PreambleGeneral Marking Guidance:

This mark scheme is designed to reward candidates for their achievements and Examiners should ensure that they use it with the same intention. Examiners should be prepared to use all of the marks available on this mark scheme judiciously; where full marks are deserved, this should always be awarded to candidates; likewise, Examiners should beprepared to award no marks to responses that are not worthy of credit. At all times, Examiners should ensure that they only use this mark scheme when assessing a response and not any other private schemes.

It should be remembered that a mark scheme is a working document that may be further developed or refined in thefuture.

General Marking Instructions:

This question paper has 75 marks.

In this mark scheme, marks are awarded in one of three forms: M marks - these are method marks and are awarded when candidates know a correct method and apply it correctly. A marks - these are accuracy marks and are awarded for the correct answer. They can only be awarded in conjunction with the relevant method mark(s). B marks - these are accuracy marks and are independent of method marks.

Common abbreviations used in this mark scheme include:

bod - benefit of doubt ft - follow through cao - correct answer onlycso - correct solution onlyisw - ignore subsequent workingawrt - answers which round tosc - special caseoe - or equivalentd... - dependent * - answer given in question

Unless stated otherwise, it should be assumed that all accuracy marks are cao (correct answer only). Do not treat accuracy marks as follow through unless explicitly instructed to do so by the mark scheme.

For a genuine misread that does not alter the nature or difficulty of the question, you should deduct two A or B marksand treat any other accuracy marks as ft. Method marks are unaffected.

If a candidate makes more than one attempt at a question, you should mark the attempt that is not crossed out. If allattempts are crossed out or no attempts are crossed out, mark all the attempts and score the highest single attempt.

Following a correct answer, any subsequent working should be ignored, including any incorrect workings or statements,unless stated otherwise.

Unless stated in the mark scheme, workings ascribed to one part of a question by a candidate can only score marks for that part of the question.

Use of a formula:We require candidates to state a formula before they use it. If a candidate states a formula before they use it, you may award method marks even if there are small errors made in the substitution of numbers, for example. If a candidate does not state a formula before they use it, method marks can be lost through errors in their working.

PreambleGeneral Principles for Core Mathematics Marking:

Method Marks:

Solving 3TQs of the form ax + bx + c = 0:Way 1: Factorisation - the method mark will be awarded for factorisation of the form (mx + p)(nx + q), where |mn| = a and |pq| = c, leading to x = ... . Way 2: Quadratic Formula - the method mark will be awarded for an attempt to use the formula with the values of a, b and c.Way 3: Completing the square - the method mark will be awarded for , leading to x = ... .

Differentiation:The method mark is awarded for decreasing the power of at least one term by one.

Integration: The method mark is awarded for increasing the power of at least one term by one.

2

a x + b2a

b2a

2

+ ca

= 0

No Working Shown:If an answer can be reasonably obtained without showing any working and it is unlikely that the correct answer may arisefrom incorrect workings, then Examiners should award full marks. It is a given that candidates are aware that an incorrectanswer obtained without working will inevitably receive no marks. In the cases where working is deemed necessary, evidence of a method is required for the candidate to be awarded any marks.

When considering whether an answer may be obtained ‘without showing any working’, Examiners should consider twothings. The first is whether or not it can be done ‘in your head’. The second is that many permitted calculators may allow candidates to solve a question directly (this is provided the exam is a calculator exam). If Examiners are in doubt,they should consult with us.

In many instances, a question may ask candidates to use a specific method or ‘show’ or ‘deduce’ a result. In such cases, candidates who provide no evidence of their workings have broken the rubric of the question and should be awardedno marks.

Question Number

Scheme Marks

1

Volume of S = ( )

2 222 2

0 0

e 1xy dx dxπ π= −∫ ∫ Correct formula with the correct

limits M1B1

( )22

4 2 4 2

00

1e 2e 1 e e4

x x x xdx xπ π ⎡ ⎤= − + = − +⎢ ⎥⎣ ⎦∫

Attempts to integrate the expression M1A1

Volume 8 4 8 41 1 1 11e e 2 1 e e4 4 4 4

π π⎛ ⎞ ⎛ ⎞= − + − + = − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ Correct volume. Equivalent forms

shown in the notes below A1 oe

5

Question 1 Notes

1st M1

B1

2nd M1

1st A1

2nd A1

Attempts to use the formula for the volume of revolution about the x-axis.

For the correct limits of integration.

Some sight of the ‘reverse chain rule’, i.e. adjusting the constant term correctly.

Correct indefinite integration.

Correct definite integration. Equivalent forms should be accepted, including but not exclusive to:

8 41 1e e 2 14 4

V π ⎛ ⎞= − + − +⎜ ⎟⎝ ⎠ scores the 2nd A1

8 41 11e e4 4

V π ⎛ ⎞= − +⎜ ⎟⎝ ⎠ scores the 2nd A1

( )8 4e 4e 114

V π= − + scores the 2nd A1

V = awrt{ } 2180 scores the 2nd A1

y = e2x −1

Question Number

Scheme Marks

2 1+ x − 14 + x

1+ x = 1+ x( )

12

Suitable form for the binomial expansion

1+ x( )12 = 1+ 1

2x + 1

212−1⎛

⎝⎜⎞⎠⎟12!

⎛⎝⎜

⎞⎠⎟ x( )2 + ...

Attempts to expand the bracket using the binomial theorem

M1A1

14 + x

= 4 + x( )−12 = 1

21+ 14x⎛

⎝⎜⎞⎠⎟−12

Suitable form for the binomial

expansion B1

4 + x( )−12 = 1

21+ − 1

2⎛⎝⎜

⎞⎠⎟14x + − 1

2⎛⎝⎜

⎞⎠⎟ − 1

2−1⎛

⎝⎜⎞⎠⎟12!

⎛⎝⎜

⎞⎠⎟14x⎛

⎝⎜⎞⎠⎟2

+ ...⎛

⎝⎜⎞

⎠⎟

Expands M1A2

1+ x − 14 + x

= 12+ 916

x − 35256

x2 + ... Combines their expansions A1

7

Question 2 Notes

1st M1

1st A1

B1

2nd M1

2nd A1

3rd A1

4th A1

Alts

Attempts to use the binomial theorem to expand (1+x)1/2.

Correct first three (ascending) terms in the binomial expansion of (1+x)1/2.

Writes (4+x)-1/2 in a form that enables use of the binomial theorem.

Attempts to use the binomial theorem to expand (4+x)-1/2.

First two terms in the binomial expansion correct. Need not be simplified.

Term in x2 correct. Need not be simplified.

Correct simplified binomial expansion with each term in its simplest form.

Attempts at a Maclaurian expansion should be sent to review.

Question Number

Scheme Marks

3 g x( ) = x1x

(a) g is defined for all positive real numbers x Correct domain oe B1 [1]

(b) This question instructs candidates to use logarithmic differentiation. Other methods are hence a rubric infringement and should receive no marks.

ln g x( )( ) = 1xln x

Takes logs M1

g' x( )g x( ) =

1x1x

⎛⎝⎜

⎞⎠⎟ + ln x − 1

x2⎛⎝⎜

⎞⎠⎟

Uses implicit differentiation and the product rule

M2A1

g' x( ) = x1x 1x2

− 1x2ln x⎛

⎝⎜⎞⎠⎟ =

x1x 1− ln x( )x2

Correct differentiation oe A1

[5]

(c) x1x ≠ 0 , ln x = 1⇒ x = e

Sets their (b) = 0 M1A1

g e( ) = e1e

Correct y coordinate of stationary point

A1 [3]

9

Question 4 Notes

(a)

(b)

(c)

B1

1st M1

2nd M1

3rd M1

1st A1

2nd A1

Alts

M1

1st A1

2nd A1

Correct description of the domain of g. Condone “g is defined for all x greater than 0”. Statements that include “less than infinity” are erroneous and should receive B0.

Takes logs to both sides of the equation.

Differentiates both sides making use of implicit differentiation on the LHS.

Uses the product rule on the RHS. Condone at most one minor error.

Completely correct differentiation of both sides of the equations.

Correct value of g’(x) oe.

Candidates are free to use logarithms in bases other than e. The same scheme holds.

Sets their (b) = 0 with some attempt to find x.

Correct x coordinate of the stationary point. There is no ft.

Correct y coordinate of the stationary point. There is no ft.

Question Number

Scheme Marks

4 2x2 − 4( ) x + 2( )

(a) Way 1

2x2 − 4( ) x + 2( ) =

Ax − 2( ) +

Bx + 2( ) +

Cx + 2( )2

Correct partitioning M1

2 = A x + 2( )2 + B x − 2( ) x + 2( ) +C x − 2( )

Multiplies through by x − 2( ) x + 2( )2 dM1

Let x = 2 , then 2 = 16A⇒ A = 18

Substitutes a value of x and attempts to use this to find either A, B or C

dM1 A1

Let x = −2 , then 2 = −4C⇒C = − 12

Uses another substitution in attempt to

find another value ddM1

When x = 0 , 2 = 4A − 4B − 2C⇒ B = − 18

Another correct value

2x2 − 4( ) x + 2( ) =

18 x − 2( ) −

18 x + 2( ) −

12 x + 2( )2

Correct partial fractions A1 [6]

(a) Way 2

2x2 − 4( ) x + 2( ) =

Ax − 2( ) +

Bx + 2( ) +

Cx + 2( )2

Correct partitioning M1

2 = A x + 2( )2 + B x − 2( ) x + 2( ) +C x − 2( ) Multiplies through by x − 2( ) x + 2( )2 dM1

2 = A + B( )x2 + 4A +C( )x + 4A − 4B − 2C( ) Collects together the like terms

A + B = 0 , 4A +C = 0 , 2A − 2B −C = 1 Compares coefficients ddM1

2A + 2A + 4A = 1⇒ A = 18

Attempts to solve the system of

equations ddM1 A1

2x2 − 4( ) x + 2( ) =

18 x − 2( ) −

18 x + 2( ) −

12 x + 2( )2

Correct partial fractions A1 [6]

(b) 1x2 − 4( ) x + 2( )

0

1

∫ dx = 12

18 x − 2( ) −

18 x + 2( ) −

12 x + 2( )2

⎛

⎝⎜⎞

⎠⎟0

1

∫ dx Attempts to integrate their

(a) – ignore constant factors here

M1

= 1218ln x − 2 − 1

8ln x + 2 + 1

2x + 2( )−1⎡

⎣⎢⎤⎦⎥0

1

Method the integrate the various terms. See notes for exemplification

dM2 A2

= 12

− 18ln3+ 1

23( )−1 − 1

22( )−1⎛

⎝⎜⎞⎠⎟

Substitutes the limits in the correct way around and performs some manipulation

ddM1

= − 148

3ln3+ 2( )

Obtains the correct answer convincingly cao

A1 [6]

8

Question 5 Notes

(a) Way 1

(a) Way 2

(b)

1st M1

2nd M1

3rd M1

1st A1

4th M1

2nd A1

Note

1st M1

2nd M1

3rd M1

1st A1

4th M1

2nd A1

1st M1

2nd M1

1st A1

3rd M1

2nd A1

4th M1 3rd A1

Partitions the fraction correctly.

Multiplies through by (x – 2)(x + 2)2. This is dependent on the 1st M1.

Substitutes a value of x and attempts to find a value. This is dependent on 1st + 2nd M1.

One correct value of x.

Substitutes another value of x to find another value. This is dependent on previous M.

Fully correct partial fractions. NB: values of A, B and C alone are not sufficient.

Some candidates may substitute other values of x and form simultaneous equations to solve. The order and nature of the 3rd and 4th M marks should be altered accordingly.

Partitions the fraction correctly.

Multiplies through by (x – 2)(x + 2)2. This is dependent on the 1st M1.

Compares coefficients. This is dependent on both previous method marks.

Attempts to solve their system of equations. This is dependent on previous M marks.

One correct value.

Fully correct partial fractions. NB: values of A, B and C alone are not sufficient.

Attempts to integrate their (a). Ignore any constant factors here.

Attempts to integrate the f `(x) / f(x) terms. This is dependent on the 1st M1.

Correct integration of these terms.

Attempts to use the reverse chain rule on the ‘last’ term in the integral. This is dependent on the 1st M1.

Correct integration of this term.

Substitutes the limits in and performs some manipulation. Correct integration cao.

Question Number

Scheme Marks

5 A 1,2,0( ) , B 2,2,0( ) , C 3, 4, 1( )

(a) (i) / (ii)

BA =300

,

BC = 5

61

Correct direction vectors

BA and

BC

B1

BA = BABA

= 3i32

= i ,

BC = BCBC

= 5i 6 j j52 + 62 +12

= 162

5i 6 j j( )

Correct method to find unit vectors in either of the two

directions

M1A1

[3]

(b)

cos x = BA BCBA BC

= ... Attempts to use this formula M1

cos x = 1532 52 + 62 +12

= 562

62 * Correct dot product and

convincing proof A1A1

[3]

(c) Area = BA BC cos 90 x( ) = BA BC sin x = awrt{ }18 sq units Correct area M1A2 [3]

9

Question 5 Notes

(a)

(b)

(c)

B1

M1

A1

M1

1st A1

2nd A1

M1

1st A1

2nd A1

Correct direction vectors.

Attempts to find the length of the segment BA and BC.

Correct unit vectors for BA and BC.

Attempts to use the dot product.

Correctly dots BA and BC.

Convincingly shows the required result. Cao.

Forms the correct expression for the area of ABCD with some angle.

Correct angle of projection.

Correct area.

Question Number

Scheme Marks

6 y = x3 16 − x2 , −4 ≤ x ≤ 4 ,

(a)

dx = 4 cosu du Use of the chain rule B1

x3 16 − x2 dx =0

4

∫ 4sinu( )3 16 − 4sinu( )2 4 cosuduπ2

0

∫ Makes use of the substitution

(ignore limits here) M1

x3 16 − x2 dx =0

4

∫ 4sinu( )3 16 1− sin2 u( )4cosuduπ2

0

∫ Manipulation

x3 16 − x2 dx =0

4

∫ 4sinu( )3 4 cosu( ) 4cosu( )duπ2

0

∫ Uses 1− sin2 x = cos2 x dM1

x3 16 − x2 dx =0

4

∫ 1024 cos2 u sin3uduπ2

0

∫ * Correct limits ; complete and

convincing proof A1A1

[5]

(b)

x3 16 − x2 dx =0

4

∫ 1024 cos2 u sinu 1− cos2 u( )duπ2

0

∫ Manipulates the integral M1

= 1024 cos2 u sinu − cos4 u sinu( )duπ2

0

∫ Correct workings so far A1

= 1024 cos3u3

− cos5 u5

⎡

⎣⎢

⎤

⎦⎥π2

0

Attempts to integrate using the reverse

chain rule dM1 A1

= 1024 13− 15

⎛⎝⎜

⎞⎠⎟ =

204815

Correct integration A1

So area of R = 2 × 204815

= 409615

= awrt{ } 273 Correct value for the area of R ft

their integration A1ft

[6]

11

Question 6 Notes

(a)

(b)

B1

1st M1

2nd M1

1st A1

2nd A1

1st M1

1st A1

2nd M1

2nd A1

3rd A1

4th A1

Note

Uses the chain rule.

Good attempt to transform the integral from x to u. Ignore the limits.

Makes use of the relevant trig identity. This is dependent on 1st M1. Ignore the limits.

Correct limits on the integral (the correct way around oe).

Convincingly obtains the required result cao.

Begins to manipulate their (a) to reduce it to a form that can be integrated (easily).

Correct workings so far.

Attempts to integrate the expression using the reverse chain rule.

Fully correct indefinite integration.

Correct definite integration.

Correctly deduces the area of R ft their definite integration.

Attempts to integrate the curve directly score NO marks. Part (b) says hence (not ‘or otherwise’), so candidates must use part (a). Accept other variations of this method. Attempts at reducing the power of sin5x by double angles or reduction formulae should be sent to review.

Question Number

Scheme Marks

7 C1 : x = s2, y = 2s , C2 : x = 7t +1, y = −3t 2

(a)

dydx

= 2 12s

⎛⎝⎜

⎞⎠⎟ =

1s

Attempts to find gradient of curve M1

⇒ dydx s=p

= 1p

, so gradient of normal = − p Correct gradient of normal at P A1

y − 2p = − p x − p2( ) Attempts to find equation of normal dM1

y + px = p3 + 2p * Convincing proof A1 [4]

(b) (i)

Q 15,−12( ) Correct coordinates of Q

−12 + p 15( ) = p3 + 2p Substitutes their coordinates of Q into (a)

dM1

p3 −13p +12 = 0 * Convincing proof A1 [2]

(ii) 3( )3 −13 3( ) +12 = 27 − 39 +12 = 0 Convincingly shows p could equal 3 (accept other methods)

B1

p − 3 p3 −13p +12p2 + 3p − 4

Attempts to find the other factors M1

p = 3 , p = 1 , p = −4 Correct values of p A1

P 9,6( ) or P 1,2( ) or P 16,−8( ) Correct possible coordinates of P A1A1 [5]

11

Question 7 Notes

(a)

(b/i)

(b/ii)

1st M1

1st A1

2nd M1

2nd A1

M1

A1

B1

M1

1st A1

2nd A1

3rd A1

Attempts to find the gradient of the curve.

Correct value for the gradient of the normal to the curve at P.

Attempts to find the equation of the normal to the curve. This is dependent on 1st M1.

Convincing proof cso.

Substitutes their coordinates of Q into (a).

Convincing proof cso.

Shows convincingly that 3 is a solution to the equation.

Uses a method to find the other values of p, i.e. long division or inspection.

Correct values of p.

One correct possible coordinate for P.

All three possible coordinates for P found.

Question Number

Scheme Marks

8 (a) 2e sin dθ θ θ∫ , (b) 2cosec edyyd

θθθ=

(a) 2 2 21 1e sin sin e e cos 2 2

d dθ θ θθ θ θ θ θ⎛ ⎞= −⎜ ⎟⎝ ⎠∫ ∫ Use of integration by parts once M1A2

2 2 2 21 1 1 1e sin e sin e cos e sin 2 2 2 2

d dθ θ θ θθ θ θ θ θ θ⎛ ⎞= − +⎜ ⎟⎝ ⎠∫ ∫ Uses IBP a second time dM1 A1

2 2 2 21 1 1e sin e sin e cos e sin 2 4 4

d A dθ θ θ θθ θ θ θ θ θ= + − −∫ ∫ Correct integration (constant not necessary)

A1 [6]

(b) 2cosec edyyd

θθθ=

Separates the variables M1

2 2 24 1 1e sin e sin e cos5 2 4

dθ θ θθ θ θ θ⎛ ⎞= −⎜ ⎟⎝ ⎠∫ From part (i)

22 24 1 1e sin e cos

2 5 2 4y cθ θθ θ⎛ ⎞= − +⎜ ⎟⎝ ⎠

Correct integration on both

sides oe. Constant necessary here

A1 oe

Attempts to find use the boundary

conditions to find the particular solution to the ODE

22 24 1 1 7e sin e cos

2 5 2 4 10y θ θθ θ⎛ ⎞= − +⎜ ⎟⎝ ⎠

Correct particular solution. ISW from

this point onwards but accept equivalent forms if given

A1 oe [3]

9

12

2= 45

− 14

⎛⎝⎜

⎞⎠⎟ + c⇒ c = ...

Question 8 Notes

(a)

(b)

1st M1

1st A1

2nd A1

2nd M1

3rd A1ft

4th A1

1st M1

1st A1

2nd A1

Attempts to use IBP with u and v’ chosen correctly.

Correct differentiation of their u and integration of their v’.

Correct expression of the integral after applying IBP once.

Attempts to apply IBP a second time. This is dependent on the 1st M1.

Correctly applies IBP a second time ft their first application.

Convincingly manipulates the integral to obtain the form required.

Attempts to solve the differential equation by separating the variables.

Correctly separates the variables and correctly integrates both sides of the DE.

Correct particular solution to the DE. Accept equivalent forms, but ignore any working once candidates have obtained the final expression given in the question.