Session 1Paper 2 Questions and

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Question 1 (Unit 1 LO1 Straight Line)

Triangle ABC has as its vertices A(-18,6) , B(2,4) and C(10,-8) .

(a) Find the equation of the median from A to BC

(b) Find the equation of the perpendicular bisector of side AC.

(c) Find the coordinates of T, the point of intersection of these

lines.

A

B

C

marks 3, 4, 3

Solution 1(a)

mid-point of BC

gradient of median

equation of line

1

),( 26

2

3

32

)18(3

16 xy

3

1

24

8

618

26

AMm

(a) ans: xy3

1

284

2102 )(

,

M

Solution 1(b)

mid-point of AC

gradient of AC

perpendicular gradient

equation of line

1

2

3

2

1

28

14

1018

86

ACm

2m

),( 14

72 xy(b) ans:

4 )4(21 xy

286

21018 )(

,

N

Solution 1(c)

solving a system of

equations

x-coordinate

y-coordinate

1

2

3

723

1 xx

3x

1

7)3(2

y

y

ans: T(-3,1)

Question 2(Unit 1 LO3 Differentiation)

The diagram below shows the parabola with equation

and a line which is a tangent to the curve at the point T(1,5). Find the size of the angle marked θ, to the nearest degree.

238 xxy

marks (4)

Solution 2

Know to differentiate

Find gradient of tangent at x = 1

Use m = tanθ

Complete calculations

1

2

3

xdx

dy68

4

2tan

2)1(68 tangentm

21tan

ans: 63

63

Question 3(Unit 2 LO4 Circle )

058422 yxyx

yx 7

The circle in the diagram has equation

.The line AB is a chord of the circle and has equation

.

(a) Show that the coordinates of A and B are (-1, -8) and (6, -1) respectively.

(b) Establish the equation of the circle which

has AB as its diameter.

marks (4,3)

x

A

B

y

O

Solution to question 3a

ans:

substituting into circle equation

multiplying brackets and tidying up

factorising and values of y

corresponding values of x

1

2

3

4

058)7(47 22 yyyy

016182 2 yy

8or1

0812

y

yy

A(-1, -8) and B(6, -1)

A(-1, -8) and B(6, -1)

Solution to question 3b

ans:

knowing to find midpoint of AB

finding radius

substituting into equation

1

2

3

(25, -45)

524)541()526( 22 r

52454y52x 22

52454y52x 22

The graph of the cubic function y = f (x) is shown in the diagram. There are turning points at (1,1) and (3,5).

Sketch the graph of y = f '(x)

Question 4(Unit 1 LO3 Differentiation)

x

y (3,5)

(1,1)

)(xfy

marks (3)

Solution 4

ans: sketch

Interpret stationary points

Parabola

Maximum TP

1

2

3

31

Session 2Paper 2 Questions and

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Harris Academy Supported Study

Question 5 (Unit 2 LO3 Trigonometry)

022cos3sin oxx

3600 x

Solve algebraically the equation

where

marks 5

Solution 5

double angle formula

re-arrange to zero and factorise

find roots

answers from

answers from

1

2

4

3

5

02sin213sin 2 xx

0)1sin2)(1sin3(

01sinsin6 2

xx

xx

21

31 xx sinsin or

5.160,5.19

330,210

ans: 1950, 16050 , 2100, 3300

31sin x

21sin x

An open box is designed in the shape of a cuboid with a square base.

The total surface area of the base andfour sides is 1200cm2 x

x

Question 6(Unit 1 LO3 Differentiation)

(a) If the length of the base is x centimetres, show that the volume V (x) is given by

3

4

1300)( xxxV

(b) Find the value of x that maximises the volume of the box.

h

marks (3,5)

Solution 6 (a)

ans: proof

Equation for surface area

Rearrange with h = …….

Find V

1

2

3

xhx 41200 2

3

22

4

1300

4

1200

xx

x

xxV

x

xh

4

1200 2

knowing to differentiate

differentiate

set derivative to zero

solve for x

nature table

Solution 6 (b)

ans: 20x

1

2

3

4

............)( xV

2

4

3030)( xxV

04

3300 2 x

5r 20

)(xV

shape

0

20x

max TP at x = 20

Question 7 (Unit 1 LO3 Differentiation Unit 2 LO1 Polynomials)

Also shown is the tangent to the curve at the point P

where

Part of the curve is shown in the diagram

xxxy 2410 23

1x

(a) Find the equation of the tangent.

(b) The tangent meets the curve again at Q. Find the coordinates of Q.

marks (4,4)

xxxy 2410 23

x

y

P

Q

O x=1

Solution 7(a)

differentiate

gradient

y-coordinate

equation

1

2

3

32

(a) ans: 87 xy

24203 2 xxdx

dy

24)1(20)1(3,1 2 mxat

15)1(24)1(10)1( 23 y

4 )1(715 xy

724203 m

Solution 7(b)

form equation

rearrange to zero

factorise

coordinates of Q

1

2

332

(a) ans: Q (8,64)

4

872410 23 xxxx

081710 23 xxx

)89)(1( 2 xxx

648)8(78 yx

1 1 -10 17 -8

1 -9 8

1 -9 8 0

)8)(1)(1( xxx

Question 8(Unit 2 LO2 Integration )

marks (4, 4)

The diagram shows the parabolas and 244 xxy

422 xxy

(a) Find the coordinates of the point A

(b) Calculate the area enclosed between the two curves.

4

A

x

y

O

244 xxy

422 xxy

Solution 8a

Form equation

Rearrange to = 0

Factorise and solve

Coordinates of A

(a) ans:

1

2

3

22 4442 xxxx

062 2 xx

30

0)3(2

xorx

xx

4 )7,3(A

)7,3(A

Solution 8b

(b) ans:

Integrate

Substitute limits

Answer

1

2

3

dxxx )62(3

0

2

4

2332 3

3

0xx

0)3(3)3( 2332

unitssquare9

27)27(32

unitssquare9

dxbottomtop )(

4

A

x

y

O

244 xxy

422 xxy

Session 3Paper 2 Questions and

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Question 9(Unit 2 LO4 Circle )

08420422 yxyx

A circle, centre C, has equation

.Show that the line with equation2y = x + 8 is a tangent to the circleand find the coordinates of the point of contact P

marks (5)

C

P

08420422 yxyx

82 xy

Solution 9

ans:

substitute into circle equation

multiply out brackets and simplify

factorise and solve for y

complete proof

point of contact

1

2

3

4

08420)82(482 22 yyyy

0180605 2 yy

60665 yyy

One point of intersection so line is a tangent

P(4, 6)

5 48)6(26 xy

Question 10(Unit 2 LO2 Integration)

The diagram shows a sketch of the graph of y = (x + 2)(x – 1)(x – 2) and the points P and Q

x

y

P Q

(0,4)

O(-2,0)

)2)(1)(2( xxxy

(a) Write down the coordinates of P and Q

(b) Find the total shaded area

marks (2,6)

Solution 10a

ans:

Coordinates of P

Coordinates of Q

1

2

)0,1(

)0,2(

)0,2()0,1( QP

Solution 10b

two integrals

multiply out brackets

integrate

integral from 0 to 1

integral from 1 to 2

total area

dxxxx )2)(1)(2(1

0

ans:

1

2

3

4

4423 xxxy

xxxx 42 23314

41

5

12111

127

6 127

12111

dxxxx )2)(1)(2(2

1

2212 units

2212 units

Question 11(Unit 2 LO3 Trigonometry)

The diagram shows a sketch of part of the graph of a trigonometric function whose equation is of the form

Find the values of a, b and c

cbxay sin

x

5

-3

0

cbxay sin

π

y

marks (3)

Solution 11

ans: 1,2,4 cba

Interpret amplitude

Interpret period

Interpret vertical displacement

1

2

3

4a

1c

2b

(b) The second diagram shows lines OA and OB. The angle between these two lines is 300

Question 12(Unit 1 LO1 Straight Line)

marks 3,1

02 yxa

.(a) The diagram shows line OA with equations

.The angle between OA and the x-axis is Find the value of a.

x

y

O

A

a

B

30

.

Calculate the gradient of line OB correct to 1 decimal place

Solution 12a

ans: 6.26

gradient of line

gradient = tan (angle) and apply

process

1

2

3

21gradient

gradient atan 0

6.262

1tan 1

Solution 12b

ans: 5.1

angle = tan-1(angle)

1 5.1)6.2630tan( m