higher revision 1, 2 & 3

8/8/2019 Higher Revision 1, 2 & 3

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HIGHER ADDITIONAL QUESTION BANK

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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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UNIT 1 :

Functions &Graphs

Straight Line

RecurrenceRelations

BasicDifferentiation

Trig Graphs& Equations

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UNIT 1 :Straight Line

You have chosen to study:

Please choose a question to attempt from the following:

1 2 3 4 5

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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).

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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).

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Reveal answer only y = -5/3x - 6

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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

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 (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

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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 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.

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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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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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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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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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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

!


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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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


16/400


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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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are P(4,-5), Q(2,3) and R(10-1).

X

Y

P(4,-5)

Q(2,3)

R(10,-1)

Find



(b) the equation of the line f, the

perpendicular bisector of QR.



y = -1(a)

y = 2x 11(b)

(5,-1)(c)


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Question 4 (a)

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are P(4,-5), Q(2,3) and R(10-1).

X

Y

P(4,-5)

Q(2,3)

R(10,-1)

Find



(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)

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(b) the equation of the line f,

the perpendicular bisector of QR.


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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are P(4,-5), Q(2,3) and R(10-1).

Find



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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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:


24/400


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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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E(6,-3), F(12,-5) and G(2,-5).Find


from vertex E.


from vertex F.



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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E(6,-3), F(12,-5) and G(2,-5).

Find


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)

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E(6,-3), F(12,-5) and G(2,-5).

Find

X

Y

G(2,-5)

E(6,-3)

F(12,-5)


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)

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E(6,-3), F(12,-5) and G(2,-5).

Find

X

Y

G(2,-5)

E(6,-3)

F(12,-5)



(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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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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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:


32/400


UNIT 1 : BasicDifferentiation



1 2 3 4 5

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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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Find the equation of the tangent to the curve (x>0)


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)


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

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Find the equation of the tangent to

the curve

y = x 16

x (x>0)


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

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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

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If x = 4 then:

= 1 + 16

24 16= + 1 = 5/4


so m =5

/4 .

We now use y b = m(x a)

this gives us y (-2) = 5/4(x 4)

or y + 2 =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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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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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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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

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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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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

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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)


45/400


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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


46/400


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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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Return to Nature Table

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 !



' dyf ( )

dxx !


dxx ! !


(We can ignore r.)


either side of 0 and p:

new y-values

x p 0 p p p

gd(x) - 0 - 0 +

Continue Comments

+

- -

x

y

a

f ( )x

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' dyf ( )

dxx !


dxx ! !

x

y

0p

y = gd(x)


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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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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y = x3 - 3x2 - 9x + 2.




y = x3 - 3x2 - 9x + 2

(-1,7) is a maximum TP and

(3,-25) is a minimum TP


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y = x3 - 3x2 - 9x + 2.




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


y = x3 - 3x2 - 9x + 2.




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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Question 4


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y = x3 - 3x2 - 9x + 2.




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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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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.


59/400


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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.


60/400


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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


61/400

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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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62/400

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S(x) = 2 - 4

x x2(x u 2)



achieve.




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

Question 5


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S(x) = 2 - 4

x x2(x u 2)



achieve.




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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


67/400


UNIT 1 : RecurrenceRelations



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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A recurrence relation is defined by the formula un+1 = aun + b,

where -1


70/400

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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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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


74/400


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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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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


76/400

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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


77/400

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un+1 = -5kun + 3.


Vn+1 = k2vn + 1.


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


78/400

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un+1 = -5kun + 3.


Vn+1 = k2vn + 1.


limit.

Since -1


79/400

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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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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


83/400


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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


84/400

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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


85/400

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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.


88/400


UNIT 1 : Trig Graphs& Equations



1 2 3

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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



90/400


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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)


91/400

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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

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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



93/400


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Solve 3tan2U + 1 = 0 ( where 0 < U < T ).



94/400

Q Q

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Solve 3tan2U + 1 = 0 ( where 0 < U < T ).

U = 5T/12

U = 11T/12

Question 23tan2U + 1 = 0


95/400

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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



97/400

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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.



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from 0 to 2T/3.

2T/3

P Q

y = 2



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




from 0 to 2T/3.

so 3 waves from 0 to 2T.

Max/min = 4 4sin()

Graph is y = 4sin3x

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Question 3

Graph is y = 4sin3x


100/400


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.


103/400


UNIT 1 : Functions& Graphs



1 2 3 4

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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)



105/400

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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


106/400

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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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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)



111/400

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y = ax

(1,a)


the function y = ax.

Make sketches of the

graphs of the functions

(I) y = a(x+2)

(II) y = 2ax - 3



112/400

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the function y = ax.



(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


(I) y = a(x+2)


113/400

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(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


(II) y = 2ax - 3

f(x) = ax


114/400

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y = ax

(1,a)



(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


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


= k

-f(x) Reflect in the x-axis




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(II) y = 2ax - 3

f(x) = ax

so 2ax - 3 = 2f(x) - 3


(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)


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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Learn Rules: Not given on

formula sheet

f(x) + k Slide k unitsparallel to y-axis


= k

-f(x) Reflect in the x-axis




(II) y = 2ax - 3

f(x) = ax

so 2ax - 3 = 2f(x) - 3


(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)



119/400

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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)) .



120/400

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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)

Question 3

Two functions f and g are defined

(a)(i) f(g(x))

2


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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


f(g(x)) = 2x2 - 1


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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!



125/400

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EXIT

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.


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.


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) For domain 3 - x { 0


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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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In finding a suitable domain it


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h(x) = (2x - 2)

. . (3 x)

(b) For domain 3 - x { 0


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.



131/400


UNIT 2 :

Integration

Polynomials

The Circle

AdditionFormulae

Quadratics

EXIT



132/400


UNIT 2 :Polynomials



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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Show that x = 3 is a root of the equation x3 + 3x2 10x 24 = 0.


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.


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

Markers Comments

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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



137/400


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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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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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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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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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 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


Given that f(x) = 6x3 + 13x2 - 4 show that (x + 2) is a factor of


144/400

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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.


Given that f(x) = 6x3 + 13x2 - 4 show that (x + 2) is a factor of


145/400

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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


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




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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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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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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.


150/400

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


151/400

(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.

Go to full solution

EXIT

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.


Go to Polynomial Menu Go to Main Menu

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)


= 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


other factor is x2 2x - 8

= (x 4)(x + 2)


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P

Q

y = -x3 + 6x2 3x 10

4


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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(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)


= 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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