higher – additional question bank

HIGHER – ADDITIONAL QUESTION BANK

UNIT 1 UNIT 2 UNIT 3

Please decide which Unit you would like to revise:

Straight LineFunctions & GraphsTrig Graphs & EquationsBasic DifferentiationRecurrence Relations

PolynomialsQuadratic Functions IntegrationAddition FormulaeThe Circle

VectorsFurther CalculusExponential / Logarithmic FunctionsThe Wave Function

UNIT 1 :

Functions & Graphs

Straight Line

Recurrence Relations

BasicDifferentiation

Trig Graphs & Equations

UNIT 1 :Straight Line

You have chosen to study:

Please choose a question to attempt from the following:

1 2 3 4 5

EXITBack to

Unit 1 Menu

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).

Go to full solution

Go to Marker’s Comments

Go to Straight Line Menu

Go to Main Menu

Reveal answer only

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).

Go to full solution

Go to Main Menu

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

Markers Comments

Begin Solution

Continue Solution

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).

Straight Line Menu

Back to Home

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

Markers Comments

Straight Line Menu

Back to Home

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

•State the gradient clearly.

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

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

m1 = 35 m2 =

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

Next Comment

Go to full solution

Go to Main Menu

Reveal answer only

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).

Go to full solution

Go to Main Menu

Reveal answer only

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

Markers Comments

Begin Solution

Continue Solution

Question 2

Straight Line Menu

Back to Home

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).

Markers Comments

Straight Line Menu

Back to Home

Next Comment

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

• State the gradient clearly.

• State the condition for parallel lines m1 = m2

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

8x + 4y – 7 = 0

4y = -8x + 7 (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

Go to full solution

Go to Main Menu

Reveal answer only

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

Go to full solution

Go to Main Menu

Reveal answer only

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

= 77.4°(b)

mAC = 3/5

mBC = - 3

Markers Comments

Begin Solution

Continue Solution

Question 3

Straight Line Menu

Back to Home

(a) Using the gradient formula:

mAC = 3 – 0

7 - 2 = 3/5

mBC = 3 – 0 7 - 8

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°

Markers Comments

Straight Line Menu

Back to Home

Next Comment

(a) Using the gradient formula:

mAC = 3 – 0

mBC = 3 – 0 7 - 8

(b) Using tan = gradient

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.

mAB = tanØ °

Ø ° = tan-1 mABso ABC = 71.6°

Go to full solution

Go to Main Menu

Reveal answer only

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

P(4,-5)

Q(2,3)

R(10,-1)

(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.

Go to full solution

Go to Main Menu

Reveal answer only

P(4,-5)

Q(2,3)

R(10,-1)

(b) the equation of the line f, the perpendicular bisector of QR.

y = -1(a)

y = 2x – 11(b)

(5,-1)(c)

Markers Comments

Begin Solution

Continue Solution

Question 4 (a)

Straight Line Menu

Back to Home

P(4,-5)

Q(2,3)

R(10,-1)

(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)

Markers Comments

Begin Solution

Continue Solution

Question 4 (b)

Straight Line Menu

Back to Home

(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

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)

Markers Comments

Begin Solution

Continue Solution

Question 4 (c)

Straight Line Menu

Back to Home

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

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)

Markers Comments

Straight Line Menu

Back to Home

Next Comment

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

Markers Comments

Straight Line Menu

Back to Home

Next Comment

(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)

Markers Comments

Straight Line Menu

Back to Home

Next Comment

(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:

Go to full solution

Go to Main Menu

Reveal answer only

In triangle EFG the vertices are E(6,-3), F(12,-5) and G(2,-5).

(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.

G(2,-5)

E(6,-3)

F(12,-5)

Go to full solution

Go to Main Menu

Reveal answer only

G(2,-5)

E(6,-3)

F(12,-5)

x = 6(a)

x + 8y + 28 = 0(b)

(6,-4.25)(c)

Markers Comments

Begin Solution

Continue Solution

Question 5(a)

Straight Line Menu

Back to Home

G(2,-5)

E(6,-3)

F(12,-5)

(a) Using the gradient formula 2 1

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)

Markers Comments

Begin Solution

Continue Solution

Question 5(b)

Straight Line Menu

Back to Home

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)

Markers Comments

Begin Solution

Continue Solution

Question 5(c)

Straight Line Menu

Back to Home

G(2,-5)

E(6,-3)

F(12,-5)

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

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

ie 8y = -34

ie y = -4.25

Point of intersection is (6,-4.25)

Markers Comments

Straight Line Menu

Back to Home

Next Comment

• Sketch the altitude and the median.

Gmedian

altitude

(a) Using the gradient formula 2 1

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)

Markers Comments

Straight Line Menu

Back to Home

Next Comment

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

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

• Look for special cases:

Markers Comments

Straight Line Menu

Back to Home

Next Comment

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

EXITBack to

Unit 1 Menu

BASIC DIFFERENTIATION : Question 1

Go to full solution

Go to Basic Differentiation Menu

Go to Main Menu

Reveal answer only

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

at the point where x = 4.

Go to full solution

Go to Main Menu

Reveal answer only

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

y = 5/4x – 7

Markers Comments

Begin Solution

Continue Solution

Question 1

Basic Differentiation Menu

Back to Home

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.

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

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

= ¼ + 1 = 5/4

Continue Solution

Markers Comments

Begin Solution

Continue Solution

Question 1

Back to Home

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

Back to Previous

Markers Comments

Differentiation Menu

Back to Home

Next Comment

• Prepare expression for differentiation. 1

y = 16x x xx

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

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

If x = 4 then:

= 1 + 16 24 16

= ¼ + 1 = 5/4

• Find gradient of the tangent

using rule:

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

• Find gradient = at x = 4.dy

Continue Comments

Markers Comments

Back to Home

Next Comment

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

or y = 5/4x – 7

16 16y 4 2

• Find y coordinate at x = 4 using:

• Use m = 5/4 and (4,-2) in

y - b = m(x - a)

Go to full solution

Go to Main Menu

Reveal answer only

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.

Go to full solution

Go to Main Menu

Reveal answer only

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.

Markers Comments

Begin Solution

Continue Solution

Question 2

Back to Home

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

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

Continue Solution

Markers Comments

Begin Solution

Continue Solution

Question 2

Back to Home

Find the coordinates of the point

on the curve y = x2 – 5x + 10

where the tangent to the curve

makes an angle of 135° with the

positive direction of the X-axis.

Back to Previous

Using y = x2 – 5x + 10 with x = 2

we get y = 22 – (5 X 2) + 10

ie y = 4

So required point is (2,4)

Markers Comments

Back to Home

Next Comment

NB: gradient of line = gradient of curve

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:

• 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

x135 °

Continue Comments

Markers Comments

Back to Home

Next Comment

• Set m =

i.e. 2x - 5 = -1

and solve for x.

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)

Go to full solution

Go to Main Menu

Reveal answer only

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)

Go to full solution

Go to Main Menu

Reveal answer only

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)

Markers Comments

Begin Solution

Continue Solution

Question 3

Back to Home

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

g(x) - 0 - 0 +

Click for graph

y = g(x)

Make a sketch of the graph of

y = g(x).

Markers Comments

Begin Solution

Continue Solution

Question 3

Back to Home

Return to Nature Table

y = g(x)

Make a sketch of the graph of

y = g(x).

y = g(x)

This now gives us the following graph

Markers Comments

Back to Home

Next Comment

To sketch the graph ofthe gradient function:

' dyf ( )

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

f ( ) 0dx

'f ( )x

y = g(x)

Continue Comments

Markers Comments

Back to Home

Next Comment

2 For each interval decide if the value of ' dy

f ( ) is - or +dx

' dyf ( )

f ( ) 0dx

(We can ignore r.)

either side of 0 and p:

new y-values

g(x) - 0 - 0 +

Continue Comments

'f ( )x

Markers Comments

Back to Home

Next Comment

' dyf ( )

f ( ) 0dx

y = g(x)

3 Draw in curve to fit information

2 For each interval decide if the value of ' dy

f ( ) is - or +dx

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

'f ( )x

Go to full solution

Go to Main Menu

Reveal answer only

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

Go to full solution

Go to Main Menu

Reveal answer only

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

y = x3 - 3x2 - 9x + 2

(-1,7) is a maximum TP and

(3,-25) is a minimum TP

Return to solution

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

y = x3 - 3x2 - 9x + 2

Markers Comments

Begin Solution

Continue Solution

Question 4

Back to Home

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)

Continue Solution

Markers Comments

Begin Solution

Continue Solution

Question 4

Back to Home

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

Back to graph

either side of -1 and 3.

x -1 3

(x + 1) - 0 + + +

(x - 3) - - - 0 +

dy/dx + 0 - 0 +

Hence (-1,7) is a maximum TP

and (3,-25) is a minimum TP

Markers Comments

Back to Home

Next Comment

• Must attempt to find

and set equal to zero

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

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

Continue Comments

Markers Comments

Back to Home

Next Comment

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”

dx + 0 -

Minimum requirement

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

Go to full solution

Go to Main Menu

Reveal answer only

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.

Go to full solution

Go to Main Menu

Reveal answer only

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.

Markers Comments

Begin Solution

Continue Solution

Question 5

Back to Home

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

Continue Solution

Markers Comments

Continue Solution

Question 5

Back to Home

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

Continue Solution

Go Back to Previous

Markers Comments

Continue Solution

Question 5

Back to Home

S(x) = 2 - 4

x x2 (x 2)

achieve.

Go Back to Previous

We now check the gradients either side of

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

Markers Comments

Back to Home

Next Comment

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

• 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.

Continue Comments

Markers Comments

Back to Home

Next Comment

• Must attempt to find

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

dyS (x)

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

• Must attempt to find and set equal to zero

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

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

Continue Comments

Markers Comments

Back to Home

Next Comment

We now check the gradients either side of

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

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

S (x) + 0 -

Minimum requirement.

• State clearly: Maximum T.P at x = 4

UNIT 1 : RecurrenceRelations

EXITBack to

Unit 1 Menu

RECURRENCE RELATIONS : Question 1

Go to full solution

Go to Main Menu

Reveal answer only

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 .

Go to full solution

Go to Main Menu

Reveal answer only

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

L = -5

Markers Comments

Begin Solution

Continue Solution

Question 1

Recurrence Relations Menu

Back to Home

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

Replacing a by 0.6 in 10a + b = 4

gives 6 + b = 4

or b = -2

a = 0.6Continue Solution

Markers Comments

Begin Solution

Continue Solution

Question 1

Back to Home

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 .

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

Markers Comments

Recurrence Menu

Back to Home

Next Comment

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

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)

Markers Comments

Recurrence Menu

Back to Home

Next Comment

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.

Go to full solution

Go to Recurrence Relations Menu

Go to Main Menu

Reveal answer only

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.

Go to full solution

Go to Main Menu

Reveal answer only

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

Markers Comments

Begin Solution

Continue Solution

Question 2

Back to Home

Continue Solution

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)

Markers Comments

Begin Solution

Continue Solution

Question 2

Back to Home

Continue Solution

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

Markers Comments

Begin Solution

Continue Solution

Question 1

Back to Home

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

Markers Comments

Recurrence Menu

Back to Home

Next Comment

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.

Continue Comments

Markers Comments

Recurrence Menu

Back to Home

Next Comment

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.

Continue Solution

Markers Comments

Recurrence Menu

Back to Home

Next Comment

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.

Go to full solution

Reveal answer only

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?

Go to full solution

Reveal answer only

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)

Markers Comments

Begin Solution

Continue Solution

Question 3

Recurrence Relation Menu

Back to Home

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

(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

Continue Solution

Markers Comments

Begin Solution

Continue Solution

Question 3

Recurrence Relation Menu

Back to Home

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?

Markers Comments

Recurrence Menu

Back to Home

Next Comment

(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.

Continue Solution

Markers Comments

Recurrence Menu

Back to Home

Next Comment

(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

EXITBack to

Unit 1 Menu

TRIG GRAPHS & EQUATIONS : Question 1

Go to full solution

Go to Trig Graphs & Equations Menu

Go to Main Menu

Reveal answer only

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

Determine the values of a, b & c.

y = acosbx + c

Go to full solution

Go to Main Menu

Reveal answer only

y = acosbx + c

c = -1

Markers Comments

Begin Solution

Continue Solution

Question 1

Trig Graphs etc. Menu

Back to Home

y = acosbx + c

a = ½(max – min)

= ½(2 – (-4))

= 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

Markers Comments

Trig Graphs Menu

Back to Home

Next Comment

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

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.

Go to full solution

Go to Main Menu

Reveal answer only

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

Go to full solution

Go to Main Menu

Reveal answer only

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

= 5/12

= 11/12

Markers Comments

Begin Solution

Continue Solution

Question 2

Back to Home

Solve 3tan2 + 1 = 0

( where 0 < < ).

3tan2 + 1 = 0

3tan2 = -1

tan2 = -1/3 Q2 or Q4

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

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

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

Markers Comments

Trig Graphs Menu

Back to Home

Next Comment

3tan2 + 1 = 0

3tan2 = -1

tan2 = -1/3 Q2 or Q4

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

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

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°

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

Go to full solution

Reveal answer only

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

Go to full solution

Reveal answer only

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).

Markers Comments

Begin Solution

Continue Solution

Question 3

Back to Home

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

so 3 waves from 0 to 2.

Max/min = ±4 4sin(…)

Graph is y = 4sin3x

Continue Solution

at points P & Q.

Find the coordinates of these two

points.

Markers Comments

Begin Solution

Continue Solution

Question 3

Back to Home

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

sin all

tan cos

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

Markers Comments

Trig Graphs Menu

Back to Home

Next Comment

(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

Markers Comments

Trig Graphs Menu

Back to Home

Next Comment

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 -

sin all

tan cos

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°

• Take care to state coordinates.

UNIT 1 : Functions& Graphs

1 2 3 4

EXITBack to

Unit 1 Menu

FUNCTIONS & GRAPHS : Question 1

Go to full solution

Go to Functions & Graphs Menu

Go to Main Menu

Reveal answer only

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

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

y = g(x)

(-p,q)

(u,-v)

Go to full solution

Go to Main Menu

Reveal answer only

y = g(x)

(-p,q)

(u,-v)

(p,-q+4)

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

(-u,v+4)

y = 4 – g(-x)

Markers Comments

Begin Solution

Continue Solution

Question 1

Functions & Graphs Menu

Back to Home

y = g(x)

(-p,q)

(u,-v)

y = 4 – g(-x) = -g(-x) + 4

Reflect in X-axis

Reflect in Y-axis

Known Points

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

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

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

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

Markers Comments

Begin Solution

Continue Solution

Question 1

Back to Home

y = g(x)

(-p,q)

(u,-v)

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

Now plot points and draw curve through

(p,-q+4)

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

(-u,v+4)

y = 4 – g(-x)

Markers Comments

Functions Menu

Back to Home

Next Comment

y = 4 – g(-x) = -g(-x) + 4

Reflect in X-axis

Reflect in Y-axis

Known Points

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

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

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

(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.

Markers Comments

Functions Menu

Back to Home

Next Comment

y = 4 – g(-x) = -g(-x) + 4

Reflect in X-axis

Reflect in Y-axis

Known Points

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

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

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

(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

Markers Comments

Functions Menu

Back to Home

Next Comment

•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

(p,-q+4)

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

(-u,v+4)

y = 4 – g(-x)

Go to full solution

Go to Main Menu

Reveal answer only

y = ax

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

Go to full solution

Reveal answer only

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)

Markers Comments

Begin Solution

Continue Solution

Question 2

Back to Home

(I) y = a(x+2)

y = ax

(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

Markers Comments

Begin Solution

Continue Solution

Question 2

Back to Home

y = ax

(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)

Markers Comments

Functions Menu

Back to Home

Next Comment

(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)

Markers Comments

Functions Menu

Back to Home

Next Comment

(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

Markers Comments

Functions Menu

Back to Home

Next Comment

(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)

Markers Comments

Functions Menu

Back to Home

Next Comment

(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)

Go to full solution

Go to Functions & Graphs

Go to Main Menu

Reveal answer only

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)) .

Go to full solution

Go to Functions & Graphs

Go to Main Menu

Reveal answer only

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)

Markers Comments

Begin Solution

Continue Solution

Question 3

Back to Home

Two functions f and g are defined

on the set of real numbers by the

formulae

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

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

(ii) g(f(x)) .

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

= f(x2)

= 2x2 - 1

(ii) g(f(x))

= g(2x-1)

= (2x – 1)2

= 4x2 – 4x + 1

Continue Solution

Markers Comments

Begin Solution

Continue Solution

Question 3

Back to Home

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

Hence only one real solution!

Markers Comments

Functions Menu

Back to Home

Next Comment

(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

Markers Comments

Functions Menu

Back to Home

Next Comment

(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

Hence only one real solution!

Go to full solution

Go to Main Menu

Reveal answer only

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.

Go to full solution

Go to Main Menu

Reveal answer only

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}

Markers Comments

Begin Solution

Continue Solution

Question 4

Back to Home

Continue Solution

(a) g(g(x)) = g( ). 2 . (x – 1)

= 2. 2 . (x – 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.

Markers Comments

Begin Solution

Continue Solution

Question 4

Back to Home

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}

Markers Comments

Functions Menu

Back to Home

Next Comment

(a) g(g(x)) = g( ). 2 . (x – 1)

= 2. 2 . (x – 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)

Markers Comments

Functions Menu

Back to Home

Next Comment

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

UNIT 2 :Polynomials

1 2 3 4

EXITBack to

Unit 2 Menu

POLYNOMIALS : Question 1

Go to full solution

Go to Polynomial Menu

Go to Main Menu

Reveal answer only

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

Hence find the other roots.

Go to full solution

Go to Main Menu

Reveal answer only

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

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

Markers Comments

Begin Solution

Continue Solution

Question 1

Polynomial Menu

Back to Home

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

Markers Comments

Polynomial Menu

Back to Home

Next Comment

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.

Go to full solution

Go to Main Menu

Reveal answer only

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.

Go to full solution

Go to Main Menu

Reveal answer only

k = -15

So full solution of equation is

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

Markers Comments

Begin Solution

Continue Solution

Question 2

Polynomial Menu

Back to Home

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

Markers Comments

Begin Solution

Continue Solution

Question 2

Polynomial Menu

Back to Home

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

Markers Comments

Polynomial Menu

Back to Home

Next Comment

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

Markers Comments

Polynomial Menu

Back to Home

Next Comment

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

Markers Comments

Polynomial Menu

Back to Home

Next Comment

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

Go to full solution

Go to Main Menu

Reveal answer only

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.

Go to full solution

Go to Main Menu

Reveal answer only

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)

Markers Comments

Begin Solution

Continue Solution

Question 3

Polynomial Menu

Back to Home

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

f(-2) = 0 so

(x + 2) is a factor

Markers Comments

Begin Solution

Continue Solution

Question 3

Polynomial Menu

Back to Home

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

Other factor is 6x2 + x – 2

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

Hence 6x3 + 13x2 - 4

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

Markers Comments

Polynomial Menu

Back to Home

Next Comment

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

f(-2) = 0 so

(x + 2) is a factor

• State clearly in solution that f(-2) = 0 x = -2 is a root

Markers Comments

Polynomial Menu

Back to Home

Next Comment

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

• 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.

Go to full solutionReveal answer onlyEXIT

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

y = -x3 + 6x2 – 3x – 10

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

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

Go to full solution

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.

Go to Polynomial Menu Go to Main Menu

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

Q is (-2,28)

Markers Comments

Begin Solution

Continue Solution

Question 4

Polynomial Menu

Back to Home

y = -x3 + 6x2 – 3x – 10

(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

Markers Comments

Begin Solution

Continue Solution

Question 4

Polynomial Menu

Back to Home

y = -x3 + 6x2 – 3x – 10

(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

Markers Comments

Begin Solution

Continue Solution

Question 4

Polynomial Menu

Back to Home

y = -x3 + 6x2 – 3x – 10

(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

Markers Comments

Begin Solution

Continue Solution

Question 4

Polynomial Menu

Back to Home

y = -x3 + 6x2 – 3x – 10

(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

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

then y = 6 + 22 = 28

Q is (-2,28)

PQ is y = -3x + 22

Markers Comments

Polynomial Menu

Back to Home

Next Comment

• Must use differentiation to find gradient. Learn rule:

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

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

equation of the curve we get …..

y = -43 + (6 X 42) – (3 X 4) - 10

= -64 + 96 – 12 - 10

= 10 ie P is (4,10)

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

When x = 4 then

dy/dx = (-3 X 16) + (12 X 4) – 3

= -48 + 48 – 3 = -3

Markers Comments

Polynomial Menu

Back to Home

Next Comment

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)

Markers Comments

Polynomial Menu

Back to Home

Next Comment

(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

Markers Comments

Polynomial Menu

Back to Home

Next Comment

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

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

then y = 6 + 22 = 28

Q is (-2,28)

• Note solution x = 4 appears twice:

Repeated root tangency

UNIT 2 :Quadratics

1 2 3 4 5

EXITBack to

Unit 2 Menu

QUADRATICS : Question 1

Go to full solution

Go to Quadratics Menu

Go to Main Menu

Reveal answer only

(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).

Go to full solution

Go to Main Menu

Reveal answer only

(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

Markers Comments

Begin Solution

Continue Solution

Question 1

Quadratics Menu

Back to Home

(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

Markers Comments

Begin Solution

Continue Solution

Question 1

Quadratics Menu

Back to Home

(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).

Markers Comments

Begin Solution

Continue Solution

Question 1

Quadratics Menu

Back to Home

(a) Express f(x) = x2 – 8x + 21

(b) Graph looks like….

(0,21)

y = x2 – 8x + 21

Markers Comments

Quadratics Menu

Back to Home

Next Comment

(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

Markers Comments

Quadratics Menu

Back to Home

Next Comment

(0,21)

• The sketch can also be obtained by calculus:

2 f(x) = x -8x 21

f (x) = 2 x-8

At stationary points 0

2 x 8 0

f(4) 4 -8.4 + 21 = 5

Markers Comments

Quadratics Menu

Back to Home

Next Comment

(0,21)

• Min. Turning Point (4,5),

coefficient of x2 is positive.

• Hence sketch.

f(0) 21

Go to full solution

Go to Main Menu

Reveal answer only

(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).

Go to full solution

Go to Main Menu

Reveal answer only

(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)

Markers Comments

Begin Solution

Continue Solution

Question 2

Quadratics Menu

Back to Home

(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

= -4[(x – 1)2 - 1] + 7

= -4(x – 1)2 + 4 + 7

= 11 - 4(x – 1)2

Markers Comments

Begin Solution

Continue Solution

Question 2

Quadratics Menu

Back to Home

(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) .

Markers Comments

Quadratics Menu

Back to Home

Next Comment

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

= - 4x2 + 8x + 7

= -4[x2 – 2x] + 7

= -4[(x2 – 2x + ) ] + 7

(-22)2

= -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

Markers Comments

Quadratics Menu

Back to Home

Next Comment

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

= - 4x2 + 8x + 7

= -4[x2 – 2x] + 7

= -4[(x2 – 2x + ) ] + 7

(-22)2

= -4[(x – 1)2 - 1] + 7

= -4(x – 1)2 + 4 + 7

= 11 - 4(x – 1)2

• Must reduce coefficient of x2 to 1

= -4(x2 - 2x) + 7

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

(a + b) 2 = 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) .

Go to full solution

Go to Main Menu

Reveal answer only

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

Go to full solution

Go to Main Menu

Reveal answer only

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

k = -4 or k = 20

Markers Comments

Begin Solution

Continue Solution

Question 3

Quadratics Menu

Back to Home

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

Markers Comments

Quadratics Menu

Back to Home

Next Comment

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

Markers Comments

Quadratics Menu

Back to Home

Next Comment

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.

Go to full solution

Go to Main Menu

Reveal answer only

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

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

Go to full solution

Go to Main Menu

Reveal answer only

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.

Markers Comments

Begin Solution

Continue Solution

Question 4

Quadratics Menu

Back to Home

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.

Markers Comments

Quadratics Menu

Back to Home

Next Comment

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.

Markers Comments

Quadratics Menu

Back to Home

Next Comment

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

Markers Comments

Quadratics Menu

Back to Home

Next Comment

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”.

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

Go to full solution

Go to Main Menu

Reveal answer only

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

then show that p = 0 or p = 48.

Go to full solution

Go to Main Menu

Reveal answer only

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

p(p - 48) = 0

ie p = 0 or p = 48

Markers Comments

Begin Solution

Continue Solution

Question 5

Quadratics Menu

Back to Home

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

Markers Comments

Quadratics Menu

Back to Home

Next Comment

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

Markers Comments

Quadratics Menu

Back to Home

Next Comment

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

Markers Comments

Quadratics Menu

Back to Home

Next Comment

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

Go to full solution

Go to Marker’s Comms

Go to Main Menu

Reveal answer only

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

Go to full solution

Go to Main Menu

Reveal answer only

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.

Markers Comments

Begin Solution

Continue Solution

Question 6

Quadratics Menu

Back to Home

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.

Markers Comments

Begin Solution

Continue Solution

Question 6

Quadratics Menu

Back to Home

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

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.

Markers Comments

Quadratics Menu

Back to Home

Next Comment

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.

Markers Comments

Quadratics Menu

Back to Home

Next Comment

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

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

Go to full solution

Go to Integration Menu

Go to Main Menu

Reveal answer only

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

Go to full solution

Go to Main Menu

Reveal answer only

EXITArea = (x2 - 8x + 18) dx

= 1/3k3 – 4k2 + 18k – 27 as required.

Markers Comments

Begin Solution

Continue Solution

Question 1

Integration Menu

Back to Home

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

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

= 1/3x3 – 4x2 + 18x[ ]k

= (1/3k3 – 4k2 + 18k)

– ((1/3 X 27) – (4 X 9) + 54)

= 1/3k3 – 4k2 + 18k – 27

as required.

Markers Comments

Integration Menu

Back to Home

Next Comment

Area = (x2 - 8x + 18) dx3

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

= 1/3x3 – 4x2 + 18x[ ]k

= (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:

f x dx

Markers Comments

Integration Menu

Back to Home

Next Comment

Area = (x2 - 8x + 18) dx3

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

= 1/3x3 – 4x2 + 18x[ ]k

= (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.”

Go to full solution

Go to Main Menu

Reveal answer only

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).

Go to full solution

Go to Main Menu

Reveal answer only

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

Markers Comments

Begin Solution

Continue Solution

Question 2

Integration Menu

Back to Home

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

Markers Comments

Integration Menu

Back to Home

Next Comment

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

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

Markers Comments

Integration Menu

Back to Home

Next Comment

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!!!

Go to full solution

Go to Main Menu

Reveal answer only

Find x2 - 4 2xx

Go to full solution

Go to Main Menu

Reveal answer only

Find x2 - 4 2xx

= xx + 4 + C 3 x

Markers Comments

Begin Solution

Continue Solution

Question 3

Integration Menu

Back to Home

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

Markers Comments

Integration Menu

Back to Home

Next Comment

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.

Go to full solution

Go to Main Menu

Reveal answer only

( )Evaluate x2 - 2 2 dx x

Go to full solution

Go to Main Menu

Reveal answer only

= 21/5

Markers Comments

Begin Solution

Continue Solution

Question 4

Integration Menu

Back to Home

= 21/5

( )x2 - 2 2 dx x

( )= x4 - 4x + 4 dx x21

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

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

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

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

Markers Comments

Integration Menu

Back to Home

Next Comment= 21/5

( )x2 - 2 2 dx x

( )= x4 - 4x + 4 dx x21

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

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

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

= (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

Go to full solution

Go to Main Menu

Reveal answer only

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

Go to full solution

Go to Main Menu

Reveal answer only

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

Markers Comments

Begin Solution

Continue Solution

Question 5

Integration Menu

Back to Home

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) .

Markers Comments

Begin Solution

Continue Solution

Question 5

Integration Menu

Back to Home

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

= (-x2 + 7x – 10) dx2

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

= (-125/3 + 175/2 – 50)

– (-8/3 +14 – 20)

= 41/2units2

Markers Comments

Integration Menu

Back to Home

Next Comment

(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.

Markers Comments

Integration Menu

Back to Home

Next Comment

(b) Curve is above line between limits so

Shaded area = (-x2 + 8x – 10 - x) dx2

= (-x2 + 7x – 10) dx2

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

= (-125/3 + 175/2 – 50)

– (-8/3 +14 – 20)

= 41/2units2

• Learn result

can be used to find the

enclosed area shown:

Area = (y y )dx

upper curve

y1area

lower curve

UNIT 2 : Addition Formulae

1 2 3 4

EXITBack to

Unit 2 Menu

ADDITION FORMULAE : Question 1

Go to full solution

Go to Add Formulae Menu

Go to Main Menu

Reveal answer only

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

Go to full solution

Go to Main Menu

Reveal answer only

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

cos(a – b) = cosacosb + sinasinb

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

Markers Comments

Begin Solution

Continue Solution

Question 1

Add Formulae Menu

Back to Home

In triangle PQR show that the

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

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

Markers Comments

Add Form Menu

Back to Home

Next Comment

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

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

cos(a - b) = cosacosb + sina sinb

• Work only with exact values when applying Pythagoras’:

Go to full solution

Go to Main Menu

Reveal answer only

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

F(4,4)

G(3,-1)

Go to full solution

Go to Main Menu

Reveal answer only

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

F(4,4)

G(3,-1)

Markers Comments

Begin Solution

Continue Solution

Question 2

Add Formulae Menu

Back to Home

Find the exact value of cos(p + q)

in the diagram below

F(4,4)

G(3,-1)

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

Markers Comments

Add Form Menu

Back to Home

Next Comment

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

Markers Comments

Add Form Menu

Back to Home

Next Comment

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

• Work only with exact values when applying Pythagoras’:

Go to full solution

Go to Main Menu

Reveal answer only

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

Go to full solution

Go to Main Menu

Reveal answer only

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

Solution = {30, 150, 199.5, 340.5}

Markers Comments

Begin Solution

Continue Solution

Question 3

Add Formulae Menu

Back to Home

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

a°(180 - a)°

(180 + a)° (360 - a)°

Markers Comments

Begin Solution

Continue Solution

Question 3

Add Formulae Menu

Back to Home

Solve the equation

where 0<x<360.

sinx° = -1/3 or sinx° = 1/2

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}

Markers Comments

Add Form Menu

Back to Home

Next Comment

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

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°

Markers Comments

Add Form Menu

Back to Home

Next Comment

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

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.

Markers Comments

Add Form Menu

Back to Home

Next Comment

sinx° = -1/3 or sinx° = 1/2

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

Go to full solution

Go to Main Menu

Reveal answer only

Solve sin2 = cos where 0 < < 2

Go to full solution

Go to Main Menu

Reveal answer only

Solve sin2 = cos where 0 < < 2

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

Markers Comments

Begin Solution

Continue Solution

Question 4

Add Formulae Menu

Back to Home

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

Markers Comments

Begin Solution

Continue Solution

Question 4

Add Formulae Menu

Back to Home

Solve sin2 = cos

where 0 < < 2

= /2 or 3/2 Q1 or Q2

sin-1(1/2) = /6

Q1: = /6

Q2: = - /6 = 5/6

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

Markers Comments

Add Form Menu

Back to Home

Next Comment

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

sin2x = 2sinxcosx

1 radians180

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

Markers Comments

Add Form Menu

Back to Home

Next Comment

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

•Only one way to solve the result:

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

Markers Comments

Add Form Menu

Back to Home

Next Comment

= /2 or 3/2 Q1 or Q2

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

y = cos x

x = , 2 3

Markers Comments

Add Form Menu

Back to Home

Next Comment

= /2 or 3/2 Q1 or Q2

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

UNIT 2 :The Circle

1 2 3 4

EXITBack to

Unit 2 Menu

(a) Find the coordinates of C and W.

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

THE CIRCLE : Question 1

Go to full solution

Go to Circle Menu

Reveal answer only

The face of a stopwatch can be modelled by the circle with equation x2 + y2 – 10x – 4y + 5 = 0.

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

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

Go to full solution

Go to Circle Menu

Reveal answer only

The face of a stopwatch can be modelled by the circle with equation x2 + y2 – 10x – 4y + 5 = 0.

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

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

Markers Comments

Begin Solution

Continue Solution

Question 1

Circle Menu

Back to Home

The face of a stopwatch can be

modelled by the circle with

equation x2 + y2 – 10x – 4y + 5 = 0.

The centre is at C and the winder

is at W.

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 – 4y + 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).

Markers Comments

Begin Solution

Continue Solution

Question 1

Circle Menu

Back to Home

The face of a stopwatch can be

modelled by the circle with

equation x2 + y2 – 10x – 4y + 5 = 0.

The centre is at C and the winder

is at W.

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

Markers Comments

Circle Menu

Back to Home

Next Comment

Large circle is x2 + y2 – 10x – 4y + 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.

Markers Comments

Circle Menu

Back to Home

Next Comment

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.

Go to full solution

Go to Circle Menu

Reveal answer only

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.

A(7,0)

Go to full solution

Go to Circle Menu

Reveal answer only

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.

PA(7,0)

Q y = -4x + 28

B is (10,-12)

(x – 18)2 + (y + 10)2 = 68

Markers Comments

Begin Solution

Continue Solution

Question 2

Circle Menu

Back to Home

(a) The small roller has centre P

and equation

x2 + y2 – 6x + 2y - 7 = 0.

(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

Markers Comments

Begin Solution

Continue Solution

Question 2

Circle Menu

Back to Home

and equation

x2 + y2 – 6x + 2y - 7 = 0.

P is (3,-1)

Gradient of PA = 0 – (-1) 7 - 3

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

Markers Comments

Begin Solution

Continue Solution

Question 2

Circle Menu

Back to Home

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)

Markers Comments

Begin Solution

Continue Solution

Question 2

Circle Menu

Back to Home

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

Markers Comments

Circle Menu

Back to Home

Next Comment

(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

Markers Comments

Circle Menu

Back to Home

Next Comment

• 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

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

Go to full solution

Go to Circle Menu

Reveal answer only

Part of the mechanism inside an alarm clock consists of

three different sized collinear cogwheels.

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

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

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

Find the equation of the large circle.

Go to full solution

Go to Circle Menu

Reveal answer only

Part of the mechanism inside an alarm clock consists of

three different sized collinear cogwheels.

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

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

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

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

Markers Comments

Begin Solution

Continue Solution

Question 3

Circle Menu

Back to Home

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

So radius of large circle = 35

C D E5 35

Markers Comments

Begin Solution

Continue Solution

Question 3

Circle Menu

Back to Home

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

It follows that CD = 4/9CE

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

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

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

Markers Comments

Circle Menu

Back to Home

Next Comment

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

• Use formula sheet to check

correct formulas and relate to

equation of the circle.

Markers Comments

Circle Menu

Back to Home

Next Comment

It follows that CD = 4/9CE

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

= 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

CD : DE = 4 : 5

5CD = 4DE

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

9d = 5c + 4e

��

(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.

Go to full solution Reveal answer onlyEXIT

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.

(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.

Go to full solution Reveal answer onlyEXIT

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

Markers Comments

Begin Solution

Continue Solution

Question 4

Circle Menu

Back to Home

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

Markers Comments

Begin Solution

Continue Solution

Question 4

Circle Menu

Back to Home

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.

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)}

Markers Comments

Begin Solution

Continue Solution

Question 4

Circle Menu

Back to Home

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)

Markers Comments

Begin Solution

Continue Solution

Question 4

Circle Menu

Back to Home

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 ….

Markers Comments

Begin Solution

Continue Solution

Question 4

Circle Menu

Back to Home

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).

Markers Comments

Circle Menu

Back to Home

Next Comment

(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

Markers Comments

Circles Menu

Back to Home

Next Comment

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)

Markers Comments

Circle Menu

Back to Home

Next Comment

(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:

b b ac

UNIT 3 :

Logs & Exponential

Wave Function

Further Calculus

Vectors

UNIT 3 :Wave Function

EXITBack to

Unit 3 Menu

WAVE FUNCTION: Question 1

Go to full solution

Go to Wave Function Menu

Go to Main Menu

Reveal answer only

(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.

Go to full solution

Go to Main Menu

Reveal answer only

(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.

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

Solution is {36.9}(b)

Markers Comments

Begin Solution

Continue Solution

Question 1

Wave Function Menu

Back to Home

(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

Markers Comments

Begin Solution

Continue Solution

Question 1

Wave Function Menu

Back to Home

(b) Hence solve

cos(x°) + 7sin (x°) = 5

for 0 x 90.

so k = 50 = 252 = 52

ksina°kcosa° =

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!

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

Markers Comments

Begin Solution

Continue Solution

Question 1

Wave Function Menu

Back to Home

(b) Hence solve

cos(x°) + 7sin (x°) = 5

for 0 x 90.

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° (**)

Markers Comments

Begin Solution

Continue Solution

Question 1

Wave Function Menu

Back to Home

(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}

Markers Comments

Wave Func Menu

Back to Home

Next Comment

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)

Markers Comments

Wave Func Menu

Back to Home

Next Comment

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°

Markers Comments

Wave Func Menu

Back to Home

Next Comment

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

Markers Comments

Wave Func Menu

Back to Home

Next Comment

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

Go to full solution

Go to Main Menu

Reveal answer only

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

where k>0 and 0 < a < 360.

Go to full solution

Go to Main Menu

Reveal answer only

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

where k>0 and 0 < a < 360.

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

Markers Comments

Begin Solution

Continue Solution

Question 2

Wave Function Menu

Back to Home

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

Markers Comments

Begin Solution

Continue Solution

Question 2

Wave Function Menu

Back to Home

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° =

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!

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

Markers Comments

Wave Func Menu

Back to Home

Next Comment

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

Markers Comments

Wave Func Menu

Back to Home

Next Comment

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

Markers Comments

Wave Func Menu

Back to Home

Next Comment

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

Markers Comments

Wave Func Menu

Back to Home

Next Comment

so k = 20 = 45 = 25

ksina°kcosa° =

tana° = (-1/2) so Q2 or Q4

tan-1(1/2) = 26.6°

Q4: a° = 360° – 26.6° = 333.4°

so a in Q4!

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.

y = 5sin(x°) - 12cos(x°)

Go to full solution

Go to Wave Funct Menu

Go to Main Menu

Reveal answer only

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.

y = 5sin(x°) - 12cos(x°)

Go to full solution

Go to Wave Funct Menu

Go to Main Menu

Reveal answer only

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

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

Max TP is at (157.4,13)

Markers Comments

Begin Solution

Continue Solution

Question 3

Wave Function Menu

Back to Home

(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

Markers Comments

Begin Solution

Continue Solution

Question 3

Wave Function Menu

Back to Home

so k = 13

ksina°kcosa° =

so a in Q4!

tana° = (-12/5) so Q2 or Q4

tan-1(12/5) = 67.4°

Q4: a° = 360° – 67.4° = 292.6°

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

Markers Comments

Begin Solution

Continue Solution

Question 3

Wave Function Menu

Back to Home

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)

Markers Comments

Wave Func Menu

Back to Home

Next Comment

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°)

Markers Comments

Wave Func Menu

Back to Home

Next Comment

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

Markers Comments

Wave Func Menu

Back to Home

Next Comment

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

Markers Comments

Wave Func Menu

Back to Home

Next Comment

so k = 13

ksina°kcosa° =

so a in Q4!

tana° = (-12/5) so Q2 or Q4

tan-1(12/5) = 67.4°

Q4: a° = 360° – 67.4° = 292.6°

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

Markers Comments

Wave Func Menu

Back to Home

Next Comment

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°

Markers Comments

Wave Func Menu

Back to Home

Next Comment

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

1 2 3 4 5

EXITBack to

Unit 3 Menu

LOGS & EXPONENTIALS: Question 1

Go to full solution

Go to Logs & Exp Menu

Go to Main Menu

Reveal answer only

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.

Go to full solution

Go to Main Menu

Reveal answer only

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

Markers Comments

Begin Solution

Continue Solution

Question 1

Logs & Exp Menu

Back to Home

(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

Markers Comments

Begin Solution

Continue Solution

Question 1

Logs & Exp Menu

Back to Home

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..)

Markers Comments

Logs & Exp Menu

Back to Home

Next Comment

(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

Markers Comments

Logs & Exp Menu

Back to Home

Next Comment

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..)

• 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

Markers Comments

Logs & Exp Menu

Back to Home

Next Comment

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..)

• 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.

Go to full solution

Go to Main Menu

Reveal answer only

log10y

log10x1

0.3The graph illustrates the law y = k

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

Go to full solution

Go to Main Menu

Reveal answer only

log10y

log10x1

0.3The graph illustrates the law y = k

Hence k = 2 and n = 0.3

Markers Comments

Begin Solution

Continue Solution

Question 2

Logs & Exp Menu

Back to Home

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

The graph illustrates the law

y = kxn

log10y

log10x1

Markers Comments

Begin Solution

Continue Solution

Question 2

Logs & Exp Menu

Back to Home

or log10y = log102x-0.3law1The graph illustrates the law

y = kxn

log10y

log10x1

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

Markers Comments

Logs & Exp Menu

Back to Home

Next Comment

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

Markers Comments

Logs & Exp Menu

Back to Home

Next Comment

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

Markers Comments

Straight Line Menu

Back to Home

Next Comment

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

• Two coordinates from the graph:

log10y

log10x1

Markers Comments

Straight Line Menu

Back to Home

Next Comment

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.

Go to full solution

Go to Main Menu

Reveal answer only

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

Go to full solution

Go to Main Menu

Reveal answer only

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

x = 22

Markers Comments

Begin Solution

Continue Solution

Question 3

Logs & Exp Menu

Back to Home

If log3(5) – log3(2x + 1) = -2

Solve the equation

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

2x + 1( ) = -2law2

(2x + 1)= 3-2

(2x + 1)= 1

9Cross mult

we get 2x + 1 = 45

or 2x = 44

ie x = 22

Markers Comments

Logs & Exp Menu

Back to Home

Next Comment

If log3(5) – log3(2x + 1) = -2

then log3 5

2x + 1( ) = -2

(2x + 1)= 3-2

(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

Markers Comments

Logs & Exp Menu

Back to Home

Next Comment

If log3(5) – log3(2x + 1) = -2

then log3 5

2x + 1( ) = -2

(2x + 1)= 3-2

(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:

Markers Comments

Logs & Exp Menu

Back to Home

Next Comment

If log3(5) – log3(2x + 1) = -2

then log3 5

2x + 1( ) = -2

(2x + 1)= 3-2

(2x + 1)= 1

9Cross mult

we get 2x + 1 = 45

or 2x = 44

ie x = 22

Drop log3 terms

= 3-2 52x+1

= 52x+1

Go to full solution

Go to Main Menu

Reveal answer only

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?

Go to full solution

Go to Main Menu

Reveal answer only

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.

1.8psi

(a) k = 0.243 to 3 dps

Markers Comments

Begin Solution

Continue Solution

Question 4

Logs & Exp Menu

Back to Home

(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

Markers Comments

Begin Solution

Continue Solution

Question 4

Logs & Exp Menu

Back to Home

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

Markers Comments

Logs & Exp Menu

Back to Home

Next Comment

(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

Markers Comments

Logs & Exp Menu

Back to Home

Next Comment

• 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

(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

Go to full solution

Go to Main Menu

Reveal answer only

Given that sin = au and cos = av show that

u – v = logatan.

Go to full solution

Go to Main Menu

Reveal answer only

Given that sin = au and cos = av show that

u – v = logatan.

u – v = logatan

Markers Comments

Begin Solution

Continue Solution

Question 5

Logs & Exp Menu

Back to Home

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)

Markers Comments

Logs & Exp Menu

Back to Home

Next Comment

• 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

Markers Comments

Logs & Exp Menu

Back to Home

Next Comment

Take logs to base a to both sides

loga tan = loga

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

1 2 3 4 5

EXITBack to

Unit 3 Menu

VECTORS: Question 1

Go to full solution

Go to Vectors Menu

Go to Main Menu

Reveal answer only

(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

Go to full solution

Go to Vectors Menu

Go to Main Menu

Reveal answer only

(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

Markers Comments

Begin Solution

Continue Solution

Question 1

Vectors Menu

Back to Home

(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

Since PQ and

PR are

Markers Comments

Begin Solution

Continue Solution

Question 1

Vectors Menu

Back to Home

(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)

Markers Comments

Vectors Menu

Back to Home

Next Comment

• Must know result:

Given coordinates of A and B

AB = b 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

Since PQ and

PR are

Markers Comments

Vectors Menu

Back to Home

Next Comment

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

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

��

Markers Comments

Vectors Menu

Back to Home

Next Comment

(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:

Markers Comments

Vectors Menu

Back to Home

Next Comment

(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

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

S(-25,7,16)

��

VECTORS: Question 2

Go to full solution

Go to Vectors Menu

Go to Main Menu

Reveal answer only

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

FE and

component form.

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

VECTORS: Question 2

Go to full solution

Go to Vectors Menu

Go to Main Menu

Reveal answer only

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

FE and

component form.

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

(a) [ ] -1 4 -3

FG [ ]

3 0 -3

so = 73.9°

Markers Comments

Begin Solution

Continue Solution

Question 2

Vectors Menu

Back to Home

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.

FE and

component form.

(b) Hence, or otherwise, find the

size of angle EFG.

Markers Comments

Begin Solution

Continue Solution

Question 2

Vectors Menu

Back to Home

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

respectively.

FE and

component form.

size of angle EFG.

(b) Let angle EFG =

FG = [ ]

-1 4 -3

[ ] 3 0 -3

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

= -3 + 0 + 9

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

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

Markers Comments

Begin Solution

Continue Solution

Question 2

Vectors Menu

Back to Home

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

respectively.

FE and

component form.

size of angle EFG.

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

then cos = FE.FG

|FE||FG|=

so = cos-1(6 26 18) = 73.9°

Markers Comments

Vectors Menu

Back to Home

Next Comment

(b) Let angle EFG =

FG = [ ]

-1 4 -3

[ ] 3 0 -3

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

= -3 + 0 + 9

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

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

• Ensure vectors are calculated

from the vertex:

Vectors FE and FG��

Markers Comments

Vectors Menu

Back to Home

Next Comment

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

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

FE.FG=a b +a b +a b��

VECTORS: Question 3

Go to full solution

Go to Vectors Menu

Reveal answer only

PQRSTUVW is a cuboid in which

PQ , PS & PW are represented by the vectors

[ ]and

-2 4 0

[ ]resp.

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

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

VECTORS: Question 3

Go to full solution

Go to Vectors Menu

Reveal answer only

PQ , PS & PW are represented by the vectors

[ ]and

-2 4 0

[ ]resp.

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

= 48.1°APB

Markers Comments

Begin Solution

Continue Solution

Question 3

Vectors Menu

Back to Home

PS + SA =

PS + 1/3ST

PS + 1/3PW

[ ]-2 4 0

+ [ ] -2 4 3

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

[ ]and

-2 4 0

[ ]resp.

Find the components of PA & PB

and hence the size of angle APB.

Markers Comments

Begin Solution

Continue Solution

Question 3

Vectors Menu

Back to Home

[ ] -2 4 3

PB = [ ]

3 4 9PQRSTUVW is a cuboid in which

[ ]and

-2 4 0

[ ]resp.

(b) Let angle APB =

[ ] -2 4 3

[ ] 3 4 9

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

= -6 + 16 + 27

Markers Comments

Begin Solution

Continue Solution

Question 3

Vectors Menu

Back to Home

[ ]and

-2 4 0

[ ]resp.

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

|PB| = (32 + 42 + 92)

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

then cos = PA.PB

|PA||PB|=

29 106

so = cos-1(37 29 106)

= 48.1°

Markers Comments

Vectors Menu

Back to Home

Next Comment

[ ] -2 4 3

PB = [ ]

(b) Let angle APB =

[ ] -2 4 3

[ ] 3 4 9

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

= -6 + 16 + 27

• Ensure vectors are calculated

from the vertex:

Vectors PB and PA��

Markers Comments

Vectors Menu

Back to Home

Next Comment

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

|PB| = (32 + 42 + 92)

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

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

PB.PA=a b +a b +a b��

VECTORS: Question 4

Go to full solution

Go to Vectors Menu

Go to Main Menu

Reveal answer only

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

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

VECTORS: Question 4

Go to full solution

Go to Vectors Menu

Go to Main Menu

Reveal answer only

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

(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

Markers Comments

Begin Solution

Continue Solution

Question 4

Vectors Menu

Back to Home

An equilateral triangle has

sides 4 units long which are

represented by the vectors

a , b & c .

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

(ii) b.(a – c)

NB: each angle is 60° but vectors should

be “tail to tail”

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

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

= 4X4Xcos60° + 4X4Xcos120°

= 16 X ½ + 16 X (-½ )

= 8 + (-8)

Dot product = 0 so

b & (a + c) are perpendicular.

Markers Comments

Begin Solution

Continue Solution

Question 4

Vectors Menu

Back to Home

An equilateral triangle has

sides 4 units long which are

represented by the vectors

a , b & c .

(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

Markers Comments

Vectors Menu

Back to Home

Next Comment

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

= 4X4Xcos60° + 4X4Xcos120°

= 16 X ½ + 16 X (-½ )

= 8 + (-8)

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.

Markers Comments

Vectors Menu

Back to Home

Next Comment

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

= 4X4Xcos60° + 4X4Xcos120°

= 16 X ½ + 16 X (-½ )

= 8 + (-8)

Dot product = 0 so

• Ensure all scalar products are

calculated from the vertex:

VECTORS: Question 5

Go to full solution

Go to Vectors Menu

Go to Main Menu

Reveal answer only

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

Go to full solution

Go to Vectors Menu

Go to Main Menu

Reveal answer only

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.

Markers Comments

Begin Solution

Continue Solution

Question 5

Vectors Menu

Back to Home

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

[ ] 4 2-6

[ ]-1 010

[ ]324

Markers Comments

Begin Solution

Continue Solution

Question 5

Vectors Menu

Back to Home

a = 2i + j – 3k, b = -i + 10k

& c = -2i + j + k.

(a) Find

(ii) | 2a + b |

are perpendicular.

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

Markers Comments

Begin Solution

Continue Solution

Question 5

Vectors Menu

Back to Home

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

Since the dot product is zero it follows

that the vectors are perpendicular.

Markers Comments

Vectors Menu

Back to Home

Next Comment

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

[ ]-1 010

[ ] 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 =

Markers Comments

Vectors Menu

Back to Home

Next Comment

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

• Must know formula for the magnitude of a vector:

n = b n a +b +c

Markers Comments

Vectors Menu

Back to Home

Next Comment

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

[ ]-2 1 1

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

= -6 + 2 + 4

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

Go to full solution

Go to Further Calculus Menu

Go to Main Menu

Reveal answer only

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

Go to full solution

Go to Main Menu

Reveal answer only

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

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

Markers Comments

Begin Solution

Continue Solution

Question 1

Further Calc Menu

Back to Home

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)

Markers Comments

Further Calc Menu

Back to Home

Next Comment

• Check formula sheet for correct result:

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

Markers Comments

Further Calc Menu

Back to Home

Next Comment

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

Go to full solution

Go to Main Menu

Reveal answer only

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

Go to full solution

Go to Main Menu

Reveal answer only

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

Markers Comments

Begin Solution

Continue Solution

Question 2

Further Calc Menu

Back to Home

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

(6x – 5)

g´(9) = 3

(6X9 – 5)

Markers Comments

Further Calc Menu

Back to Home

Next Comment

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

(6x – 5)

g´(9) = 3

(6X9 – 5)

49= 3/7

• Apply the laws of indices to replace the sign

• Apply the chain rule

• Learn the rule for differentiation

Multiply by the power then reduce the power by 1

Markers Comments

Further Calc Menu

Back to Home

Next Comment

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

(6x – 5)

g´(9) = 3

(6X9 – 5)

49= 3/7

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

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

Go to full solution

Go to Main Menu

Reveal answer only

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

Go to full solution

Go to Main Menu

Reveal answer only

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

Markers Comments

Begin Solution

Continue Solution

Question 3

Further Calc Menu

Back to Home

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

Markers Comments

Further Calc Menu

Back to Home

Next Comment

= 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

Markers Comments

Further Calc Menu

Back to Home

Next Comment

= 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)

cos(ax) sin(ax) a

• Do not forget the constant of integration.

Go to full solution

Go to Main Menu

Reveal answer only

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

(b) Hence find x2

(2x3 + 1)1/3

Go to full solution

Go to Main Menu

Reveal answer only

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

(b) Hence find x2

(2x3 + 1)1/3

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

Markers Comments

Begin Solution

Continue Solution

Question 4

Further Calc Menu

Back to Home

(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

(2x3 + 1)1/3

Markers Comments

Begin Solution

Continue Solution

Question 4

Further Calc Menu

Back to Home

(b) Hence find

(2x3 + 1)1/3

(b) From (a) it follows that

(2x3 + 1)1/3

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

(2x3 + 1)1/3

= ¼ X

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

Markers Comments

Further Calc Menu

Back to Home

Next Comment

(2x3 + 1)2/3

outer inner

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

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

(2x3 + 1)1/3

•. Apply the chain rule“Peel an onion”

(2x3 + 1)2/3

outer inner

3 32( ) ( )

3 22x +1 6x

Markers Comments

Further Calc Menu

Back to Home

Next Comment

(b) From (a) it follows that

(2x3 + 1)1/3

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

(2x3 + 1)1/3

= ¼ X

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

i.e. given

higher – additional question bank

gradient of line

form y

straight line menugo

menustraight line

straight line menu8x

straight line menu3x

straight lineyou

original equation

Documents

higher psychology question paper 2015 candidate 2 · pdf...

intermediate 2 – additional question bank

additional science 1 h unit 5 · 6 gcse additional science...

gcse - ysgol john bright · gcse 239/02 additional science...

intermediate 2 – additional question bank

unit 1 higher question paper january 2012

2019 modern studies higher finalised marking...

higher business management question paper 2015 … ›...

higher music specimen question paper

higher human biology specimen question paper

mathematicsmaths777.weebly.com/.../ah_aqb_selective_csys_1996-2000.pdfmathematics...

higher geography specimen question paper

gcse biology specimen question paper higher specimen ... ·...

question leave number blank higher tier

higher media question paper 2015 - q2 candidate f€¦ ·...

nick lacey repertoire additional to question 1

question leave higher tier - edexcel

additional examplary multiple-choice question for assessment

b3 higher question paper

gcse additional science higher tier … in additional...