(2 marks)

States thatcos3

cot3sin3

xx

x

M1

Makes an attempt to find

cos3d

sin 3

xx

x

Writing

f 'd ln f

f

xx x

x or writing ln (sin x) constitutes an attempt.

M1

States a fully correct answer 1

ln sin33

x C A1

TOTAL: 3 marks

Begins the proof by assuming the opposite is true.

‘Assumption: there exists a rational numbera

bsuch that

a

bis the greatest positive rational

number.’

B1

Makes an attempt to consider a number that is clearly greater thana

b:

‘Consider the number 1a

b , which must be greater than

a

b’

M1

Simplifies 1a

b and concludes that this is a rational number.

1a a b a b

b b b b

By definition,a b

b

is a rational number.

M1

Makes a valid conclusion.

This contradicts the assumption that there exists a greatest positive rational number,

so we can conclude that there is not a greatest positive rational number.

B1

TOTAL: 4 marks

1

PURE MATHEMATICS

A level Practice Papers

PAPER O

MARK SCHEME

2

Makes an attempt to find

3sin 4 1 cos4 dx x x

Raising the power by 1 would constitute an attempt.

M1

States a fully correct answe r 3 41

sin 4 1 cos4 d 1 cos416

x x x x C M1

Makes an attempt to substitute the limits 4

41 11 0 1

16 2

M1 ft

Correctly states answer is 15

256

A1 ft

TOTAL: 4 marks

Makes an attempt to factor all the quadratics on the left-hand side of the identity. M1

Correctly factors each expression on the left-hand side of the identity:

(x - 6)(x + 6)

(x - 5)(x - 6)´

(5- x)(5+ x)

Ax2 + Bx + C´

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

(3x -1)(x + 6)

A1

Successfully cancels common factors:

(–1)(5+ x)(2x + 3)

Ax2 + Bx + Cº

x + 5

(-1)(x - 6)

M1

States that Ax2 + Bx + C º (2x + 3)(x - 6) M1

States or implies that A = 2, B = −9 and C = −18 A1

TOTAL: 5 marks

NOTES: Alternative method

Makes an attempt to substitute x = 0 (M1)

Finds C = −18 (A1)

Substitutes x = 1 to give A + B = −7 (M1)

Substitutes x = −1 to give A − B = 11 (M1)

Solves to get A = 2, B = −9 and C = −18 (A1)

3

4

Begins the proof by assuming the opposite is true.

‘Assumption: there exists a number n such that n is odd and n3 + 1 is also odd.’

B1

Defines an odd number: ‘Let 2k + 1 be an odd number.’ B1

Successfully calculates (2k +1)3 +1

(2k +1)3 +1º 8k3 +12k2 + 6k +1( ) +1º 8k3 +12k2 + 6k + 2

M1

Factors the expression and concludes that this number must be even.

8k3 +12k2 + 6k + 2 º 2 4k3 + 6k2 + 3k +1( )

2 4k3 + 6k2 + 3k +1( ) is even.

M1

Makes a valid conclusion.

This contradicts the assumption that there exists a number n such that n is odd and n3 + 1 is also

odd, so if n is odd, then n3 + 1 is even.

B1

TOTAL: 5 marks

NOTES: Alternative method

Assume the opposite is true: there exists a number n such that n is odd and n3 + 1 is also odd. (B1)

If n3 + 1 is odd, then n3 is even. (B1)

So 2 is a factor of n3. (M1)

This implies 2 is a factor of n. (M1)

This contradicts the statement n is odd. (B1)

Understands that for the series to be convergent 1r or states 4 1x M1

Correctly concludes that 1

4x . Accept

1 1

4 4x A1

(2 marks)

Understands to use the sum to infinity formula. For example, states 1

41 4x

M1

Makes an attempt to solve for x. For example, 3

44

x is seen. M1

States 3

16x A1

(3 marks)

TOTAL: 5 marks

5

6a

°

6b

°

States 10a b and 7 5 2a b M1

Makes an attempt to solve the pair of simultaneous equations.

Attempt could include making a substitution or multiplying the first equation by 5 or by 7.

M1

Finds a = −4 A1

Finds b = 6 A1

States −2abc = −96 M1

Finds c = −2 A1

TOTAL: 6 marks

Understands the need to complete the square, and makes an attempt to do this.

For example, 2

4x is seen.

M1

Correctly writes 2

g( ) 4 9x x A1

Demonstrates an understanding of the method for finding the inverse is to switch the x and y.

For example, 2

4 9x y is seen.

B1

Makes an attempt to rearrange to make y the subject.

Attempt must include taking the square root.

M1

Correctly states 1g ( ) 9 4x x A1

Correctly states domain is x > −9 and range is y > 4 B1

TOTAL: 6 marks

7

8

Makes an attempt to set up a long division.

For example: 9x2 + 0x -16 9x2 + 25x +16 is seen.

The ‘0x’ being seen is not necessary to award the mark.

M1

Long division completed so that a ‘1’ is seen in the quotient

and a remainder of 25x + 32 is also seen.

9x2 + 0x -16 9x2 + 25x +16

9x2 + 0x -16

25x + 32

1

M1

States B(3x + 4) + C(3x - 4) º 25x - 32 M1

Equates the various terms.

Equating the coefficients of x: 3B+3C = 25

Equating constant terms: 4B- 4C = 32

M1

Multiplies one or both of the equations in an effort to equate one of the two variables. M1

Finds 49

6B

A1

Finds 1

6C

A1

TOTAL: 7 marks

NOTES: Alternative method

Writes A+

B

3x - 4+

C

3x + 4 as

A(3x - 4)(3x + 4)

9x2 -16+

B(3x + 4)

9x2 -16+

C(3x - 4)

9x2 -16

States A(3x - 4)(3x + 4) + B(3x + 4) + C(3x - 4) º 9x2 + 25x +16

Substitutes x =

4

3 to obtain:

8B =

196

3Þ B =

49

6

Substitutes x = -

4

3 to obtain:

-8C = -

4

3Þ C =

1

6

Equating the coefficients of x2: 9A = 9 Þ A =1

9

States that sin1

BD and concludes that BD = sinq

M1

States that cos1

AD and concludes that AD = cosq

M1

States that ÐDBC =q M1

States that tansin

DC

and concludes that

2sin

cosDC

oe.

M1

States thatsin

cosBC

and concludes that BC = tanq oe.

M1

Recognises the need to use Pythagoras’ theorem. For example, AB2 + BC2 = AC2 M1

Makes substitutions and begins to manipulate the equation:

22

2 cos sin1 tan

1 cos

22 2

2 cos sin1 tan

cos

M1

Uses a clear algebraic progression to arrive at the final answer:

2

2 11 tan

cos

1+ tan2q = sec2q

A1

TOTAL: 8 marks

10

States4

sin7

xt

and

3cos

7

yt

M1

Recognises that the identity 2 2sin cos 1t t can be used to find the cartesian equation. M1

Makes the substitution to find 2 2 24 3 7x y A1

(3 marks)

States or implies that the curve is a circle with centre (−4, 3) and radius 7 M1 ft

Substitutesπ

2t to find x = −11 and y = 3 (−11, 3)

Substitutesπ

3t to find x ≈ 2.06 and y = 6.5 (2.06, 6.5)

Could also substitute t = 0 to find x = −4 and y = 10 (−4, 10)

M1 ft

A1 ft

(3 marks)

Makes an attempt to find the length of the curve by recognising that the length is part of the

circumference. Must at least attempt to find the circumference to award method mark.

2 π 7 14πC

M1 ft

Uses the fact that the arc is5

12of the circumference to write: arc length =

35π

6 A1 ft

(2 marks)

TOTAL: 8 marks

NOTES:

11b Award ft marks for correct sketch using incorrect values from part a.

11c Award ft marks for correct answer using incorrect values from part a.

11c Alternative method: use s r , with 7r and5

6

Award one mark for the attempt and one for the correct answer.

11a

11c

11b

Draws fully correct

curve.

Finds d

2sin 2d

xt

t and

dcos

d

yt

t

M1

Writes −2sin 2t = − 4sin t cos t M1

Calculates d cos 1

cosecd 4sin cos 4

y tt

x t t

A1

(3 marks)

Evaluatesd

d

y

xat

5

6t

dy

dx=

-1

4sin -5p

6

æ

èçö

ø÷

=1

2

A1 ft

Understands that the gradient of the tangent is1

2, and then the gradient of the normal is −2

M1 ft

Finds the values of x and y at5

6t

5 1

cos 26 2

x

and 5 1

sin6 2

y

M1 ft

Attempts to substitute values into 1 1( )y y m x x

For example,1 1

22 2

y x

is seen.

M1 ft

Shows logical progression to simplify algebra, arriving at:

1

22

y x or 4 2 1 0x y

A1

(5 marks)

TOTAL: 8 marks

NOTES: 12b

Award ft marks for a correct answer using an incorrect answer from part a.

12a

12b

M1

States that y =

2

x -1meets

y = ex in just one place, therefore

2

x -1= ex has just one root

g(x) = 0 has just one root

A1

(2 marks)

Makes an attempt to rearrange the equation. For example,

2

x -1- ex = 0 Þ xex - ex = 2

M1

Shows logical progression to state x = 2e-x +1

For example, x =

2 + ex

exis seen.

A1

(2 marks)

Attempts to use iterative procedure to find subsequent values. M1

Correctly finds: x1 = 1.4463 x2 = 1.4709 x3 = 1.4594 x4 = 1.4647 A1

(2 marks)

Correctly finds

¢g (x) = -2

(x -1)2- ex

A1

Finds g(1.5) = -0.4816...and ¢g (1.5) = -12.4816... M1

Attempts to find x

1: 0

1 0 1

0

g( ) 0.4816...1.5

g ( ) 12.4816...

xx x x

x

M1

Finds x

1= 1.461 A1

(4 marks)

TOTAL: 10 marks

13a

13b

13c

13d

Attempts to sketch

both y =

2

x -1and

y = ex

NOTES:

13a

Uses their graphing calculator to sketch g(x) =

2

x -1- ex (M1)

States that as g(x) only intersects the x-axis in one place, there is only one solution. (A1)

13c

Award M1 if finds at least one correct answer.

Writes:

1+ x

1- 2x as

(1+ x)(1- 2x)

-1

2 M1

Uses the binomial expansion to write:

21

2

1 3( 2 )

1 2 2(1 2 ) 1 ( 2 ) ...

2 2

x

x x

M1

Simplifies to obtain:

(1+ x)(1- 2x)-

1

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

2x2æ

èçö

ø÷ M1

Writes the correct final answer: 251 2

2x x … A1 ft

(4 marks)

Either states1

2x or states

-

1

2< x <

1

2 B1

(1 mark)

Makes an attempt to substitute 1

100x into

1+ x

1- 2x

For example

1 101

100 100

981

100100

1

1 2

M1

Continues to simplify the expression:

101

100´

100

98 And states the correct final answer:

101 2

140

A1

(2 marks)

Substitutes 1

100x into 25

1 22

x x Obtains:

1+ 21

100

æ

èçö

ø÷+

5

2

1

100

æ

èçö

ø÷

2

»1.02025 M1 ft

States that 1.02025 »

101 2

140

M1 ft

Deduces that140 1.02025

2 1.41421101

A1 ft

(3 marks)

TOTAL: 10 marks

NOTES:

14a Award 3 marks if a student has used an incorrect expansion but worked out all the other steps correctly.

14d Award all three marks if a student provided an incorrect answer in part a, but accurately works out

an approximation for root 2 consistent with this incorrect answer.

14a

14d

14c

14b

Correctly substitutes x = 1.5 into 3 214

2y x x and obtains 2.2323… A1

(1 mark)

States or implies formula for the trapezium rule: 0 1 2 3 422

hA y y y y y

M1

Makes an attempt to substitute into the formula 0.5

0 2 0.12103 0.86603 2.23235 02

A M1

States correct final answer 1.610 (4 s.f.) A1

(3 marks)

Recognises the need to make a substitution.

Method 1

24u x is seen.

M1

Correctly statesd

2d

ux

x and finds new

limits 0 4x u and 2 0x u

M1

Correctly transforms the integral

23 2

0

14 d

2

x

xx x x

into

1 30

2 2

4

14 d

4

u

uu u u

M1

Correctly finds the integral

03 5

2 2

4

1 8 2

4 3 5u u

M1

Makes an attempt to substitute the limits

3 5

2 2

3 5

2 2

8 20 0

3 51

4 8 24 4

3 5

M1

Correctly finds answer32

15

A1

(6 marks)

Using more strips would improve the accuracy of the answer. B1

(1 mark)

TOTAL: 11 marks

(TOTAL: 100 MARKS)

15a

15b

15c

15d

Method 2

1

2 24u x is seen

States 2 24u x and finds d

2 2d

uu x

x and finds

new limits 0 2x u and 2 0x u

into

02 4

2

14 d

2

u

uu u u

0

3 5

2

1 4 1

2 3 5u u

3 5

3 5

4 10 0

3 51

2 4 12 2

3 5

15c Either method is acceptable.