HIGHER – ADDITIONAL QUESTION BANK EXIT UNIT 1UNIT 2UNIT 3 Please decide which Unit you would like...

HIGHER – ADDITIONAL QUESTION BANK

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UNIT 1 UNIT 2 UNIT 3

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Straight LineFunctions & GraphsTrig Graphs & EquationsBasic DifferentiationRecurrence Relations

PolynomialsQuadratic Functions IntegrationAddition FormulaeThe Circle

VectorsFurther CalculusExponential / Logarithmic FunctionsThe Wave Function


UNIT 1 :

Functions & Graphs

Straight Line

Recurrence Relations

BasicDifferentiation

Trig Graphs & Equations

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UNIT 1 :Straight Line

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STRAIGHT LINE : Question 1

Find the equation of the straight line which is perpendicular to the line with equation 3x – 5y = 4 and which passes through the point (-6,4).

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Find the equation of the straight line which is perpendicular to the line with equation 3x – 5y = 4 and which passes through the point (-6,4).

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

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3x – 5y = 4

3x - 4 = 5y

5y = 3x - 4 (5)

y = 3/5x - 4/5

Using y = mx + c , gradient of line is 3/5

So required gradient = -5/3 , ( m1m2 = -1)

We now have (a,b) = (-6,4) & m = -5/3 .

Using y – b = m(x – a)

We get y – 4 = -5/3 (x – (-6))

y – 4 = -5/3 (x + 6)

y – 4 = -5/3x - 10

y = -5/3x - 6

Question 1

Find the equation of the

straight line which is

perpendicular to the line with

equation 3x – 5y = 4 and

which passes through

the point (-6,4).

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3x – 5y = 4

3x - 4 = 5y

5y = 3x - 4 (5)

y = 3/5x - 4/5

Using y = mx + c , gradient of line is 3/5

So required gradient = -5/3 , ( m1m2 = -1)

We now have (a,b) = (-6,4) & m = -5/3 .


We get y – 4 = -5/3 (x – (-6))

y – 4 = -5/3 (x + 6)

y – 4 = -5/3x - 10

y = -5/3x - 6

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

•State the gradient clearly.

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

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

m1 = 35 m2 =

-5 3

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

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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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Find the equation of the straight line which is parallel to the line with equation 8x + 4y – 7 = 0 and which passes through the point (5,-3).

y = -2x + 7

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Question 2

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8x + 4y – 7 = 0

4y = -8x + 7 (4)

y = -2x + 7/4

y = -2x + 7

Using y = mx + c , gradient of line is -2

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

We now have (a,b) = (5,-3) & m = -2.


We get y – (-3) = -2(x – 5)

y + 3 = -2x + 10


straight line which is parallel

to the line with equation

8x + 4y – 7 = 0 and which

passes through the point

(5,-3).

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

• State the gradient clearly.

• State the condition for parallel lines m1 = m2

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

8x + 4y – 7 = 0

4y = -8x + 7 (4)

y = -2x + 7/4

y = -2x + 7

Using y = mx + c , gradient of line is -2

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

We now have (a,b) = (5,-3) & m = -2.


We get y – (-3) = -2(x – 5)

y + 3 = -2x + 10

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In triangle ABC, A is (2,0),

B is (8,0) and C is (7,3).

(a) Find the gradients of AC and BC.

(b) Hence find the size of ACB. X

Y

A B

C

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B is (8,0) and C is (7,3).

(a) Find the gradients of AC and BC.

(b) Hence find the size of ACB. X

Y

A B

C

= 77.4°(b)

mAC = 3/5

mBC = - 3

(a)

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(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 of AC

and BC.

(b) Hence find the size of ACB. (b) Using tan = gradient

If tan = 3/5 then CAB = 31.0°

If tan = -3 then CBX = (180-71.6)°

= 108.4 o

Hence :

ACB = 180° – 31.0° – 71.6°

= 77.4°

so ABC = 71.6°

X

Y

A B

C

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(a) Using the gradient formula:

mAC = 3 – 0

7 - 2

mBC = 3 – 0 7 - 8

= - 3

2 1

2 1

y ym

x x

(b) Using tan = gradient

= 3/5

If tan = 3/5 then CAB = 31.0°

then CBX = (180-71.6)°

= 108.4 o

Hence :

ACB = 180° – 31.0° – 71.6°

= 77.4°

If tan = -3

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

• In calculating gradients state the gradient formula.

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

A

B

Ø °

mAB = tanØ °

Ø ° = tan-1 mABso ABC = 71.6°


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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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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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X

Y

P(4,-5)

Q(2,3)

R(10,-1)

Find


(a) Midpoint of PQ is (3,-1): let’s call this S

Using the gradient formula m = y2 – y1

x2 – x1

mSR = -1 – (-1)

10 - 3

Since it passes through (3,-1)

equation of e is y = -1

= 0 (ie line is horizontal)

Solution to 4 (b)

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Question 4 (b)

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(b) the equation of the line f,

the perpendicular bisector of QR.

In triangle PQR the vertices are P(4,-5), Q(2,3) and R(10-1). Find

X

Y

P(4,-5)

Q(2,3)

R(10,-1)

(b) Midpoint of QR is (6,1)

mQR = 3 – (-1)

2 - 10 = 4/-8 = - 1/2

required gradient = 2 (m1m2 = -1)

Using y – b = m(x – a) with (a,b) = (6,1)

& m = 2

we get y – 1 = 2(x – 6)

so f is y = 2x – 11

Solution to 4 (c)

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Question 4 (c)

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In triangle PQR the vertices are P(4,-5), Q(2,3) and R(10-1). Find


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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Q

P

R

y

x

median

Perpendicular bisector

(a) Midpoint of PQ is (3,-1): let’s call this S

Using the gradient formula m = y2 – y1

x2 – x1

mSR = -1 – (-1)

10 - 3

Since it passes through (3,-1)

equation of e is y = -1

(ie line is horizontal)

Comments for 4 (b)

•Sketch the median and the perpendicular bisector

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Q

P

R

y

x

(b) Midpoint of QR is (6,1)

mQR = 3 – (-1)

2 - 10 = 4/-8

required gradient = 2 (m1m2 = -1)

Using y – b = m(x – a) with (a,b) = (6,1)

& m = 2

we get y – 1 = 2(x – 6)

so f is y = 2x – 11

= - 1/2

• To find midpoint of QR

2 + 10 3 + (-1) 2 2

,

• Look for special cases:

Horizontal lines in the form y = kVertical lines in the form x = k

Comments for 4 (c)

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(c) e & f meet when y = -1 & y = 2x -11

so 2x – 11 = -1

ie 2x = 10

ie x = 5

Point of intersection is (5,-1)

y = -1y = 2x - 11

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


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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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Find




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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Find


XY

G(2,-5)

E(6,-3)

F(12,-5)

(a) Using the gradient formula 2 1

2 1

y ym

x x

mFG = -5 – (-5)

12 - 2 = 0

(ie line is horizontal so altitude is vertical)

Altitude is vertical line through (6,-3)

ie x = 6

Solution to 5 (b)

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Question 5(b)

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Find

XY

G(2,-5)

E(6,-3)

F(12,-5)


(b) Midpoint of EG is (4,-4)- let’s call this H

mFH = -5 – (-4)

12 - 4 = -1/8

Using y – b = m(x – a) with (a,b) = (4,-4)

& m = -1/8

we get y – (-4) = -1/8(x – 4) (X8)

or 8y + 32 = -x + 4

Median is x + 8y + 28 = 0

Solution to 5 (c)

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Question 5(c)

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Find

XY

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 = 0

ie 8y = -34

ie y = -4.25

Point of intersection is (6,-4.25)

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• Sketch the altitude and the median.

y

x

F

E

Gmedian

altitude

(a) Using the gradient formula 2 1

2 1

y ym

x x

mFG = -5 – (-5)

12 - 2 = 0

(ie line is horizontal so altitude is vertical)

Altitude is vertical line through (6,-3)

ie x = 6

Comments for 5 (b)

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y

x

F

E

G

Comments for 5 (c)

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

mFH = -5 – (-4)

12 - 4 = -1/8

Using y – b = m(x – a) with (a,b) = (4,-4)

& m = -1/8

we get y – (-4) = -1/8(x – 4) (X8)

or 8y + 32 = -x + 4

Median is x + 8y + 28 = 0

• To find midpoint of EG

2 + 6 -3 + (-5) 2 2

,H

Horizontal lines in the form y = kVertical lines in the form x = k

• Look for special cases:

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(c)

Lines meet when x = 6 & x + 8y + 28 = 0

put x =6 in 2nd equation 8y + 34 = 0

ie 8y = -34

ie y = -4.25

Point of intersection is (6,-4.25)

x = 6x + 8y = -28

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


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.

Point

If x = 4 then y = 4 – 16 4

= 2 – 4 = -2

so (a,b) = (4,-2)

Gradient: y = x – 16 x

= x1/2 – 16x -1

dy/dx = 1/2x-1/2 + 16x-2 = 1 + 16

2x x2

If x = 4 then: dy/dx = 1 + 16

24 16

= ¼ + 1 = 5/4

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Question 1


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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 24 16

= ¼ + 1 = 5/4

Gradient of tangent = gradient of curve

so m = 5/4 .

We now use y – b = m(x – a)

this gives us y – (-2) = 5/4(x – 4)

or y + 2 = 5/4x – 5

or y = 5/4x – 7

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• Prepare expression for differentiation. 1

1216

y = 16x x xx

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

Point

If x = 4 then y = 4 – 16 4

= 2 – 4 = -2

so (a,b) = (4,-2)

Gradient: y = x – 16 x

= x1/2 – 16x -1

dy/dx = 1/2x-1/2 + 16x-2 = 1 + 16

2x x2

If x = 4 then:

= 1 + 16 24 16

= ¼ + 1 = 5/4

• Find gradient of the tangent

using rule:

dy

dx

“multiply by the power and reduce the power by 1”

dy/dx

• Find gradient = at x = 4.dy

dx

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If x = 4 then:

= 1 + 16 24 16

= ¼ + 1 = 5/4


so m = 5/4 .

We now use y – b = m(x – a)

this gives us y – (-2) = 5/4(x – 4)

or y + 2 = 5/4x – 5

dy/dx

or y = 5/4x – 7

16 16y 4 2

4x

x

• Find y coordinate at x = 4 using:

• Use m = 5/4 and (4,-2) in

y - b = m(x - a)


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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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Find the coordinates of the point on the curve y = x2 – 5x + 10 where the tangent to the curve makes an angle of 135° with the positive direction of the X-axis.

(2,4)

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Find the coordinates of the point

on the curve y = x2 – 5x + 10

where the tangent to the curve

makes an angle of 135° with the

positive direction of the X-axis.

NB: gradient of line = gradient of curve

Line

Using gradient = tan

we get gradient of line = tan135°

= -tan45°

= -1Curve

Gradient of curve = dy/dx = 2x - 5

It now follows that

2x – 5 = -1

Or 2x = 4

Or x = 2

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Question 2


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Find the coordinates of the point

on the curve y = x2 – 5x + 10

where the tangent to the curve

makes an angle of 135° with the

positive direction of the X-axis.

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Using y = x2 – 5x + 10 with x = 2

we get y = 22 – (5 X 2) + 10

ie y = 4

So required point is (2,4)

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NB: gradient of line = gradient of curve

Line

Using gradient = tan

we get gradient of line = tan135°

= -tan45°

= -1Curve

Gradient of curve = dy/dx = 2x - 5

It now follows that

2x – 5 = -1

Or 2x = 4

Or x = 2

• Find gradient of the tangent

using rule:

dy

dx


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

m = tan135 ° = -1

y

x135 °

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• Set m =

i.e. 2x - 5 = -1

and solve for x.

dy

dxIt now follows that

2x – 5 = -1

Or 2x = 4

Or x = 2

Using y = x2 – 5x + 10 with x = 2

we get y = 22 – (5 X 2) + 10

ie y = 4

So required point is (2,4)


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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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The graph of y = g(x) is shown here.

There is a point of inflection at the origin, a minimum turning point at (p,q) and the graph also cuts the X-axis at r.

Make a sketch of the graph of y = g(x).

y = g(x)

(p,q)

r

x

y

y = g(x)

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Stationary points occur at x = 0 and x = p.

(We can ignore r.)

We now consider the sign of the gradient

either side of 0 and p:

new y-values

x 0 p

g(x) - 0 - 0 +

Click for graph

y = g(x)

(p,q)

r

Make a sketch of the graph of

y = g(x).

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Return to Nature Table

y = g(x)

(p,q)

r

Make a sketch of the graph of

y = g(x).

x

y

0p

y = g(x)

This now gives us the following graph

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To sketch the graph ofthe gradient function:

' dyf ( )

dxx

1 Mark the stationary points on the x axis i.e. ' dy

f ( ) 0dx

x

x

y

a

'f ( )x

x

y

0p

y = g(x)


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2 For each interval decide if the value of ' dy

f ( ) is - or +dx

x


' dyf ( )

dxx


f ( ) 0dx

x


(We can ignore r.)


either side of 0 and p:

new y-values

x 0 p

g(x) - 0 - 0 +

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+

- -x

y

a

'f ( )x

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' dyf ( )

dxx


f ( ) 0dx

x

x

y

0p

y = g(x)


3 Draw in curve to fit information

2 For each interval decide if the value of ' dy

f ( ) is - or +dx

x

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

+

- -

x

y

a

'f ( )x


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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

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


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either side of -1 and 3.

x -1 3

(x + 1) - 0 + + +

(x - 3) - - - 0 +

dy/dx + 0 - 0 +

Hence (-1,7) is a maximum TP

and (3,-25) is a minimum TP

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• Must attempt to find

and set equal to zero

dy

dx

SPs occur where dy/dx = 0

ie 3x2 – 6x – 9 = 0

ie 3(x2 – 2x – 3) = 0

ie 3(x – 3)(x + 1) = 0

ie x = -1 or x = 3

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

when x = -1

y = -1 – 3 + 9 + 2 = 7

& when x = 3

y = 27 – 27 - 27 + 2 = -25

So stationary points are at

(-1,7) and (3,-25)

“multiply by the power andreduce the power by 1”

• Make the statement:“At stationary points “dy

0dx

• Find the value of y from y = x3 -3x2-9x+2 not from dy

dx

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either side of -1 and 3.

x -1 3

(x + 1) - 0 + + +

(x - 3) - - - 0 +

dy/dx + 0 - 0 +

Hence (-1,7) is a maximum TP

and (3,-25) is a minimum TP

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

x -1

dy

dx + 0 -

Minimum requirement

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


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When a company launches a new product its share of the market after x months is calculated by the formula

S(x) = 2 - 4

x x2 (x 2)

So after 5 months the share is S(5) = 2/5 – 4/25 = 6/25

Find the maximum share of the market that the company can achieve.


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When a company launches a new product its share of the market after x months is calculated by the formula

S(x) = 2 - 4

x x2 (x 2)

So after 5 months the share is S(5) = 2/5 – 4/25 = 6/25

Find the maximum share of the market that the company can achieve.

= 1/4

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End points

S(2) = 1 – 1 = 0

There is no upper limit but as x

S(x) 0. S(x) = 2 - 4

x x2 (x 2)

Find the maximum share of the

market that the company can

achieve.

When a company launches a new

product its share of the market

after x months is calculated as:

Stationary Points

S(x) = 2 - 4

x x2 = 2x-1 – 4x-2

So S (x) = -2x-2 + 8x-3 = - 2 + 8

x2 x3 = 8 - 2

x3 x2

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S(x) = 2 - 4

x x2 (x 2)



achieve.




SPs occur where S (x) = 0

8 - 2 x3 x2

or8 = 2 x3 x2

( cross mult!)

8x2 = 2x3

8x2 - 2x3 = 0

2x2(4 – x) = 0

x = 0 or x = 4

NB: x 2In required interval

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= 0

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S(x) = 2 - 4

x x2 (x 2)



achieve.




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We now check the gradients either side of

X = 4

x 4

S (x) + 0 -

S (3.9 ) = 0.00337…

S (4.1) = -0.0029…

Hence max TP at x = 4

So max share of market

= S(4) = 2/4 – 4/16

= 1/2 – 1/4

= 1/4

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End points

S(2) = 1 – 1 = 0

There is no upper limit but as x

S(x) 0.

Stationary Points

S(x) = 2 - 4

x x2 = 2x-1 – 4x-2

So S (x) = -2x-2 + 8x-3 = - 2 + 8

x2 x3 = 8 - 2

x3 x2

• Must consider end points and stationary points.

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

Maximum, minimum, greatest , least etc.

• Prepare expression for differentiation.

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• Must attempt to find

( Note: No marks are allocated for trial and error solution.)

dyS (x)

dx

SPs occur where S (x) = 0

8 - 2 x3 x2

or8 = 2 x3 x2

( cross mult!)

8x2 = 2x3

= 0

8x2 - 2x3 = 0

2x2(4 – x) = 0

x = 0 or x = 4

NB: x 2In required interval


• Must attempt to find and set equal to zero

dy

dx

• Usually easier to solve resulting equation using cross-multiplication.

• Take care to reject “solutions” outwith the domain.

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We now check the gradients either side of

X = 4

x 4

S (x) + 0 -

S (3.9 ) = 0.00337…

S (4.1) = -0.0029…

Hence max TP at x = 4

So max share of market

= S(4) = 2/4 – 4/16

= 1/2 – 1/4

= 1/4

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

x 4

S (x) + 0 -

Minimum requirement.

• State clearly: Maximum T.P at x = 4


UNIT 1 : RecurrenceRelations



1 2 3

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A recurrence relation is defined by the formula un+1 = aun + b, where -1<a<1 and u0 = 20.

(a) If u1 = 10 and u2 = 4 then find the values of a and b.

(b) Find the limit of this recurrence relation as n .


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A recurrence relation is defined by the formula un+1 = aun + b, where -1<a<1 and u0 = 20.

(a) If u1 = 10 and u2 = 4 then find the values of a and b.

(b) Find the limit of this recurrence relation as n .

a = 0.6

b = -2

(a)

L = -5

(b)

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A recurrence relation is defined

by the formula un+1 = aun + b,

where -1<a<1 and u0 = 20.

(a) If u1 = 10 and u2 = 4 then

find the values of a and b.

(b) Find the limit of this recurrence

relation as n .

Using un+1 = aun + b

we get u1 = au0 + b

and u2 = au1 + b

Replacing u0 by 20, u1 by 10 & u2 by 4

gives us 20a + b = 10

and 10a + b = 4

subtract 10a = 6

or

(a)

Replacing a by 0.6 in 10a + b = 4

gives 6 + b = 4

or b = -2

a = 0.6Continue Solution

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A recurrence relation is defined

by the formula un+1 = aun + b,

where -1<a<1 and u0 = 20.

(a) If u1 = 10 and u2 = 4 then

find the values of a and b.

(b) Find the limit of this recurrence

relation as n .

(b)

L = -5

un+1 = aun + b is now un+1 = 0.6un - 2

This has a limit since -1<0.6<1

At this limit, L, un+1 = un = L

So we now have L = 0.6 L - 2

or 0.4L = -2

or L = -2 0.4

or L = -20 4so

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Using un+1 = aun + b

we get u1 = au0 + b

and u2 = au1 + b

Replacing u0 by 20, u1 by 10 & u2 by 4

gives us 20a + b = 10

and 10a + b = 4

subtract 10a = 6

or

(a)

Replacing a by 0.6 in 10a + b = 4

gives 6 + b = 4

or b = -2

a = 0.6

• Must form the two simultaneous equations and solve.

• u1 is obtained from u0 and u2 is obtained from u1 .

• A trial and error solution would only score 1 mark.

Comments for 1(b)

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(b)

un+1 = aun + b is now un+1 = 0.6un - 2

This has a limit since -1<0.6<1

At this limit, L, un+1 = un = L

So we now have L = 0.6 L - 2

or 0.4L = -2

or L = -2 0.4

or L = -20 4so L = -5

• Must state condition for limit i.e. -1 < 0.6 < 1

• At limit L, state un+1 = un = L

• Substitute L for un+1 and un

and solve for L.


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Two different recurrence relations are known to have the same limit as n . The first is defined by the formula un+1 = -5kun + 3. The second is defined by the formula vn+1 = k2vn + 1. Find the value of k and hence this limit.


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Two different recurrence relations are known to have the same limit as n . The first is defined by the formula un+1 = -5kun + 3. The second is defined by the formula vn+1 = k2vn + 1. Find the value of k and hence this limit.

k = 1/3

L = 9/8

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Two different recurrence relations

are known to have the same limit

as n . The first is defined by the formula

un+1 = -5kun + 3.

The second is defined by

Vn+1 = k2vn + 1.

Find the value of k and hence this

limit.

If the limit is L then as n we have

un+1 = un = L and vn+1 = vn = L

First Sequence

un+1 = -5kun + 3

becomes

L + 5kL = 3

L(1 + 5k) = 3

L = 3 . . (1 + 5k)

L = -5kL + 3

Second Sequence

vn+1 = k2vn + 1

becomesL = k2L + 1 L - k2L = 1

L(1 - k2) = 1

L = 1 . . (1 - k2)

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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 – k2)

Cross multiply to get 1 + 5k = 3 – 3k2

This becomes 3k2 + 5k – 2 = 0

Or (3k – 1)(k + 2) = 0

So k = 1/3 or k = -2

Since -1<k<1 then k = 1/3

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un+1 = -5kun + 3.


Vn+1 = k2vn + 1.


limit.


Using L = 1 . . (1 - k2)

gives us L = . 1 . . (1 – 1/9)

or L = 1 8/9

ie L = 9/8

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If the limit is L then as n we have

un+1 = un = L and vn+1 = vn = L

First Sequence

un+1 = -5kun + 3

becomes

L + 5kL = 3

L(1 + 5k) = 3

L = 3 . . (1 + 5k)

L = -5kL + 3

Second Sequence

vn+1 = k2vn + 1

becomesL = k2L + 1 L - k2L = 1

L(1 - k2) = 1

L = 1 . . (1 - k2)

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

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L = 1 . . (1 - k2)

It follows that

L = 3 . . (1 + 5k)

. 3 . = . 1 . . (1 + 5k) (1 – k2)

Cross multiply to get 1 + 5k = 3 – 3k2

This becomes 3k2 + 5k – 2 = 0

Or (3k – 1)(k + 2) = 0

So k = 1/3 or k = -2


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

• Only one way to solve resulting equation i.e. terms to the left, form the quadratic and factorise.

• State clearly the condition for the recurrence relation to approach a limit. -1< k < 1.

• Take care to reject the “solution” which is outwith the range.

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Using L = 1 . . (1 - k2)

gives us L = . 1 . . (1 – 1/9)

or L = 1 8/9

ie L = 9/8

Find L from either formula.


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A man plants a row of fast growing trees between his own house and his neighbour’s. These trees are known grow at a rate of 1m per annum so cannot be allowed to become too high. He therefore decides to prune 30% from their height at the beginning of each year.

(a) Using the 30% pruning scheme what height should he expect the trees to grow to in the long run?

(b) The neighbour is concerned at the growth rate and asks that the trees be kept to a maximum height of 3m. What percentage should be pruned to ensure that this happens?


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A man plants a row of fast growing trees between his own house and his neighbour’s. These trees are known grow at a rate of 1m per annum so cannot be allowed to become too high. He therefore decides to prune 30% from their height at the beginning of each year.

(a) Using the 30% pruning scheme what height should he expect the trees to grow to in the long run?

(b) The neighbour is concerned at the growth rate and asks that the trees be kept to a maximum height of 3m. What percentage should be pruned to ensure that this happens?

Height of trees in long run is 31/3m.

331/3% (b)

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The trees are known grow at a rate

of 1m per annum. He therefore

decides to prune 30% from their

height at the beginning of each year.

(a) Using the 30% pruning scheme

what height should he expect

the trees to grow to in the long

run?

(a) Removing 30% leaves 70% or 0.7

If Hn is the tree height in year n then

Hn+1 = 0.7Hn + 1

Since -1<0.7<1 this sequence has a limit, L.

At the limit Hn+1 = Hn = L

So L= 0.7L + 1

or 0.3 L = 1

ie L = 1 0.3 = 10 3 = 31/3


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The trees are known grow at a rate

of 1m per annum. He therefore

decides to prune 30% from their

height at the beginning of each year.

(b) If fraction left after pruning is a and

we need the limit to be 3

then we have 3 = a X 3 + 1

or 3a = 2

or a = 2/3

This means that the fraction pruned is

1/3 or 331/3%

(b) The neighbour asks that

the trees be kept to a maximum

height of 3m. What percentage should be pruned to ensure that this happens?

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(a) Removing 30% leaves 70% or 0.7

If Hn is the tree height in year n then

Hn+1 = 0.7Hn + 1

Since -1<0.7<1 this sequence has a limit, L.

At the limit Hn+1 = Hn = L

So L= 0.7L + 1

or 0.3 L = 1

ie L = 1 0.3 = 10 3


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

H0 = 1 (any value acceptable)H1 = 0.7 x1 + 1 = 1.7 H2 = 0.7 x1.7 + 1 = 2.19 etc.

• State the recurrence relation, with the starting value. Hn+1 = 0.7 Hn + 1, H0 = 1

• State the condition for the limit-1< 0.7< 1

• At limit L, state Hn+1 = Hn = L

• Substitute L for Hn+1 and Hn and solve for L.

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(b) If fraction left after pruning is a and

we need the limit to be 3

then we have 3 = a X 3 + 1

or 3a = 2

or a = 2/3

This means that the fraction pruned is

1/3 or 331/3%

• Since we know the limit we are working backwards to %.

L = 0.7L + 1

New limit, L = 3 and multiplier a

3 = a x3 + 1 etc.

• Take care to subtract from 1 to get fraction pruned.


UNIT 1 : Trig Graphs& Equations



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TRIG GRAPHS & EQUATIONS : Question 1

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This diagram shows the graph of y = acosbx + c.

Determine the values of a, b & c.

/2

y = acosbx + c


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/2

y = acosbx + c

a = 3

b = 2

c = -1

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/2

y = acosbx + c

a = ½(max – min)

= ½(2 – (-4))

= 3

= ½ X 6

Period of graph = so two complete

sections between 0 & 2

ie b = 2

For 3cos(…) max = 3 & min = -3.

This graph: max = 2 & min = -4.

ie 1 lessso c = -1

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a = ½(max – min)

= ½(2 – (-4))

= ½ X 6

Period of graph = so two complete

sections between 0 & 2

For 3cos(…) max = 3 & min = -3.

This graph: max = 2 & min = -4.

ie 1 lessso c = -1

ie b = 2

= 3

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


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Solve 3tan2 + 1 = 0 ( where 0 < < ).


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Solve 3tan2 + 1 = 0 ( where 0 < < ).

= 5/12

= 11/12

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Solve 3tan2 + 1 = 0

( where 0 < < ).

3tan2 + 1 = 0

3tan2 = -1

tan2 = -1/3 Q2 or Q4

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

+ 2 -

sin all

tan cos

Q2: angle = - /6

so 2 = 5/6 ie = 5/12

Q4: angle = 2 - /6

so 2 = 11/6

ie = 11/12

1

23/6

tan2 repeats every /2 radians but repeat values are not in interval.

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3tan2 + 1 = 0

3tan2 = -1

tan2 = -1/3 Q2 or Q4

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

+ 2 -

sin all

tan cos

1

23/6

Q2: angle = - /6

so 2 = 5/6 ie = 5/12

Q4: angle = 2 - /6

so 2 = 11/6

ie = 11/12

tan2 repeats every /2 radians but repeat values are not in interval.

• Solve the equation for tan .2• Use the positive value when finding tan-1.

• Use the quadrant rule to find the solutions.

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

3145°

1 260°

30°

12• Take care to reject ”solutions”

outwith domain.

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

radians 180

1 /180 radians


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The diagram shows a the graph of a sine function from 0 to 2/3.

2/3

P Qy = 2

(a) State the equation of the graph.

(b) The line y = 2 meets the graph

at points P & Q.

Find the coordinates of these two points.


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2/3

P Qy = 2



at points P & Q.

Find the coordinates of these two points.

Graph is y = 4sin3x

P is (/18, 2) and Q is (5/18, 2).

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2/3

P Q

y = 2



(a) One complete wave from 0 to 2/3

so 3 waves from 0 to 2.

Max/min = ±4 4sin(…)

Graph is y = 4sin3x

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at points P & Q.

Find the coordinates of these two

points.

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2/3

P Q

y = 2

Graph is y = 4sin3x

so 4sin3x = 2

or sin3x = 1/2

(b) At P & Q y = 4sin3x and y = 2

Q1 or Q2

sin-1(1/2) = /6

Q1: angle = /6

so 3x = /6

ie x = /18

Q2: angle = - /6

so 3x = 5/6

ie x = 5/18

-

+ 2 -

sin all

tan cos

1

23/6

P is (/18, 2) and Q is (5/18, 2).

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(a) One complete wave from 0 to 2/3

so 3 waves from 0 to 2.

Max/min = ±4

Graph is y = 4sin3x

4sin(…)

• Identify graph is of the form y = asinbx.

• Must justify choice of a and b.

• Possible justification of a

Max = 4, Min = -4 4sin(…)y = sinx stretched by a factor of 4

• Possible justification for b

Period = 3 waves from 0 to

2 / 32

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Graph is y = 4sin3x

so 4sin3x = 2

or sin3x = 1/2

(b) At P & Q y = 4sin3x and y = 2

Q1 or Q2

sin-1(1/2) = /6 -

+ 2 -

sin all

tan cos

1

23/6

Q1: angle = /6

so 3x = /6

ie x = /18

Q2: angle = - /6

so 3x = 5/6

ie x = 5/18

P is (/18, 2) and Q is (5/18, 2).

• At intersection y1 = y2

4sin3x = 2

• Solve for sin3x

• Use the quadrant rule to find the solutions.

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

3145°

1 260°

30°

1 2

• Take care to state coordinates.


UNIT 1 : Functions& Graphs



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This graph shows the the function y = g(x).

Make a sketch of the graph of the function y = 4 – g(-x).

y = g(x)

-8 12

(-p,q)

(u,-v)


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y = g(x)

-8 12

(-p,q)

(u,-v)

(p,-q+4)

(8,4) (-12,4) (0,4)

(-u,v+4)

y = 4 – g(-x)

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y = g(x)

-8 12

(-p,q)

(u,-v)

y = 4 – g(-x) = -g(-x) + 4

Reflect in X-axis

Reflect in Y-axis

Slide 4 up

A

B C

Known Points

(-8,0), (-p,q), (0,0), (u,-v), (12,0)

(-8,0), (-p,-q), (0,0), (u,v), (12,0)A

B

(8,0), (p,-q), (0,0), (-u,v), (-12,0)

C

(8,4), (p,-q+4), (0,4), (-u,v+4), (-12,4)

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y = g(x)

-8 12

(-p,q)

(u,-v)

(8,4), (p,-q+4), (0,4), (-u,v+4), (-12,4)

Now plot points and draw curve through

them.

(p,-q+4)

(8,4) (-12,4) (0,4)

(-u,v+4)

y = 4 – g(-x)

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y = 4 – g(-x) = -g(-x) + 4

Reflect in X-axis

Reflect in Y-axis

Slide 4 up

A

B C

Known Points

(-8,0), (-p,q), (0,0), (u,-v), (12,0)

(-8,0), (-p,-q), (0,0), (u,v), (12,0)A

B

(8,0), (p,-q), (0,0), (-u,v), (-12,0)

C

(8,4), (p,-q+4), (0,4), (-u,v+4), (-12,4)

•Change order to give form: y = k.g(x) + c

•When the function is being changed by more than one related function take each change one at a time either listing the coordinates or sketching the steps to final solution.

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y = 4 – g(-x) = -g(-x) + 4

Reflect in X-axis

Reflect in Y-axis

Slide 4 up

A

B C

Known Points

(-8,0), (-p,q), (0,0), (u,-v), (12,0)

(-8,0), (-p,-q), (0,0), (u,v), (12,0)A

B

(8,0), (p,-q), (0,0), (-u,v), (-12,0)

C

(8,4), (p,-q+4), (0,4), (-u,v+4), (-12,4)

•Learn Rules: Not given on formula sheet

f(x) + k Slide k units parallel to y-axis

kf(x) Stretch by a factor = k

-f(x) Reflect in the x-axis

f(x-k) Slide k units parallel to the x-axis

f(-x) Reflect in y-axis

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

(8,4), (p,-q+4), (0,4), (-u,v+4), (-12,4)

Now plot points and draw curve through

them.

(p,-q+4)

(8,4) (-12,4) (0,4)

(-u,v+4)

y = 4 – g(-x)


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y = ax

(1,a)

This graph shows the the function y = ax.

Make sketches of the graphs of the functions

(I) y = a(x+2)

(II) y = 2ax - 3


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This graph shows the the function y = ax.


(I) y = a(x+2)

(II) y = 2ax - 3

(-1,a)

(-2,1)

y = a(x+2)

y = ax

ANSWER TO PART (I)

y = 2ax - 3

y = ax

(0,-1)

(1,2a-3)

ANSWER to PART (II)

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(I) y = a(x+2)

y = ax

(1,a)

(I) y = a(x+2)

f(x) = ax

so a(x+2) = f(x+2)

move f(x) 2 to left

(0,1)(-2,1) & (1,a) (-1,a)

(-1,a)

(-2,1)

y = a(x+2)

y = ax

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y = ax

(1,a)


(II) y = 2ax - 3

(II) y = 2ax - 3

f(x) = ax

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

double y-coords slide 3 down

(0,1)(0,2)(0,-1)

(1,a) (1,2a) (1,2a-3)

y = 2ax - 3

y = ax

(0,-1)

(1,2a-3)

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(I) y = a(x+2)

f(x) = ax

so a(x+2) = f(x+2)

move f(x) 2 to left

(0,1)(-2,1) & (1,a) (-1,a)

(-1,a)

(-2,1)

y = a(x+2)

y = ax

• When the problem is given in terms of a specific function rather in terms of the general f(x), change back to f(x) eg. y = 2x, y = log3x, y = x2 + 3x, each becomes y = f(x)


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(I) y = a(x+2)

f(x) = ax

so a(x+2) = f(x+2)

move f(x) 2 to left

(0,1)(-2,1) & (1,a) (-1,a)

(-1,a)

(-2,1)

y = a(x+2)

y = ax







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(II) y = 2ax - 3

f(x) = ax

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


(0,1)(0,2)(0,-1)

(1,a) (1,2a) (1,2a-3)

y = 2ax - 3

y = ax

(0,-1)

(1,2a-3)

• When the problem is given in terms of a specific function rather in terms of the general f(x), change back to f(x) eg. y = 2x, y = log3x, y = x2 + 3x, each becomes y = f(x)


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(II) y = 2ax - 3

f(x) = ax

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


(0,1)(0,2)(0,-1)

(1,a) (1,2a) (1,2a-3)

y = 2ax - 3

y = ax

(0,-1)

(1,2a-3)


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Two functions f and g are defined on the set of real numbers by the formulae f(x) = 2x - 1 and g(x) = x2 .

(b) Hence show that the equation g(f(x)) = f(g(x)) has only one real solution.

(a) Find formulae for (i) f(g(x)) (ii) g(f(x)) .


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Two functions f and g are defined on the set of real numbers by the formulae f(x) = 2x - 1 and g(x) = x2 .

(b) Hence show that the equation g(f(x)) = f(g(x)) has only one real solution.

(a) Find formulae for (i) f(g(x)) (ii) g(f(x)) .

= 2x2 - 1(a) (i)

= 4x2 – 4x + 1(ii)

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Two functions f and g are defined

on the set of real numbers by the

formulae

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

(a) Find formulae for (i) f(g(x))

(ii) g(f(x)) .

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

= f(x2)

= 2x2 - 1

(ii) g(f(x))

= g(2x-1)

= (2x – 1)2

= 4x2 – 4x + 1

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Two functions f and g are defined

on the set of real numbers by the

formulae

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

g(f(x)) = 4x2 – 4x + 1

(b) Hence show that the equation

g(f(x)) = f(g(x)) has only one

real solution.

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

(b) g(f(x)) = f(g(x))

4x2 – 4x + 1 = 2x2 - 1

2x2 – 4x + 2 = 0

x2 – 2x + 1 = 0

(x – 1)(x – 1) = 0

x = 1

Hence only one real solution!

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

= f(x2)

= 2x2 - 1

(ii) g(f(x))

= g(2x-1)

= (2x – 1)2

= 4x2 – 4x + 1

(a) • In composite function problems take at least 3 lines to answer the problem:

State required composite function: f(g(x))Replace g(x) without simplifying: f(x2)In f(x) replace each x by g(x): 2 x2 - 1

(II) State required composite function: g(f(x))Replace f(x) without simplifying: g(2x-1)In g(x) replace each x by f(x): (2x – 1)2

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(b)• Only one way to solve resulting equation:

Terms to the left, simplify and factorise.

g(f(x)) = 4x2 – 4x + 1

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

(b) g(f(x)) = f(g(x))

4x2 – 4x + 1 = 2x2 - 1

2x2 – 4x + 2 = 0

x2 – 2x + 1 = 0

(x – 1)(x – 1) = 0

x = 1

Hence only one real solution!


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A function g is defined by the formula g(x) = . 2 (x1) (x – 1)

(a) Find a formula for h(x) = g(g(x)) in its simplest form.

(b) State a suitable domain for h.


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A function g is defined by the formula g(x) = . 2 (x1) (x – 1)

(a) Find a formula for h(x) = g(g(x)) in its simplest form.


h(x) = (2x - 2) . . (3 – x)

Domain = {x R: x 3}

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(a) g(g(x)) = g( ). 2 . (x – 1)

= 2. 2 . (x – 1)

- 1

= 2 2 - (x – 1) . (x – 1)

= 2 (3 - x) .(x – 1)

= 2 (x - 1) .(3 – x)

= (2x - 2) . . (3 – x)

A function g is defined by the

formula g(x) = . 2 (x1) (x – 1)

(a) Find a formula for

h(x) = g(g(x))

in its simplest form.

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h(x) = (2x - 2) . . (3 – x)A function g is defined by the

formula g(x) = . 2 (x1) (x – 1)

(a) Find a formula for

h(x) = g(g(x))

in its simplest form.


(b) For domain 3 - x 0

Domain = {x R: x 3}

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(a) g(g(x)) = g( ). 2 . (x – 1)

= 2. 2 . (x – 1)

- 1

= 2 2 - (x – 1) . (x – 1)

= 2 (3 - x) .(x – 1)

= 2 (x - 1) .(3 – x)

= (2x - 2) . . (3 – x)

(a)• In composite function problems take at least 3 lines to answer the problem:

State required composite function: g(g(x))Replace g(x) without simplifying: g(2/(x-1))

In g(x) replace each x by g(x): 2(x-1)

2 - 1

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h(x) = (2x - 2) . . (3 – x)

(b) For domain 3 - x 0

Domain = {x R: x 3}

(b)• In finding a suitable domain it is often necessary to restrict R to prevent either division by zero

or the root of a negative number: In this case: 3 - x = 0

i.e. preventing division by zero.


UNIT 2 :

Integration

Polynomials

The Circle

AdditionFormulae

Quadratics

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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

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Show that x = 3 is a root of the

equation x3 + 3x2 – 10x – 24 = 0.


Using the nested method -

coefficients are 1, 3, -10, -24

f(3) = 3 1 3 -10 -24

3 18 24

1 6 8 0

f(3) = 0 so x = 3 is a root.

Also (x – 3) is a factor.

Other factor: x2 + 6x + 8 or (x + 4)(x + 2)

If (x + 4)(x + 2) = 0 then x = -4 or x = -2

Hence other roots are x = -4 & x = -2

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coefficients are 1, 3, -10, -24

f(3) = 3 1 3 -10 -24

3 18 24

1 6 8 0

Other factor: x2 + 6x + 8 or (x + 4)(x + 2)

If (x + 4)(x + 2) = 0 then x = -4 or x = -2

Hence other roots are x = -4 & x = -2

f(3) = 0 so x = 3 is a root.

Also (x – 3) is a factor.

• State clearly in solution that f(3) = 0 x = 3 is a root

• Show completed factorisation of cubic i.e.(x - 3)(x + 4)(x + 2) = 0

•Take care to set factorised expression = 0

•List all the roots of the polynomial

x = 3, x = -4, x = -2

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

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

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Given that (x + 4) is a factor of

the polynomial

f(x) = 3x3 + 8x2 + kx + 4

find the value of k.

Hence solve the equation

3x3 + 8x2 + kx + 4 = 0

for this value of k.

Since (x + 4) a factor then f(-4) = 0 .

Now using the nested method -

coefficients are 3, 8, k, 4

f(-4) = -4 3 8 k 4

-12 16 (-4k – 64)

3 -4 (k + 16) (-4k – 60)

Since -4k – 60 = 0

then -4k = 60

so k = -15

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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.

If k = -15 then we now have

f(-4) = -4 3 8 -15 4

-12 16 -4

Other factor is 3x2 – 4x + 1

or (3x - 1)(x – 1)

3 -4 1 0

If (3x - 1)(x – 1) = 0 then x = 1/3 or x = 1

So full solution of equation is:

x = -4 or x = 1/3 or x = 1

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f(-4) = -4 3 8 k 4

-12 16 (-4k – 64)

3 -4 (k + 16) (-4k – 60)

Since -4k – 60 = 0

then -4k = 60

so k = -15

• The working in the nested solution can sometimes be eased by working in both directions toward the variable:

-4 3 8 k 4 -12 16 -4

3 -4 1 0

k + 16 = 1 k = -15

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f(-4) = -4 3 8 k 4

-12 16 (-4k – 64)

3 -4 (k + 16) (-4k – 60)

Since -4k – 60 = 0

then -4k = 60

so k = -15

• Simply making f(-4) = 0 will also yield k i.e. 3(-4)3 + 8(-4)2 + k(-4) + 4 = 0

k = -15

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If k = -15 then we now have

f(-4) = -4

-12 16 -4

Other factor is 3x2 – 4x + 1

or (3x - 1)(x – 1)

3 -4 1 0

If (3x - 1)(x – 1) = 0

So full solution of equation is:

x = -4 or x = 1/3 or x = 1

3 8 -15 4

• Show completed factorisation of the cubic:

(x + 4)(3x - 1)(x - 1) = 0


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

Hence express f(x) in its fully factorised form.


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

Hence express f(x) in its fully factorised form.

6x3 + 13x2 - 4 = (3x + 2)(2x - 1)(x + 2)

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Given that f(x) = 6x3 + 13x2 - 4

show that (x + 2) is a factor

of f(x).

Hence express f(x) in its fully

factorised form.


coefficients are 6, 13, 0, -4

f(-2) = -2 6 13 0 -4

6

-12

1

-2

-2

4

0

f(-2) = 0 so

(x + 2) is a factor

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Given that f(x) = 6x3 + 13x2 - 4

show that (x + 2) is a factor

of f(x).

Hence express f(x) in its fully

factorised form.



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

• State clearly in solution that 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

• Show completed factorisation of cubic i.e.

(3x + 2)(2x - 1)(x +2).

Other factor is 6x2 + x – 2

or (3x + 2)(2x - 1)

Hence 6x3 + 13x2 - 4

= (3x + 2)(2x - 1)(x + 2)

•Can show (x + 2) is a factor by showing f(-2) = 0 but still need nested method for quadratic factor.

(a) Find the coordinates of P and the equation of the bypass PQ.

(b) Hence find the coordinates of Q – the point where the bypass rejoins the original road.


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A busy road passes through several small villages so it is decided to build a by-pass to reduce the volume of traffic. Relative to a set of coordinate axes the road can be modelled by the curve y = -x3 + 6x2 – 3x – 10. The by-pass is a tangent to this curve at point P and rejoins the original road at Q as shown below.

bypass

P

Q

y = -x3 + 6x2 – 3x – 10

4

(a) Find the coordinates of P and the equation of the bypass PQ.

(b) Hence find the coordinates of Q – the point where the bypass rejoins the original road.


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EXIT

A busy road passes through several small villages so it is decided to build a by-pass to reduce the volume of traffic. Relative to a set of coordinate axes the road can be modelled by the curve y = -x3 + 6x2 – 3x – 10. The by-pass is a tangent to this curve at point P and rejoins the original road at Q as shown below.


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P is (4,10) PQ is y = -3x + 22

Q is (-2,28)

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y = -x3 + 6x2 – 3x – 10

P

Q

y = -x3 + 6x2 – 3x – 10

4

(a) Find the coordinates of P

and the equation of the

bypass PQ.

(a) At point P, x = 4 so using the

equation of the curve we get …..

y = -43 + (6 X 42) – (3 X 4) - 10

= -64 + 96 – 12 - 10

= 10 ie P is (4,10)


= dy/dx= -3x2 + 12x - 3

When x = 4 then

dy/dx = (-3 X 16) + (12 X 4) – 3

= -48 + 48 – 3 = -3

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y = -x3 + 6x2 – 3x – 10

P

Q

y = -x3 + 6x2 – 3x – 10

4

(a) Find the coordinates of P

and the equation of the

bypass PQ.

P is (4,10) dy/dx = -3

Now using : y – b = m(x – a)

where (a,b) = (4,10) & m = -3

We get y – 10 = -3(x – 4)

or y – 10 = -3x + 12

So PQ is y = -3x + 22

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y = -x3 + 6x2 – 3x – 10

P

Q

y = -x3 + 6x2 – 3x – 10

4

(b) The tangent & curve meet whenever

y = -3x + 22 and y = -x3 + 6x2 – 3x – 10

ie -3x + 22 = -x3 + 6x2 – 3x – 10

or x3 - 6x2 + 32 = 0

(b)Hence find the coordinates of Q – the point where the bypass rejoinsthe original road.

We already know that x = 4 is one

solution to this so using the nested

method we get …..

f(4) = 4 1 -6 0 32

4 -8 -32

1 -2 -8 0

Other factor is x2 – 2x - 8

PQ is y = -3x + 22

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y = -x3 + 6x2 – 3x – 10

P

Q

y = -x3 + 6x2 – 3x – 10

4

(b) The

(b)Hence find the coordinates of Q – the point where the bypass rejoinsthe original road.

other factor is x2 – 2x - 8

= (x – 4)(x + 2)

Solving (x – 4)(x + 2) = 0

we get x = 4 or x = -2

It now follows that Q has an x-coordinate

of -2

Using y = -3x + 22 if x = -2

then y = 6 + 22 = 28

Hence

Q is (-2,28)

PQ is y = -3x + 22

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• Must use differentiation to find gradient. Learn rule:

“Multiply by the power then reduce the power by 1”

(a)

(a) At point P, x = 4 so using the

equation of the curve we get …..

y = -43 + (6 X 42) – (3 X 4) - 10

= -64 + 96 – 12 - 10

= 10 ie P is (4,10)


= dy/dx= -3x2 + 12x - 3

When x = 4 then

dy/dx = (-3 X 16) + (12 X 4) – 3

= -48 + 48 – 3 = -3

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(a)

P is (4,10) dy/dx = -3

Now using : y – b = m(x – a)

where (a,b) = (4,10) & m = -3

We get y – 10 = -3(x – 4)

or y – 10 = -3x + 12

So PQ is y = -3x + 22

• Use :

1. the point of contact (4,10)&

2. Gradient of curve at this point (m = -3) in equation

y - b = m(x - a)

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(b)

(b) The tangent & curve meet whenever

y = -3x + 22 and y = -x3 + 6x2 – 3x – 10

ie -3x + 22 = -x3 + 6x2 – 3x – 10

or x3 - 6x2 + 32 = 0

We already know that x = 4 is one

solution to this so using the nested

method we get …..

f(4) = 4 1 -6 0 32

4 -8 -32

1 -2 -8 0

Other factor is x2 – 2x - 8

• At intersection y1 = y2

Terms to the left, simplify and factorise

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(b)

other factor is x2 – 2x - 8

= (x – 4)(x + 2)

Solving (x – 4)(x + 2) = 0

we get x = 4 or x = -2

It now follows that Q has an x-coordinate

of -2

Using y = -3x + 22 if x = -2

then y = 6 + 22 = 28

Hence

Q is (-2,28)

• Note solution x = 4 appears twice:

Repeated root tangency


UNIT 2 :Quadratics



1 2 3 4 5

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QUADRATICS : Question 1

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(a) Express f(x) = x2 – 8x + 21 in the form (x – a)2 + b.

(b) Hence, or otherwise, sketch the graph of y = f(x).


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(a) Express f(x) = x2 – 8x + 21 in the form (x – a)2 + b.

(b) Hence, or otherwise, sketch the graph of y = f(x).

= (x – 4)2 + 5(a)

y = x2 – 8x + 21

(4,5)

(b)

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(a) Express f(x) = x2 – 8x + 21

in the form (x – a)2 + b.

(b) Hence, or otherwise,

sketch the graph of y = f(x).

(a) f(x) = x2 – 8x + 21

= (x2 – 8x + ) + 2116 - 16

(-82)2

= (x – 4)2 + 5

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(a) Express f(x) = x2 – 8x + 21




(b) f(x) = (x – 4)2 + 5 has a minimum

value of 5 when (x – 4)2 = 0 ie x = 4

So the graph has a minimum

turning point at (4,5).

When x = 0 , y = 21 (from original formula!)

so Y-intercept is (0,21).

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(a) Express f(x) = x2 – 8x + 21




(b) Graph looks like….

(4,5)

(0,21)

y = x2 – 8x + 21

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(a) f(x) = x2 – 8x + 21

= (x2 – 8x + ) + 2116 - 16

(-82)2

= (x – 4)2 + 5

• Move towards desired form in stages:

f(x) = x2 - 8x + 21 = (x2 - 8x) + 21 = (x2 - 8x +16) + 21 - 16

Find the number to complete the perfect square and balance the expression.(a + b) 2 = a2 + 2ab + b2

= (x - 4)2 + 5

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(4,5)

(0,21)

• The sketch can also be obtained by calculus:

2 f(x) = x -8x 21

f (x) = 2 x-8

2

At stationary points 0

2 x 8 0

x 4

f(4) 4 -8.4 + 21 = 5

dy

dx

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(4,5)

(0,21)

• Min. Turning Point (4,5),

coefficient of x2 is positive.

• Hence sketch.

f(0) 21

21

(4,5)


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(a)Express f(x) = 7 + 8x - 4x2 in the form a - b(x - c)2.

(b) Hence find the maximum turning point on the graph of y = f(x).


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(a)Express f(x) = 7 + 8x - 4x2 in the form a - b(x - c)2.

(b) Hence find the maximum turning point on the graph of y = f(x).

= 11 - 4(x – 1)2 (a)

maximum t p is at (1,11) .(b)

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(a) f(x) = 7 + 8x - 4x2

= - 4x2 + 8x + 7

= -4[x2 – 2x] + 7

(a) Express f(x) = 7 + 8x - 4x2 in

the form a - b(x - c)2.

(b) Hence find the maximum

turning point on the graph of

y = f(x).

= -4[(x2 – 2x + ) ] + 7

(-22)2

1 - 1

= -4[(x – 1)2 - 1] + 7

= -4(x – 1)2 + 4 + 7

= 11 - 4(x – 1)2

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(a) Express f(x) = 7 + 8x - 4x2 in

the form a - b(x - c)2.

(b) Hence find the maximum

turning point on the graph of

y = f(x).

(b) Maximum value is 11 when

(x – 1)2 = 0 ie x = 1.

= 11 - 4(x – 1)2

so maximum turning point is at (1,11) .

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(a) f(x) = 7 + 8x - 4x2

= - 4x2 + 8x + 7

= -4[x2 – 2x] + 7

= -4[(x2 – 2x + ) ] + 7

(-22)2

1 - 1

= -4[(x – 1)2 - 1] + 7

= -4(x – 1)2 + 4 + 7

= 11 - 4(x – 1)2

• Move towards desired form in stages:

f(x) = 7 + 8x - 4x2

= - 4x2 + 8x + 7 = (- 4x2 + 8x) + 7

• Must reduce coefficient of x2 to 1

= -4(x2 - 2x) + 7

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(a) f(x) = 7 + 8x - 4x2

= - 4x2 + 8x + 7

= -4[x2 – 2x] + 7

= -4[(x2 – 2x + ) ] + 7

(-22)2

1 - 1

= -4[(x – 1)2 - 1] + 7

= -4(x – 1)2 + 4 + 7

= 11 - 4(x – 1)2

• Must reduce coefficient of x2 to 1

= -4(x2 - 2x) + 7

Find the number to complete the perfect square and balance the expression.

(a + b) 2 = a2 + 2ab + b2= -4[(x2 - 2x +1) -1] +7 )= -4(x-1)2 + 7 + 4= 11 - 4(x-1)2

Max. TP at (1,11) (b) Maximum value is 11 when

(x – 1)2 = 0 ie x = 1.

so maximum turning point is at (1,11) .


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For what value(s) of k does the equation 4x2 – kx + (k + 5) = 0 have equal roots.


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For what value(s) of k does the equation 4x2 – kx + (k + 5) = 0 have equal roots.

k = -4 or k = 20

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Question 3

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For what value(s) of k does the

equation 4x2 – kx + (k + 5) = 0

have equal roots.

Let 4x2 – kx + (k + 5) = ax2 + bx + c

then a = 4, b = -k & c = (k + 5)

For equal roots we need discriminant = 0

ie b2 – 4ac = 0

(-k)2 – (4 X 4 X (k + 5)) = 0

k2 – 16(k + 5) = 0

k2 – 16k - 80 = 0

(k + 4)(k – 20) = 0

k = -4 or k = 20

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Let 4x2 – kx + (k + 5) = ax2 + bx + c

then a = 4, b = -k & c = (k + 5)


ie b2 – 4ac = 0

(-k)2 – (4 X 4 X (k + 5)) = 0

k2 – 16(k + 5) = 0

k2 – 16k - 80 = 0

(k + 4)(k – 20) = 0

k = -4 or k = 20

• Learn Rules relating to discriminant b2- 4ac

b2- 4ac = 0 Equal rootsb2- 4ac > 0 2 Real, distinct roots b2- 4ac < 0 No real roots b2- 4ac 0 Real roots, equal or distinct

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Let 4x2 – kx + (k + 5) = ax2 + bx + c

then a = 4, b = -k & c = (k + 5)


ie b2 – 4ac = 0

(-k)2 – (4 X 4 X (k + 5)) = 0

k2 – 16(k + 5) = 0

k2 – 16k - 80 = 0

(k + 4)(k – 20) = 0

k = -4 or k = 20

• Must use factorisation to solve resulting quadratic.

Trial and error receives no credit.


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The equation of a parabola is f(x) = px2 + 5x – 2p .

Prove that the equation f(x) = 0 always has two distinct roots.


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The equation of a parabola is f(x) = px2 + 5x – 2p .

Prove that the equation f(x) = 0 always has two distinct roots.

discriminant = b2 – 4ac

= 8p2 + 25

Since p2 0 for all values of p

then 8p2 + 25 > 0.

The discriminant is always positive

so there are always two distinct roots.

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The equation of a parabola is

f(x) = px2 + 5x – 2p .

Prove that the equation f(x) = 0

always has two distinct roots.

Let px2 + 5x – 2p = ax2 + bx + c

then a = p, b = 5 & c = -2p.

So discriminant = b2 – 4ac

= 52 – (4 X p X (-2p))

= 25 – (-8p2)

= 8p2 + 25


then 8p2 + 25 > 0.



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Let px2 + 5x – 2p = ax2 + bx + c

then a = p, b = 5 & c = -2p.


= 52 – (4 X p X (-2p))

= 25 – (-8p2)

= 8p2 + 25


then 8p2 + 25 > 0.





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Let px2 + 5x – 2p = ax2 + bx + c

then a = p, b = 5 & c = -2p.


= 52 – (4 X p X (-2p))

= 25 – (-8p2)

= 8p2 + 25


then 8p2 + 25 > 0.



• State condition you require to show true explicitly:

For two distinct roots b2- 4ac > 0

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Let px2 + 5x – 2p = ax2 + bx + c

then a = p, b = 5 & c = -2p.


= 52 – (4 X p X (-2p))

= 25 – (-8p2)

= 8p2 + 25


then 8p2 + 25 > 0.



•To show 8p2 + 25> 0 use algebraic logic or show the graph of 8p2 + 25 is always above the “x-axis”.

25

Min. T.P. at (0,25) hence graph always above the “x - axis.”


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Given that the roots of 3x(x + p) = 4p(x – 1) are equal

then show that p = 0 or p = 48.


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Given that the roots of 3x(x + p) = 4p(x – 1) are equal



p(p - 48) = 0

ie p = 0 or p = 48

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Given that the roots of

3x(x + p) = 4p(x – 1) are equal


Rearranging 3x(x + p) = 4p(x – 1) 3x2 + 3px = 4px - 4p

3x2 - px + 4p = 0

Let 3x2 - px + 4p = ax2 + bx + c

then a = 3, b = -p & c = 4p


ie b2 – 4ac = 0

(-p)2 - (4 X 3 X 4p) = 0

p2 - 48p = 0

p(p - 48) = 0

ie p = 0 or p = 48

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3x2 - px + 4p = 0

Let 3x2 - px + 4p = ax2 + bx + c

then a = 3, b = -p & c = 4p


ie b2 – 4ac = 0

(-p)2 - (4 X 3 X 4p) = 0

p2 - 48p = 0

p(p - 48) = 0

ie p = 0 or p = 48

• Must put the equation into standard quadratic form before reading off a,b and c.

i.e. ax2 + bx +c = 0

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3x2 - px + 4p = 0

Let 3x2 - px + 4p = ax2 + bx + c

then a = 3, b = -p & c = 4p


ie b2 – 4ac = 0

(-p)2 - (4 X 3 X 4p) = 0

p2 - 48p = 0

p(p - 48) = 0

ie p = 0 or p = 48



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3x2 - px + 4p = 0

Let 3x2 - px + 4p = ax2 + bx + c

then a = 3, b = -p & c = 4p


ie b2 – 4ac = 0

(-p)2 - (4 X 3 X 4p) = 0

p2 - 48p = 0

p(p - 48) = 0

ie p = 0 or p = 48

•State condition you require explicitly:

For two equal roots b2- 4ac = 0

• Must use factorisation to solve resulting quadratic.

y = -2x2 + 3x + 2

x + y – 4 = 0


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The diagram below shows the parabola y = -2x2 + 3x + 2 and the line x + y – 4 = 0.

Prove that the line is a tangent to the curve.

y = -2x2 + 3x + 2

x + y – 4 = 0


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The diagram below shows the parabola y = -2x2 + 3x + 2 and the line x + y – 4 = 0.

Prove that the line is a tangent to the curve.

Since the discriminant = 0 then there is only one solution to the equation

so only one point of contact and it follows that the line is a tangent.

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y = -2x2 + 3x + 2

x + y – 4 = 0

Prove that the line is a tangent.

Linear equation can be changed from

x + y – 4 = 0 to y = -x + 4.

The line and curve meet when

y = -x + 4 and y = -2x2 + 3x + 2 .

So -x + 4 = -2x2 + 3x + 2

Or 2x2 - 4x + 2 = 0

Let 2x2 - 4x + 2 = ax2 + bx + c

then a = 2, b = -4 & c = 2.

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Question 6

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y = -2x2 + 3x + 2

x + y – 4 = 0

Prove that the line is a tangent.

then a = 2, b = -4 & c = 2.

So discriminant = b2 – 4ac = (-4)2 – (4 X 2 X 2)

= 16 - 16

= 0

Since the discriminant = 0 then there is

only one solution to the equation so only

one point of contact and it follows that

the line is a tangent.

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Linear equation can be changed from

x + y – 4 = 0 to y = -x + 4.

The line and curve meet when

y = -x + 4 and y = -2x2 + 3x + 2 .

So -x + 4 = -2x2 + 3x + 2

Or 2x2 - 4x + 2 = 0

Let 2x2 - 4x + 2 = ax2 + bx + c

then a = 2, b = -4 & c = 2.

• For intersection of line and polynomial

yy11 = y = y22

Terms to the left, simplify and Terms to the left, simplify and factorise.factorise.

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Or -x + 4 = -2x2 + 3x + 2

2x2 - 4x + 2 = 0

2(x2 - 2x + 1) = 0

2(x - 1)(x - 1) = 0

x = 1 (twice)

Equal roots tangency

then a = 2, b = -4 & c = 2.

So discriminant = b2 – 4ac = (-4)2 – (4 X 2 X 2)

= 16 - 16

= 0

Since the discriminant = 0 then there is

only one solution to the equation so only

one point of contact and it follows that

the line is a tangent.

• To prove tangency To prove tangency “ “equal roots” may be used inequal roots” may be used in place of the discriminant. place of the discriminant. The statement must be madeThe statement must be made explicitlyexplicitly..


UNIT 2 :Integration



1 2 3 4 5

EXITBack to

Unit 2 Menu

y = x2 - 8x + 18

x = 3 x = k

Show that the shaded area is given by

1/3k3 – 4k2 + 18k - 27

INTEGRATION : Question 1

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The diagram below shows the curve y = x2 - 8x + 18 and the lines x = 3 and x = k.

The diagram below shows the curve y = x2 - 8x + 18 and the lines x = 3 and x = k.

y = x2 - 8x + 18

x = 3 x = k

Show that the shaded area is given by

1/3k3 – 4k2 + 18k - 27


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EXITArea = (x2 - 8x + 18) dx

3

k

= 1/3k3 – 4k2 + 18k – 27 as required.

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Question 1

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The diagram shows the

curve y = x2 - 8x + 18 and the

lines x = 3 and x = k.

Show that the shaded area is

given by 1/3k3 – 4k2 + 18k - 27

Area = (x2 - 8x + 18) dx3

k

= x3 - 8x2 + 18x [ ]3 2

k

3

= 1/3x3 – 4x2 + 18x[ ]k

3

= (1/3k3 – 4k2 + 18k)

– ((1/3 X 27) – (4 X 9) + 54)

= 1/3k3 – 4k2 + 18k – 27

as required.

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Area = (x2 - 8x + 18) dx3

k

= x3 - 8x2 + 18x [ ]3 2

k

3

= 1/3x3 – 4x2 + 18x[ ]k

3

= (1/3k3 – 4k2 + 18k)

– ((1/3 X 27) – (4 X 9) + 54)

= 1/3k3 – 4k2 + 18k – 27

as required.

• Learn result

can be used to find the

enclosed area shown:

( )b

a

f x dx

a b

f(x)

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Area = (x2 - 8x + 18) dx3

k

= x3 - 8x2 + 18x [ ]3 2

k

3

= 1/3x3 – 4x2 + 18x[ ]k

3

= (1/3k3 – 4k2 + 18k)

– ((1/3 X 27) – (4 X 9) + 54)

= 1/3k3 – 4k2 + 18k – 27

as required.

• Learn result for integration:

“Add 1 to the power and divide by the new power.”


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Given that dy/dx = 12x2 – 6x and the curve y = f(x) passes

through the point (2,15) then find the equation of the curve

y = f(x).


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Given that dy/dx = 12x2 – 6x and the curve y = f(x) passes

through the point (2,15) then find the equation of the curve

y = f(x).

Equation of curve is y = 4x3 – 3x2 - 5

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Given that dy/dx = 12x2 – 6x and

the curve y = f(x) passes through

the point (2,15) then find the

equation of the curve y = f(x).

dy/dx = 12x2 – 6x

So 2(12 6 )y x x dx

= 12x3 – 6x2 + C 3 2

= 4x3 – 3x2 + C

Substituting (2,15) into y = 4x3 – 3x2 + C

We get 15 = (4 X 8) – (3 X 4) + C

So C + 20 = 15

ie C = -5

Equation of curve is

y = 4x3 – 3x2 - 5

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dy/dx = 12x2 – 6x

So 2(12 6 )y x x dx

= 12x3 – 6x2 + C 3 2

= 4x3 – 3x2 + C


We get 15 = (4 X 8) – (3 X 4) + C

So C + 20 = 15

ie C = -5


y = 4x3 – 3x2 - 5

• Learn the result that integration undoes differentiation:

i.e. given

= f(x) y = f(x) dx dy

dx


“Add 1 to the power and divide by the new power”.

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dy/dx = 12x2 – 6x

So 2(12 6 )y x x dx

= 12x3 – 6x2 + C 3 2

= 4x3 – 3x2 + C


We get 15 = (4 X 8) – (3 X 4) + C

So C + 20 = 15

ie C = -5


y = 4x3 – 3x2 - 5

• Do not forget the constant of integration!!!


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Find x2 - 4 2xx

dx


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Find x2 - 4 2xx

dx

= xx + 4 + C 3 x

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Find x2 - 4 2xx

dxx2 - 4 2xx

dx= x2 - 4

2x3/2 2x3/2dx

= 1/2x1/2 - 2x-3/2 dx

= 2/3 X 1/2x3/2 - (-2) X 2x-1/2 + C

= 1/3x3/2 + 4x-1/2 + C

= xx + 4 + C 3 x

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x2 - 4 2xx

dx= x2 - 4

2x3/2 2x3/2dx

= 1/2x1/2 - 2x-3/2 dx

= 2/3 X 1/2x3/2 - (-2) X 2x-1/2 + C

= 1/3x3/2 + 4x-1/2 + C

= xx + 4 + C 3 x

• Prepare expression by:

1 Dividing out the fraction. 2 Applying the laws of indices.


Add 1 to the power and divide by the new power.

• Do not forget the constant of integration.


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1

2

( )Evaluate x2 - 2 2 dx x


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1

2


= 21/5

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Question 4

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1

2


= 21/5

1

2

( )x2 - 2 2 dx x

( )= x4 - 4x + 4 dx x21

2

( )= x4 - 4x + 4x-2 dx1

2

[ ]= x5 - 4x2 + 4x-1 5 2 -1 1

2

= x5 - 2x2 - 4 5 x[ ]2

1

= (32/5 - 8 - 2) - (1/5 - 2 - 4)

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Next Comment= 21/5

1

2

( )x2 - 2 2 dx x

( )= x4 - 4x + 4 dx x21

2

( )= x4 - 4x + 4x-2 dx1

2

[ ]= x5 - 4x2 + 4x-1 5 2 -1 1

2

= x5 - 2x2 - 4 5 x[ ]2

1

= (32/5 - 8 - 2) - (1/5 - 2 - 4)

• Prepare expression by:

1 Expanding the bracket2 Applying the laws of indices.


“Add 1 to the power and divide by the new power”.

• When applying limits show substitution clearly.

(a) Find the coordinates of A and B.

(b)Hence find the shaded area between the

curves.

y = -x2 + 8x - 10

y = x

A

B


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The diagram below shows the parabola y = -x2 + 8x - 10 and the line y = x. They meet at the points A and B.


(b)Hence find the shaded area between the

curves.

y = -x2 + 8x - 10

y = x

A

B


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The diagram below shows the parabola y = -x2 + 8x - 10 and the line y = x. They meet at the points A and B.

A is (2,2) and B is (5,5) .

= 41/2units2

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Question 5

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The diagram shows the parabola

y = -x2 + 8x - 10 and the line y = x.

They meet at the points A and B.


(b) Hence find the shaded area

between the curves.

(a) Line & curve meet when

y = x and y = -x2 + 8x - 10 .

So x = -x2 + 8x - 10

or x2 - 7x + 10 = 0

ie (x – 2)(x – 5) = 0

ie x = 2 or x = 5

Since points lie on y = x then

A is (2,2) and B is (5,5) .

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Question 5

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The diagram shows the parabola

y = -x2 + 8x - 10 and the line y = x.

They meet at the points A and B.


(b) Hence find the shaded area

between the curves.

A is (2,2) and B is (5,5) .

(b) Curve is above line between limits so

Shaded area = (-x2 + 8x – 10 - x) dx2

5

= (-x2 + 7x – 10) dx2

5

= -x3 + 7x2 - 10x3 2[ ] 5

2

= (-125/3 + 175/2 – 50)

– (-8/3 +14 – 20)

= 41/2units2

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(a) Line & curve meet when

y = x and y = -x2 + 8x - 10 .

So x = -x2 + 8x - 10

or x2 - 7x + 10 = 0

ie (x – 2)(x – 5) = 0

ie x = 2 or x = 5

Since points lie on y = x then

A is (2,2) and B is (5,5) .

• At intersection of line and curve

yy11 = y = y22

Terms to the left, simplify and Terms to the left, simplify and factorise.factorise.

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(b) Curve is above line between limits so

Shaded area = (-x2 + 8x – 10 - x) dx2

5

= (-x2 + 7x – 10) dx2

5

= -x3 + 7x2 - 10x3 2[ ] 5

2

= (-125/3 + 175/2 – 50)

– (-8/3 +14 – 20)

= 41/2units2

• Learn result

can be used to find the

enclosed area shown:

b

2 1

a

Area = (y y )dx

a b

upper curve

y1area

y2

lower curve


UNIT 2 : Addition Formulae



1 2 3 4

EXITBack to

Unit 2 Menu

ADDITION FORMULAE : Question 1

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In triangle PQR show that the exact value of cos(a - b) is 4/5.

P

Q

R1 4

2ab


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In triangle PQR show that the exact value of cos(a - b) is 4/5.

P

Q

R1 4

2ab

cos(a – b) = cosacosb + sinasinb

= (2/5 X 1/5 ) + (1/5 X 2/5 )

= 4/5

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In triangle PQR show that the

exact value of cos(a - b) is 4/5.

P

Q

R1 4

2ab

PQ2 = 12 + 22 = 5

so PQ = 5

QR2 = 42 + 22 = 20

so QR = 20 = 45 = 25

sina = 2/25 = 1/5 & sinb = 2/5

cosa = 4/25 = 2/5 & cosb = 1/5


= (2/5 X 1/5 ) + (1/5 X 2/5 )

= 2/5 + 2/5

= 4/5

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PQ2 = 12 + 22 = 5

so PQ = 5

QR2 = 42 + 22 = 20

so QR = 20 = 45 = 25

sina = 2/25 = 1/5 & sinb = 2/5

cosa = 4/25 = 2/5 & cosb = 1/5


= (2/5 X 1/5 ) + (1/5 X 2/5 )

= 2/5 + 2/5

= 4/5

• Use formula sheet to check correct expansion and relate to given variables:

cos(a - b) = cosacosb + sina sinb

112

a

a = 7

• Work only with exact values when applying Pythagoras’:


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Find the exact value of cos(p + q) in the diagram below

Y

XO

F(4,4)

G(3,-1)

pq


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Find the exact value of cos(p + q) in the diagram below

Y

XO

F(4,4)

G(3,-1)

pq

= 1/5

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Find the exact value of cos(p + q)

in the diagram below

Y

XO

F(4,4)

G(3,-1)

pq

OF2 = 42 + 42 = 32

so OF = 32 = 162 = 42

OG2 = 32 + 12 = 10

so OG = 10

sinp = 4/42 = 1/2 & sinq = 1/10

cosp = 4/42 = 1/2 & cosq = 3/10

cos(p + q) = cospcosq - sinpsinq

= (1/2 X 3/10 ) - (1/2 X 1/10 )

= 3/20 - 1/20

= 2/20

= 2/4 5

= 2/2 5

= 1/5

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cos(a + b) = cosacosb - sina sinb


Formula Sheet:

becomes:

OF2 = 42 + 42 = 32

so OF = 32 = 162 = 42

OG2 = 32 + 12 = 10

so OG = 10

sinp = 4/42 = 1/2 & sinq = 1/10

cosp = 4/42 = 1/2 & cosq = 3/10


= (1/2 X 3/10 ) - (1/2 X 1/10 )

= 3/20 - 1/20

= 2/20

= 2/4 5

= 2/2 5

= 1/5

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OF2 = 42 + 42 = 32

so OF = 32 = 162 = 42

OG2 = 32 + 12 = 10

so OG = 10

sinp = 4/42 = 1/2 & sinq = 1/10

cosp = 4/42 = 1/2 & cosq = 3/10


= (1/2 X 3/10 ) - (1/2 X 1/10 )

= 3/20 - 1/20

= 2/20

= 2/4 5

= 2/2 5

= 1/5

112

a

a = 7

• Work only with exact values when applying Pythagoras’:


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Solve the equation 2 - sinx° = 3cos2x° where 0<x<360.


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Solve the equation 2 - sinx° = 3cos2x° where 0<x<360.

Solution = {30, 150, 199.5, 340.5}

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Solve the equation

2 - sinx° = 3cos2x°

where 0<x<360.


2 - sinx° = 3(1 – 2sin2x°)

2 - sinx° = 3 – 6sin2x°

Rearrange into quadratic form

6sin2x° - sinx° - 1 = 0

(3sinx° + 1)(2sinx° - 1) = 0

6s2 – s – 1 = (3s + 1)(2s – 1)

sinx° = -1/3 or sinx° = 1/2

Q3 or Q4 Q1 or Q2

AS

T C

a°(180 - a)°

(180 + a)° (360 - a)°

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Question 3

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Solve the equation


where 0<x<360.

sinx° = -1/3 or sinx° = 1/2

AS

T C

a°(180 - a)°

(180 + a)° (360 - a)°

sin-1 (1/2) = 30°

Q1: x = 30

Q2: x = 180 – 30 = 150

sin-1 (1/3) = 19.5°

Q3: x = 180 + 19.5 = 199.5

Q4: x = 360 - 19.5 = 340.5

Solution = {30, 150, 199.5, 340.5}

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2 - sinx° = 3(1 – 2sin2x°)

2 - sinx° = 3 – 6sin2x°


6sin2x° - sinx° - 1 = 0

(3sinx° + 1)(2sinx° - 1) = 0

6s2 – s – 1 = (3s + 1)(2s – 1)

sinx° = -1/3 or sinx° = 1/2

Q3 or Q4 Q1 or Q2

AS

T C

a°(180 - a)°

(180 + a)° (360 - a)°

•Use formula sheet to check correct expansion and relate to given variables:

cos2x° =1 - 2sin2x°

•Formula must be consistent with the rest of the equation.

2 - sinx° i.e. choose formula with sinx°

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2 - sinx° = 3(1 – 2sin2x°)

2 - sinx° = 3 – 6sin2x°


6sin2x° - sinx° - 1 = 0

(3sinx° + 1)(2sinx° - 1) = 0

6s2 – s – 1 = (3s + 1)(2s – 1)

sinx° = -1/3 or sinx° = 1/2

Q3 or Q4 Q1 or Q2

AS

T C

a°(180 - a)°

(180 + a)° (360 - a)°

•Only one way to solve theresulting quadratic:

Terms to the left, put in standard quadratic form and factorise.

•Take care to relate “solutions” to the given domain. Since all 4 values are included.

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sinx° = -1/3 or sinx° = 1/2

AS

T C

a°(180 - a)°

(180 + a)° (360 - a)°

sin-1 (1/2) = 30°

Q1: x = 30

Q2: x = 180 – 30 = 150

sin-1 (1/3) = 19.5°

Q3: x = 180 + 19.5 = 199.5

Q4: x = 360 - 19.5 = 340.5

Solution = {30, 150, 199.5, 340.5}

0 360x


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Solve sin2 = cos where 0 < < 2


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Solve sin2 = cos where 0 < < 2

Soltn = {/6 , /2 , 5/6 , 3/2}

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sin2 = cos

2sin cos - cos = 0

sin2 - cos = 0

(common factor cos)

Solve sin2 = cos

where 0 < < 2

cos(2sin - 1) = 0

cos = 0 or (2sin - 1) = 0

cos = 0 or sin = 1/2

(roller-coaster graph)

= /2 or 3/2

Q1 or Q2

-

+ 2 -

AS

T C

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Solve sin2 = cos

where 0 < < 2


= /2 or 3/2 Q1 or Q2

-

+ 2 -

AS

T C

sin-1(1/2) = /6

Q1: = /6

Q2: = - /6 = 5/6

Soltn = {/6 , /2 , 5/6 , 3/2}

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sin2 = cos

2sin cos - cos = 0

sin2 - cos = 0

(common factor cos)

cos(2sin - 1) = 0

cos = 0 or (2sin - 1) = 0



= /2 or 3/2

Q1 or Q2

-

+ 2 -

AS

T C


sin2x = 2sinxcosx

1 radians180

•Although equation is in radians possible to work in degreesand convert to radians using:

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sin2 = cos

2sin cos - cos = 0

sin2 - cos = 0

(common factor cos)

cos(2sin - 1) = 0

cos = 0 or (2sin - 1) = 0



= /2 or 3/2

Q1 or Q2

-

+ 2 -

AS

T C

•Only one way to solve the result:

Terms to the left, put in standard quadratic form and factorise.

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= /2 or 3/2 Q1 or Q2

-

+ 2 -

AS

T C

sin-1(1/2) = /6

Q1: = /6

Q2: = - /6 = 5/6

Soltn = {/6 , /2 , 5/6 , 3/2}

• For trig. equations

sinx = 0 or 1 or cosx = 0 or 1

use sketch of graph to obtain

solutions: cosx = 0 y

x

y = cos x

2 3

2

x = , 2 3

2

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= /2 or 3/2 Q1 or Q2

-

+ 2 -

AS

T C

sin-1(1/2) = /6

Q1: = /6

Q2: = - /6 = 5/6

Soltn = {/6 , /2 , 5/6 , 3/2}

• Take care to relate “solutions” to the given domain.

Since all 4 values are included

0 2


UNIT 2 :The Circle



1 2 3 4

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(a) Find the coordinates of C and W.

(b) Hence find the equation of the dial for the second hand.

W

C

THE CIRCLE : Question 1

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The face of a stopwatch can be modelled by the circle with equation x2 + y2 – 10x – 8y + 5 = 0.

The centre is at C and the winder is at W.

The dial for the second hand is 1/3 the size of the face and is located half way between C and W.



W

C


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The face of a stopwatch can be modelled by the circle with equation x2 + y2 – 10x – 8y + 5 = 0.

The centre is at C and the winder is at W.

The dial for the second hand is 1/3 the size of the face and is located half way between C and W. C is (5,4) W is (5,10).

(x – 5)2 + (y – 7)2 = 4

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The face of a stopwatch can be

modelled by the circle with

equation x2 + y2 – 10x – 8y + 5 = 0.

The centre is at C and the winder

is at W.

The dial for the second hand is 1/3

the size of the face and is located

half way between C and W.


Large circle is x2 + y2 – 10x – 8y + 5 = 0.

x2 + y2 + 2gx + 2fy + c = 0Comparing Coefficients

2g = -10, 2f = -8 and c = 5

So g = -5, f = -4 and c = 5

Centre is (-g,-f) and r = (g2 + f2 – c)

C is (5,4) r = (25 + 16 – 5)

r = 36 = 6

W is 6 units above C so W is (5,10).

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Question 1

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The face of a stopwatch can be

modelled by the circle with

equation x2 + y2 – 10x – 8y + 5 = 0.

The centre is at C and the winder

is at W.

The dial for the second hand is 1/3

the size of the face and is located

half way between C and W.


Radius of small circle = 6 3 = 2.

Centre is midpoint of CW ie (5,7).

Using (x – a)2 + (y – b)2 = r2 we get

(x – 5)2 + (y – 7)2 = 4

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Large circle is x2 + y2 – 10x – 8y + 5 = 0.

x2 + y2 + 2gx + 2fy + c = 0Comparing Coefficients

2g = -10, 2f = -8 and c = 5

So g = -5, f = -4 and c = 5


C is (5,4) r = (25 + 16 – 5)

r = 36 = 6

W is 6 units above C so W is (5,10).

• Use formula sheet to check correct formulas and relate to general equation of the circle.

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Radius of small circle = 6 3 = 2.

Centre is midpoint of CW ie (5,7).


(x – 5)2 + (y – 7)2 = 4

• Identify centre and radius before putting values into circle equation:

Centre (5,7), radius = 2(x - a)2 + (y - b)2= r2

(x - 5)2 + (y - 7)2= 22

There is no need to expand this form of the circle equation.


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In a factory a conveyor belt passes through several rollers. Part of this mechanism is shown below.

(a) The small roller has centre P and equation x2 + y2 – 6x + 2y - 7 = 0.

(i) The belt is a common tangent which meets the small roller at A(7,0)and the large roller at B. Find the equation of the tangent.

(ii) B has an x-coordinate of 10. Find its y-coordinate.

(b) Find the equation of the larger roller given that its diameter is twice that of the smaller roller, and has centre Q.

belt

P

A(7,0)

B

Q


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In a factory a conveyor belt passes through several rollers. Part of this mechanism is shown below.

(a) The small roller has centre P and equation x2 + y2 – 6x + 2y - 7 = 0.



(b) Find the equation of the larger roller given that its diameter is twice that of the smaller roller, and has centre Q.

belt

PA(7,0)

B

Q y = -4x + 28

B is (10,-12)

(x – 18)2 + (y + 10)2 = 68

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(a) The small roller has centre P

and equation

x2 + y2 – 6x + 2y - 7 = 0.



(a)

(i) Small circle is x2 + y2 – 6x + 2y - 7 = 0.

x2 + y2 + 2gx + 2fy + c = 0Comparingcoefficients

2g = -6, 2f = 2 and c = -7

So g = -3, f = 1 and c = -7


P is (3,-1) r = (9 + 1 + 7)

r = 17

For equation of tangent:

Gradient of PA = 0 – (-1) 7 - 3

= 1/4

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and equation

x2 + y2 – 6x + 2y - 7 = 0.



P is (3,-1)


Gradient of PA = 0 – (-1) 7 - 3

= 1/4

Gradient of tangent = -4 (as m1m2 = -1)


with (a,b) = (7,0) & m = -4 we get ….

y – 0 = -4(x – 7) or y = -4x + 28

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and equation

x2 + y2 – 6x + 2y - 7 = 0.



(a)(ii) At B x = 10

so y = (-4 X 10) + 28 = -12.

ie B is (10,-12)

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and equation

x2 + y2 – 6x + 2y - 7 = 0.

(b) Find the equation of the larger

roller given that its diameter is

twice that of the smaller roller,

and has centre Q.

r = 17 Small Circle:

(b) From P to A is 4 along and 1 up.

So from B to Q is 8 along and 2 up.

Q is at (10+8, -12+2) ie (18,-10)

P is (3,-1) A(7,0)

Radius of larger circle is 217

so r2 = (217)2 = 4 X 17 = 68


(x – 18)2 + (y + 10)2 = 68

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(a)

(i) Small circle is x2 + y2 – 6x + 2y - 7 = 0.


2g = -6, 2f = 2 and c = -7

So g = -3, f = 1 and c = -7


P is (3,-1) r = (9 + 1 + 7)

r = 17


Gradient of PA = 0 – (-1) 7 - 3

= 1/4


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• Identify centre and radius before putting values into circle equation: Centre (18,-10), radius = 2 (x - a)2 + (y - b)2= r2

(x - 18)2 + (y + 10)2= (2 ) 2

17

17

There is no need to expand thisform of the circle equation.

If the radius is “squared out”it must be left as an exact value.

(b) From P to A is 4 along and 1 up.

So from B to Q is 8 along and 2 up.

Q is at (10+8, -12+2) ie (18,-10)

Radius of larger circle is 217

so r2 = (217)2 = 4 X 17 = 68


(x – 18)2 + (y + 10)2 = 68


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Part of the mechanism inside an alarm clock consists of

three different sized collinear cogwheels.

CD

E

The wheels can be represented by circles with centres C, D and E as shown.

The small circle with centre C has equation x2 + (y –12)2 = 5.

The medium circle with centre E has equation (x - 18)2 + (y –3)2 = 20.

Find the equation of the large circle.


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Part of the mechanism inside an alarm clock consists of

three different sized collinear cogwheels.

CD

E

The wheels can be represented by circles with centres C, D and E as shown.

The small circle with centre C has equation x2 + (y –12)2 = 5.

The medium circle with centre E has equation (x - 18)2 + (y –3)2 = 20.

Find the equation of the large circle. (x – 8)2 + (y – 8)2 = 45

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Question 3

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Small circle has centre (0,12) and radius 5 The small circle with centre C

has equation x2 + (y –12)2 = 5.

The medium circle with centre E

has equation

(x - 18)2 + (y –3)2 = 20.


large circle.

Med. circle has centre (18,3) and radius 20

= 4 X 5 = 25

CE2 = (18 – 0)2 + (3 – 12)2

= 182 + (-9)2 = 324 + 81 = 405

(Distance form!!)

CE = 405 = 81 X 5 = 95

Diameter of large circle = 95 - 25 - 5

= 65

So radius of large circle = 35

C D E5 35

95

25

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Question 3

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The small circle with centre C

has equation x2 + (y –12)2 = 5.

The medium circle with centre E

has equation

(x - 18)2 + (y –3)2 = 20.


large circle.

C D E5 35

95

25

It follows that CD = 4/9CE

= 4/9[( ) - ( )]18 3

012

= 4/9( )18-9 = ( ) 8

-4

So D is the point (0 + 8,12- 4) or (8,8)


(x – 8)2 + (y – 8)2 = (35)2

or (x – 8)2 + (y – 8)2 = 45

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Small circle has centre (0,12) and radius 5

Med. circle has centre (18,3) and radius 20

= 4 X 5 = 25

CE2 = (18 – 0)2 + (3 – 12)2

= 182 + (-9)2 = 324 + 81 = 405

(Distance form!!)

CE = 405 = 81 X 5 = 95

Diameter of large circle = 95 - 25 - 5

= 65C D E5 35

95

25

• Use formula sheet to check

correct formulas and relate to

equation of the circle.

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It follows that CD = 4/9CE

= 4/9[( ) - ( )]18 3

012

= 4/9( )18-9

So D is the point (0 + 8,12- 4) or (8,8)


(x – 8)2 + (y – 8)2 = (35)2

or (x – 8)2 + (y – 8)2 = 45

• The section formula is used to find Centre D

C

E

D4 5

5 5

CD : DE = 4 : 5

5CD = 4DE

5(d - c) = 4(e - d)

9d = 5c + 4e

8d =

8

��

��

(b) The central wheel has equation x2 + y2 – 28x – 10y + 196 = 0 and the midpoint of DE is M(13.5,1.5). Find the equation of DE and hence find the coordinates of D & E.


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The central driving wheels on a steam locomotive are linked by a piston rod.

A B C

ED

(a) The rear wheel has centre A & equation x2 + y2 – 4x – 10y + 4 = 0

the front wheel has centre C & equation x2 + y2 – 52x – 10y + 676 = 0.

If one unit is 5cm then find the size of the minimum gap between the wheels given that the gaps are equal and the wheels are identical.

(b) The central wheel has equation x2 + y2 – 28x – 10y + 196 = 0 and the midpoint of DE is M(13.5,1.5). Find the equation of DE and hence find the coordinates of D & E.


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The central driving wheels on a steam locomotive are linked by a piston rod.

(a) The rear wheel has centre A & equation x2 + y2 – 4x – 10y + 4 = 0

the front wheel has centre C & equation x2 + y2 – 52x – 10y + 676 = 0.

If one unit is 5cm then find the size of the minimum gap between the wheels given that the gaps are equal and the wheels are identical.

Each gap = 2 units = 10cm.

Hence D is (10,2) and E is (17,1). Equation of DE x + 7y = 24

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The rear wheel has centre A &

equation x2 + y2 – 4x – 10y + 4 = 0

the front wheel has centre C &

equation

x2 + y2 – 52x – 10y + 676 = 0

If one unit is 5cm then find the

size of the minimum gap between

the wheels.

(a)(a) Rear wheel is x2 + y2 – 4x – 10y + 4 = 0 .


2g = -4, 2f = -10 and c = 4

So g = -2, f = -5 and c = 4


A is (2,5) r = (4 + 25 – 4)

r = 25 = 5

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Question 4

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The rear wheel has centre A &

equation x2 + y2 – 4x – 10y + 4 = 0

the front wheel has centre C &

equation

x2 + y2 – 52x – 10y + 676 = 0

If one unit is 5cm then find the

size of the minimum gap between

the wheels.

(a)

Comparingcoefficients

Front wheel is x2 + y2 – 52x – 10y + 676 = 0

x2 + y2 + 2gx + 2fy + c = 0

2g = -52, 2f = -10 and c = 676

So g = -26, f = -5 and c = 676

Centre is (-g,-f)

C is (26,5)

r = 5 as wheels are identical

AC = 24 units

Both gaps = AC – 2 radii - diameter

= 24 – 5 – 5 – 10 = 4 units

Each gap = 2 units = 10cm.

{A is (2,5)}

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(b)

The central wheel has equation

x2 + y2 – 28x – 10y + 196 = 0

and the midpoint of DE is

M(13.5,1.5).

Find the equation of DE and hence

find the coordinates of D & E.

Middle wheel is x2 + y2 – 28x – 10y + 196 = 0 .

x2 + y2 + 2gx + 2fy + c = 0

2g = -28, 2f = -10 and c = 196

So g = -14, f = -5 and c = 196

Centre is (-g,-f)

B is (14,5)

Gradient of BM =5 – 1.5

14 – 13.5

Equation of DE

= 3.5/0.5 = 7

Gradient of DE = -1/7 (m1m2 = -1)

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(b)


x2 + y2 – 28x – 10y + 196 = 0


M(13.5,1.5).

Find the equation of DE and hence

find the coordinates of D & E.

Equation of DE


we get y – 1.5 = -1/7 (x – 13.5) ( X 7)

Or 7y – 10.5 = -x + 13.5

So DE is x + 7y = 24

Line & circle meet when

x2 + y2 – 28x – 10y + 196 = 0 & x = 24 – 7y.

x + 7y = 24 can be written as x = 24 – 7y

Substituting (24 – 7y) for x in the

circle equation we get ….

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(b)


x2 + y2 – 28x – 10y + 196 = 0


M(13.5,1.5).

Find the equation of DE and

Hence find the coordinates

of D & E.

(24 – 7y)2 + y2 – 28(24 – 7y) – 10y + 196 = 0

576 – 336y + 49y2 + y2 – 672 + 196y – 10y + 196 = 0

50y2 – 150y + 100 = 0 (50)

y2 – 3y + 2 = 0

(y – 1)(y – 2) = 0

So y = 1 or y = 2

Using x = 24 – 7y

If y = 1 then x = 17 If y = 2 then x = 10

Hence D is (10,2) and E is (17,1).

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(a) Rear wheel is x2 + y2 – 4x – 10y + 4 = 0 .


2g = -4, 2f = -10 and c = 4

So g = -2, f = -5 and c = 4


A is (2,5) r = (4 + 25 – 4)

r = 25 = 5


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Equation of DE


we get y – 1.5 = -1/7 (x – 13.5)

Or 7y – 10.5 = -x + 13.5

So DE is x + 7y = 24

Line & circle meet when

x2 + y2 – 28x – 10y + 196 = 0 & x = 24 – 7y.

x + 7y = 24 can be written as x = 24 – 7y

Substituting (24 – 7y) for x in the

circle equation we get ….

• In part b) avoid fractions when substituting into the circle equation:

Use x = (24 - 7y) not y = (24 -x)

1

7

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(24 – 7y)2 + y2 – 28(24 – 7y) – 10y + 196 = 0

576 – 336y + 49y2 + y2 – 672 + 196y – 10y + 196 = 0

50y2 – 150y + 100 = 0 (50)

y2 – 3y + 2 = 0

(y – 1)(y – 2) = 0

So y = 1 or y = 2

Using x = 24 – 7y

If y = 1 then x = 17 If y = 2 then x = 10

Hence D is (10,2) and E is (17,1).

• Even if it appears an error has occurred continue thesolution even if it means applying the quadratic formula:

2 4

2

b b ac

a

x =


UNIT 3 :

Logs & Exponential

Wave Function

Further Calculus

Vectors

EXIT


UNIT 3 :Wave Function



1 2 3

EXITBack to

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WAVE FUNCTION: Question 1

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(a) Express cos(x°) + 7sin (x°) in the form kcos(x° - a°) where

k>0 and 0 a 90.

(b) Hence solve cos(x°) + 7sin (x°) = 5 for 0 x 90.


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(a) Express cos(x°) + 7sin (x°) in the form kcos(x° - a°) where

k>0 and 0 a 90.

(b) Hence solve cos(x°) + 7sin (x°) = 5 for 0 x 90.

Hence

cos(x°) + 7sin (x°) = 52cos(x° - 81.9°)

Solution is {36.9}(b)

(a)

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(a) Express cos(x°) + 7sin (x°)

in the form kcos(x° - a°)

where k>0 and 0 a 90.

(b) Hence solve

cos(x°) + 7sin (x°) = 5

for 0 x 90.

Let cos(x°) + 7sin (x°) = kcos(x° - a°)

= k(cosx°cosa° + sinx°sina° )

= (kcosa°)cosx° + (ksina°)sinx°

kcosa° = 1 & ksina° = 7Comparing coefficients

So (kcosa°)2 + (ksina°)2 = 12 + 72

or k2cos2a° + k2sin2a° = 1 + 49

or k2(cos2a° + sin2a°) = 50

or k2 = 50 (cos2a° + sin2a°) = 1

so k = 50 = 252 = 52

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(b) Hence solve

cos(x°) + 7sin (x°) = 5

for 0 x 90.

so k = 50 = 252 = 52

ksina°kcosa° =

71

tana° = 7

a° = tan-1(7) = 81.9°

ksina° > 0 so a in Q1 or Q2

kcosa° > 0 so a in Q1 or Q4

so a in Q1!

Hence

cos(x°) + 7sin (x°) = 52cos(x° - 81.9°)

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(b) Hence solve

cos(x°) + 7sin (x°) = 5

for 0 x 90.

Hence

cos(x°) + 7sin (x°) = 52cos(x° - 81.9°)

(b) If cos(x°) + 7sin (x°) = 5

then 52cos(x° - 81.9°) = 5

or cos(x° - 81.9°) = 1/2

Q1 or Q4 & cos-1(1/2) = 45°

Q1: angle = 45° so x° - 81.9° = 45°

so x° = 126.9° not in range.

Q4: angle = 360° - 45° so x° - 81.9° = 315°

so x° = 396.9° (**)

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(b) Hence solve

cos(x°) + 7sin (x°) = 5

for 0 x 90.

Q4: angle = 360° - 45° so x° - 81.9° = 315°

so x° = 396.9° (**)

(**) function repeats every 360°

& 396.9° - 360° = 36.9°

Solution is {36.9}

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So (kcosa°)2 + (ksina°)2 = 12 + 72

or k2cos2a° + k2sin2a° = 1 + 49


or k2 = 50 (cos2a° + sin2a°) = 1

so k = 50 = 252 = 52

• Use the formula sheet for the correct expansion:

kcos(x° - a°)

= kcosx°cosa° + ksinx°sina°

(Take care not to omit the “k” term)

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So (kcosa°)2 + (ksina°)2 = 12 + 72

or k2cos2a° + k2sin2a° = 1 + 49


or k2 = 50 (cos2a° + sin2a°) = 1

so k = 50 = 252 = 52

• When equating coefficients “square” and “ring” corresponding coefficients:

1cosx° + 7sinx° = kcosx°cosa° + ksinx°sina°

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So (kcosa°)2 + (ksina°)2 = 12 + 72

or k2cos2a° + k2sin2a° = 1 + 49


or k2 = 50 (cos2a° + sin2a°) = 1

so k = 50 = 252 = 52

• State the resulting equations explicitly:

kcosa° = 1 ksina° = 7

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So (kcosa°)2 + (ksina°)2 = 12 + 72

or k2cos2a° + k2sin2a° = 1 + 49


or k2 = 50 (cos2a° + sin2a°) = 1

so k = 50 = 252 = 52

• k can be found directly:

2 2k 1 7 50

Note: k is always positive


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Express 4sin(x°)–2cos(x°) in the form ksin(x° + a°)

where k>0 and 0 < a < 360.


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Express 4sin(x°)–2cos(x°) in the form ksin(x° + a°)

where k>0 and 0 < a < 360.

Hence

4sin(x°) – 2cos(x°) = 25sin(x° + 333.4°)

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Let 4sin(x°)–2cos(x°) = ksin(x° + a°)

Express 4sin(x°)–2cos(x°)

in the form ksin(x° + a°)

where k>0 and 0 < a < 360.

= k(sinx°cosa° + cosx°sina° )

= (kcosa°)sinx° + (ksina°)cosx°

Comparing coefficients kcosa° = 4 &

ksina° = -2

So (kcosa°)2 + (ksina°)2 = 42 + (-2)2

or k2cos2a° + k2sin2a° = 16 + 4


or k2 = 20 (cos2a° + sin2a°) = 1

so k = 20 = 45 = 25

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Express 4sin(x°)–2cos(x°)

in the form ksin(x° + a°)

where k>0 and 0 < a < 360.

so k = 20 = 45 = 25

ksina°kcosa° =

-2 4

tana° = (-1/2) so Q2 or Q4

tan-1(1/2) = 26.6°

Q4: a° = 360° – 26.6° = 333.4°

ksina° < 0 so a in Q3 or Q4


so a in Q4!

Hence

4sin(x°) – 2cos(x°) = 25sin(x° + 333.4°)

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ksin(x° + a°)

= ksinx°cosa° + kcosx°sina°






ksina° = -2

So (kcosa°)2 + (ksina°)2 = 42 + (-2)2

or k2cos2a° + k2sin2a° = 16 + 4


or k2 = 20 (cos2a° + sin2a°) = 1

so k = 20 = 45 = 25

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ksina° = -2

So (kcosa°)2 + (ksina°)2 = 42 + (-2)2

or k2cos2a° + k2sin2a° = 16 + 4


or k2 = 20 (cos2a° + sin2a°) = 1

so k = 20 = 45 = 25


4sinx° - 2cosx°



kcosa° = 4 ksina° = -2

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ksina° = -2

So (kcosa°)2 + (ksina°)2 = 42 + (-2)2

or k2cos2a° + k2sin2a° = 16 + 4


or k2 = 20 (cos2a° + sin2a°) = 1

so k = 20 = 45 = 25



2 2k 4 ( 2) 20

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so k = 20 = 45 = 25

ksina°kcosa° =

-2 4

tana° = (-1/2) so Q2 or Q4

tan-1(1/2) = 26.6°

Q4: a° = 360° – 26.6° = 333.4°



so a in Q4!

Hence

4sin(x°) – 2cos(x°) = 25sin(x° + 333.4°)

• Use the sign of the equations to determine the correct quadrant:

kcosa° = 4 (cos +ve)

ksina° = -2 (sin -ve)

cos +ve&sin -ve

(a) Express y = 5sin(x°) - 12cos(x°) in the form y = ksin(x° + a°) where k > 0 and 0 a < 360

(b) Hence find the coordinates of the maximum turning point at P.

P

y = 5sin(x°) - 12cos(x°)


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The graph below is that of y = 5sin(x°) - 12cos(x°)

(a) Express y = 5sin(x°) - 12cos(x°) in the form y = ksin(x° + a°) where k > 0 and 0 a < 360

(b) Hence find the coordinates of the maximum turning point at P.

P

y = 5sin(x°) - 12cos(x°)


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The graph below is that of y = 5sin(x°) - 12cos(x°)

Hence

5sin(x°) – 12cos(x°) = 13sin(x° + 292.6°)

Max TP is at (157.4,13)

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(a) Express y = 5sin(x°) - 12cos(x°)

in the form y = ksin(x° + a°)

where k > 0 and 0 a < 360

(b) Hence find the coordinates of

the maximum turning point at P.

(a) Let 5sin(x°) – 12cos(x°) = ksin(x° + a°)



kcosa° = 5 & ksina° = -12Comparing coefficients

So (kcosa°)2 + (ksina°)2 = 52 + (-12)2

or k2cos2a° + k2sin2a° = 25 + 144

or k2(cos2a° + sin2a°) = 169

or k2 = 169 (cos2a° + sin2a°) = 1

so k = 13

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so k = 13

ksina°kcosa° =

-12 5



so a in Q4!

tana° = (-12/5) so Q2 or Q4

tan-1(12/5) = 67.4°

Q4: a° = 360° – 67.4° = 292.6°

Hence

5sin(x°) – 12cos(x°) = 13sin(x° + 292.6°)

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Hence

5sin(x°) – 12cos(x°) = 13sin(x° + 292.6°)

(b) The maximum value of

13sin(x° + 292.6°) is 13.

Maximum on a sin graph occurs

when angle = 90°.

ie x + 292.6 = 90

or x = -202.6 (**)

This is not in the desired range but the

function repeats every 360°.

Taking -202.6 + 360 = 157.4

Max TP is at (157.4,13)

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So (kcosa°)2 + (ksina°)2 = 52 + (-12)2

or k2cos2a° + k2sin2a° = 25 + 144

or k2(cos2a° + sin2a°) = 169

or k2 = 169 (cos2a° + sin2a°) = 1

so k = 13


ksin(x° + a°)



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So (kcosa°)2 + (ksina°)2 = 52 + (-12)2

or k2cos2a° + k2sin2a° = 25 + 144

or k2(cos2a° + sin2a°) = 169

or k2 = 169 (cos2a° + sin2a°) = 1

so k = 13


5sinx° - 12cosx°



kcosa° = 5 ksina° = -12

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So (kcosa°)2 + (ksina°)2 = 52 + (-12)2

or k2cos2a° + k2sin2a° = 25 + 144

or k2(cos2a° + sin2a°) = 169

or k2 = 169 (cos2a° + sin2a°) = 1

so k = 13



2 2k 5 ( 12)

169 13

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so k = 13

ksina°kcosa° =

-12 5



so a in Q4!

tana° = (-12/5) so Q2 or Q4

tan-1(12/5) = 67.4°

Q4: a° = 360° – 67.4° = 292.6°

Hence

5sin(x°) – 12cos(x°) = 13sin(x° + 292.6°)

• Use the sign of the equations to determine the correct quadrant:

kcosa° = 5 (cos +ve)

ksina° = -12 (sin -ve)

cos +ve&sin -ve

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13sin(x° + 292.6°) is 13.


when angle = 90°.

ie x + 292.6 = 90

or x = -202.6 (**)



Taking -202.6 + 360 = 157.4

Max TP is at (157.4,13)

• The maximum value of 13sin(x°+292.6°) can also be found by considering rules for related functions:

13sin(x° + 292.6°) slides 13sinx° graph 292.6° to the left. Maximum value of sinx° occurs at 90°

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13sin(x° + 292.6°) is 13.


when angle = 90°.

ie x + 292.6 = 90

or x = -202.6 (**)



Taking -202.6 + 360 = 157.4

Max TP is at (157.4,13)

• Maximum value of sin(x°+292.6°) occurs at (90°- 292.6°)

i.e at x = - 202.6°

Add 360° to bring into domain x = 157.4°


UNIT 3 : Logs &Exponentials



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LOGS & EXPONENTIALS: Question 1

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As a radioactive substance decays the amount of radioactive material remaining after t hours, At , is given by the formula

At = A0e-0.161t

where A0 is the original amount of material.

(a) If 400mg of material remain after 10 hours then determine how much material there was at the start.

(b) The half life of the substance is the time required for exactly half of the initial amount to decay. Find the half life to the nearest minute.


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As a radioactive substance decays the amount of radioactive material remaining after t hours, At , is given by the formula

At = A0e-0.161t

where A0 is the original amount of material.


(b) The half life of the substance is the time required for exactly half of the initial amount to decay. Find the half life to the nearest minute.

Initial amount of material = 2000mg

t = 4hrs 18mins

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(a) When t = 10, A10 = 400 so At = A0e-0.161t


0.1610

ttA A e

becomes A10 = A0e-0.161X10

or 400 = A0e-1.61

and A0 = 400 e-1.61

ie A0 = 2001.12…

or 2000 to 3 sfs


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0.1610

ttA A e

(b) The half life of the substance is

the time required for exactly

half of the initial amount to

decay. Find the half life to the

nearest minute.

(b) For half life At = 1/2A0

so At = A0e-0.161t

becomes 1/2A0 = A0e-0.161t

or e-0.161t = 0.5

or ln(e-0.161t ) = ln0.5

ie -0.161t = ln0.5

so t = ln0.5 (-0.161)

ie t = 4.3052…hrs

or t = 4hrs 18mins

( 0.3052 X 60 = 18.3..)

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(a) When t = 10, A10 = 400 so At = A0e-0.161t

becomes A10 = A0e-0.161X10

or 400 = A0e-1.61

and A0 = 400 e-1.61

ie A0 = 2001.12…

or 2000 to 3 sfs


• The ex is found on the calculator:

2nd lnex

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so At = A0e-0.161t


or e-0.161t = 0.5

or ln(e-0.161t ) = ln0.5

ie -0.161t = ln0.5

so t = ln0.5 (-0.161)

ie t = 4.3052…hrs

or t = 4hrs 18mins

( 0.3052 X 60 = 18.3..)

(b)

• The half life can be found using any real value:

e.g. A0 = 2000 At = 1000

• This results in the equation

At = A0e-0.161t

1000 = 2000e-0.161t

etc.

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so At = A0e-0.161t


or e-0.161t = 0.5

or ln(e-0.161t ) = ln0.5

ie -0.161t = ln0.5

so t = ln0.5 (-0.161)

ie t = 4.3052…hrs

or t = 4hrs 18mins

( 0.3052 X 60 = 18.3..)

(b)

• To solve an exponential equation must use logs. A trial and error solution will only be given minimum credit:

e.g. 12 = e2x

take logs. to both sides ln12 = ln (e2x)

log and exponential are inverse functions

ln 12 = 2xetc.


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log10y

log10x1

0.3The graph illustrates the law y = k

xn

The line passes through (0,0.3) and (1,0). Find the values of k and n.


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log10y

log10x1

0.3The graph illustrates the law y = k

xn


Hence k = 2 and n = 0.3

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Intercept = (0,0.3)

gradient = 0 – 0.31 - 0

Straight line so in form Y = mX + c

with Y = log10y and X = log10x

This becomes log10y = -0.3log10x + 0.3

Or log10y = -0.3log10x + log10100.3

= -0.3

or log10y = -0.3log10x + log102

or log10y = log10x-0.3 + log102

or log10y = log102x-0.3

law3

law1

The graph illustrates the law

y = kxn


log10y

log10x1

0.3

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or log10y = log102x-0.3law1The graph illustrates the law

y = kxn


log10y

log10x1

0.3

so y = 2x-0.3 = 2x0.3


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Intercept = (0,0.3)

gradient = 0 – 0.31 - 0




Or log10y = -0.3log10x + log10100.3




• It is also possible to find the values of k and n by applying the laws of logs to the given equation and substituting two coordinates from the graph:

e.g. y = kxn Take logs to both sides

log y = log kxn Apply Law 1:

log AB = log A + logB

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Intercept = (0,0.3)

gradient = 0 – 0.31 - 0




Or log10y = -0.3log10x + log10100.3




log y = logk + logxn

Apply Law 3: logxn = nlogx

log y = logk + nlogx

log y = nlogx + logk

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so y = 2x-0.3 = 2x0.3


• Two coordinates from the graph:

log10y

log10x1

0.3

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so y = 2x-0.3 = 2x0.3


• Two coordinates from the graph:

log y = nlogx + logk(0,0.3) 0.3 = n.0 + logk - 1

(1,0) 0 = n.1 + logk - 2

Hence solve 1 and 2 to find k and n

logk = 0.3 hence k = 2, and n = - logk hence n= -0.3.


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Solve the equation log3(5) – log3(2x + 1) = -2


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Solve the equation log3(5) – log3(2x + 1) = -2

x = 22

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If log3(5) – log3(2x + 1) = -2

Solve the equation

log3(5) – log3(2x + 1) = -2then log3 5

2x + 1( ) = -2law2

so 5

(2x + 1)= 3-2

or 5

(2x + 1)= 1

9Cross mult

we get 2x + 1 = 45

or 2x = 44

ie x = 22

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If log3(5) – log3(2x + 1) = -2

then log3 5

2x + 1( ) = -2

so 5

(2x + 1)= 3-2

or 5

(2x + 1)= 1

9Cross mult

we get 2x + 1 = 45

or 2x = 44

ie x = 22

• To solve an equation involving logs apply the laws so that the equation is reduced to log=log and the log can be removed and the equation solved:

log35 - log3(2x+1) = -2

Law 2: log = logA - logBAB

Must know: log33 = 1

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If log3(5) – log3(2x + 1) = -2

then log3 5

2x + 1( ) = -2

so 5

(2x + 1)= 3-2

or 5

(2x + 1)= 1

9Cross mult

we get 2x + 1 = 45

or 2x = 44

ie x = 22

log3 = -2 log33 52x+1

log3 = log33-2 52x+1

Law 3: log = nlogxxn

The log terms can now be dropped from both sides of theequation and the equation solved:

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If log3(5) – log3(2x + 1) = -2

then log3 5

2x + 1( ) = -2

so 5

(2x + 1)= 3-2

or 5

(2x + 1)= 1

9Cross mult

we get 2x + 1 = 45

or 2x = 44

ie x = 22

Drop log3 terms

= 3-2 52x+1

etc.

= 52x+1

19


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The pressure in a leaky tyre drops according to the formula

Pt = P0e-kt

where P0 is the initial tyre pressure and Pt is the pressure after t hours.

(a) If the tyre is inflated to 35psi but this drops to 31 psi in 30 minutes then find the value of k to 3 decimal places.

(b) By how many more psi will the pressure drop in the next 15 mins?

sssss


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The pressure in a leaky tyre drops according to the formula

Pt = P0e-kt

where P0 is the initial tyre pressure and Pt is the pressure after t hours.

(a) If the tyre is inflated to 35psi but this drops to 31 psi in 30 minutes then find the value of k to 3 decimal places.


sssss

1.8psi

(a) k = 0.243 to 3 dps

(b)

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0kt

tP Pe

(a) If the tyre is inflated to 35psi

but this drops to 31 psi in 30

minutes then find the value of

k to 3 decimal places.

(a) We have P0 = 35, Pt = 31 and t = 0.5

(30mins = ½ hr)

So Pt = P0e-kt

becomes 31 = 35e(-0.5k)

this becomes e(-0.5k) = 31/35

or lne(-0.5k) = ln(31/35)

or -0.5k = ln(31/35)

or k = ln(31/35) (-0.5)

ie k = 0.243 to 3 dps

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0kt

tP Peie k = 0.243 to 3 dps


(b) We now have k = 0.243,

P0 = 31 and t = 0.25 (15mins = 1/4hr)

using Pt = P0e-0.243t

we get P0.25 = 31e(-0.243X0.25)

so P0.25 = 29.2

Pressure drop in next 15 mins is

31 – 29.2 or 1.8psi

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(a) We have P0 = 35, Pt = 31 and t = 0.5

(30mins = ½ hr)

So Pt = P0e-kt

becomes 31 = 35e(-0.5k)


or lne(-0.5k) = ln(31/35)

or -0.5k = ln(31/35)

or k = ln(31/35) (-0.5)

ie k = 0.243 to 3 dps

• The ex is found on the calculator:

2nd lnex

(a)

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• To solve an exponential equation must use logs.

A trial and error solution will only be given minimum credit:

e.g. e(-0.5k) = 31/35

take logs. to both sides ln e(-0.5k) = ln 31/35

log and exponential are inverse functions

-0.5k = ln31/35

etc.

(a) We have P0 = 35, Pt = 31 and t = 0.5

(30mins = ½ hr)

So Pt = P0e-kt

becomes 31 = 35e(-0.5k)


or lne(-0.5k) = ln(31/35)

or -0.5k = ln(31/35)

or k = ln(31/35) (-0.5)

ie k = 0.243 to 3 dps


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Given that sin = au and cos = av show that

u – v = logatan.


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Given that sin = au and cos = av show that

u – v = logatan.

u – v = logatan

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Given that sin = au

and cos = av show that

u – v = logatan.

Since sin = au and cos = av

then u = logasin and v = logacos

so u – v = logasin - logacos

or u – v = loga(sin/cos)

hence u – v = logatan

logx – logy = log(x/y)

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• There are different routes to the same result but all involve correct application of formulas and laws of logs:

e.g. tan = sin cos

Should be known from

standard grade

tan = au

av

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Take logs to base a to both sides

loga tan = loga

au

av

loga tan = loga au - loga av

Law2: log =logA-logBAB

loga tan = uloga a - vloga a

Law 3: log = nlogxxn

logaa=1

loga tan = u - v


UNIT 3 :Vectors



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(a) P, Q & R are the points (3,0,-5), (-1,1,-2) & (-13,4,7) respectively.

Prove that these points are collinear.

(b) Given that PS = 7

PQ then find

the coordinates of S.

VECTORS: Question 1

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(a) P, Q & R are the points (3,0,-5), (-1,1,-2) & (-13,4,7) respectively.

Prove that these points are collinear.

(b) Given that PS = 7

PQ then find


Since PQ and

PR are

multiples of the same

common point then vector and have P as a

it follows that P, Q & R are collinear.

= (-25, 7, 16)(b) S

(a)

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(a) P, Q & R are the points (3,0,-5),

(-1,1,-2) & (-13,4,7) respectively.

Prove that these points are

collinear.

(b) Given that

PS = 7

PQ then find


PQ = q - p =

(a) [ ]-1 1-2

[ ] 3 0-2

-

[ ] -4 1 3

=

PR = r - p = [ ]

-13 4 7

[ ] 3 0-2

-

[ ]-16 4 9

[ ] -4 1 3

= 4=

Since PQ and

PR are




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(a) P, Q & R are the points (3,0,-5),

(-1,1,-2) & (-13,4,7) respectively.

Prove that these points are

collinear.

(b) Given that

PS = 7

PQ then find


(b) PS = 7

PQ [ ]

-4 1 3

= 7 [ ]-28 7 21

=

S is (3,0,-5) + [ ]-28 7 21

= (3-28, 0+7, -5+21)

= (-25, 7, 16)

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• Must know result:

Given coordinates of A and B

AB = b a��

a)

PQ = q - p =

(a) [ ]-1 1-2

[ ] 3 0-2

-

[ ] -4 1 3

=

PR = r - p = [ ]

-13 4 7

[ ] 3 0-2

-

[ ]-16 4 9

[ ] -4 1 3

= 4=

Since PQ and

PR are




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PQ = q - p =

(a) [ ]-1 1-2

[ ] 3 0-2

-

[ ] -4 1 3

=

PR = r - p = [ ]

-13 4 7

[ ] 3 0-2

-

[ ]-16 4 9

[ ] -4 1 3

= 4=

Since PQ and

PR are




• Must be able to state explicitly the result for collinearity:

Since PR = 4PQ with common

point P P,Q and R are collinear

Beware common error 4PR = PQ

��

��

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(b) PS = 7

PQ [ ]

-4 1 3

= 7 [ ]-28 7 21

=

S is (3,0,-5) + [ ]-28 7 21

= (3-28, 0+7, -5+21)

= (-25, 7, 16)

• An alternative approach is to

form a vector equation and

solve it:

b)

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(b) PS = 7

PQ [ ]

-4 1 3

= 7 [ ]-28 7 21

=

S is (3,0,-5) + [ ]-28 7 21

= (3-28, 0+7, -5+21)

= (-25, 7, 16)

b) PS = 7PQ

s - p = 7(q - p)

s = 7 q - 7p + p

-1 3

s = 7 q - 6p = 7 1 6 0

-2 5

25

7

16

S(-25,7,16)

��

VECTORS: Question 2

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(a) E, F & G are the points (4,3,-1), (5,-1,2) & (8,-1,-1) respectively.

Find

FE and

FG in

component form.

(b) Hence, or otherwise, find the size of angle EFG.

VECTORS: Question 2

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(a) E, F & G are the points (4,3,-1), (5,-1,2) & (8,-1,-1) respectively.

Find

FE and

FG in

component form.

(b) Hence, or otherwise, find the size of angle EFG.

FE

(a) [ ] -1 4 -3

=

FG [ ]

3 0 -3

=

so = 73.9°

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FE = e - f =

(a) [ ] 4 3-1

[ ] 5-1 2

-

[ ] -1 4 -3

=

FG = g - f = [ ]

8 -1 -1

[ ] 5 -1 2

-

[ ] 3 0 -3

=

(a) E, F & G are the points

(4,3,-1), (5,-1,2) & (8,-1,-1)

respectively.

Find

FE and

FG in

component form.

(b) Hence, or otherwise, find the

size of angle EFG.

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(4,3,-1), (5,-1,2) & (8,-1,-1)

respectively.

Find

FE and

FG in

component form.


size of angle EFG.

(b) Let angle EFG =

E

F

G

ie

FE .

FG = [ ]

-1 4 -3

[ ] 3 0 -3

.

= (-1 X 3) + (4 X 0) + (-3 X (-3))

= -3 + 0 + 9

= 6

|FE| = ((-1)2 + 42 + (-3)2)

|FG| = (32 + 02 + (-3)2)

= 26

= 18

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(4,3,-1), (5,-1,2) & (8,-1,-1)

respectively.

Find

FE and

FG in

component form.


size of angle EFG.

Given that FE.FG = |FE||FG|cos

then cos = FE.FG

|FE||FG|=

6

26 18

so = cos-1(6 26 18) = 73.9°

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(b) Let angle EFG =

E

F

G

ie

FE .

FG = [ ]

-1 4 -3

[ ] 3 0 -3

.

= (-1 X 3) + (4 X 0) + (-3 X (-3))

= -3 + 0 + 9

= 6

|FE| = ((-1)2 + 42 + (-3)2)

|FG| = (32 + 02 + (-3)2)

= 26

= 18

• Ensure vectors are calculated

from the vertex:

F G

E

Vectors FE and FG��

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Given that FE.FG = |FE||FG|cos

then cos = FE.FG

|FE||FG|

so = cos-1(6 26 18) = 73.9°

• Refer to formula sheet and relate formula to given variables:

a.b= a . b .cosθ

FE.FG= FE . FG .cosθ��

1 1 2 2 3 3

1 1 2 2 3 3

a.b=a b +a b +a b

FE.FG=a b +a b +a b��

VECTORS: Question 3

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P

QR

S

T

U V

W

A

B

PQRSTUVW is a cuboid in which

PQ , PS & PW are represented by the vectors

[ ],

4 2 0

[ ]and

-2 4 0

[ ]resp.

0 0 9

A is 1/3 of the way up ST & B is the midpoint of UV.ie SA:AT = 1:2 & VB:BU = 1:1.

Find the components of PA & PB and hence the size of angle APB.

VECTORS: Question 3

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P

QR

S

T

U V

W

A

B


PQ , PS & PW are represented by the vectors

[ ],

4 2 0

[ ]and

-2 4 0

[ ]resp.

0 0 9


Find the components of PA & PB and hence the size of angle APB.

|PA|

= 29

|PB|

= 106

= 48.1°APB

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PA =

PS + SA =

PS + 1/3ST

PS + 1/3PW

=

[ ]-2 4 0

[ ] =

0 0 3

+ [ ] -2 4 3

=

PB =

PQ + QV + VB

PQ + PW + 1/2PS

=

[ ] 4 2 0

[ ]+ 0 0 9

+ [ ]= -1 2 0

[ ] 3 4 9

=


PQ , PS & PW are represented by vectors

[ ],

4 2 0

[ ]and

-2 4 0

[ ]resp.

0 0 9


Find the components of PA & PB

and hence the size of angle APB.

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PA =

[ ] -2 4 3

PB = [ ]

3 4 9PQRSTUVW is a cuboid in which


[ ],

4 2 0

[ ]and

-2 4 0

[ ]resp.

0 0 9




(b) Let angle APB =

A

P

B

ie

PA .

PB =

[ ] -2 4 3

[ ] 3 4 9

.

= (-2 X 3) + (4 X 4) + (3 X 9)

= -6 + 16 + 27

= 37

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[ ],

4 2 0

[ ]and

-2 4 0

[ ]resp.

0 0 9




PA .

PB =

37

|PA| = ((-2)2 + 42 + 32)

= 29

|PB| = (32 + 42 + 92)

= 106

Given that PA.PB = |PA||PB|cos

then cos = PA.PB

|PA||PB|=

37

29 106

so = cos-1(37 29 106)

= 48.1°

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PA =

[ ] -2 4 3

PB = [ ]

3 4 9

(b) Let angle APB =

A

P

B

ie

PA .

PB =

[ ] -2 4 3

[ ] 3 4 9

.

= (-2 X 3) + (4 X 4) + (3 X 9)

= -6 + 16 + 27

= 37

• Ensure vectors are calculated

from the vertex:

P A

B

Vectors PB and PA��

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PA .

PB =

|PA| = ((-2)2 + 42 + 32)

= 29

|PB| = (32 + 42 + 92)

= 106

Given that PA.PB = |PA||PB|cos

then cos = PA.PB

|PA||PB|

so = cos-1(37 29 106)

= 48.1°

• Refer to formula sheet and relate formula to given variables:

a.b= a . b .cosθ

PB.PA= PB . PA .cosθ��

1 1 2 2 3 3

1 1 2 2 3 3

a.b=a b +a b +a b

PB.PA=a b +a b +a b��

VECTORS: Question 4

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An equilateral triangle has sides 4 units long which are represented by the vectors a , b & c as shown.

b c

aFind

(i) b.(a + c) & comment on your answer (ii) b.(a – c)

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An equilateral triangle has sides 4 units long which are represented by the vectors a , b & c as shown.

b c

aFind

(i) b.(a + c) & comment on your answer (ii) b.(a – c)

Dot product = 0 so

b & (a + c) are perpendicular.(ii) b.(a – c) = b.b

= 42 = 16

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An equilateral triangle has

sides 4 units long which are

represented by the vectors

a , b & c .

Find

(i) b.(a + c) & comment

(ii) b.(a – c)

NB: each angle is 60° but vectors should

be “tail to tail”

b

b

c60°

120°

(i) b.(a + c) = b.a + b.c

= |b||a|cos1 + |b||c|cos2

= 4X4Xcos60° + 4X4Xcos120°

= 16 X ½ + 16 X (-½ )

= 8 + (-8)

= 0

Dot product = 0 so

b & (a + c) are perpendicular.

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An equilateral triangle has

sides 4 units long which are

represented by the vectors

a , b & c .

Find

(i) b.(a + c) & comment

(ii) b.(a – c)

NB: a – c = a + (-c) or b .

(ii) b.(a – c) = b.b

= |b|2

= 42 = 16

= |b||b|cos0

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b

b

c60°

120°

(i) b.(a + c) = b.a + b.c


= 4X4Xcos60° + 4X4Xcos120°

= 16 X ½ + 16 X (-½ )

= 8 + (-8)

= 0

Dot product = 0 so


• This geometric question is based on the scalar product definition and the distributive law:

a.b= a . b .cosθ

and a(b+c) = ab + ac

No other results should be applied.

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b

b

c60°

120°

(i) b.(a + c) = b.a + b.c


= 4X4Xcos60° + 4X4Xcos120°

= 16 X ½ + 16 X (-½ )

= 8 + (-8)

= 0

Dot product = 0 so


• Ensure all scalar products are

calculated from the vertex:

b

aθ

VECTORS: Question 5

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a = 2i + j – 3k, b = -i + 10k & c = -2i + j + k.

(a) Find (i) 2a + b in component form (ii) | 2a + b |

(b) Show that 2a + b and c are perpendicular.

VECTORS: Question 5

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a = 2i + j – 3k, b = -i + 10k & c = -2i + j + k.

(a) Find (i) 2a + b in component form (ii) | 2a + b |

(b) Show that 2a + b and c are perpendicular.

(a)(i) 2a + b [ ]324

=

= 29 (ii) | 2a + b |

Since the dot product is zero it

follows that the vectors are perpendicular.

(b)

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a = 2i + j – 3k, b = -i + 10k

& c = -2i + j + k.

(a) Find

(i) 2a + b in component form

(ii) | 2a + b |

(b)Show that 2a + b and c

are perpendicular.

(a)(i) 2a + b = [ ] 2 1-3

[ ]-1 010

+2

[ ] 4 2-6

[ ]-1 010

+=

[ ]324

=

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a = 2i + j – 3k, b = -i + 10k

& c = -2i + j + k.

(a) Find


(ii) | 2a + b |


are perpendicular.

(ii) | 2a + b | = (32 + 22 + 42)

= 29

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a = 2i + j – 3k, b = -i + 10k

& c = -2i + j + k.

(a) Find


(ii) | 2a + b |


are perpendicular.

(b) (2a + b ).c = [ ]324

[ ]-2 1 1

.

= (3 X (-2))+(2 X 1)+(4 X 1)

= -6 + 2 + 4

= 0

Since the dot product is zero it follows

that the vectors are perpendicular.

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(a)(i) 2a + b = [ ] 2 1-3

[ ]-1 010

+2

[ ] 4 2-6

[ ]-1 010

+=

[ ]324

=

• When a vector is given in i , j , k form change to component form:

e.g. 5i - 2j + k =

5

-2

1

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(ii) | 2a + b | = (32 + 22 + 42)

= 29

• Must know formula for the magnitude of a vector:

2 2 2

a

n = b n a +b +c

c

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(b) (2a + b ).c = [ ]324

[ ]-2 1 1

.

= (3 X (-2))+(2 X 1)+(4 X 1)

= -6 + 2 + 4

= 0

Since the dot product is zero it follows

that the vectors are perpendicular.

• Must know condition for perpendicular vectors:

a.b = 0 a is perpendicular to b


UNIT 3 :Further Calculus



1 2 3 4

EXITBack to

Unit 3 Menu

FURTHER CALCULUS : Question 1

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Given that y = 3sin(2x) – 1/2cos(4x) then find dy/dx .


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Given that y = 3sin(2x) – 1/2cos(4x) then find dy/dx .

= 6cos(2x) + 2sin(4x)

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Given that

y = 3sin(2x) – 1/2cos(4x)

then find dy/dx .

3sin(2x) 1/2cos(4x)

OUTER / INNER

Differentiate outer then inner

y = 3sin(2x) – 1/2cos(4x)

dy/dx = 3cos(2x) X 2 - (-1/2sin(4x)) X 4

= 6cos(2x) + 2sin(4x)

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• Check formula sheet for correct result:

ydy

dx

sin(ax) acos(ax)cos(ax) -asin(ax)

3sin(2x) 1/2cos(4x)

OUTER / INNER


y = 3sin(2x) – 1/2cos(4x)

dy/dx = 3cos(2x) X 2 - (-1/2sin(4x)) X 4

= 6cos(2x) + 2sin(4x)

• Relate formula to given variables

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3sin(2x) 1/2cos(4x)

OUTER / INNER


y = 3sin(2x) – 1/2cos(4x)

dy/dx = 3cos(2x) X 2 - (-1/2sin(4x)) X 4

= 6cos(2x) + 2sin(4x)

• When applying the “chain rule” “Peel an onion”

3sin(2x) 1/2cos(4x)

OUTER / INNER


e.g. 3sin - 3cos 2x 2


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Given that g(x) = (6x – 5) then evaluate g´(9) .


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Given that g(x) = (6x – 5) then evaluate g´(9) .

= 3/7

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Given that g(x) = (6x – 5)

then evaluate g´(9) .

g(x) = (6x – 5) = (6x – 5)1/2

(6x – 5)1/2

outer / innerdiff outer then inner

g´(x) = 1/2(6x – 5)-1/2 X 6

= 3(6x – 5)-1/2

= 3

(6x – 5)

g´(9) = 3

(6X9 – 5)

= 3

49

= 3/7

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g(x) = (6x – 5) = (6x – 5)1/2

(6x – 5)1/2


g´(x) = 1/2(6x – 5)-1/2 X 6

= 3(6x – 5)-1/2

= 3

(6x – 5)

g´(9) = 3

(6X9 – 5)

= 3

49= 3/7

• Apply the laws of indices to replace the sign

1

2x x

• Apply the chain rule

• Learn the rule for differentiation

Multiply by the power then reduce the power by 1

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g(x) = (6x – 5) = (6x – 5)1/2

(6x – 5)1/2


g´(x) = 1/2(6x – 5)-1/2 X 6

= 3(6x – 5)-1/2

= 3

(6x – 5)

g´(9) = 3

(6X9 – 5)

= 3

49= 3/7

• Apply the laws of indices to return power to a positive value then the root

1

21

2

1 1x

xx

• Will usually work out to an exact value without need to use calculator.


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A curve for which dy/dx = -12sin(3x) passes through the point (/3,-2). Express y in terms of x.


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A curve for which dy/dx = -12sin(3x) passes through the point (/3,-2). Express y in terms of x.

So y = 4cos(3x) + 2

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A curve for which

dy/dx = -12sin(3x) passes

through the point (/3,-2).

Express y in terms of x.

if dy/dx = -12sin(3x)

then y = -12sin(3x) dx

= 1/3 X 12cos(3x) + C

= 4cos(3x) + C

At the point (/3,-2) y = 4cos(3x) + C

becomes -2 = 4cos(3 X /3) + C

or -2 = 4cos + C

or -2 = -4 + C

ie C = 2

So y = 4cos(3x) + 2

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= 1/3 X 12cos(3x) + C

= 4cos(3x) + C



or -2 = 4cos + C

or -2 = -4 + C

ie C = 2

So y = 4cos(3x) + 2


i.e. given


dx

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= 1/3 X 12cos(3x) + C

= 4cos(3x) + C



or -2 = 4cos + C

or -2 = -4 + C

ie C = 2

So y = 4cos(3x) + 2

• Check formula sheet for correct result:

y sinax -cos(ax)

a

cos(ax) sin(ax) a

+ c

+ c

• Do not forget the constant of integration.


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(a) Find the derivative of y = (2x3 + 1)2/3 where x > 0.

(b) Hence find x2

(2x3 + 1)1/3

dx


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(a) Find the derivative of y = (2x3 + 1)2/3 where x > 0.

(b) Hence find x2

(2x3 + 1)1/3

dx

= 4x2

(2x3 + 1)1/3

= 1/4(2x3 + 1)2/3 + C

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(2x3 + 1)2/3

outer inner

diff outer then inner

(a) if y = (2x3 + 1)2/3

(a) Find the derivative of

y = (2x3 + 1)2/3 where x > 0.

then dy/dx = 2/3 (2x3 + 1)-1/3 X 6x2

= 4x2

(2x3 + 1)1/3

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(b) Hence find

x2

(2x3 + 1)1/3

dx

(b) From (a) it follows that

4x2

(2x3 + 1)1/3

dx = (2x3 + 1)2/3 + C

4x2

(2x3 + 1)1/3

dx

now

x2

(2x3 + 1)1/3

dx

= ¼ X

= 1/4(2x3 + 1)2/3 + C

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(2x3 + 1)2/3

outer inner


(a) if y = (2x3 + 1)2/3

then dy/dx = 2/3 (2x3 + 1)-1/3 X 6x2

= 4x2

(2x3 + 1)1/3

•. Apply the chain rule“Peel an onion”

(2x3 + 1)2/3

outer inner


2 1-

3 32( ) ( )

3e.g.

a)

3 22x +1 6x

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(b) From (a) it follows that

4x2

(2x3 + 1)1/3

dx = (2x3 + 1)2/3 + C

4x2

(2x3 + 1)1/3

dx

now

x2

(2x3 + 1)1/3

dx

= ¼ X

= 1/4(2x3 + 1)2/3 + C


i.e. given


dx

b)

HIGHER – ADDITIONAL QUESTION BANK EXIT UNIT 1UNIT 2UNIT 3 Please decide which Unit you would like...

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