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8/8/2019 Higher Revision 1, 2 & 3
1/400
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 Differentiation
Recurrence Relations
PolynomialsQuadratic FunctionsIntegration
Addition Formulae
The Circle
VectorsFurther CalculusExponential /Logarithmic FunctionsThe Wave Function
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HIGHER ADDITIONAL QUESTION BANK
UNIT 1 :
Functions &Graphs
Straight Line
RecurrenceRelations
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
EXIT Back toUnit 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 thepoint (-6,4).
Go to full solution
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Go to Straight Line Menu
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Reveal answer only
EXIT
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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 thepoint (-6,4).
Go to full solution
Go to Markers Comments
Go to Straight Line Menu
Go to Main Menu
Reveal answer only y = -5/3x - 6
EXIT
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Markers Comments
Begin Solution
Continue Solution
3x 5y = 4
3x - 4 = 5y
5y = 3x - 4 (z5)
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).
Straight Line Menu
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3x 5y = 4
3x - 4 = 5y
5y = 3x - 4 (z5)
y = 3/5x -4/5
Using y = mx + c , gradient of line is 3/5
So required gradient = -5/3 , ( m1m2 = -1)
We now have (a,b) = (-6,4) & m = -5/3 .
Using y b = m(x a)
We get y 4 = -5/3 (x (-6))
y 4 = -5/3 (x + 6)
y 4 = -5/3x - 10
y =-5
/3x - 6
Markers Comments
Straight Line Menu
Back to Home
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 m1m2 = -1.
When finding m2simply invert
and change the sign on m1
m1 =3
5 m2 =-5
3
U
se the y - b = m(x - a) formto obtain the equation of the
line.
Next Comment
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Go to full solution
Go to Markers Comments
Go to Straight Line Menu
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Reveal answer only
EXIT
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 thepoint (5,-3).
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Go to full solution
Go to Markers Comments
Go to Straight Line Menu
Go to Main Menu
Reveal answer only
EXIT
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 thepoint (5,-3).
y = -2x + 7
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Question 2
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8x + 4y 7 = 0
4y = -8x + 7 (z4)
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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Markers Comments
Straight Line Menu
Back to Home
Next Comment
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 (z4)
y = -2x + 7/4
y = -2x + 7
sing y = mx + c , gradient of line is -2
o required gradient = -2 as parallel linesave equal gradients.
We now have (a,b) = (5,-3) & m = -2.
Using y b = m(x a)
e get y (-3) = -2(x 5)
y + 3 = -2x + 10
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Go to full solution
Go to Markers Comments
Go to Straight Line Menu
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Reveal answer only
EXIT
STRAIGHT LINE : Question 3
In triangleABC, A is (2,0),
B is (8,0) and C is (7,3).
(a) Find the gradients ofAC and BC.
(b) Hence find the size ofACB.
X
Y
A B
C
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Go to full solution
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Reveal answer only
EXIT
STRAIGHT LINE : Question 3
In triangleABC, A is (2,0),
B is (8,0) and C is (7,3).
(a) Find the gradients ofAC and BC.
(b) Hence find the size ofACB.
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 triangleABC, A is (2,0),
B is (8,0) and C is (7,3).
(a) Find the gradients ofAC
and BC.
(b) Hence find the size ofACB. (b) Using tanU = gradient
If tanU = 3/5 then CAB = 31.0
If tanU = -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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Markers Comments
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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
!
) Using tanU = gradient
= 3/5
If tanU = 3/5 then CAB = 31.0
then CBX = (180-71.6)
= 108.4 o
Hence :
ACB = 180 31.0 71.6
= 77.4
If tanU = -3
If no diagram is given draw a
neat labelled diagram.
In calculating gradients statethe gradient formula.
Must use the result that the
gradient of the line is equal
to the tangent of the angle
the line makes with thepositive direction of the x-axis.
Not given on the formula sheet.
A
B
mAB
= tan
= tan-1 mABso ABC = 71.6
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STRAIGHT LINE : Question 4
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Reveal answer only
EXIT
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
Go to full solution
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Reveal answer only
EXIT
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)
Straight Line Menu
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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): lets call this S
Using the gradient formula m = y2 y1x
2
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)
Straight Line Menu
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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)
Straight Line Menu
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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): lets call this S
Using the gradient formula m = y2 y1x2 x1
mSR= -1 (-1)
10 - 3
ince it passes through (3,-1)
quation of e is y = -1
(ie line is horizontal)
Comments for4 (b)
Sketch the median and the
perpendicular bisector
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Markers Comments
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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 = k
Vertical lines in the form x = k
Comments for4 (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 = -1
y = 2x - 11
To find the point of
intersection of the two
lines solve the twoequations:
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STRAIGHT LINE : Question 5
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Reveal answer only
EXIT
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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Reveal answer only
EXIT
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)
Straight Line Menu
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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.
X
Y
G(2,-5)
E(6,-3)
F(12,-5)
(a) Using the gradient formula2 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)
Straight Line Menu
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In triangle EFG the vertices are
E(6,-3), F(12,-5) and G(2,-5).
Find
X
Y
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)- lets 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)
Straight Line Menu
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In triangle EFG the vertices are
E(6,-3), F(12,-5) and G(2,-5).
Find
X
Y
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 = 0ie 8y = -34
ie y = -4.25
Point of intersection is (6,-4.25)
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Markers Comments
Straight Line Menu
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Next Comment
If no diagram is given draw a
neat labelled diagram.
Sketch the altitude andthe median.
y
x
F
E
G
median
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 for5 (b)
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y
x
F
E
G
Comments for5 (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 = k
Vertical lines in the form x = k
Look for special cases:
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Straight Line Menu
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c)
ines 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 = 6
x + 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
You have chosen to study:
Please choose a question to attempt from the following:
1 2 3 4 5
EXIT Back toUnit 1 Menu
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BASIC DIFFERENTIATION : Question 1
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EXIT
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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EXIT
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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Question 1
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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.
PointIf x = 4 then y = 4 16
4= 2 4 = -2
so (a,b) = (4,-2)
Gradient: y = x 16x
= x1/2 16x -1
dy/dx =1/2x
-1/2 + 16x-2 = 1 + 16
2x x2
If x = 4 then:dy
/dx = 1 + 1624 16
= + 1 = 5/4
Continue Solution
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Question 1
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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
2
4 16
= + 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
Back to Previous
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Prepare expression for differentiation.1
1216
y 16 x x xx
!
NB: a tangent is a line so we need a
point of contact and a gradient.
Point
If x = 4 then y = 4 164
= 2 4 = -2
so (a,b) = (4,-2)
Gradient: y = x 16x
= x1/2 16x -1
dy/dx =1/2x
-1/2 + 16x-2 = 1 + 16
2
x x2
If x = 4 then:
= 1 + 16
24 16
= + 1 = 5/4
Find gradient of the tangent
using rule:
dy
dx
multiply by the power and reducethe power by 1
dy/dx
Find gradient = at x = 4.dy
dx
Continue Comments
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If x = 4 then:
= 1 + 16
24 16= + 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 24x 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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EXIT
Find the coordinates of the point on the curve y = x2 5x + 10
where the tangent to the curve makes an angle of135 with thepositive direction of the X-axis.
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BASIC DIFFERENTIATION : Question 2
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EXIT
Find the coordinates of the point on the curve y = x2 5x + 10
where the tangent to the curve makes an angle of135 with thepositive direction of the X-axis.
(2,4)
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olutionContinue Solution
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.
NB: gradient of line = gradient of curve
Line
Using gradient = tanU
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
Continue Solution
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olutionContinue Solution
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.
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)
M k C t
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NB: gradient of line = gradient of curve
Line
Using gradient = tanU
we get gradient of line = tan135
= -tan45
= -1
Curve
Gradient of curve = dy/dx = 2x - 5
It now follows that2x 5 = -1
Or 2x = 4
Or x = 2
Find gradient of the tangent
using rule:
dy
dx
multiply by the power and reducethe 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
x
135
Continue Comments
M k C t
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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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EXIT
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
y
y =
g (x)
Q i 3
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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 p 0 p p p
gd(x) - 0 - 0 +
Click for graph
y = g(x)
(p,q)
r
Make a sketch of the graph of
y = gd(x).
Q i 3
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y = g(x)
(p,q)
r
Make a sketch of the graph of
y = gd(x).
y
0p
y = gd(x)
This now gives us the following graph
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To sketch the graph of
the gradient function:
' dyf ( )
dxx !
1 Mark the stationary points on thex axis i.e. ' dy( ) 0
dxx ! !
x
y
a
'( )x
x
y
0p
y = gd(x)
tationary points occur at x = 0 and x = p.
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2 For each interval decide if the
value of dyf ( ) is - or +
dx
x !
To sketch the graph of
the gradient function:
' dyf ( )
dxx !
1 Mark the stationary points on thex axis i.e. ' dy( ) 0
dxx ! !
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 p 0 p p p
gd(x) - 0 - 0 +
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+
- -
x
y
a
f ( )x
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To sketch the graph of
the gradient function:
' dyf ( )
dxx !
1 Mark the stationary points on thex axis i.e. ' dy( ) 0
dxx ! !
x
y
0p
y = gd(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 dyf ( ) 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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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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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
Question 4
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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 p -1 p 3 p
(x + 1) - 0 + + +
(x - 3) - - - 0 +
dy/dx + 0 - 0 +
Hence (-1,7) is a maximum TPand (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 and
reduce 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 p -1 p 3 p
(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
ofsigns
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 u 2)
So after 5 months the share is S(5) = 2/54/25 =
6/25
Find the maximum share of the market that the company can achieve.
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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 u 2)
So after 5 months the share is S(5) = 2/54/25 =
6/25
Find the maximum share of the market that the company can achieve.
= 1/4
Question 5
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End points
S(2) = 1 1 = 0
There is no upper limit but as x p g
S(x) p 0.S(x) = 2 - 4
x x2(x u 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 d(x) = -2x-2 + 8x-3
= - 2 + 8
x2 x3= 8 - 2
x3 x2
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S(x) = 2 - 4
x x2(x u 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 d(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 u 2
In required interval
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= 0
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S(x) = 2 - 4
x x2(x u 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 p 4 pS d(x) + 0 -
S d(3.9 ) = 0.00337
S d(4.1) = -0.0029
Hence max TP at x = 4
So max share of market
= S(4) = 2/44/16
= 1/21/4
= 1/4
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End points
S(2) = 1 1 = 0
There is no upper limit but as x p g
S(x) p 0.
Stationary Points
S(x) = 2 - 4
x x2= 2x-1 4x-2
So S d(x) = -2x-2 + 8x-3
= - 2 + 8
x2 x3= 8 - 2
x3 x2
Must consider end points
and stationary points.
Must look forkey 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.)
dy(x)
dxd!
SPs occur where S d(x) = 0
8 - 2x3 x2
or8 = 2
x3 x2( cross mult!)
8x2 = 2x3
= 0
8x2 - 2x3 = 0
2x2(4 x) = 0
x = 0 or x = 4
NB: x u 2
In 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 solutionsoutwith the domain.
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We now check the gradients either side of
X = 4
x p 4 p
S d(x) + 0 -
S d(3.9 ) = 0.00337
S d(4.1) = -0.0029
Hence max TP at x = 4
So max share of market
= S(4) = 2/44/16
= 1/21/4
= 1/4
Must show a maximum value
using a table ofsigns.
x 4
(x)d + 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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1 2 3
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RECURRENCERELATIONS : Question 1
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A recurrence relation is defined by the formula un+1 = aun + b,
where -1
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RECURRENCERELATIONS : Question 1
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A recurrence relation is defined by the formula un+1 = aun + b,
where -1
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Q
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A recurrence relation is defined
by the formula un+1 = aun + b,
where -1
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A recurrence relation is defined
by the formula un+1 = aun + b,
where -1
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Using un+1 = aun + bwe 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 q 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.
Comments for1(b)
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(b)
un+1 = aun + b is now un+1 = 0.6un - 2
This has a limit since -1
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RECURRENCERELATIONS : Question 2
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Two different recurrence relations are known to have the same
limit as n p g.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.
RECURRENCE RELATIONS Q ti 2
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RECURRENCERELATIONS : Question 2
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Two different recurrence relations are known to have the same
limit as n p g.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
Question 2If the limit is L then as n p g we have
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Two different recurrence relations
are known to have the same limit
as n p g.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 p g 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
becomes
L = k2L + 1
L - k2L = 1
L(1 - k2) = 1
L = 1 .
. (1 - k2)
Question 2
L = 1L = 3
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Two different recurrence relations
are known to have the same limit
as n p g.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 k 2)
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
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Two different recurrence relations
are known to have the same limit
as n p g.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
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If the limit is L then as n p g 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 k 2)
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
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Since -1
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RECURRENCERELATIONS : Question 3
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A man plants a row of fast growing trees between his own house
and his neighbours. These trees are known grow at a rate of 1m
per annum so cannot be allowed to become too high. Hetherefore 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 thatthe trees be kept to a maximum height of 3m. What
percentage should be pruned to ensure that this happens?
RECURRENCE RELATIONS : Question 3
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RECURRENCERELATIONS : Question 3
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A man plants a row of fast growing trees between his own house
and his neighbours. These trees are known grow at a rate of 1m
per annum so cannot be allowed to become too high. Hetherefore 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)
Question 3(a) Removing 30% leaves 70% or 0.7
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The trees are known grow at a rate
of1m 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 expectthe trees to grow to in the long
run?
( ) g % %
If Hn is the tree height in year n then
Hn+1 = 0.7Hn + 1
Since -1
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The trees are known grow at a rate
of1m per annum. He therefore
decides to prune 30% from their
height at the beginning of each year.
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
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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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TRIG GRAPHS & EQUATIONS : Question 1
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TRIGGRAPHS & EQUATIONS : Question 1
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This diagram shows the
graph of y = acosbx + c.Determine the values of a, b
& c.
T/2T
y = acosbx + c
TRIG GRAPHS & EQUATIONS : Question 1
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TRIGGRAPHS & EQUATIONS : Question 1
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This diagram shows the
graph of y = acosbx + c.Determine the values of a, b
& c.
T/2T
y = acosbx + c
a = 3
b = 2
c = -1
Question 1
a = (max min)
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This diagram shows the
graph of y = acosbx + c.
Determine the values of a, b& c.
T/2T
y = acosbx
+ c
a = (max min)
= (2 (-4))
= 3=
X6
Period of graph = T so two complete
sections between 0 & 2T ie b = 2
For 3cos() max = 3 & min = -3.
This graph: max = 2 & min = -4.
ie 1 lessso c = -1
Markers CommentsThe values chosen for a,b and c
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a = (max min)
= (2 (-4))
= X 6
Period of graph = T so two complete
sections between 0 & 2T
For 3cos() max = 3 & min = -3.
This graph: max = 2 & min = -4.
ie 1 lessso c = -1
ie b = 2
= 3
must be justified.
Possible justification of a = 3
a = 1/2(max - min)
y = cosx graph stretched by afactor of3 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 down1 unit etc.
T
T
T
T T
TRIG GRAPHS & EQUATIONS : Question 2
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TRIGGRAPHS & EQUATIONS : Question 2
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Solve 3tan2U + 1 = 0 ( where 0 < U < T ).
TRIGGRAPHS & EQUATIONS : Question 2
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Solve 3tan2U + 1 = 0 ( where 0 < U < T ).
U = 5T/12
U = 11T/12
Question 23tan2U + 1 = 0
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Solve 3tan2U + 1 = 0
( where 0 < U < T ).
3tan2U = -1
tan2U = -1/3 Q2 or Q4
tan -1(1/3) =T/6
T - U U
T + U 2T - U
sin all
tan cos
Q2: angle = T - T/6
so 2U = 5T/6
ie U = 5T/12
Q4: angle = 2T - T/6
so 2U = 11T/6
ie U = 11T/12
1
23
T/6
tan2U repeats every T/2 radians but
repeat values are not in interval.
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3tan2U + 1 = 0
3tan2U = -1tan2U = -1/3 Q2 or Q4
tan -1(1/3) =T/6 T - U U
T + U 2T - U
sin all
tan cos
1
23
T
/6
Q2: angle = T - T/6
so 2U = 5T/6
ie U = 5T/12
Q4: angle = 2T
-
T
/6
so 2U = 11T/6
ie U = 11T/12
tan2U repeats every T/2 radians but
repeat values are not in interval.
Solve the equation for tan .2U
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 fromtriangles.
31
45
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
T
T
!
!
o
o
TRIGGRAPHS & EQUATIONS : Question 3
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The diagram shows a the
graph of a sine function
from 0 to 2T/3.
2T/3
P Q
y = 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.
TRIGGRAPHS & EQUATIONS : Question 3
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The diagram shows a the
graph of a sine function
from 0 to 2T/3.
2T/3
P Q
y = 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 (T
/18, 2) and Q is (5T
/18, 2).
Question 3
Th di h th(a) One complete wave from 0 to 2T/3
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2T/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 2T/3.
so 3 waves from 0 to 2T.
Max/min = 4 4sin()
Graph is y = 4sin3x
Continue Solution
(b) The line y = 2 meets the graph
Question 3
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 twopoints.
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2T/3
P
Q
y =
2
so 4sin3x = 2
or sin3x = 1/2
(b) At P & Q y = 4sin3x and y = 2
Q1 or Q2
sin-1(1/2) =T/6
Q1: angle = T/6
so 3x = T/6
ie x = T/18
Q2: angle = T - T/6
so 3x = 5T/6
ie x = 5T/18
T - U U
T+ U 2
T- U
sin all
tan cos
1
23T/6
P is (T/18, 2) and Q is (5T/18, 2).
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(a) One complete wave from 0 to 2T/3
so 3 waves from 0 to 2T.
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 of4
Possible justification for b
Period =
3 waves from 0 to
2 / 3T
2T
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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) =T/6
T - U U
T + U 2T - U
sin all
tan cos
1
23T/6
Q1: angle = T/6
so 3x = T/6
ie x = T/18
Q2: angle = T - T/6
so 3x = 5T/6
ie x = 5T/18
P is (
T
/18, 2) and Q is (
5T
/18, 2).
At intersection y1 = y24sin3x = 2
Solve for sin3x
Use the quadrant rule to find
the solutions.
Must learn special angles or be
able to calculate from triangles
31
45
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
EXIT
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Unit 1 Menu
FUNCTIONS & GRAPHS : Question 1
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This graph shows the
the function y = g(x).
Make a sketch of thegraph of the function
y = 4 g(-x).
y = g(x)
-8 12
(-p,q)
(u,-v)
FUNCTIONS & GRAPHS : Question 1
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This graph shows the
the function y = g(x).
Make a sketch of thegraph 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)
Question 1
Thi h h th
y = 4 g(-x) = -g(-x) + 4
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This graph shows the
the function y = g(x).
Make a sketch of thegraph of the function
y = 4 g(-x).
y = g(x)
-8 12
(-p,q)
(u,-v)
Reflect in
X-axisReflect 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)
Question 1
Thi h h th
(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 thegraph of the function
y = 4 g(-x).
y = g(x)
-8 12
(-p,q)
(u,-v)
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-axisReflect 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-axisReflect 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 onformula 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 throughthem.
(p,-q+4)
(8,4)12,4) (0,4)
(-u,v+4)
y = 4 g(-x)
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
FUNCTIONS & GRAPHS : Question 2
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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
ANSWERTOPART (I)
y = 2ax
- 3
y = ax
(0,-1)
(1,2a-3)
ANSWER to PART (II)
Question 2
Make sketches of the
(I) y = a(x+2)
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graphs of the functions
(I) y = a(x+2)
y = ax
(1,a)
f(x) = ax
so a(x+2) = f(x+2)
move f(x) 2 to left
(0,1)p(-2,1) & (1,a) p(-1,a)
(-1,a)
(-2,1)
y = a(x+2)
y = a
x
Question 2
Make sketches of the
(II) y = 2ax - 3
f(x) = ax
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y = ax
(1,a)
Make sketches of the
graphs of the functions
(II) y = 2ax - 3
f(x) = a
so 2ax - 3 = 2f(x) - 3
double y-coords slide 3 down
(0,1)p(0,2)p(0,-1)
(1,a) p(1,2a) p(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)p(-2,1) & (1,a) p(-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 = x
2 + 3x,
each becomes y = f(x)
In any curve sketchingquestion use a rulerand
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)p(-2,1) & (1,a) p(-1,a)
(-1,a)
(-2,1)
y = a(x+2)
y = ax
Learn Rules: Not given on
formula sheet
f(x) + k Slide k unitsparallel 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
Markers Comments
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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)p(0,2)p(0,-1)
(1,a) p(1,2a) p(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 = x
2 + 3x,
each becomes y = f(x)
In any curve sketchingquestion use a rulerand
annotate the sketch
i.e.label all known
coordinates.
Markers Comments
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Learn Rules: Not given on
formula sheet
f(x) + k Slide k unitsparallel 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)p(0,2)p(0,-1)
(1,a) p(1,2a) p(1,2a-3)
y = 2ax
- 3
y = ax
(0,-1)
(1,2a-3)
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)) .
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)) .
= 2x2 - 1(a) (i)
= 4x2 4x + 1(ii)
Question 3
Two functions f and g are defined
(a)(i) f(g(x))
2
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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)) .
= f(x2)
= 2x2 - 1
(ii) g(f(x))
= g(2x-1)
= (2x 1)2
= 4x2 4x + 1
Continue Solution
Question 3
Two functions f and g are defined
f(g(x)) = 2x2 - 1
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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.
(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)
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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
( )
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): 2x2 - 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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( )
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!
FUNCTIONS & GRAPHS : Question 4
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A function g is defined by the formula g(x) = . 2 (x{1)
(x 1)
(a) Find a formula for h(x) = g(g(x)) in its simplest form.
(b) State a suitable domain for h.
FUNCTIONS & GRAPHS : Question 4
f i i fi f ( ) 2 ( 1)
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A function g is defined by the formula g(x) = . 2 (x{1)
(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}
Question 4(a) g(g(x)) = g( )
. 2 .
(x 1)
A function g is defined by the
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= 2. 2 .
(x 1)
- 1
= 22 - (x 1)
. (x 1)
= 2(3 - x)
.(x 1)
= 2 (x - 1)
.(3 x)
= (2x - 2). . (3 x)
g y
formula g(x) = . 2 (x{1)
(x 1)
(a) Find a formula for
h(x) = g(g(x))
in its simplest form.
Question 4 h(x) = (
2x - 2)
. . (3 x)A function g is defined by the
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formula g(x) = . 2 (x{1)
(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}
Markers Comments
2
(a)
I it f ti bl
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(a) g(g(x)) = g( ). 2 .
(x 1)
= 2. 2 .
(x 1)- 1
= 22 - (x 1)
. (x 1)
= 2(3 - x)
.(x 1)
= 2 (x - 1)
.(3 x)
= (2x - 2)
. . (3 x)
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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(b)
In finding a suitable domain it
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h(x) = (2x - 2)
. . (3 x)
(b) For domain 3 - x { 0
Domain = {x R: x { 3}
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.
HIGHER ADDITIONAL QUESTION BANK
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HIGHER ADDITIONAL QUESTION BANK
UNIT 2 :
Integration
Polynomials
The Circle
AdditionFormulae
Quadratics
EXIT
HIGHER ADDITIONAL QUESTION BANK
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HIGHER ADDITIONAL QUESTION BANK
UNIT 2 :Polynomials
You have chosen to study:
Please choose a question to attempt from the following:
1 2 3 4
EXITBack to
Unit 2 Menu
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.
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
Question 1
Show that x = 3 is a root of theUsing the nested method -
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equation x3 + 3x2 10x 24 = 0.
Hence find the other roots.
coefficients are 1, 3, -10, -24
f(3) = 3 1 3 -10 -243 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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State clearly in solution that
f( ) i
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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: x
2
+ 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.
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 factorisedexpression = 0
List all the roots of the
polynomial
x = 3, x = -4, x = -2
Given that (x + 4) is a factor of the polynomial
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.
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Given that (x + 4) is a factor of the polynomial
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.
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Reveal answer only
EXIT
k = -15
So full solution of equation is
x = -4 or x = 1/3 or x = 1
Question 2
Given that (x + 4) is a factor of
Since (x + 4) a factor then f(-4) = 0 .
N i th t d th d
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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.
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
Question 2
Given that (x + 4) is a factor of
If k = -15 then we now have
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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.
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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The working in the nested
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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
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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Simply making f(-4) = 0
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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
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 h h Show completed factorisation
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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
p
of the cubic:
(x + 4)(3x - 1)(x - 1) = 0
POLYNOMIALS : Question 3
Given that f(x) = 6x3 + 13x2 - 4 show that (x + 2) is a factor of
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EXIT
Given that f(x) 6x 13x 4 show that (x 2) is a factor of
f(x).
Hence express f(x) in its fully factorised form.
POLYNOMIALS : Question 3
Given that f(x) = 6x3 + 13x2 - 4 show that (x + 2) is a factor of
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( ) ( )
f(x).
Hence express f(x) in its fully factorised form.
6x3 + 13x2 - 4 = (3x + 2)(2x - 1)(x + 2)
Question 3
Given that f(x) = 6x3 + 13x2 - 4
Using the nested method -
coefficients are 6, 13, 0, -4
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show that (x + 2) is a factor
of f(x).
Hence express f(x) in its fully
factorised form.
f(-2) = -2 6 13 0 -4
6
-12
1
-2
-2
4
0
f(-2) = 0 so
(x + 2) is a factor
Question 3
Given that f(x) = 6x3 + 13x2 - 4
Using the nested method -
coefficients are 6, 13, 0, -4
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show that (x + 2) is a factor
of f(x).
Hence express f(x) in its fully
factorised form.
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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State clearly in solution that
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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
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
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 stillneed nested method for
quadratic factor.
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POLYNOMIALS : Question 4A busy road passes through several small villages so it is
decided to build a by pass to reduce the volume of traffic
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(a) Find the coordinates ofP and the equation of the bypass
PQ.
(b) Hence find the coordinates of Q the point where the
bypass rejoins the original road.
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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 thiscurve 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)
Question 4
y = -x3 + 6x2 3x 10
(a) At point P, x = 4 so using the
equation of the curve we get ..
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P
Q
y = -x3 + 6x2 3x 10
4
(a) Find the coordinates ofP
and the equation of the
bypass PQ.
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
Question 4
y = -x3 + 6x2 3x 10
P is (4,10) dy/dx = -3
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P
Q
y = -x3 + 6x2 3x 10
4
(a) Find the coordinates ofP
and the equation of the
bypass PQ.
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
Question 4
y = -x3 + 6x2 3x 10
(b) The tangent & curve meet whenever
y = -3x + 22 and y = -x3 + 6x2 3x 10(b)Hence find the coordinates of Q
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P
Q
y = -x3 + 6x2 3x 10
4
ie -3x + 22 = -x3 + 6x2 3x 10
or x3 - 6x2 + 32 = 0
(b)Hence find the coordinates of Q
the point where the bypass rejoins
the 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
Question 4
y = -x3 + 6x2 3x 10
(b) The
(b)Hence find the coordinates of Q
other factor is x2 2x - 8
= (x 4)(x + 2)
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P
Q
y = -x3 + 6x2 3x 10
4
(b)Hence find the coordinates of Q
the point where the bypass rejoins
the original road. Solving (x 4)(x + 2) = 0we 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
(a)
(a) At point P x = 4 so using the
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find gradient.
Learn rule:
Multiply by the power
then reduce the power by 1
(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 Use :
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( , ) 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 At intersection y = y
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