gradients of AC and BC.
(b) Hence find the size of ACB. X
Y
A B
C
= 77.4°(b)
mAC = 3/5
mBC = - 3
(a)
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Question 3
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(a) Using the gradient formula:
mAC = 3 – 0
7 - 2 = 3/5
mBC = 3 – 0 7 - 8
= - 3
2 1
2 1
y ym
x x
In triangle ABC, A is (2,0),
B is (8,0) and C is (7,3).
(a)Find the gradients of AC
and BC.
(b) Hence find the size of ACB. (b) Using tan = gradient
If tan = 3/5 then CAB = 31.0°
If tan = -3 then CBX = (180-71.6)°
= 108.4 o
Hence :
ACB = 180° – 31.0° – 71.6°
= 77.4°
so ABC = 71.6°
X
Y
A B
C
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(a) Using the gradient formula:
mAC = 3 – 0
7 - 2
mBC = 3 – 0 7 - 8
= - 3
2 1
2 1
y ym
x x
(b) Using tan = gradient
= 3/5
If tan = 3/5 then CAB = 31.0°
then CBX = (180-71.6)°
= 108.4 o
Hence :
ACB = 180° – 31.0° – 71.6°
= 77.4°
If tan = -3
• If no diagram is given draw a neat labelled diagram.
• 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.
A
B
Ø °
mAB = tanØ °
Ø ° = tan-1 mABso ABC = 71.6°
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STRAIGHT LINE : Question 4
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In triangle PQR the vertices are P(4,-5), Q(2,3) and R(10-1).
X
Y
P(4,-5)
Q(2,3)
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.
(c) The coordinates of the point of intersection of lines e & f.
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STRAIGHT LINE : Question 4
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In triangle PQR the vertices are P(4,-5), Q(2,3) and R(10-1).
X
Y
P(4,-5)
Q(2,3)
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.
(c) The coordinates of the point of intersection of lines e & f.
y = -1(a)
y = 2x – 11(b)
(5,-1)(c)
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Question 4 (a)
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In triangle PQR the vertices are P(4,-5), Q(2,3) and R(10-1).
X
Y
P(4,-5)
Q(2,3)
R(10,-1)
Find
(a) the equation of the line e, the median from R of triangle PQR.
(a) Midpoint of PQ is (3,-1): let’s call this S
Using the gradient formula m = y2 – y1
x2 – x1
mSR = -1 – (-1)
10 - 3
Since it passes through (3,-1)
equation of e is y = -1
= 0 (ie line is horizontal)
Solution to 4 (b)
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Question 4 (b)
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(b) the equation of the line f,
the perpendicular bisector of QR.
In triangle PQR the vertices are P(4,-5), Q(2,3) and R(10-1). Find
X
Y
P(4,-5)
Q(2,3)
R(10,-1)
(b) Midpoint of QR is (6,1)
mQR = 3 – (-1)
2 - 10 = 4/-8 = - 1/2
required gradient = 2 (m1m2 = -1)
Using y – b = m(x – a) with (a,b) = (6,1)
& m = 2
we get y – 1 = 2(x – 6)
so f is y = 2x – 11
Solution to 4 (c)
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Question 4 (c)
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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.
X
Y
P(4,-5)
Q(2,3)
R(10,-1)
(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)
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• If no diagram is given draw a neat labelled diagram.
Q
P
R
y
x
median
Perpendicular bisector
(a) Midpoint of PQ is (3,-1): let’s call this S
Using the gradient formula m = y2 – y1
x2 – x1
mSR = -1 – (-1)
10 - 3
Since it passes through (3,-1)
equation of e is y = -1
(ie line is horizontal)
Comments for 4 (b)
•Sketch the median and the perpendicular bisector
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Q
P
R
y
x
(b) Midpoint of QR is (6,1)
mQR = 3 – (-1)
2 - 10 = 4/-8
required gradient = 2 (m1m2 = -1)
Using y – b = m(x – a) with (a,b) = (6,1)
& m = 2
we get y – 1 = 2(x – 6)
so f is y = 2x – 11
= - 1/2
• To find midpoint of QR
2 + 10 3 + (-1) 2 2
,
• Look for special cases:
Horizontal lines in the form y = kVertical lines in the form x = k
Comments for 4 (c)
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(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)
y = -1y = 2x - 11
• To find the point of intersection of the two lines solve the two equations:
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STRAIGHT LINE : Question 5
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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.
X
Y
G(2,-5)
E(6,-3)
F(12,-5)
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STRAIGHT LINE : Question 5
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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.
X
Y
G(2,-5)
E(6,-3)
F(12,-5)
x = 6(a)
x + 8y + 28 = 0(b)
(6,-4.25)(c)
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Question 5(a)
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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.
XY
G(2,-5)
E(6,-3)
F(12,-5)
(a) Using the gradient formula 2 1
2 1
y ym
x x
mFG = -5 – (-5)
12 - 2 = 0
(ie line is horizontal so altitude is vertical)
Altitude is vertical line through (6,-3)
ie x = 6
Solution to 5 (b)
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Question 5(b)
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In triangle EFG the vertices are E(6,-3), F(12,-5) and G(2,-5).
Find
XY
G(2,-5)
E(6,-3)
F(12,-5)
(b) the equation of the median from vertex F.
(b) Midpoint of EG is (4,-4)- let’s 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) (X8)
or 8y + 32 = -x + 4
Median is x + 8y + 28 = 0
Solution to 5 (c)
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Question 5(c)
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In triangle EFG the vertices are E(6,-3), F(12,-5) and G(2,-5).
Find
XY
G(2,-5)
E(6,-3)
F(12,-5)
(c) The point of intersection of the altitude and median.
(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
Point of intersection is (6,-4.25)
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• If no diagram is given draw a neat labelled diagram.
• Sketch the altitude and the median.
y
x
F
E
Gmedian
altitude
(a) Using the gradient formula 2 1
2 1
y ym
x x
mFG = -5 – (-5)
12 - 2 = 0
(ie line is horizontal so altitude is vertical)
Altitude is vertical line through (6,-3)
ie x = 6
Comments for 5 (b)
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y
x
F
E
G
Comments for 5 (c)
(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) (X8)
or 8y + 32 = -x + 4
Median is x + 8y + 28 = 0
• To find midpoint of EG
2 + 6 -3 + (-5) 2 2
,H
Horizontal lines in the form y = kVertical lines in the form x = k
• Look for special cases:
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(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
Point of intersection is (6,-4.25)
x = 6x + 8y = -28
• To find the point of intersection of the two lines solve the two equations:
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HIGHER – ADDITIONAL QUESTION BANK
UNIT 1 : BasicDifferentiation
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BASIC DIFFERENTIATION : Question 1
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Find the equation of the tangent to the curve (x>0)
at the point where x = 4.
16y x
x
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BASIC DIFFERENTIATION : Question 1
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Find the equation of the tangent to the curve (x>0)
at the point where x = 4.
y = 5/4x – 7
16y x
x
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Find the equation of the tangent to
the curve
y = x – 16
x (x>0)
at the point where x = 4.
NB: a tangent is a line so we need a point of contact and a gradient.
Point
If x = 4 then y = 4 – 16 4
= 2 – 4 = -2
so (a,b) = (4,-2)
Gradient: y = x – 16 x
= x1/2 – 16x -1
dy/dx = 1/2x-1/2 + 16x-2 = 1 + 16
2x x2
If x = 4 then: dy/dx = 1 + 16
24 16
= ¼ + 1 = 5/4
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Find the equation of the tangent to
the curve
y = x – 16
x (x>0)
at the point where x = 4.
If x = 4 then:
dy/dx = 1 + 16 24 16
= ¼ + 1 = 5/4
Gradient of tangent = gradient of curve
so m = 5/4 .
We now use y – b = m(x – a)
this gives us y – (-2) = 5/4(x – 4)
or y + 2 = 5/4x – 5
or y = 5/4x – 7
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• Prepare expression for differentiation. 1
1216
y = 16x x xx
NB: a tangent is a line so we need a point of contact and a gradient.
Point
If x = 4 then y = 4 – 16 4
= 2 – 4 = -2
so (a,b) = (4,-2)
Gradient: y = x – 16 x
= x1/2 – 16x -1
dy/dx = 1/2x-1/2 + 16x-2 = 1 + 16
2x x2
If x = 4 then:
= 1 + 16 24 16
= ¼ + 1 = 5/4
• Find gradient of the tangent
using rule:
dy
dx
“multiply by the power and reduce the power by 1”
dy/dx
• Find gradient = at x = 4.dy
dx
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If x = 4 then:
= 1 + 16 24 16
= ¼ + 1 = 5/4
Gradient of tangent = gradient of curve
so m = 5/4 .
We now use y – b = m(x – a)
this gives us y – (-2) = 5/4(x – 4)
or y + 2 = 5/4x – 5
dy/dx
or y = 5/4x – 7
16 16y 4 2
4x
x
• Find y coordinate at x = 4 using:
• Use m = 5/4 and (4,-2) in
y - b = m(x - a)
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BASIC DIFFERENTIATION : Question 2
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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.
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BASIC DIFFERENTIATION : Question 2
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Find the coordinates of the point on the curve y = x2 – 5x + 10 where the tangent to the curve makes an angle of 135° with the positive direction of the X-axis.
(2,4)
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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.
NB: gradient of line = gradient of curve
Line
Using gradient = tan
we get gradient of line = tan135°
= -tan45°
= -1Curve
Gradient of curve = dy/dx = 2x - 5
It now follows that
2x – 5 = -1
Or 2x = 4
Or x = 2
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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.
Back to Previous
Using y = x2 – 5x + 10 with x = 2
we get y = 22 – (5 X 2) + 10
ie y = 4
So required point is (2,4)
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NB: gradient of line = gradient of curve
Line
Using gradient = tan
we get gradient of line = tan135°
= -tan45°
= -1Curve
Gradient of curve = dy/dx = 2x - 5
It now follows that
2x – 5 = -1
Or 2x = 4
Or x = 2
• Find gradient of the tangent
using rule:
dy
dx
“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.
m = tan135 ° = -1
y
x135 °
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• Set m =
i.e. 2x - 5 = -1
and solve for x.
dy
dxIt now follows that
2x – 5 = -1
Or 2x = 4
Or x = 2
Using y = x2 – 5x + 10 with x = 2
we get y = 22 – (5 X 2) + 10
ie y = 4
So required point is (2,4)
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BASIC DIFFERENTIATION : Question 3
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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.
Make a sketch of the graph of y = g(x).
y = g(x)
(p,q)
r
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BASIC DIFFERENTIATION : Question 3
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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.
Make a sketch of the graph of y = g(x).
y = g(x)
(p,q)
r
x
y
y = g(x)
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Stationary points occur at x = 0 and x = p.
(We can ignore r.)
We now consider the sign of the gradient
either side of 0 and p:
new y-values
x 0 p
g(x) - 0 - 0 +
Click for graph
y = g(x)
(p,q)
r
Make a sketch of the graph of
y = g(x).
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Return to Nature Table
y = g(x)
(p,q)
r
Make a sketch of the graph of
y = g(x).
x
y
0p
y = g(x)
This now gives us the following graph
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To sketch the graph ofthe gradient function:
' dyf ( )
dxx
1 Mark the stationary points on the x axis i.e. ' dy
f ( ) 0dx
x
x
y
a
'f ( )x
x
y
0p
y = g(x)
Stationary points occur at x = 0 and x = p.
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2 For each interval decide if the value of ' dy
f ( ) is - or +dx
x
To sketch the graph ofthe gradient function:
' dyf ( )
dxx
1 Mark the stationary points on the x axis i.e. ' dy
f ( ) 0dx
x
Stationary points occur at x = 0 and x = p.
(We can ignore r.)
We now consider the sign of the gradient
either side of 0 and p:
new y-values
x 0 p
g(x) - 0 - 0 +
Continue Comments
+
- -x
y
a
'f ( )x
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To sketch the graph ofthe gradient function:
' dyf ( )
dxx
1 Mark the stationary points on the x axis i.e. ' dy
f ( ) 0dx
x
x
y
0p
y = g(x)
Stationary points occur at x = 0 and x = p.
3 Draw in curve to fit information
2 For each interval decide if the value of ' dy
f ( ) is - or +dx
x
• In any curve sketching question use a ruler and annotate the sketch i.e. label all known coordinates.
+
- -
x
y
a
'f ( )x
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BASIC DIFFERENTIATION : Question 4
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EXIT
Here is part of the graph of
y = x3 - 3x2 - 9x + 2.
Find the coordinates of the stationary points and determine their nature algebraically.
y = x3 - 3x2 - 9x + 2
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BASIC DIFFERENTIATION : Question 4
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Here is part of the graph of
y = x3 - 3x2 - 9x + 2.
Find the coordinates of the stationary points and determine their nature algebraically.
y = x3 - 3x2 - 9x + 2
(-1,7) is a maximum TP and
(3,-25) is a minimum TP
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BASIC DIFFERENTIATION : Question 4
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EXIT
Here is part of the graph of
y = x3 - 3x2 - 9x + 2.
Find the coordinates of the stationary points and determine their nature algebraically.
y = x3 - 3x2 - 9x + 2
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SPs occur where dy/dx = 0
ie 3x2 – 6x – 9 = 0
ie 3(x2 – 2x – 3) = 0
ie 3(x – 3)(x + 1) = 0
ie x = -1 or x = 3
Here is part of the graph of
y = x3 - 3x2 - 9x + 2.
Find the coordinates of the stationary points and determine their nature algebraically.
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)
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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
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
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• Must attempt to find
and set equal to zero
dy
dx
SPs occur where dy/dx = 0
ie 3x2 – 6x – 9 = 0
ie 3(x2 – 2x – 3) = 0
ie 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)
“multiply by the power andreduce the power by 1”
• Make the statement:“At stationary points “dy
0dx
• Find the value of y from y = x3 -3x2-9x+2 not from dy
dx
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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
• Justify the nature of each stationary point using a table of “signs”
x -1
dy
dx + 0 -
Minimum requirement
• State the nature of the stationary point i.e. Maximum T.P.
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BASIC DIFFERENTIATION : Question 5
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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 x2 (x 2)
So after 5 months the share is S(5) = 2/5 – 4/25 = 6/25
Find the maximum share of the market that the company can achieve.
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EXIT
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 x2 (x 2)
So after 5 months the share is S(5) = 2/5 – 4/25 = 6/25
Find the maximum share of the market that the company can achieve.
= 1/4
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End points
S(2) = 1 – 1 = 0
There is no upper limit but as x
S(x) 0. S(x) = 2 - 4
x x2 (x 2)
Find the maximum share of the
market that the company can
achieve.
When a company launches a new
product its share of the market
after x months is calculated as:
Stationary Points
S(x) = 2 - 4
x x2 = 2x-1 – 4x-2
So S (x) = -2x-2 + 8x-3 = - 2 + 8
x2 x3 = 8 - 2
x3 x2
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S(x) = 2 - 4
x x2 (x 2)
Find the maximum share of the
market that the company can
achieve.
When a company launches a new
product its share of the market
after x months is calculated as:
SPs occur where S (x) = 0
8 - 2 x3 x2
or8 = 2 x3 x2
( cross mult!)
8x2 = 2x3
8x2 - 2x3 = 0
2x2(4 – x) = 0
x = 0 or x = 4
NB: x 2In required interval
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= 0
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S(x) = 2 - 4
x x2 (x 2)
Find the maximum share of the
market that the company can
achieve.
When a company launches a new
product its share of the market
after x months is calculated as:
Go Back to Previous
We now check the gradients either side of
X = 4
x 4
S (x) + 0 -
S (3.9 ) = 0.00337…
S (4.1) = -0.0029…
Hence max TP at x = 4
So max share of market
= S(4) = 2/4 – 4/16
= 1/2 – 1/4
= 1/4
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End points
S(2) = 1 – 1 = 0
There is no upper limit but as x
S(x) 0.
Stationary Points
S(x) = 2 - 4
x x2 = 2x-1 – 4x-2
So S (x) = -2x-2 + 8x-3 = - 2 + 8
x2 x3 = 8 - 2
x3 x2
• Must consider end points and stationary points.
• Must look for key word to spot the optimisation question i.e.
Maximum, minimum, greatest , least etc.
• Prepare expression for differentiation.
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• Must attempt to find
( Note: No marks are allocated for trial and error solution.)
dyS (x)
dx
SPs occur where S (x) = 0
8 - 2 x3 x2
or8 = 2 x3 x2
( cross mult!)
8x2 = 2x3
= 0
8x2 - 2x3 = 0
2x2(4 – x) = 0
x = 0 or x = 4
NB: x 2In required interval
“multiply by the power and reduce the power by 1”
• Must attempt to find and set equal to zero
dy
dx
• Usually easier to solve resulting equation using cross-multiplication.
• Take care to reject “solutions” outwith the domain.
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We now check the gradients either side of
X = 4
x 4
S (x) + 0 -
S (3.9 ) = 0.00337…
S (4.1) = -0.0029…
Hence max TP at x = 4
So max share of market
= S(4) = 2/4 – 4/16
= 1/2 – 1/4
= 1/4
•Must show a maximum value using a table of “signs”.
x 4
S (x) + 0 -
Minimum requirement.
• State clearly: Maximum T.P at x = 4
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HIGHER – ADDITIONAL QUESTION BANK
UNIT 1 : RecurrenceRelations
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RECURRENCE RELATIONS : Question 1
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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 .
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RECURRENCE RELATIONS : Question 1
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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 = 0.6
b = -2
(a)
L = -5
(b)
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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 .
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 10a = 6
or
(a)
Replacing a by 0.6 in 10a + b = 4
gives 6 + b = 4
or b = -2
a = 0.6Continue Solution
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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 .
(b)
L = -5
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 0.4L = -2
or L = -2 0.4
or L = -20 4so
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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 10a = 6
or
(a)
Replacing a by 0.6 in 10a + b = 4
gives 6 + b = 4
or b = -2
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.
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(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 0.4L = -2
or L = -2 0.4
or L = -20 4so L = -5
• Must state condition for limit i.e. -1 < 0.6 < 1
• At limit L, state un+1 = un = L
• Substitute L for un+1 and un
and solve for L.
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RECURRENCE RELATIONS : Question 2
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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.
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RECURRENCE RELATIONS : Question 2
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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.
k = 1/3
L = 9/8
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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
Vn+1 = k2vn + 1.
Find the value of k and hence this
limit.
If the limit is L then as n we have
un+1 = un = L and vn+1 = vn = L
First Sequence
un+1 = -5kun + 3
becomes
L + 5kL = 3
L(1 + 5k) = 3
L = 3 . . (1 + 5k)
L = -5kL + 3
Second Sequence
vn+1 = k2vn + 1
becomesL = k2L + 1 L - k2L = 1
L(1 - k2) = 1
L = 1 . . (1 - k2)
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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
Vn+1 = k2vn + 1.
Find the value of k and hence this
limit.
L = 1 . . (1 - k2)
It follows that
L = 3 . . (1 + 5k)
. 3 . = . 1 . . (1 + 5k) (1 – k2)
Cross multiply to get 1 + 5k = 3 – 3k2
This becomes 3k2 + 5k – 2 = 0
Or (3k – 1)(k + 2) = 0
So k = 1/3 or k = -2
Since -1<k<1 then k = 1/3
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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
Vn+1 = k2vn + 1.
Find the value of k and hence this
limit.
Since -1<k<1 then k = 1/3
Using L = 1 . . (1 - k2)
gives us L = . 1 . . (1 – 1/9)
or L = 1 8/9
ie L = 9/8
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If the limit is L then as n we have
un+1 = un = L and vn+1 = vn = L
First Sequence
un+1 = -5kun + 3
becomes
L + 5kL = 3
L(1 + 5k) = 3
L = 3 . . (1 + 5k)
L = -5kL + 3
Second Sequence
vn+1 = k2vn + 1
becomesL = k2L + 1 L - k2L = 1
L(1 - k2) = 1
L = 1 . . (1 - k2)
• Since both recurrence relations have the same limit, L, find the limit for both and set equal.
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L = 1 . . (1 - k2)
It follows that
L = 3 . . (1 + 5k)
. 3 . = . 1 . . (1 + 5k) (1 – k2)
Cross multiply to get 1 + 5k = 3 – 3k2
This becomes 3k2 + 5k – 2 = 0
Or (3k – 1)(k + 2) = 0
So k = 1/3 or k = -2
Since -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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Since -1<k<1 then k = 1/3
Using L = 1 . . (1 - k2)
gives us L = . 1 . . (1 – 1/9)
or L = 1 8/9
ie L = 9/8
Find L from either formula.
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RECURRENCE RELATIONS : Question 3
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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?
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RECURRENCE RELATIONS : Question 3
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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?
Height of trees in long run is 31/3m.
331/3% (b)
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The trees are known grow at a rate
of 1m per annum. He therefore
decides to prune 30% from their
height at the beginning of each year.
(a) Using the 30% pruning scheme
what height should he expect
the trees to grow to in the long
run?
(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 L = 1 0.3 = 10 3 = 31/3
Height of trees in long run is 31/3m.
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The trees are known grow at a rate
of 1m per annum. He therefore
decides to prune 30% from their
height at the beginning of each year.
(b) If fraction left after pruning is a and
we need the limit to be 3
then we have 3 = a X 3 + 1
or 3a = 2
or a = 2/3
This means that the fraction pruned is
1/3 or 331/3%
(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?
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(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 L = 1 0.3 = 10 3
Height of trees in long run is 31/3m.
• 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.
• 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
• Substitute L for Hn+1 and Hn and solve for L.
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(b) If fraction left after pruning is a and
we need the limit to be 3
then we have 3 = a X 3 + 1
or 3a = 2
or a = 2/3
This means that the fraction pruned is
1/3 or 331/3%
• Since we know the limit we are working backwards to %.
L = 0.7L + 1
New limit, L = 3 and multiplier a
3 = a x3 + 1 etc.
• Take care to subtract from 1 to get fraction pruned.
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HIGHER – ADDITIONAL QUESTION BANK
UNIT 1 : Trig Graphs& Equations
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1 2 3
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TRIG GRAPHS & EQUATIONS : Question 1
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This diagram shows the graph of y = acosbx + c.
Determine the values of a, b & c.
/2
y = acosbx + c
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This diagram shows the graph of y = acosbx + c.
Determine the values of a, b & c.
/2
y = acosbx + c
a = 3
b = 2
c = -1
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This diagram shows the graph of y = acosbx + c.
Determine the values of a, b & c.
/2
y = acosbx + c
a = ½(max – min)
= ½(2 – (-4))
= 3
= ½ X 6
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 lessso c = -1
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a = ½(max – min)
= ½(2 – (-4))
= ½ X 6
Period of graph = so two complete
sections between 0 & 2
For 3cos(…) max = 3 & min = -3.
This graph: max = 2 & min = -4.
ie 1 lessso c = -1
ie b = 2
= 3
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.
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TRIG GRAPHS & EQUATIONS : Question 2
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Solve 3tan2 + 1 = 0 ( where 0 < < ).
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TRIG GRAPHS & EQUATIONS : Question 2
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Solve 3tan2 + 1 = 0 ( where 0 < < ).
= 5/12
= 11/12
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Solve 3tan2 + 1 = 0
( where 0 < < ).
3tan2 + 1 = 0
3tan2 = -1
tan2 = -1/3 Q2 or Q4
tan -1(1/3) = /6 -
+ 2 -
sin all
tan cos
Q2: angle = - /6
so 2 = 5/6 ie = 5/12
Q4: angle = 2 - /6
so 2 = 11/6
ie = 11/12
1
23/6
tan2 repeats every /2 radians but repeat values are not in interval.
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3tan2 + 1 = 0
3tan2 = -1
tan2 = -1/3 Q2 or Q4
tan -1(1/3) = /6 -
+ 2 -
sin all
tan cos
1
23/6
Q2: angle = - /6
so 2 = 5/6 ie = 5/12
Q4: angle = 2 - /6
so 2 = 11/6
ie = 11/12
tan2 repeats every /2 radians but repeat values are not in interval.
• Solve the equation for tan .2• Use the positive value when finding tan-1.
• Use the quadrant rule to find the solutions.
• Must learn special angles or be able to calculate from triangles.
3145°
1 260°
30°
12• Take care to reject ”solutions”
outwith domain.
• Full marks can be obtained by working in degrees and changing final answers back to radians.
radians 180
1 /180 radians
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TRIG GRAPHS & EQUATIONS : Question 3
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EXIT
The diagram shows a the graph of a sine function from 0 to 2/3.
2/3
P Qy = 2
(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.
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TRIG GRAPHS & EQUATIONS : Question 3
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The diagram shows a the graph of a sine function from 0 to 2/3.
2/3
P Qy = 2
(a) State the equation of the graph.
(b) The line y = 2 meets the graph
at points P & Q.
Find the coordinates of these two points.
Graph is y = 4sin3x
P is (/18, 2) and Q is (5/18, 2).
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2/3
P Q
y = 2
(a) State the equation of the graph.
The diagram shows a the graph of a sine function from 0 to 2/3.
(a) One complete wave from 0 to 2/3
so 3 waves from 0 to 2.
Max/min = ±4 4sin(…)
Graph is y = 4sin3x
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(b) The line y = 2 meets the graph
at points P & Q.
Find the coordinates of these two
points.
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2/3
P Q
y = 2
Graph is y = 4sin3x
so 4sin3x = 2
or sin3x = 1/2
(b) At P & Q y = 4sin3x and y = 2
Q1 or Q2
sin-1(1/2) = /6
Q1: angle = /6
so 3x = /6
ie x = /18
Q2: angle = - /6
so 3x = 5/6
ie x = 5/18
-
+ 2 -
sin all
tan cos
1
23/6
P is (/18, 2) and Q is (5/18, 2).
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(a) One complete wave from 0 to 2/3
so 3 waves from 0 to 2.
Max/min = ±4
Graph is y = 4sin3x
4sin(…)
• Identify graph is of the form y = asinbx.
• Must justify choice of a and b.
• Possible justification of a
Max = 4, Min = -4 4sin(…)y = sinx stretched by a factor of 4
• Possible justification for b
Period = 3 waves from 0 to
2 / 32
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Graph is y = 4sin3x
so 4sin3x = 2
or sin3x = 1/2
(b) At P & Q y = 4sin3x and y = 2
Q1 or Q2
sin-1(1/2) = /6 -
+ 2 -
sin all
tan cos
1
23/6
Q1: angle = /6
so 3x = /6
ie x = /18
Q2: angle = - /6
so 3x = 5/6
ie x = 5/18
P is (/18, 2) and Q is (5/18, 2).
• At intersection y1 = y2
4sin3x = 2
• Solve for sin3x
• Use the quadrant rule to find the solutions.
• Must learn special angles or be able to calculate from triangles.
3145°
1 260°
30°
1 2
• Take care to state coordinates.
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HIGHER – ADDITIONAL QUESTION BANK
UNIT 1 : Functions& Graphs
You have chosen to study:
Please choose a question to attempt from the following:
1 2 3 4
EXITBack to
Unit 1 Menu
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FUNCTIONS & GRAPHS : Question 1
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EXIT
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)
-8 12
(-p,q)
(u,-v)
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FUNCTIONS & GRAPHS : Question 1
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EXIT
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)
-8 12
(-p,q)
(u,-v)
(p,-q+4)
(8,4) (-12,4) (0,4)
(-u,v+4)
y = 4 – g(-x)
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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)
-8 12
(-p,q)
(u,-v)
y = 4 – g(-x) = -g(-x) + 4
Reflect in X-axis
Reflect in Y-axis
Slide 4 up
A
B C
Known Points
(-8,0), (-p,q), (0,0), (u,-v), (12,0)
(-8,0), (-p,-q), (0,0), (u,v), (12,0)A
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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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)
-8 12
(-p,q)
(u,-v)
(8,4), (p,-q+4), (0,4), (-u,v+4), (-12,4)
Now plot points and draw curve through
them.
(p,-q+4)
(8,4) (-12,4) (0,4)
(-u,v+4)
y = 4 – g(-x)
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y = 4 – g(-x) = -g(-x) + 4
Reflect in X-axis
Reflect in Y-axis
Slide 4 up
A
B C
Known Points
(-8,0), (-p,q), (0,0), (u,-v), (12,0)
(-8,0), (-p,-q), (0,0), (u,v), (12,0)A
B
(8,0), (p,-q), (0,0), (-u,v), (-12,0)
C
(8,4), (p,-q+4), (0,4), (-u,v+4), (-12,4)
•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.
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y = 4 – g(-x) = -g(-x) + 4
Reflect in X-axis
Reflect in Y-axis
Slide 4 up
A
B C
Known Points
(-8,0), (-p,q), (0,0), (u,-v), (12,0)
(-8,0), (-p,-q), (0,0), (u,v), (12,0)A
B
(8,0), (p,-q), (0,0), (-u,v), (-12,0)
C
(8,4), (p,-q+4), (0,4), (-u,v+4), (-12,4)
•Learn Rules: Not given on formula sheet
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
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•In any curve sketching question use a ruler and annotate the sketch i.e. label all known coordinates.
(8,4), (p,-q+4), (0,4), (-u,v+4), (-12,4)
Now plot points and draw curve through
them.
(p,-q+4)
(8,4) (-12,4) (0,4)
(-u,v+4)
y = 4 – g(-x)
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FUNCTIONS & GRAPHS : Question 2
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y = ax
(1,a)
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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FUNCTIONS & GRAPHS : Question 2
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EXIT
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
(-1,a)
(-2,1)
y = a(x+2)
y = ax
ANSWER TO PART (I)
y = 2ax - 3
y = ax
(0,-1)
(1,2a-3)
ANSWER to PART (II)
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Make sketches of the graphs of the functions
(I) y = a(x+2)
y = ax
(1,a)
(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)
(-1,a)
(-2,1)
y = a(x+2)
y = ax
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y = ax
(1,a)
Make sketches of the graphs of the functions
(II) y = 2ax - 3
(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 = 2ax - 3
y = ax
(0,-1)
(1,2a-3)
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(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)
(-1,a)
(-2,1)
y = a(x+2)
y = ax
• 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)
• In any curve sketching question use a ruler and annotate the sketch i.e. label all known coordinates.
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(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)
(-1,a)
(-2,1)
y = a(x+2)
y = ax
•Learn Rules: Not given on formula sheet
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
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(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 = 2ax - 3
y = ax
(0,-1)
(1,2a-3)
• 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)
• In any curve sketching question use a ruler and annotate the sketch i.e. label all known coordinates.
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•Learn Rules: Not given on formula sheet
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
(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 = 2ax - 3
y = ax
(0,-1)
(1,2a-3)
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FUNCTIONS & GRAPHS : Question 3
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EXIT
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.
(a) Find formulae for (i) f(g(x)) (ii) g(f(x)) .
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FUNCTIONS & GRAPHS : Question 3
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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.
(a) Find formulae for (i) f(g(x)) (ii) g(f(x)) .
= 2x2 - 1(a) (i)
= 4x2 – 4x + 1(ii)
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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)) .
(a)(i) f(g(x))
= f(x2)
= 2x2 - 1
(ii) g(f(x))
= g(2x-1)
= (2x – 1)2
= 4x2 – 4x + 1
Continue Solution
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Two functions f and g are defined
on the set of real numbers by the
formulae
f(x) = 2x - 1 and g(x) = x2 .
g(f(x)) = 4x2 – 4x + 1
(b) Hence show that the equation
g(f(x)) = f(g(x)) has only one
real solution.
f(g(x)) = 2x2 - 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!
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(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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(b)• Only one way to solve resulting equation:
Terms to the left, simplify and factorise.
g(f(x)) = 4x2 – 4x + 1
f(g(x)) = 2x2 - 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!
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FUNCTIONS & GRAPHS : Question 4
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A function g is defined by the formula g(x) = . 2 (x1) (x – 1)
(a) Find a formula for h(x) = g(g(x)) in its simplest form.
(b) State a suitable domain for h.
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FUNCTIONS & GRAPHS : Question 4
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A function g is defined by the formula g(x) = . 2 (x1) (x – 1)
(a) Find a formula for h(x) = g(g(x)) in its simplest form.
(b) State a suitable domain for h.
h(x) = (2x - 2) . . (3 – x)
Domain = {x R: x 3}
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(a) g(g(x)) = g( ). 2 . (x – 1)
= 2. 2 . (x – 1)
- 1
= 2 2 - (x – 1) . (x – 1)
= 2 (3 - x) .(x – 1)
= 2 (x - 1) .(3 – x)
= (2x - 2) . . (3 – x)
A function g is defined by the
formula g(x) = . 2 (x1) (x – 1)
(a) Find a formula for
h(x) = g(g(x))
in its simplest form.
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h(x) = (2x - 2) . . (3 – x)A function g is defined by the
formula g(x) = . 2 (x1) (x – 1)
(a) Find a formula for
h(x) = g(g(x))
in its simplest form.
(b) State a suitable domain for h.
(b) For domain 3 - x 0
Domain = {x R: x 3}
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(a) g(g(x)) = g( ). 2 . (x – 1)
= 2. 2 . (x – 1)
- 1
= 2 2 - (x – 1) . (x – 1)
= 2 (3 - x) .(x – 1)
= 2 (x - 1) .(3 – x)
= (2x - 2) . . (3 – x)
(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(x-1)
2 - 1
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h(x) = (2x - 2) . . (3 – x)
(b) 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 :
Integration
Polynomials
The Circle
AdditionFormulae
Quadratics
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UNIT 2 :Polynomials
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Please choose a question to attempt from the following:
1 2 3 4
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POLYNOMIALS : Question 1
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Show that x = 3 is a root of the equation x3 + 3x2 – 10x – 24 = 0.
Hence find the other roots.
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POLYNOMIALS : Question 1
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Show that x = 3 is a root of the equation x3 + 3x2 – 10x – 24 = 0.
Hence find the other roots.
other roots are x = -4 & x = -2
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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 -10 -24
3 18 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
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Using the nested method -
coefficients are 1, 3, -10, -24
f(3) = 3 1 3 -10 -24
3 18 24
1 6 8 0
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
f(3) = 0 so x = 3 is a root.
Also (x – 3) is a factor.
• State clearly in solution that f(3) = 0 x = 3 is a root
• 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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Given that (x + 4) is a factor of the polynomial
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.
POLYNOMIALS : Question 2
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Given that (x + 4) is a factor of the polynomial
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.
POLYNOMIALS : Question 2
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k = -15
So full solution of equation is
x = -4 or x = 1/3 or x = 1
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Given that (x + 4) is a factor of
the polynomial
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.
Since (x + 4) a factor then f(-4) = 0 .
Now using the nested method -
coefficients are 3, 8, k, 4
f(-4) = -4 3 8 k 4
-12 16 (-4k – 64)
3 -4 (k + 16) (-4k – 60)
Since -4k – 60 = 0
then -4k = 60
so k = -15
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Given that (x + 4) is a factor of
the polynomial
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.
If k = -15 then we now have
f(-4) = -4 3 8 -15 4
-12 16 -4
Other factor is 3x2 – 4x + 1
or (3x - 1)(x – 1)
3 -4 1 0
If (3x - 1)(x – 1) = 0 then x = 1/3 or x = 1
So full solution of equation is:
x = -4 or x = 1/3 or x = 1
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Since (x + 4) a factor then f(-4) = 0 .
Now using the nested method -
coefficients are 3, 8, k, 4
f(-4) = -4 3 8 k 4
-12 16 (-4k – 64)
3 -4 (k + 16) (-4k – 60)
Since -4k – 60 = 0
then -4k = 60
so k = -15
• The working in the nested solution can sometimes be eased by working in both directions toward the variable:
-4 3 8 k 4 -12 16 -4
3 -4 1 0
k + 16 = 1 k = -15
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Since (x + 4) a factor then f(-4) = 0 .
Now using the nested method -
coefficients are 3, 8, k, 4
f(-4) = -4 3 8 k 4
-12 16 (-4k – 64)
3 -4 (k + 16) (-4k – 60)
Since -4k – 60 = 0
then -4k = 60
so k = -15
• Simply making f(-4) = 0 will also yield k i.e. 3(-4)3 + 8(-4)2 + k(-4) + 4 = 0
k = -15
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If k = -15 then we now have
f(-4) = -4
-12 16 -4
Other factor is 3x2 – 4x + 1
or (3x - 1)(x – 1)
3 -4 1 0
If (3x - 1)(x – 1) = 0
So full solution of equation is:
x = -4 or x = 1/3 or x = 1
3 8 -15 4
• Show completed factorisation of the cubic:
(x + 4)(3x - 1)(x - 1) = 0
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POLYNOMIALS : Question 3
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Given that f(x) = 6x3 + 13x2 - 4 show that (x + 2) is a factor of f(x).
Hence express f(x) in its fully factorised form.
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POLYNOMIALS : Question 3
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Given that f(x) = 6x3 + 13x2 - 4 show that (x + 2) is a factor of f(x).
Hence express f(x) in its fully factorised form.
6x3 + 13x2 - 4 = (3x + 2)(2x - 1)(x + 2)
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Given that f(x) = 6x3 + 13x2 - 4
show that (x + 2) is a factor
of f(x).
Hence express f(x) in its fully
factorised form.
Using the nested method -
coefficients are 6, 13, 0, -4
f(-2) = -2 6 13 0 -4
6
-12
1
-2
-2
4
0
f(-2) = 0 so
(x + 2) is a factor
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Given that f(x) = 6x3 + 13x2 - 4
show that (x + 2) is a factor
of f(x).
Hence express f(x) in its fully
factorised form.
Using the nested method -
coefficients are 6, 13, 0, -4
f(-2) = -2 6 13 0 -4
6
-12
1
-2
-2
4
0
Other factor is 6x2 + x – 2
or (3x + 2)(2x - 1)
Hence 6x3 + 13x2 - 4
= (3x + 2)(2x - 1)(x + 2)
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Using the nested method -
coefficients are 6, 13, 0, -4
f(-2) = -2 6 13 0 -4
6
-12
1
-2
-2
4
0
f(-2) = 0 so
(x + 2) is a factor
• State clearly in solution that f(-2) = 0 x = -2 is a root
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Using the nested method -
coefficients are 6, 13, 0, -4
f(-2) = -2 6 13 0 -4
6
-12
1
-2
-2
4
0
• Show completed factorisation of cubic i.e.
(3x + 2)(2x - 1)(x +2).
Other factor is 6x2 + x – 2
or (3x + 2)(2x - 1)
Hence 6x3 + 13x2 - 4
= (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.
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(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.
POLYNOMIALS : Question 4
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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
P
Q
y = -x3 + 6x2 – 3x – 10
4
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(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.
POLYNOMIALS : Question 4
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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.
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P is (4,10) PQ is y = -3x + 22
Q is (-2,28)
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y = -x3 + 6x2 – 3x – 10
P
Q
y = -x3 + 6x2 – 3x – 10
4
(a) Find the coordinates of P
and the equation of the
bypass PQ.
(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)
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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y = -x3 + 6x2 – 3x – 10
P
Q
y = -x3 + 6x2 – 3x – 10
4
(a) Find the coordinates of P
and the equation of the
bypass PQ.
P is (4,10) dy/dx = -3
Now using : 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
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y = -x3 + 6x2 – 3x – 10
P
Q
y = -x3 + 6x2 – 3x – 10
4
(b) The tangent & curve meet whenever
y = -3x + 22 and y = -x3 + 6x2 – 3x – 10
ie -3x + 22 = -x3 + 6x2 – 3x – 10
or x3 - 6x2 + 32 = 0
(b)Hence find the coordinates of Q – the point where the bypass rejoinsthe original road.
We already know that x = 4 is one
solution to this so using the nested
method we get …..
f(4) = 4 1 -6 0 32
4 -8 -32
1 -2 -8 0
Other factor is x2 – 2x - 8
PQ is y = -3x + 22
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y = -x3 + 6x2 – 3x – 10
P
Q
y = -x3 + 6x2 – 3x – 10
4
(b) The
(b)Hence find the coordinates of Q – the point where the bypass rejoinsthe original road.
other factor is x2 – 2x - 8
= (x – 4)(x + 2)
Solving (x – 4)(x + 2) = 0
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
Q is (-2,28)
PQ is y = -3x + 22
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• Must use differentiation to find gradient. Learn rule:
“Multiply by the power then reduce the power by 1”
(a)
(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)
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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(a)
P is (4,10) dy/dx = -3
Now using : 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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(b)
(b) The tangent & curve meet whenever
y = -3x + 22 and y = -x3 + 6x2 – 3x – 10
ie -3x + 22 = -x3 + 6x2 – 3x – 10
or 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 0 32
4 -8 -32
1 -2 -8 0
Other factor is x2 – 2x - 8
• At intersection y1 = y2
Terms to the left, simplify and factorise
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(b)
other factor is x2 – 2x - 8
= (x – 4)(x + 2)
Solving (x – 4)(x + 2) = 0
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
Q is (-2,28)
• Note solution x = 4 appears twice:
Repeated root tangency
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HIGHER – ADDITIONAL QUESTION BANK
UNIT 2 :Quadratics
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1 2 3 4 5
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QUADRATICS : Question 1
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(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).
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(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).
= (x – 4)2 + 5(a)
y = x2 – 8x + 21
(4,5)
(b)
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(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) f(x) = x2 – 8x + 21
= (x2 – 8x + ) + 2116 - 16
(-82)2
= (x – 4)2 + 5
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(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).
(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
turning point at (4,5).
When x = 0 , y = 21 (from original formula!)
so Y-intercept is (0,21).
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(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).
(b) Graph looks like….
(4,5)
(0,21)
y = x2 – 8x + 21
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(a) f(x) = x2 – 8x + 21
= (x2 – 8x + ) + 2116 - 16
(-82)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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(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
turning point at (4,5).
When x = 0 , y = 21 (from original formula!)
so Y-intercept is (0,21).
(4,5)
(0,21)
• The sketch can also be obtained by calculus:
2 f(x) = x -8x 21
f (x) = 2 x-8
2
At stationary points 0
2 x 8 0
x 4
f(4) 4 -8.4 + 21 = 5
dy
dx
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(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
turning point at (4,5).
When x = 0 , y = 21 (from original formula!)
so Y-intercept is (0,21).
(4,5)
(0,21)
• Min. Turning Point (4,5),
coefficient of x2 is positive.
• Hence sketch.
f(0) 21
21
(4,5)
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QUADRATICS : Question 2
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EXIT
(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).
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QUADRATICS : Question 2
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EXIT
(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).
= 11 - 4(x – 1)2 (a)
maximum t p is at (1,11) .(b)
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(a) f(x) = 7 + 8x - 4x2
= - 4x2 + 8x + 7
= -4[x2 – 2x] + 7
(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).
= -4[(x2 – 2x + ) ] + 7
(-22)2
1 - 1
= -4[(x – 1)2 - 1] + 7
= -4(x – 1)2 + 4 + 7
= 11 - 4(x – 1)2
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Question 2
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(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).
(b) Maximum value is 11 when
(x – 1)2 = 0 ie x = 1.
= 11 - 4(x – 1)2
so maximum turning point is at (1,11) .
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(a) f(x) = 7 + 8x - 4x2
= - 4x2 + 8x + 7
= -4[x2 – 2x] + 7
= -4[(x2 – 2x + ) ] + 7
(-22)2
1 - 1
= -4[(x – 1)2 - 1] + 7
= -4(x – 1)2 + 4 + 7
= 11 - 4(x – 1)2
• 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
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(a) f(x) = 7 + 8x - 4x2
= - 4x2 + 8x + 7
= -4[x2 – 2x] + 7
= -4[(x2 – 2x + ) ] + 7
(-22)2
1 - 1
= -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 = a2 + 2ab + b2= -4[(x2 - 2x +1) -1] +7 )= -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.
so maximum turning point is at (1,11) .
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QUADRATICS : Question 3
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For what value(s) of k does the equation 4x2 – kx + (k + 5) = 0 have equal roots.
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For what value(s) of k does the equation 4x2 – kx + (k + 5) = 0 have equal roots.
k = -4 or k = 20
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For what value(s) of k does the
equation 4x2 – kx + (k + 5) = 0
have equal roots.
Let 4x2 – kx + (k + 5) = ax2 + bx + c
then 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
(k + 4)(k – 20) = 0
k = -4 or k = 20
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Let 4x2 – kx + (k + 5) = ax2 + bx + c
then 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
(k + 4)(k – 20) = 0
k = -4 or k = 20
• Learn Rules relating to discriminant b2- 4ac
b2- 4ac = 0 Equal rootsb2- 4ac > 0 2 Real, distinct roots b2- 4ac < 0 No real roots b2- 4ac 0 Real roots, equal or distinct
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Let 4x2 – kx + (k + 5) = ax2 + bx + c
then 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
(k + 4)(k – 20) = 0
k = -4 or k = 20
• Must use factorisation to solve resulting quadratic.
Trial and error receives no credit.
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QUADRATICS : Question 4
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The equation of a parabola is f(x) = px2 + 5x – 2p .
Prove that the equation f(x) = 0 always has two distinct roots.
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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
= 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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The equation of a parabola is
f(x) = px2 + 5x – 2p .
Prove that the equation f(x) = 0
always has two distinct roots.
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
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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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
Since p2 0 for all values of p
then 8p2 + 25 > 0.
The discriminant is always positive
so there are always two distinct roots.
• Learn Rules relating to discriminant b2- 4ac
b2- 4ac = 0 Equal rootsb2- 4ac > 0 2 Real, distinct roots b2- 4ac < 0 No real roots b2- 4ac 0 Real roots, equal or distinct
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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
Since p2 0 for all values of p
then 8p2 + 25 > 0.
The discriminant is always positive
so there are always two distinct roots.
• State condition you require to show true explicitly:
For two distinct roots b2- 4ac > 0
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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
Since p2 0 for all values of p
then 8p2 + 25 > 0.
The discriminant is always positive
so there are always two distinct roots.
•To show 8p2 + 25> 0 use algebraic logic or show the graph of 8p2 + 25 is always above the “x-axis”.
25
Min. T.P. at (0,25) hence graph always above the “x - axis.”
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QUADRATICS : Question 5
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Given that the roots of 3x(x + p) = 4p(x – 1) are equal
then show that p = 0 or p = 48.
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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
ie p = 0 or p = 48
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Given that the roots of
3x(x + p) = 4p(x – 1) are equal
then show that p = 0 or p = 48.
Rearranging 3x(x + p) = 4p(x – 1) 3x2 + 3px = 4px - 4p
3x2 - px + 4p = 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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Rearranging 3x(x + p) = 4p(x – 1) 3x2 + 3px = 4px - 4p
3x2 - px + 4p = 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
• Must put the equation into standard quadratic form before reading off a,b and c.
i.e. ax2 + bx +c = 0
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Rearranging 3x(x + p) = 4p(x – 1) 3x2 + 3px = 4px - 4p
3x2 - px + 4p = 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
• Learn Rules relating to discriminant b2- 4ac
b2- 4ac = 0 Equal rootsb2- 4ac > 0 2 Real, distinct roots b2- 4ac < 0 No real roots b2- 4ac 0 Real roots, equal or distinct
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Rearranging 3x(x + p) = 4p(x – 1) 3x2 + 3px = 4px - 4p
3x2 - px + 4p = 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
•State condition you require explicitly:
For two equal roots b2- 4ac = 0
• Must use factorisation to solve resulting quadratic.
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y = -2x2 + 3x + 2
x + y – 4 = 0
QUADRATICS : Question 6
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The diagram below shows the parabola y = -2x2 + 3x + 2 and the line x + y – 4 = 0.
Prove that the line is a tangent to the curve.
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y = -2x2 + 3x + 2
x + y – 4 = 0
QUADRATICS : Question 6
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The diagram below shows the parabola y = -2x2 + 3x + 2 and the line x + y – 4 = 0.
Prove that the line is a tangent to the curve.
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.
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y = -2x2 + 3x + 2
x + y – 4 = 0
Prove that the line is a tangent.
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 .
So -x + 4 = -2x2 + 3x + 2
Or 2x2 - 4x + 2 = 0
Let 2x2 - 4x + 2 = ax2 + bx + c
then a = 2, b = -4 & c = 2.
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y = -2x2 + 3x + 2
x + y – 4 = 0
Prove that the line is a tangent.
then a = 2, b = -4 & c = 2.
So discriminant = b2 – 4ac = (-4)2 – (4 X 2 X 2)
= 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.
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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 .
So -x + 4 = -2x2 + 3x + 2
Or 2x2 - 4x + 2 = 0
Let 2x2 - 4x + 2 = ax2 + bx + c
then a = 2, b = -4 & c = 2.
• For intersection of line and polynomial
yy11 = y = y22
Terms to the left, simplify and Terms to the left, simplify and factorise.factorise.
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Or -x + 4 = -2x2 + 3x + 2
2x2 - 4x + 2 = 0
2(x2 - 2x + 1) = 0
2(x - 1)(x - 1) = 0
x = 1 (twice)
Equal roots tangency
then a = 2, b = -4 & c = 2.
So discriminant = b2 – 4ac = (-4)2 – (4 X 2 X 2)
= 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.
• To prove tangency To prove tangency “ “equal roots” may be used inequal roots” may be used in place of the discriminant. place of the discriminant. The statement must be madeThe statement must be made explicitlyexplicitly..
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HIGHER – ADDITIONAL QUESTION BANK
UNIT 2 :Integration
You have chosen to study:
Please choose a question to attempt from the following:
1 2 3 4 5
EXITBack to
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y = x2 - 8x + 18
x = 3 x = k
Show that the shaded area is given by
1/3k3 – 4k2 + 18k - 27
INTEGRATION : Question 1
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The diagram below shows the curve y = x2 - 8x + 18 and the lines x = 3 and x = k.
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The diagram below shows the curve y = x2 - 8x + 18 and the lines x = 3 and x = k.
y = x2 - 8x + 18
x = 3 x = k
Show that the shaded area is given by
1/3k3 – 4k2 + 18k - 27
INTEGRATION : Question 1
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EXITArea = (x2 - 8x + 18) dx
3
k
= 1/3k3 – 4k2 + 18k – 27 as required.
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The diagram shows the
curve y = x2 - 8x + 18 and the
lines x = 3 and x = k.
Show that the shaded area is
given by 1/3k3 – 4k2 + 18k - 27
Area = (x2 - 8x + 18) dx3
k
= x3 - 8x2 + 18x [ ]3 2
k
3
= 1/3x3 – 4x2 + 18x[ ]k
3
= (1/3k3 – 4k2 + 18k)
– ((1/3 X 27) – (4 X 9) + 54)
= 1/3k3 – 4k2 + 18k – 27
as required.
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Area = (x2 - 8x + 18) dx3
k
= x3 - 8x2 + 18x [ ]3 2
k
3
= 1/3x3 – 4x2 + 18x[ ]k
3
= (1/3k3 – 4k2 + 18k)
– ((1/3 X 27) – (4 X 9) + 54)
= 1/3k3 – 4k2 + 18k – 27
as required.
• Learn result
can be used to find the
enclosed area shown:
( )b
a
f x dx
a b
f(x)
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Area = (x2 - 8x + 18) dx3
k
= x3 - 8x2 + 18x [ ]3 2
k
3
= 1/3x3 – 4x2 + 18x[ ]k
3
= (1/3k3 – 4k2 + 18k)
– ((1/3 X 27) – (4 X 9) + 54)
= 1/3k3 – 4k2 + 18k – 27
as required.
• Learn result for integration:
“Add 1 to the power and divide by the new power.”
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INTEGRATION : Question 2
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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).
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INTEGRATION : Question 2
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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 y = 4x3 – 3x2 - 5
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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).
dy/dx = 12x2 – 6x
So 2(12 6 )y x 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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dy/dx = 12x2 – 6x
So 2(12 6 )y x 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
= f(x) y = f(x) dx dy
dx
• Learn result for integration:
“Add 1 to the power and divide by the new power”.
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dy/dx = 12x2 – 6x
So 2(12 6 )y x 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
• Do not forget the constant of integration!!!
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INTEGRATION : Question 3
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Find x2 - 4 2xx
dx
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INTEGRATION : Question 3
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Find x2 - 4 2xx
dx
= xx + 4 + C 3 x
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Find x2 - 4 2xx
dxx2 - 4 2xx
dx= x2 - 4
2x3/2 2x3/2dx
= 1/2x1/2 - 2x-3/2 dx
= 2/3 X 1/2x3/2 - (-2) X 2x-1/2 + C
= 1/3x3/2 + 4x-1/2 + C
= xx + 4 + C 3 x
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x2 - 4 2xx
dx= x2 - 4
2x3/2 2x3/2dx
= 1/2x1/2 - 2x-3/2 dx
= 2/3 X 1/2x3/2 - (-2) X 2x-1/2 + C
= 1/3x3/2 + 4x-1/2 + C
= xx + 4 + C 3 x
• Prepare expression by:
1 Dividing out the fraction. 2 Applying the laws of indices.
• 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
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EXIT
1
2
( )Evaluate x2 - 2 2 dx x
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INTEGRATION : Question 4
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1
2
( )Evaluate x2 - 2 2 dx x
= 21/5
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1
2
( )Evaluate x2 - 2 2 dx x
= 21/5
1
2
( )x2 - 2 2 dx x
( )= x4 - 4x + 4 dx x21
2
( )= x4 - 4x + 4x-2 dx1
2
[ ]= x5 - 4x2 + 4x-1 5 2 -1 1
2
= x5 - 2x2 - 4 5 x[ ]2
1
= (32/5 - 8 - 2) - (1/5 - 2 - 4)
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Next Comment= 21/5
1
2
( )x2 - 2 2 dx x
( )= x4 - 4x + 4 dx x21
2
( )= x4 - 4x + 4x-2 dx1
2
[ ]= x5 - 4x2 + 4x-1 5 2 -1 1
2
= x5 - 2x2 - 4 5 x[ ]2
1
= (32/5 - 8 - 2) - (1/5 - 2 - 4)
• Prepare expression by:
1 Expanding the bracket2 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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(a) Find the coordinates of A and B.
(b)Hence find the shaded area between the
curves.
y = -x2 + 8x - 10
y = x
A
B
INTEGRATION : Question 5
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The diagram below shows the parabola y = -x2 + 8x - 10 and the line y = x. They meet at the points A and B.
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(a) Find the coordinates of A and B.
(b)Hence find the shaded area between the
curves.
y = -x2 + 8x - 10
y = x
A
B
INTEGRATION : Question 5
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The diagram below shows the parabola y = -x2 + 8x - 10 and the line y = x. They meet at the points A and B.
A is (2,2) and B is (5,5) .
= 41/2units2
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The diagram shows the parabola
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
between the curves.
(a) Line & curve meet when
y = x and y = -x2 + 8x - 10 .
So x = -x2 + 8x - 10
or x2 - 7x + 10 = 0
ie (x – 2)(x – 5) = 0
ie x = 2 or x = 5
Since points lie on y = x then
A is (2,2) and B is (5,5) .
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The diagram shows the parabola
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
between the curves.
A is (2,2) and B is (5,5) .
(b) Curve is above line between limits so
Shaded area = (-x2 + 8x – 10 - x) dx2
5
= (-x2 + 7x – 10) dx2
5
= -x3 + 7x2 - 10x3 2[ ] 5
2
= (-125/3 + 175/2 – 50)
– (-8/3 +14 – 20)
= 41/2units2
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(a) Line & curve meet when
y = x and y = -x2 + 8x - 10 .
So x = -x2 + 8x - 10
or x2 - 7x + 10 = 0
ie (x – 2)(x – 5) = 0
ie x = 2 or x = 5
Since points lie on y = x then
A is (2,2) and B is (5,5) .
• At intersection of line and curve
yy11 = y = y22
Terms to the left, simplify and Terms to the left, simplify and factorise.factorise.
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(b) Curve is above line between limits so
Shaded area = (-x2 + 8x – 10 - x) dx2
5
= (-x2 + 7x – 10) dx2
5
= -x3 + 7x2 - 10x3 2[ ] 5
2
= (-125/3 + 175/2 – 50)
– (-8/3 +14 – 20)
= 41/2units2
• Learn result
can be used to find the
enclosed area shown:
b
2 1
a
Area = (y y )dx
a b
upper curve
y1area
y2
lower curve
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HIGHER – ADDITIONAL QUESTION BANK
UNIT 2 : Addition Formulae
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Please choose a question to attempt from the following:
1 2 3 4
EXITBack to
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ADDITION FORMULAE : Question 1
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EXIT
In triangle PQR show that the exact value of cos(a - b) is 4/5.
P
Q
R1 4
2ab
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ADDITION FORMULAE : Question 1
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EXIT
In triangle PQR show that the exact value of cos(a - b) is 4/5.
P
Q
R1 4
2ab
cos(a – b) = cosacosb + sinasinb
= (2/5 X 1/5 ) + (1/5 X 2/5 )
= 4/5
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In triangle PQR show that the
exact value of cos(a - b) is 4/5.
P
Q
R1 4
2ab
PQ2 = 12 + 22 = 5
so PQ = 5
QR2 = 42 + 22 = 20
so QR = 20 = 45 = 25
sina = 2/25 = 1/5 & sinb = 2/5
cosa = 4/25 = 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
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PQ2 = 12 + 22 = 5
so PQ = 5
QR2 = 42 + 22 = 20
so QR = 20 = 45 = 25
sina = 2/25 = 1/5 & sinb = 2/5
cosa = 4/25 = 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
• Use formula sheet to check correct expansion and relate to given variables:
cos(a - b) = cosacosb + sina sinb
112
a
a = 7
• Work only with exact values when applying Pythagoras’:
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ADDITION FORMULAE : Question 2
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Find the exact value of cos(p + q) in the diagram below
Y
XO
F(4,4)
G(3,-1)
pq
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ADDITION FORMULAE : Question 2
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Find the exact value of cos(p + q) in the diagram below
Y
XO
F(4,4)
G(3,-1)
pq
= 1/5
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Find the exact value of cos(p + q)
in the diagram below
Y
XO
F(4,4)
G(3,-1)
pq
OF2 = 42 + 42 = 32
so OF = 32 = 162 = 42
OG2 = 32 + 12 = 10
so OG = 10
sinp = 4/42 = 1/2 & sinq = 1/10
cosp = 4/42 = 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
= 2/2 5
= 1/5
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• Use formula sheet to check correct expansion and relate to given variables:
cos(a + b) = cosacosb - sina sinb
cos(p + q) = cospcosq - sinpsinq
Formula Sheet:
becomes:
OF2 = 42 + 42 = 32
so OF = 32 = 162 = 42
OG2 = 32 + 12 = 10
so OG = 10
sinp = 4/42 = 1/2 & sinq = 1/10
cosp = 4/42 = 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
= 2/2 5
= 1/5
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OF2 = 42 + 42 = 32
so OF = 32 = 162 = 42
OG2 = 32 + 12 = 10
so OG = 10
sinp = 4/42 = 1/2 & sinq = 1/10
cosp = 4/42 = 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
= 2/2 5
= 1/5
112
a
a = 7
• Work only with exact values when applying Pythagoras’:
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ADDITION FORMULAE : Question 3
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EXIT
Solve the equation 2 - sinx° = 3cos2x° where 0<x<360.
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ADDITION FORMULAE : Question 3
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Solve the equation 2 - sinx° = 3cos2x° where 0<x<360.
Solution = {30, 150, 199.5, 340.5}
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Solve the equation
2 - sinx° = 3cos2x°
where 0<x<360.
2 - sinx° = 3cos2x°
2 - sinx° = 3(1 – 2sin2x°)
2 - sinx° = 3 – 6sin2x°
Rearrange into quadratic form
6sin2x° - sinx° - 1 = 0
(3sinx° + 1)(2sinx° - 1) = 0
6s2 – s – 1 = (3s + 1)(2s – 1)
sinx° = -1/3 or sinx° = 1/2
Q3 or Q4 Q1 or Q2
AS
T C
a°(180 - a)°
(180 + a)° (360 - a)°
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Solve the equation
2 - sinx° = 3cos2x°
where 0<x<360.
sinx° = -1/3 or sinx° = 1/2
AS
T C
a°(180 - a)°
(180 + a)° (360 - a)°
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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2 - sinx° = 3cos2x°
2 - sinx° = 3(1 – 2sin2x°)
2 - sinx° = 3 – 6sin2x°
Rearrange into quadratic form
6sin2x° - sinx° - 1 = 0
(3sinx° + 1)(2sinx° - 1) = 0
6s2 – s – 1 = (3s + 1)(2s – 1)
sinx° = -1/3 or sinx° = 1/2
Q3 or Q4 Q1 or Q2
AS
T C
a°(180 - a)°
(180 + a)° (360 - a)°
•Use formula sheet to check correct expansion and relate to given variables:
cos2x° =1 - 2sin2x°
•Formula must be consistent with the rest of the equation.
2 - sinx° i.e. choose formula with sinx°
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2 - sinx° = 3cos2x°
2 - sinx° = 3(1 – 2sin2x°)
2 - sinx° = 3 – 6sin2x°
Rearrange into quadratic form
6sin2x° - sinx° - 1 = 0
(3sinx° + 1)(2sinx° - 1) = 0
6s2 – s – 1 = (3s + 1)(2s – 1)
sinx° = -1/3 or sinx° = 1/2
Q3 or Q4 Q1 or Q2
AS
T C
a°(180 - a)°
(180 + a)° (360 - a)°
•Only one way to solve theresulting quadratic:
Terms to the left, put in standard quadratic form and factorise.
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•Take care to relate “solutions” to the given domain. Since all 4 values are included.
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sinx° = -1/3 or sinx° = 1/2
AS
T C
a°(180 - a)°
(180 + a)° (360 - a)°
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}
0 360x
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ADDITION FORMULAE : Question 4
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EXIT
Solve sin2 = cos where 0 < < 2
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ADDITION FORMULAE : Question 4
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EXIT
Solve sin2 = cos where 0 < < 2
Soltn = {/6 , /2 , 5/6 , 3/2}
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sin2 = cos
2sin cos - cos = 0
sin2 - cos = 0
(common factor cos)
Solve sin2 = cos
where 0 < < 2
cos(2sin - 1) = 0
cos = 0 or (2sin - 1) = 0
cos = 0 or sin = 1/2
(roller-coaster graph)
= /2 or 3/2
Q1 or Q2
-
+ 2 -
AS
T C
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Solve sin2 = cos
where 0 < < 2
cos = 0 or sin = 1/2
= /2 or 3/2 Q1 or Q2
-
+ 2 -
AS
T C
sin-1(1/2) = /6
Q1: = /6
Q2: = - /6 = 5/6
Soltn = {/6 , /2 , 5/6 , 3/2}
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sin2 = cos
2sin cos - cos = 0
sin2 - 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 3/2
Q1 or Q2
-
+ 2 -
AS
T C
• Use formula sheet to check correct expansion and relate to given variables:
sin2x = 2sinxcosx
1 radians180
•Although equation is in radians possible to work in degreesand convert to radians using:
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sin2 = cos
2sin cos - cos = 0
sin2 - 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 3/2
Q1 or Q2
-
+ 2 -
AS
T C
•Only one way to solve the result:
Terms to the left, put in standard quadratic form and factorise.
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cos = 0 or sin = 1/2
= /2 or 3/2 Q1 or Q2
-
+ 2 -
AS
T C
sin-1(1/2) = /6
Q1: = /6
Q2: = - /6 = 5/6
Soltn = {/6 , /2 , 5/6 , 3/2}
• For trig. equations
sinx = 0 or 1 or cosx = 0 or 1
use sketch of graph to obtain
solutions: cosx = 0 y
x
y = cos x
2 3
2
x = , 2 3
2
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cos = 0 or sin = 1/2
= /2 or 3/2 Q1 or Q2
-
+ 2 -
AS
T C
sin-1(1/2) = /6
Q1: = /6
Q2: = - /6 = 5/6
Soltn = {/6 , /2 , 5/6 , 3/2}
• Take care to relate “solutions” to the given domain.
Since all 4 values are included
0 2
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HIGHER – ADDITIONAL QUESTION BANK
UNIT 2 :The Circle
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Please choose a question to attempt from the following:
1 2 3 4
EXITBack to
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(a) Find the coordinates of C and W.
(b) Hence find the equation of the dial for the second hand.
W
C
THE CIRCLE : Question 1
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The face of a stopwatch can be modelled by the circle with equation x2 + y2 – 10x – 8y + 5 = 0.
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.
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(a) Find the coordinates of C and W.
(b) Hence find the equation of the dial for the second hand.
W
C
THE CIRCLE : Question 1
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EXIT
The face of a stopwatch can be modelled by the circle with equation x2 + y2 – 10x – 8y + 5 = 0.
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. C is (5,4) W is (5,10).
(x – 5)2 + (y – 7)2 = 4
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The face of a stopwatch can be
modelled by the circle with
equation x2 + y2 – 10x – 8y + 5 = 0.
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.
(a) Find the coordinates of C and W.
Large circle is x2 + y2 – 10x – 8y + 5 = 0.
x2 + y2 + 2gx + 2fy + c = 0Comparing Coefficients
2g = -10, 2f = -8 and c = 5
So g = -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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The face of a stopwatch can be
modelled by the circle with
equation x2 + y2 – 10x – 8y + 5 = 0.
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.
(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
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(a) Find the coordinates of C and W.
Large circle is x2 + y2 – 10x – 8y + 5 = 0.
x2 + y2 + 2gx + 2fy + c = 0Comparing Coefficients
2g = -10, 2f = -8 and c = 5
So g = -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).
• Use formula sheet to check correct formulas and relate to general equation of the circle.
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(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
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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.
(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.
(b) Find the equation of the larger roller given that its diameter is twice that of the smaller roller, and has centre Q.
belt
P
A(7,0)
B
Q
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THE CIRCLE : Question 2
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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.
(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.
(b) Find the equation of the larger roller given that its diameter is twice that of the smaller roller, and has centre Q.
belt
PA(7,0)
B
Q y = -4x + 28
B is (10,-12)
(x – 18)2 + (y + 10)2 = 68
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(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.
(a)
(i) Small circle is x2 + y2 – 6x + 2y - 7 = 0.
x2 + y2 + 2gx + 2fy + c = 0Comparingcoefficients
2g = -6, 2f = 2 and c = -7
So g = -3, f = 1 and 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
= 1/4
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(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.
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
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(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.
(a)(ii) At B x = 10
so y = (-4 X 10) + 28 = -12.
ie B is (10,-12)
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(a) The small roller has centre P
and equation
x2 + y2 – 6x + 2y - 7 = 0.
(b) Find the equation of the larger
roller given that its diameter is
twice that of the smaller roller,
and has centre Q.
r = 17 Small Circle:
(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)
P is (3,-1) A(7,0)
Radius of larger circle is 217
so r2 = (217)2 = 4 X 17 = 68
Using (x – a)2 + (y – b)2 = r2 we get
(x – 18)2 + (y + 10)2 = 68
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(a)
(i) Small circle is x2 + y2 – 6x + 2y - 7 = 0.
x2 + y2 + 2gx + 2fy + c = 0Comparingcoefficients
2g = -6, 2f = 2 and c = -7
So g = -3, f = 1 and 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
= 1/4
• Use formula sheet to check correct formulas and relate to general equation of the circle.
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• Identify centre and radius before putting values into circle equation: Centre (18,-10), radius = 2 (x - a)2 + (y - b)2= r2
(x - 18)2 + (y + 10)2= (2 ) 2
17
17
There is no need to expand thisform of the circle equation.
If the radius is “squared out”it must be left as an exact value.
(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 217
so r2 = (217)2 = 4 X 17 = 68
Using (x – a)2 + (y – b)2 = r2 we get
(x – 18)2 + (y + 10)2 = 68
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THE CIRCLE : Question 3
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Part of the mechanism inside an alarm clock consists of
three different sized collinear cogwheels.
CD
E
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.
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THE CIRCLE : Question 3
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Part of the mechanism inside an alarm clock consists of
three different sized collinear cogwheels.
CD
E
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. (x – 8)2 + (y – 8)2 = 45
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Small circle has centre (0,12) and radius 5 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.
Med. circle has centre (18,3) and radius 20
= 4 X 5 = 25
CE2 = (18 – 0)2 + (3 – 12)2
= 182 + (-9)2 = 324 + 81 = 405
(Distance form!!)
CE = 405 = 81 X 5 = 95
Diameter of large circle = 95 - 25 - 5
= 65
So radius of large circle = 35
C D E5 35
95
25
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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.
C D E5 35
95
25
It follows that CD = 4/9CE
= 4/9[( ) - ( )]18 3
012
= 4/9( )18-9 = ( ) 8
-4
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 = (35)2
or (x – 8)2 + (y – 8)2 = 45
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Small circle has centre (0,12) and radius 5
Med. circle has centre (18,3) and radius 20
= 4 X 5 = 25
CE2 = (18 – 0)2 + (3 – 12)2
= 182 + (-9)2 = 324 + 81 = 405
(Distance form!!)
CE = 405 = 81 X 5 = 95
Diameter of large circle = 95 - 25 - 5
= 65C D E5 35
95
25
• Use formula sheet to check
correct formulas and relate to
equation of the circle.
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It follows that CD = 4/9CE
= 4/9[( ) - ( )]18 3
012
= 4/9( )18-9
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 = (35)2
or (x – 8)2 + (y – 8)2 = 45
• The section formula is used to find Centre D
C
E
D4 5
5 5
CD : DE = 4 : 5
5CD = 4DE
5(d - c) = 4(e - d)
9d = 5c + 4e
8d =
8
����������������������������
����������������������������
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(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.
THE CIRCLE : Question 4
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The central driving wheels on a steam locomotive are linked by a piston rod.
A B C
ED
(a) The rear wheel has centre A & equation x2 + y2 – 4x – 10y + 4 = 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 given that the gaps are equal and the wheels are identical.
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(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.
THE CIRCLE : Question 4
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The central driving wheels on a steam locomotive are linked by a piston rod.
(a) The rear wheel has centre A & equation x2 + y2 – 4x – 10y + 4 = 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 given that the gaps are equal and the wheels are identical.
Each gap = 2 units = 10cm.
Hence D is (10,2) and E is (17,1). Equation of DE x + 7y = 24
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The rear wheel has centre A &
equation x2 + y2 – 4x – 10y + 4 = 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.
(a)(a) Rear wheel is x2 + y2 – 4x – 10y + 4 = 0 .
x2 + y2 + 2gx + 2fy + c = 0Comparingcoefficients
2g = -4, 2f = -10 and c = 4
So g = -2, f = -5 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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The rear wheel has centre A &
equation x2 + y2 – 4x – 10y + 4 = 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.
(a)
Comparingcoefficients
Front wheel is x2 + y2 – 52x – 10y + 676 = 0
x2 + y2 + 2gx + 2fy + c = 0
2g = -52, 2f = -10 and c = 676
So g = -26, f = -5 and c = 676
Centre is (-g,-f)
C is (26,5)
r = 5 as wheels are identical
AC = 24 units
Both gaps = AC – 2 radii - diameter
= 24 – 5 – 5 – 10 = 4 units
Each gap = 2 units = 10cm.
{A is (2,5)}
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(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.
Middle wheel is x2 + y2 – 28x – 10y + 196 = 0 .
x2 + y2 + 2gx + 2fy + c = 0
2g = -28, 2f = -10 and c = 196
So g = -14, f = -5 and c = 196
Centre is (-g,-f)
B is (14,5)
Gradient of BM =5 – 1.5
14 – 13.5
Equation of DE
= 3.5/0.5 = 7
Gradient of DE = -1/7 (m1m2 = -1)
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(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
Using y – b = m(x – a)
we get y – 1.5 = -1/7 (x – 13.5) ( X 7)
Or 7y – 10.5 = -x + 13.5
So DE is x + 7y = 24
Line & circle meet when
x2 + y2 – 28x – 10y + 196 = 0 & x = 24 – 7y.
x + 7y = 24 can be written as x = 24 – 7y
Substituting (24 – 7y) for x in the
circle equation we get ….
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(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.
(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 If y = 2 then x = 10
Hence D is (10,2) and E is (17,1).
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(a) Rear wheel is x2 + y2 – 4x – 10y + 4 = 0 .
x2 + y2 + 2gx + 2fy + c = 0Comparingcoefficients
2g = -4, 2f = -10 and c = 4
So g = -2, f = -5 and c = 4
Centre is (-g,-f) and r = (g2 + f2 – c)
A is (2,5) r = (4 + 25 – 4)
r = 25 = 5
• Use formula sheet to check correct formulas and relate to general equation of the circle.
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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 x + 7y = 24
Line & circle meet when
x2 + y2 – 28x – 10y + 196 = 0 & x = 24 – 7y.
x + 7y = 24 can be written as x = 24 – 7y
Substituting (24 – 7y) for x in the
circle equation we get ….
• In part b) avoid fractions when substituting into the circle equation:
Use x = (24 - 7y) not y = (24 -x)
1
7
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(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 If y = 2 then x = 10
Hence D is (10,2) and E is (17,1).
• Even if it appears an error has occurred continue thesolution even if it means applying the quadratic formula:
2 4
2
b b ac
a
x =
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HIGHER – ADDITIONAL QUESTION BANK
UNIT 3 :
Logs & Exponential
Wave Function
Further Calculus
Vectors
EXIT
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HIGHER – ADDITIONAL QUESTION BANK
UNIT 3 :Wave Function
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Please choose a question to attempt from the following:
1 2 3
EXITBack to
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WAVE FUNCTION: Question 1
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(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.
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WAVE FUNCTION: Question 1
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(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
cos(x°) + 7sin (x°) = 52cos(x° - 81.9°)
Solution is {36.9}(b)
(a)
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(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.
Let cos(x°) + 7sin (x°) = kcos(x° - a°)
= k(cosx°cosa° + sinx°sina° )
= (kcosa°)cosx° + (ksina°)sinx°
kcosa° = 1 & ksina° = 7Comparing coefficients
So (kcosa°)2 + (ksina°)2 = 12 + 72
or k2cos2a° + k2sin2a° = 1 + 49
or k2(cos2a° + sin2a°) = 50
or k2 = 50 (cos2a° + sin2a°) = 1
so k = 50 = 252 = 52
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(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.
so k = 50 = 252 = 52
ksina°kcosa° =
71
tana° = 7
a° = tan-1(7) = 81.9°
ksina° > 0 so a in Q1 or Q2
kcosa° > 0 so a in Q1 or Q4
so a in Q1!
Hence
cos(x°) + 7sin (x°) = 52cos(x° - 81.9°)
![Page 285: HIGHER – ADDITIONAL QUESTION BANK EXIT UNIT 1UNIT 2UNIT 3 Please decide which Unit you would like to revise: Straight Line Functions & Graphs Trig Graphs](https://reader038.vdocument.in/reader038/viewer/2022102906/56649d045503460f949d762a/html5/thumbnails/285.jpg)
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(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
cos(x°) + 7sin (x°) = 52cos(x° - 81.9°)
(b) If cos(x°) + 7sin (x°) = 5
then 52cos(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°
so x° = 126.9° not in range.
Q4: angle = 360° - 45° so x° - 81.9° = 315°
so x° = 396.9° (**)
![Page 286: HIGHER – ADDITIONAL QUESTION BANK EXIT UNIT 1UNIT 2UNIT 3 Please decide which Unit you would like to revise: Straight Line Functions & Graphs Trig Graphs](https://reader038.vdocument.in/reader038/viewer/2022102906/56649d045503460f949d762a/html5/thumbnails/286.jpg)
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(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.
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}
![Page 287: HIGHER – ADDITIONAL QUESTION BANK EXIT UNIT 1UNIT 2UNIT 3 Please decide which Unit you would like to revise: Straight Line Functions & Graphs Trig Graphs](https://reader038.vdocument.in/reader038/viewer/2022102906/56649d045503460f949d762a/html5/thumbnails/287.jpg)
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Let cos(x°) + 7sin (x°) = kcos(x° - a°)
= k(cosx°cosa° + sinx°sina° )
= (kcosa°)cosx° + (ksina°)sinx°
kcosa° = 1 & ksina° = 7Comparing coefficients
So (kcosa°)2 + (ksina°)2 = 12 + 72
or k2cos2a° + k2sin2a° = 1 + 49
or k2(cos2a° + sin2a°) = 50
or k2 = 50 (cos2a° + sin2a°) = 1
so k = 50 = 252 = 52
• Use the formula sheet for the correct expansion:
kcos(x° - a°)
= kcosx°cosa° + ksinx°sina°
(Take care not to omit the “k” term)
![Page 288: HIGHER – ADDITIONAL QUESTION BANK EXIT UNIT 1UNIT 2UNIT 3 Please decide which Unit you would like to revise: Straight Line Functions & Graphs Trig Graphs](https://reader038.vdocument.in/reader038/viewer/2022102906/56649d045503460f949d762a/html5/thumbnails/288.jpg)
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Let cos(x°) + 7sin (x°) = kcos(x° - a°)
= k(cosx°cosa° + sinx°sina° )
= (kcosa°)cosx° + (ksina°)sinx°
kcosa° = 1 & ksina° = 7Comparing coefficients
So (kcosa°)2 + (ksina°)2 = 12 + 72
or k2cos2a° + k2sin2a° = 1 + 49
or k2(cos2a° + sin2a°) = 50
or k2 = 50 (cos2a° + sin2a°) = 1
so k = 50 = 252 = 52
• When equating coefficients “square” and “ring” corresponding coefficients:
1cosx° + 7sinx° = kcosx°cosa° + ksinx°sina°
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Let cos(x°) + 7sin (x°) = kcos(x° - a°)
= k(cosx°cosa° + sinx°sina° )
= (kcosa°)cosx° + (ksina°)sinx°
kcosa° = 1 & ksina° = 7Comparing coefficients
So (kcosa°)2 + (ksina°)2 = 12 + 72
or k2cos2a° + k2sin2a° = 1 + 49
or k2(cos2a° + sin2a°) = 50
or k2 = 50 (cos2a° + sin2a°) = 1
so k = 50 = 252 = 52
• State the resulting equations explicitly:
kcosa° = 1 ksina° = 7
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Let cos(x°) + 7sin (x°) = kcos(x° - a°)
= k(cosx°cosa° + sinx°sina° )
= (kcosa°)cosx° + (ksina°)sinx°
kcosa° = 1 & ksina° = 7Comparing coefficients
So (kcosa°)2 + (ksina°)2 = 12 + 72
or k2cos2a° + k2sin2a° = 1 + 49
or k2(cos2a° + sin2a°) = 50
or k2 = 50 (cos2a° + sin2a°) = 1
so k = 50 = 252 = 52
• k can be found directly:
2 2k 1 7 50
Note: k is always positive
![Page 291: HIGHER – ADDITIONAL QUESTION BANK EXIT UNIT 1UNIT 2UNIT 3 Please decide which Unit you would like to revise: Straight Line Functions & Graphs Trig Graphs](https://reader038.vdocument.in/reader038/viewer/2022102906/56649d045503460f949d762a/html5/thumbnails/291.jpg)
WAVE FUNCTION: Question 2
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Express 4sin(x°)–2cos(x°) in the form ksin(x° + a°)
where k>0 and 0 < a < 360.
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WAVE FUNCTION: Question 2
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Express 4sin(x°)–2cos(x°) in the form ksin(x° + a°)
where k>0 and 0 < a < 360.
Hence
4sin(x°) – 2cos(x°) = 25sin(x° + 333.4°)
![Page 293: HIGHER – ADDITIONAL QUESTION BANK EXIT UNIT 1UNIT 2UNIT 3 Please decide which Unit you would like to revise: Straight Line Functions & Graphs Trig Graphs](https://reader038.vdocument.in/reader038/viewer/2022102906/56649d045503460f949d762a/html5/thumbnails/293.jpg)
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Let 4sin(x°)–2cos(x°) = ksin(x° + a°)
Express 4sin(x°)–2cos(x°)
in the form ksin(x° + a°)
where k>0 and 0 < a < 360.
= k(sinx°cosa° + cosx°sina° )
= (kcosa°)sinx° + (ksina°)cosx°
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 (cos2a° + sin2a°) = 1
so k = 20 = 45 = 25
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Express 4sin(x°)–2cos(x°)
in the form ksin(x° + a°)
where k>0 and 0 < a < 360.
so k = 20 = 45 = 25
ksina°kcosa° =
-2 4
tana° = (-1/2) so Q2 or Q4
tan-1(1/2) = 26.6°
Q4: a° = 360° – 26.6° = 333.4°
ksina° < 0 so a in Q3 or Q4
kcosa° > 0 so a in Q1 or Q4
so a in Q4!
Hence
4sin(x°) – 2cos(x°) = 25sin(x° + 333.4°)
![Page 295: HIGHER – ADDITIONAL QUESTION BANK EXIT UNIT 1UNIT 2UNIT 3 Please decide which Unit you would like to revise: Straight Line Functions & Graphs Trig Graphs](https://reader038.vdocument.in/reader038/viewer/2022102906/56649d045503460f949d762a/html5/thumbnails/295.jpg)
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• Use the formula sheet for the correct expansion:
ksin(x° + a°)
= ksinx°cosa° + kcosx°sina°
(Take care not to omit the “k” term)
Let 4sin(x°)–2cos(x°) = ksin(x° + a°)
= k(sinx°cosa° + cosx°sina° )
= (kcosa°)sinx° + (ksina°)cosx°
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 (cos2a° + sin2a°) = 1
so k = 20 = 45 = 25
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Let 4sin(x°)–2cos(x°) = ksin(x° + a°)
= k(sinx°cosa° + cosx°sina° )
= (kcosa°)sinx° + (ksina°)cosx°
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 (cos2a° + sin2a°) = 1
so k = 20 = 45 = 25
• When equating coefficients “square” and “ring” corresponding coefficients:
4sinx° - 2cosx°
= ksinx°cosa° + kcosx°sina°
• State the resulting equations explicitly:
kcosa° = 4 ksina° = -2
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Let 4sin(x°)–2cos(x°) = ksin(x° + a°)
= k(sinx°cosa° + cosx°sina° )
= (kcosa°)sinx° + (ksina°)cosx°
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 (cos2a° + sin2a°) = 1
so k = 20 = 45 = 25
• k can be found directly:
Note: k is always positive
2 2k 4 ( 2) 20
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so k = 20 = 45 = 25
ksina°kcosa° =
-2 4
tana° = (-1/2) so Q2 or Q4
tan-1(1/2) = 26.6°
Q4: a° = 360° – 26.6° = 333.4°
ksina° < 0 so a in Q3 or Q4
kcosa° > 0 so a in Q1 or Q4
so a in Q4!
Hence
4sin(x°) – 2cos(x°) = 25sin(x° + 333.4°)
• Use the sign of the equations to determine the correct quadrant:
kcosa° = 4 (cos +ve)
ksina° = -2 (sin -ve)
cos +ve&sin -ve
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(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.
P
y = 5sin(x°) - 12cos(x°)
WAVE FUNCTION: Question 3
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The graph below is that of y = 5sin(x°) - 12cos(x°)
![Page 300: HIGHER – ADDITIONAL QUESTION BANK EXIT UNIT 1UNIT 2UNIT 3 Please decide which Unit you would like to revise: Straight Line Functions & Graphs Trig Graphs](https://reader038.vdocument.in/reader038/viewer/2022102906/56649d045503460f949d762a/html5/thumbnails/300.jpg)
(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.
P
y = 5sin(x°) - 12cos(x°)
WAVE FUNCTION: Question 3
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The graph below is that of y = 5sin(x°) - 12cos(x°)
Hence
5sin(x°) – 12cos(x°) = 13sin(x° + 292.6°)
Max TP is at (157.4,13)
![Page 301: HIGHER – ADDITIONAL QUESTION BANK EXIT UNIT 1UNIT 2UNIT 3 Please decide which Unit you would like to revise: Straight Line Functions & Graphs Trig Graphs](https://reader038.vdocument.in/reader038/viewer/2022102906/56649d045503460f949d762a/html5/thumbnails/301.jpg)
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(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.
(a) Let 5sin(x°) – 12cos(x°) = ksin(x° + a°)
= k(sinx°cosa° + cosx°sina° )
= (kcosa°)sinx° + (ksina°)cosx°
kcosa° = 5 & ksina° = -12Comparing coefficients
So (kcosa°)2 + (ksina°)2 = 52 + (-12)2
or k2cos2a° + k2sin2a° = 25 + 144
or k2(cos2a° + sin2a°) = 169
or k2 = 169 (cos2a° + sin2a°) = 1
so k = 13
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(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.
so k = 13
ksina°kcosa° =
-12 5
ksina° < 0 so a in Q3 or Q4
kcosa° > 0 so a in Q1 or Q4
so a in Q4!
tana° = (-12/5) so Q2 or Q4
tan-1(12/5) = 67.4°
Q4: a° = 360° – 67.4° = 292.6°
Hence
5sin(x°) – 12cos(x°) = 13sin(x° + 292.6°)
![Page 303: HIGHER – ADDITIONAL QUESTION BANK EXIT UNIT 1UNIT 2UNIT 3 Please decide which Unit you would like to revise: Straight Line Functions & Graphs Trig Graphs](https://reader038.vdocument.in/reader038/viewer/2022102906/56649d045503460f949d762a/html5/thumbnails/303.jpg)
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(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.
Hence
5sin(x°) – 12cos(x°) = 13sin(x° + 292.6°)
(b) The maximum value of
13sin(x° + 292.6°) is 13.
Maximum on a sin graph occurs
when angle = 90°.
ie x + 292.6 = 90
or 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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(a) Let 5sin(x°) – 12cos(x°) = ksin(x° + a°)
= k(sinx°cosa° + cosx°sina° )
= (kcosa°)sinx° + (ksina°)cosx°
kcosa° = 5 & ksina° = -12Comparing coefficients
So (kcosa°)2 + (ksina°)2 = 52 + (-12)2
or k2cos2a° + k2sin2a° = 25 + 144
or k2(cos2a° + sin2a°) = 169
or k2 = 169 (cos2a° + sin2a°) = 1
so k = 13
• Use the formula sheet for the correct expansion:
ksin(x° + a°)
= ksinx°cosa° + kcosx°sina°
(Take care not to omit the “k” term)
![Page 305: HIGHER – ADDITIONAL QUESTION BANK EXIT UNIT 1UNIT 2UNIT 3 Please decide which Unit you would like to revise: Straight Line Functions & Graphs Trig Graphs](https://reader038.vdocument.in/reader038/viewer/2022102906/56649d045503460f949d762a/html5/thumbnails/305.jpg)
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(a) Let 5sin(x°) – 12cos(x°) = ksin(x° + a°)
= k(sinx°cosa° + cosx°sina° )
= (kcosa°)sinx° + (ksina°)cosx°
kcosa° = 5 & ksina° = -12Comparing coefficients
So (kcosa°)2 + (ksina°)2 = 52 + (-12)2
or k2cos2a° + k2sin2a° = 25 + 144
or k2(cos2a° + sin2a°) = 169
or k2 = 169 (cos2a° + sin2a°) = 1
so k = 13
• When equating coefficients “square” and “ring” corresponding coefficients:
5sinx° - 12cosx°
= ksinx°cosa° + kcosx°sina°
• State the resulting equations explicitly:
kcosa° = 5 ksina° = -12
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(a) Let 5sin(x°) – 12cos(x°) = ksin(x° + a°)
= k(sinx°cosa° + cosx°sina° )
= (kcosa°)sinx° + (ksina°)cosx°
kcosa° = 5 & ksina° = -12Comparing coefficients
So (kcosa°)2 + (ksina°)2 = 52 + (-12)2
or k2cos2a° + k2sin2a° = 25 + 144
or k2(cos2a° + sin2a°) = 169
or k2 = 169 (cos2a° + sin2a°) = 1
so k = 13
• k can be found directly:
Note: k is always positive
2 2k 5 ( 12)
169 13
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so k = 13
ksina°kcosa° =
-12 5
ksina° < 0 so a in Q3 or Q4
kcosa° > 0 so a in Q1 or Q4
so a in Q4!
tana° = (-12/5) so Q2 or Q4
tan-1(12/5) = 67.4°
Q4: a° = 360° – 67.4° = 292.6°
Hence
5sin(x°) – 12cos(x°) = 13sin(x° + 292.6°)
• Use the sign of the equations to determine the correct quadrant:
kcosa° = 5 (cos +ve)
ksina° = -12 (sin -ve)
cos +ve&sin -ve
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(b) The maximum value of
13sin(x° + 292.6°) is 13.
Maximum on a sin graph occurs
when angle = 90°.
ie x + 292.6 = 90
or 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)
• The maximum value of 13sin(x°+292.6°) can also be found by considering rules for related functions:
13sin(x° + 292.6°) slides 13sinx° graph 292.6° to the left. Maximum value of sinx° occurs at 90°
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(b) The maximum value of
13sin(x° + 292.6°) is 13.
Maximum on a sin graph occurs
when angle = 90°.
ie x + 292.6 = 90
or 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)
• Maximum value of sin(x°+292.6°) occurs at (90°- 292.6°)
i.e at x = - 202.6°
Add 360° to bring into domain x = 157.4°
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HIGHER – ADDITIONAL QUESTION BANK
UNIT 3 : Logs &Exponentials
You have chosen to study:
Please choose a question to attempt from the following:
1 2 3 4 5
EXITBack to
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LOGS & EXPONENTIALS: Question 1
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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.
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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.
Initial amount of material = 2000mg
t = 4hrs 18mins
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(a) When t = 10, A10 = 400 so At = A0e-0.161t
(a) If 400mg of material remain after 10 hours then determine how much material there was at the start.
0.1610
ttA A e
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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0.1610
ttA A e
(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.
(b) For half life At = 1/2A0
so At = A0e-0.161t
becomes 1/2A0 = 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
or t = 4hrs 18mins
( 0.3052 X 60 = 18.3..)
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(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
• The ex is found on the calculator:
2nd lnex
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(b) For half life At = 1/2A0
so At = A0e-0.161t
becomes 1/2A0 = 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
or t = 4hrs 18mins
( 0.3052 X 60 = 18.3..)
(b)
• 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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(b) For half life At = 1/2A0
so At = A0e-0.161t
becomes 1/2A0 = 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
or t = 4hrs 18mins
( 0.3052 X 60 = 18.3..)
(b)
• 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 = 2xetc.
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LOGS & EXPONENTIALS: Question 2
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log10y
log10x1
0.3The graph illustrates the law y = k
xn
The line passes through (0,0.3) and (1,0). Find the values of k and n.
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LOGS & EXPONENTIALS: Question 2
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log10y
log10x1
0.3The graph illustrates the law y = k
xn
The line passes through (0,0.3) and (1,0). Find the values of k and n.
Hence k = 2 and n = 0.3
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Intercept = (0,0.3)
gradient = 0 – 0.31 - 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
= -0.3
or log10y = -0.3log10x + log102
or log10y = log10x-0.3 + log102
or log10y = log102x-0.3
law3
law1
The graph illustrates the law
y = kxn
The line passes through (0,0.3) and (1,0). Find the values of k and n.
log10y
log10x1
0.3
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or log10y = log102x-0.3law1The graph illustrates the law
y = kxn
The line passes through (0,0.3) and (1,0). Find the values of k and n.
log10y
log10x1
0.3
so y = 2x-0.3 = 2x0.3
Hence k = 2 and n = 0.3
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Intercept = (0,0.3)
gradient = 0 – 0.31 - 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
or log10y = log102x-0.3
• 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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Intercept = (0,0.3)
gradient = 0 – 0.31 - 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
or log10y = log102x-0.3
log y = logk + logxn
Apply Law 3: logxn = nlogx
log y = logk + nlogx
log y = nlogx + logk
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or log10y = log102x-0.3
so y = 2x-0.3 = 2x0.3
Hence k = 2 and n = 0.3
• Two coordinates from the graph:
log10y
log10x1
0.3
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or log10y = log102x-0.3
so y = 2x-0.3 = 2x0.3
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
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Solve the equation log3(5) – log3(2x + 1) = -2
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LOGS & EXPONENTIALS: Question 3
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Solve the equation log3(5) – log3(2x + 1) = -2
x = 22
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If log3(5) – log3(2x + 1) = -2
Solve the equation
log3(5) – log3(2x + 1) = -2then log3 5
2x + 1( ) = -2law2
so 5
(2x + 1)= 3-2
or 5
(2x + 1)= 1
9Cross mult
we get 2x + 1 = 45
or 2x = 44
ie x = 22
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If log3(5) – log3(2x + 1) = -2
then log3 5
2x + 1( ) = -2
so 5
(2x + 1)= 3-2
or 5
(2x + 1)= 1
9Cross mult
we get 2x + 1 = 45
or 2x = 44
ie x = 22
• 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
Law 2: log = logA - logBAB
Must know: log33 = 1
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If log3(5) – log3(2x + 1) = -2
then log3 5
2x + 1( ) = -2
so 5
(2x + 1)= 3-2
or 5
(2x + 1)= 1
9Cross mult
we get 2x + 1 = 45
or 2x = 44
ie x = 22
log3 = -2 log33 52x+1
log3 = log33-2 52x+1
Law 3: log = nlogxxn
The log terms can now be dropped from both sides of theequation and the equation solved:
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If log3(5) – log3(2x + 1) = -2
then log3 5
2x + 1( ) = -2
so 5
(2x + 1)= 3-2
or 5
(2x + 1)= 1
9Cross mult
we get 2x + 1 = 45
or 2x = 44
ie x = 22
Drop log3 terms
= 3-2 52x+1
etc.
= 52x+1
19
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LOGS & EXPONENTIALS: Question 4
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The pressure in a leaky tyre drops according to the formula
Pt = P0e-kt
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?
sssss
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LOGS & EXPONENTIALS: Question 4
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EXIT
The pressure in a leaky tyre drops according to the formula
Pt = P0e-kt
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?
sssss
1.8psi
(a) k = 0.243 to 3 dps
(b)
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0kt
tP Pe
(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.
(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
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0kt
tP Peie k = 0.243 to 3 dps
(b) By how many more psi will the pressure drop in the next 15 mins?
(b) We now have k = 0.243,
P0 = 31 and t = 0.25 (15mins = 1/4hr)
using Pt = P0e-0.243t
we get P0.25 = 31e(-0.243X0.25)
so P0.25 = 29.2
Pressure drop in next 15 mins is
31 – 29.2 or 1.8psi
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(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
• The ex is found on the calculator:
2nd lnex
(a)
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• 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.
(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
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LOGS & EXPONENTIALS: Question 5
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EXIT
Given that sin = au and cos = av show that
u – v = logatan.
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LOGS & EXPONENTIALS: Question 5
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Given that sin = au and cos = av show that
u – v = logatan.
u – v = logatan
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Given that sin = au
and cos = av show that
u – v = logatan.
Since sin = au and cos = av
then u = logasin and v = logacos
so u – v = logasin - logacos
or u – v = loga(sin/cos)
hence u – v = logatan
logx – logy = log(x/y)
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Since sin = au and cos = av
then u = logasin and v = logacos
so u – v = logasin - logacos
or u – v = loga(sin/cos)
hence u – v = logatan
logx – logy = log(x/y)
• There are different routes to the same result but all involve correct application of formulas and laws of logs:
e.g. tan = sin cos
Should be known from
standard grade
tan = au
av
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Since sin = au and cos = av
then u = logasin and v = logacos
so u – v = logasin - logacos
or u – v = loga(sin/cos)
hence u – v = logatan
logx – logy = log(x/y)
Take logs to base a to both sides
loga tan = loga
au
av
loga tan = loga au - loga av
Law2: log =logA-logBAB
loga tan = uloga a - vloga a
Law 3: log = nlogxxn
logaa=1
loga tan = u - v
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HIGHER – ADDITIONAL QUESTION BANK
UNIT 3 :Vectors
You have chosen to study:
Please choose a question to attempt from the following:
1 2 3 4 5
EXITBack to
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VECTORS: Question 1
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(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.
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VECTORS: Question 1
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(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.
Since PQ and
PR are
multiples of the same
common point then vector and have P as a
it follows that P, Q & R are collinear.
= (-25, 7, 16)(b) S
(a)
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(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.
PQ = q - p =
(a) [ ]-1 1-2
[ ] 3 0-2
-
[ ] -4 1 3
=
PR = r - p = [ ]
-13 4 7
[ ] 3 0-2
-
[ ]-16 4 9
[ ] -4 1 3
= 4=
Since PQ and
PR are
multiples of the same
common point then vector and have P as a
it follows that P, Q & R are collinear.
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(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.
(b) PS = 7
PQ [ ]
-4 1 3
= 7 [ ]-28 7 21
=
S is (3,0,-5) + [ ]-28 7 21
= (3-28, 0+7, -5+21)
= (-25, 7, 16)
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• Must know result:
Given coordinates of A and B
AB = b a��������������
a)
PQ = q - p =
(a) [ ]-1 1-2
[ ] 3 0-2
-
[ ] -4 1 3
=
PR = r - p = [ ]
-13 4 7
[ ] 3 0-2
-
[ ]-16 4 9
[ ] -4 1 3
= 4=
Since PQ and
PR are
common point then vector and have P as a
it follows that P, Q & R are collinear.
multiples of the same
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PQ = q - p =
(a) [ ]-1 1-2
[ ] 3 0-2
-
[ ] -4 1 3
=
PR = r - p = [ ]
-13 4 7
[ ] 3 0-2
-
[ ]-16 4 9
[ ] -4 1 3
= 4=
Since PQ and
PR are
common point then vector and have P as a
it follows that P, Q & R are collinear.
multiples of the same
• Must be able to state explicitly the result for collinearity:
Since PR = 4PQ with common
point P P,Q and R are collinear
Beware common error 4PR = PQ
����������������������������
����������������������������
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(b) PS = 7
PQ [ ]
-4 1 3
= 7 [ ]-28 7 21
=
S is (3,0,-5) + [ ]-28 7 21
= (3-28, 0+7, -5+21)
= (-25, 7, 16)
• An alternative approach is to
form a vector equation and
solve it:
b)
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(b) PS = 7
PQ [ ]
-4 1 3
= 7 [ ]-28 7 21
=
S is (3,0,-5) + [ ]-28 7 21
= (3-28, 0+7, -5+21)
= (-25, 7, 16)
b) 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
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(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.
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VECTORS: Question 2
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(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.
FE
(a) [ ] -1 4 -3
=
FG [ ]
3 0 -3
=
so = 73.9°
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FE = e - f =
(a) [ ] 4 3-1
[ ] 5-1 2
-
[ ] -1 4 -3
=
FG = g - f = [ ]
8 -1 -1
[ ] 5 -1 2
-
[ ] 3 0 -3
=
(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.
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(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.
(b) Let angle EFG =
E
F
G
ie
FE .
FG = [ ]
-1 4 -3
[ ] 3 0 -3
.
= (-1 X 3) + (4 X 0) + (-3 X (-3))
= -3 + 0 + 9
= 6
|FE| = ((-1)2 + 42 + (-3)2)
|FG| = (32 + 02 + (-3)2)
= 26
= 18
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(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.
Given that FE.FG = |FE||FG|cos
then cos = FE.FG
|FE||FG|=
6
26 18
so = cos-1(6 26 18) = 73.9°
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(b) Let angle EFG =
E
F
G
ie
FE .
FG = [ ]
-1 4 -3
[ ] 3 0 -3
.
= (-1 X 3) + (4 X 0) + (-3 X (-3))
= -3 + 0 + 9
= 6
|FE| = ((-1)2 + 42 + (-3)2)
|FG| = (32 + 02 + (-3)2)
= 26
= 18
• Ensure vectors are calculated
from the vertex:
F G
E
Vectors FE and FG����������������������������
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Given that FE.FG = |FE||FG|cos
then cos = FE.FG
|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 .cos�������������������������������������������������������
1 1 2 2 3 3
1 1 2 2 3 3
a.b=a b +a b +a b
FE.FG=a b +a b +a b����������������������������
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VECTORS: Question 3
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EXIT
P
QR
S
T
U V
W
A
B
PQRSTUVW is a cuboid in which
PQ , PS & PW are represented by the vectors
[ ],
4 2 0
[ ]and
-2 4 0
[ ]resp.
0 0 9
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.
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VECTORS: Question 3
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P
QR
S
T
U V
W
A
B
PQRSTUVW is a cuboid in which
PQ , PS & PW are represented by the vectors
[ ],
4 2 0
[ ]and
-2 4 0
[ ]resp.
0 0 9
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.
|PA|
= 29
|PB|
= 106
= 48.1°APB
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PA =
PS + SA =
PS + 1/3ST
PS + 1/3PW
=
[ ]-2 4 0
[ ] =
0 0 3
+ [ ] -2 4 3
=
PB =
PQ + QV + VB
PQ + PW + 1/2PS
=
[ ] 4 2 0
[ ]+ 0 0 9
+ [ ]= -1 2 0
[ ] 3 4 9
=
PQRSTUVW is a cuboid in which
PQ , PS & PW are represented by vectors
[ ],
4 2 0
[ ]and
-2 4 0
[ ]resp.
0 0 9
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.
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PA =
[ ] -2 4 3
PB = [ ]
3 4 9PQRSTUVW is a cuboid in which
PQ , PS & PW are represented by vectors
[ ],
4 2 0
[ ]and
-2 4 0
[ ]resp.
0 0 9
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.
(b) Let angle APB =
A
P
B
ie
PA .
PB =
[ ] -2 4 3
[ ] 3 4 9
.
= (-2 X 3) + (4 X 4) + (3 X 9)
= -6 + 16 + 27
= 37
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PQRSTUVW is a cuboid in which
PQ , PS & PW are represented by vectors
[ ],
4 2 0
[ ]and
-2 4 0
[ ]resp.
0 0 9
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.
PA .
PB =
37
|PA| = ((-2)2 + 42 + 32)
= 29
|PB| = (32 + 42 + 92)
= 106
Given that PA.PB = |PA||PB|cos
then cos = PA.PB
|PA||PB|=
37
29 106
so = cos-1(37 29 106)
= 48.1°
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PA =
[ ] -2 4 3
PB = [ ]
3 4 9
(b) Let angle APB =
A
P
B
ie
PA .
PB =
[ ] -2 4 3
[ ] 3 4 9
.
= (-2 X 3) + (4 X 4) + (3 X 9)
= -6 + 16 + 27
= 37
• Ensure vectors are calculated
from the vertex:
P A
B
Vectors PB and PA����������������������������
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PA .
PB =
|PA| = ((-2)2 + 42 + 32)
= 29
|PB| = (32 + 42 + 92)
= 106
Given that PA.PB = |PA||PB|cos
then cos = PA.PB
|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 .cos�������������������������������������������������������
1 1 2 2 3 3
1 1 2 2 3 3
a.b=a b +a b +a b
PB.PA=a b +a b +a b����������������������������
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VECTORS: Question 4
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An equilateral triangle has sides 4 units long which are represented by the vectors a , b & c as shown.
b c
aFind
(i) b.(a + c) & comment on your answer (ii) b.(a – c)
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VECTORS: Question 4
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An equilateral triangle has sides 4 units long which are represented by the vectors a , b & c as shown.
b c
aFind
(i) b.(a + c) & comment on your answer (ii) b.(a – c)
Dot product = 0 so
b & (a + c) are perpendicular.(ii) b.(a – c) = b.b
= 42 = 16
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An equilateral triangle has
sides 4 units long which are
represented by the vectors
a , b & c .
Find
(i) b.(a + c) & comment
(ii) b.(a – c)
NB: each angle is 60° but vectors should
be “tail to tail”
b
b
c60°
120°
(i) b.(a + c) = b.a + b.c
= |b||a|cos1 + |b||c|cos2
= 4X4Xcos60° + 4X4Xcos120°
= 16 X ½ + 16 X (-½ )
= 8 + (-8)
= 0
Dot product = 0 so
b & (a + c) are perpendicular.
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Question 4
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An equilateral triangle has
sides 4 units long which are
represented by the vectors
a , b & c .
Find
(i) b.(a + c) & comment
(ii) b.(a – c)
NB: a – c = a + (-c) or b .
(ii) b.(a – c) = b.b
= |b|2
= 42 = 16
= |b||b|cos0
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NB: each angle is 60° but vectors should
be “tail to tail”
b
b
c60°
120°
(i) b.(a + c) = b.a + b.c
= |b||a|cos1 + |b||c|cos2
= 4X4Xcos60° + 4X4Xcos120°
= 16 X ½ + 16 X (-½ )
= 8 + (-8)
= 0
Dot product = 0 so
b & (a + c) are perpendicular.
• This geometric question is based on the scalar product definition and the distributive law:
a.b= a . b .cosθ
and a(b+c) = ab + ac
No other results should be applied.
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NB: each angle is 60° but vectors should
be “tail to tail”
b
b
c60°
120°
(i) b.(a + c) = b.a + b.c
= |b||a|cos1 + |b||c|cos2
= 4X4Xcos60° + 4X4Xcos120°
= 16 X ½ + 16 X (-½ )
= 8 + (-8)
= 0
Dot product = 0 so
b & (a + c) are perpendicular.
• Ensure all scalar products are
calculated from the vertex:
b
aθ
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VECTORS: Question 5
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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.
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VECTORS: Question 5
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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) 2a + b [ ]324
=
= 29 (ii) | 2a + b |
Since the dot product is zero it
follows that the vectors are perpendicular.
(b)
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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) 2a + b = [ ] 2 1-3
[ ]-1 010
+2
[ ] 4 2-6
[ ]-1 010
+=
[ ]324
=
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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.
(ii) | 2a + b | = (32 + 22 + 42)
= 29
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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.
(b) (2a + b ).c = [ ]324
[ ]-2 1 1
.
= (3 X (-2))+(2 X 1)+(4 X 1)
= -6 + 2 + 4
= 0
Since the dot product is zero it follows
that the vectors are perpendicular.
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(a)(i) 2a + b = [ ] 2 1-3
[ ]-1 010
+2
[ ] 4 2-6
[ ]-1 010
+=
[ ]324
=
• When a vector is given in i , j , k form change to component form:
e.g. 5i - 2j + k =
5
-2
1
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(ii) | 2a + b | = (32 + 22 + 42)
= 29
• Must know formula for the magnitude of a vector:
2 2 2
a
n = b n a +b +c
c
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(b) (2a + b ).c = [ ]324
[ ]-2 1 1
.
= (3 X (-2))+(2 X 1)+(4 X 1)
= -6 + 2 + 4
= 0
Since the dot product is zero it follows
that the vectors are perpendicular.
• Must know condition for perpendicular vectors:
a.b = 0 a is perpendicular to b
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HIGHER – ADDITIONAL QUESTION BANK
UNIT 3 :Further Calculus
You have chosen to study:
Please choose a question to attempt from the following:
1 2 3 4
EXITBack to
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FURTHER CALCULUS : Question 1
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Given that y = 3sin(2x) – 1/2cos(4x) then find dy/dx .
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FURTHER CALCULUS : Question 1
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Given that y = 3sin(2x) – 1/2cos(4x) then find dy/dx .
= 6cos(2x) + 2sin(4x)
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Given that
y = 3sin(2x) – 1/2cos(4x)
then find dy/dx .
3sin(2x) 1/2cos(4x)
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)
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• Check formula sheet for correct result:
ydy
dx
sin(ax) acos(ax)cos(ax) -asin(ax)
3sin(2x) 1/2cos(4x)
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)
• Relate formula to given variables
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3sin(2x) 1/2cos(4x)
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)
• When applying the “chain rule” “Peel an onion”
3sin(2x) 1/2cos(4x)
OUTER / INNER
Differentiate outer then inner
e.g. 3sin - 3cos 2x 2
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FURTHER CALCULUS : Question 2
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Given that g(x) = (6x – 5) then evaluate g´(9) .
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FURTHER CALCULUS : Question 2
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Given that g(x) = (6x – 5) then evaluate g´(9) .
= 3/7
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Given that g(x) = (6x – 5)
then evaluate g´(9) .
g(x) = (6x – 5) = (6x – 5)1/2
(6x – 5)1/2
outer / innerdiff outer then inner
g´(x) = 1/2(6x – 5)-1/2 X 6
= 3(6x – 5)-1/2
= 3
(6x – 5)
g´(9) = 3
(6X9 – 5)
= 3
49
= 3/7
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g(x) = (6x – 5) = (6x – 5)1/2
(6x – 5)1/2
outer / innerdiff outer then inner
g´(x) = 1/2(6x – 5)-1/2 X 6
= 3(6x – 5)-1/2
= 3
(6x – 5)
g´(9) = 3
(6X9 – 5)
= 3
49= 3/7
• Apply the laws of indices to replace the sign
1
2x x
• Apply the chain rule
• Learn the rule for differentiation
Multiply by the power then reduce the power by 1
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g(x) = (6x – 5) = (6x – 5)1/2
(6x – 5)1/2
outer / innerdiff outer then inner
g´(x) = 1/2(6x – 5)-1/2 X 6
= 3(6x – 5)-1/2
= 3
(6x – 5)
g´(9) = 3
(6X9 – 5)
= 3
49= 3/7
• Apply the laws of indices to return power to a positive value then the root
1
21
2
1 1x
xx
• Will usually work out to an exact value without need to use calculator.
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FURTHER CALCULUS : Question 3
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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
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A curve for which dy/dx = -12sin(3x) passes through the point (/3,-2). Express y in terms of x.
So y = 4cos(3x) + 2
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A curve for which
dy/dx = -12sin(3x) passes
through the point (/3,-2).
Express y in terms of x.
if dy/dx = -12sin(3x)
then 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 -2 = -4 + C
ie C = 2
So y = 4cos(3x) + 2
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if dy/dx = -12sin(3x)
then 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 -2 = -4 + C
ie C = 2
So y = 4cos(3x) + 2
• Learn the result that integration undoes differentiation:
i.e. given
= f(x) y = f(x) dx dy
dx
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if dy/dx = -12sin(3x)
then 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 -2 = -4 + C
ie C = 2
So y = 4cos(3x) + 2
• Check formula sheet for correct result:
y sinax -cos(ax)
a
cos(ax) sin(ax) a
+ c
+ c
• Do not forget the constant of integration.
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FURTHER CALCULUS : Question 4
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(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
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(a) Find the derivative of y = (2x3 + 1)2/3 where x > 0.
(b) Hence find x2
(2x3 + 1)1/3
dx
= 4x2
(2x3 + 1)1/3
= 1/4(2x3 + 1)2/3 + C
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(2x3 + 1)2/3
outer inner
diff outer then inner
(a) if y = (2x3 + 1)2/3
(a) Find the derivative of
y = (2x3 + 1)2/3 where x > 0.
then dy/dx = 2/3 (2x3 + 1)-1/3 X 6x2
= 4x2
(2x3 + 1)1/3
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(b) Hence find
x2
(2x3 + 1)1/3
dx
(b) From (a) it follows that
4x2
(2x3 + 1)1/3
dx = (2x3 + 1)2/3 + C
4x2
(2x3 + 1)1/3
dx
now
x2
(2x3 + 1)1/3
dx
= ¼ X
= 1/4(2x3 + 1)2/3 + C
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(2x3 + 1)2/3
outer inner
diff outer then inner
(a) if y = (2x3 + 1)2/3
then dy/dx = 2/3 (2x3 + 1)-1/3 X 6x2
= 4x2
(2x3 + 1)1/3
•. Apply the chain rule“Peel an onion”
(2x3 + 1)2/3
outer inner
diff outer then inner
2 1-
3 32( ) ( )
3e.g.
a)
3 22x +1 6x
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(b) From (a) it follows that
4x2
(2x3 + 1)1/3
dx = (2x3 + 1)2/3 + C
4x2
(2x3 + 1)1/3
dx
now
x2
(2x3 + 1)1/3
dx
= ¼ X
= 1/4(2x3 + 1)2/3 + C
• Learn the result that integration undoes differentiation:
i.e. given
= f(x) y = f(x) dx dy
dx
b)