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Mark schemeQuestio

nAnswer/Indicative content

Mark

sPart marks and guidance

1 y = x2 tan 2x M1 product rule u × their v′ + v × their u′ attempted

M1 d / du(tan u) = sec2u soi M0 if d / dx (tan 2x) = (2) sec2x

⇒ dy / dx = 2x2sec22x + 2xtan 2xA1ca

oor 2x2 / cos22x + 2xtan 2x isw

M1 product rulesee additional notes for complete solution

u × their v′ + v × their u′ attempted

A1 correct expression

= … = 2x2sec22x + 2xtan 2xA1ca

oor 2x2 / cos22x + 2xtan 2x (isw)

or (2x2 + 2xsin2xcos2x) / cos22x

or 2x2 / cos22x + 2xsin2x / cos2x

M1 quotient rule (v × their (u′ − u × their v′) / v2 attempted

A1 correct expression

= … = 2x2sec22x + 2xtan 2x A1ca

o

or 2x2 / cos22x + 2xtan 2x (isw)

Examiner's Comments

The derivative of tan x was usually

familiar, but those candidates who

started with sin 2x/cos 2x usually got lost

in algebraic complexity. A surprising

number lost marks through giving the

derivative of tan 2x as sec2x, or omitting

the ‘2’ in 2 sec22x.

However, better candidates just wrote

the result down.

or (2x2 + 2xsin2xcos2x) / cos22x

or 2x2 / cos22x + 2xsin2x / cos2x

Additional notes and solutions

© OCR 2017. You may photocopy this page. of 23 Created in ExamBuilder

Total 3

2 i at A y = 3 B1

i B1

itheir

M1* must follow from attempt at differentiation

i grad of normal = −1/their4

M1d

ep*

i y − 3 = (−¼) × (x − 4) oe isw A1

isubstitution of y = 0 and completion to given

result with at least 1 correct intierim step wwwA1

or substitution of x = 16 to obtain y = 0

Examiner's Comments

This was done extremely well, with the

majority of even the weakest candidates

scoring full marks. A few wrote 2x − 4 = 0

to incorrectly obtain m = 2 and made no

further progress, and a very small

minority tried to answer the question

without using calculus and working

backwards.

correct interim step may occur before

substitution

i

iat B, x = 3 B1 may be embedded

i

iM1*

condone one error, must be three terms,

ignore + c

i

iF[4] − F[their 3]

M1*d

epdependent on integration attempted

i

iarea of triangle = 18 soi B1 may be embedded in final answer

i

i Area of region = oe iswA1 19.3 or better

Examiner's Comments

Nearly all candidates identified the

coordinates of B correctly. However,

most — as if by rote — subtracted the

equation of the line from the equation of

the curve and then integrated. Some

candidates integrated the equation of the

curve correctly, but used the wrong limits

(usually 3 to 16) and made no further


progress, and of those that did adopt the

correct approach, a large number were

unable to find the area of the triangle

correctly (½ × 12 × 4 was common).

Total 11

3 i M1Rearranging for y and differentiating explicitly

is M0

i A1 correct equationIgnore superfluous dy/dx = … unless used

subsequently

i A1

o.e., but mark final answer

Examiner's Comments

This relatively simple implicit

differentiation was very well done by

almost all candidates.

i

i

B1de

pdep correct derivative

i

i⇒ 4 + 2y2 = 8 ⇒ y2 = 2, y = √2 or −√2

B1B

1

√2, −√2

Examiner's Comments

Most candidates scored two out of three

for the point (2, √2), but missed the y

= .√2 solution. In a few cases, the

denominator was set to zero, giving y =

0.

can isw, penalise inexact answers of ±1.41 or

better once only

−1 for extra solutions found from using y = 0

Total 6

4 i At P(a, a) g(a) = a so ½(ea −1) = a

i ⇒ ea = 1 + 2a* B1

NB AG

Examiner's Comments

This mark was usually earned.

i

iM1 correct integral and limits limits can be implied from subsequent work

i

iB1 integral of ex − 1 is ex − x

i

i= ½ (ea −a e0) A1

i

i= ½ (1 + 2a −a −1) = ½ a* A1 NB AG


i

iarea of triangle = ½ a2 B1

i

iarea between curve and line = ½ a2−½ a

B1ca

o

mark final answer

Examiner's Comments

Virtually everyone scored M1 for writing

down the correct integral and limits, but

many candidates made a meal of trying

to integrate ½ (ex – 1) , with ¼ (ex – 1)2

not an uncommon wrong answer. Having

successfully negotiated this hurdle, using

part (i) to derive ½ a was spotted by

about 50% of the candidates. Quite a few

candidates managed to recover to earn

the final 2 marks for ½ (a2 – a) (without

incorrectly simplifying this to½ a!).

i

i

i

y = ½(ex − 1) swap x and y

x = ½ (ey − 1)

i

i

i

⇒ 2x = ey − 1 M1 Attempt to invert — one valid step merely swapping x and y is not ‘one step’

i

i

i

⇒ 2x + 1 = ey A1

i

i

i

⇒ ln(2x + 1) = y*

⇒ g(x) = ln(2x + 1)

A1

y = ln(2x + 1) or

g(x) = ln(2x + 1) AGapply a similar scheme if they start with g(x)

and invert to get f(x).

or g f(x) = g((ex − 1)/2) M1

i

i

i

Sketch: recognisable attempt to reflect in

i

i

i

y = x M1 through O and (a, a) = ln(1 + ex − 1) = ln(ex) A1 = x A1

i

i

i

Good shape A1 no obvious inflexion or TP, extends to

third quadrant, without gradient

becoming too negative

Examiner's Comments

Finding the inverse function proved to be

an easy 3 marks for most candidates –

candidates are clearly well practiced in

this. The graphs were usually

recognisable reflections in y = x, but only

well drawn examples – without

unnecessary maxima or inflections –

similar scheme for fg

See appendix for examples


were awarded the ‘A’ mark.

i

vf ′(x) = ½ ex B1

i

vg ′(x) = 2/(2x + 1) M1

1/(2x + 1) (or 1/u with

u = 2x + 1) …

i

vA1 … × 2 to get 2/(2x + 1)

i

vg ′(a) = 2/(2a + 1) , f ′(a) = ½ ea B1 either g′(a) or f ′(a) correct soi

i

vso g ′(a) = 2/ea or f ′(a) = ½ (2a + 1) M1 substituting ea = 1 + 2a

i

v

= 1/(½ea) = (2a + 1)/2

[= 1/f ′(a)] [= 1/g ′(a])A1 establishing f ′(a) = 1/g ′(a) either way round

i

vtangents are reflections in y = x B1

must mention tangents

Examiner's Comments

This proved to be more difficult, as

intended for the final question in the

paper. As with the integral, many

candidates struggled to differentiate ½

(ex – 1) correctly, and equally many

omitted the ‘2’ in the numerator of the

derivative of ln(1 + 2x). Once these were

established correctly the substitution of x

= a and establishing of f'(a) = 1/g'(a) was

generally done well, though sometimes

the arguments using the result in part (i)

were either inconclusive or done

‘backwards’. The final mark proved to be

elusive for most, as we needed the word

‘tangent’ used here to provide a

geometric interpretation of the reciprocal

gradients.

Total 19

5 6x2 + 18x – 24 B1

their 6x2 + 18x − 24 = 0 or >0 or ⩽ 0 M1or sketch of y = 6x2 + 18x − 24 with attempt to

find x-intercepts

− 4 and + 1 identified oe A1

x < − 4 and x > 1 cao A1 or x ⩽ − 4 and x ≤ 1

Examiner's Comments

Most candidates differentiated correctly

and identified the correct values of x. The

final mark was often lost, either due to a

if B0M0 then SC2 for fully correct answer


misunderstanding of what had been

found — answer given as −4 < x < 1 or

poor notation — answer given as −4 > x

> 1. Those who used a graphical

approach with the derivative generally

scored full marks. A few candidates

missed the last term out, converted the

first plus sign to a minus sign or failed to

multiply 2 by 3 correctly, and lost the first

mark.

Total 4

6u = x, du / dx = 1, dv / dx = cos ½ x, v = 2sin

½ xM1 correct u, u′, v, v′

but allow v to be any multiple of sin ½ x

M0 if u = cos ½ x, v′ = x

A1ft consistent with their u, v

A1 2x sin ½ x + 4 cos ½ x oe (no ft)

M1substituting correct limits into correct

expression

can be implied by one correct intermediate

step

A1ca

o

NB AG

Examiner's Comments

There was a mixed response to the

question, with plenty of faultless answers,

but others with errors in v = 2sin ½ x, e.g.

v = sin ½ x or –2 sin ½ x or ½ sin ½

x. Occasionally there was insufficient

working to show that the given result had

been established: candidates are well

advised to include ample working.

Total 5

7 i h = 20, stops growing B1

AG need interpretation

Examiner's Comments

Most candidates correctly wrote down the

value of h but quite a number failed to

give the interpretation that the tree

stopped growing when its height was

20m.

i

i

h = 20 − 20e−t/10

dh/dt = 2e−t/10

M1A

1

differentiation (for M1 need ke−t/10, k

const)

i

i

20e−t/10 = 20 − 20(1 − e−t/10) = 20 − h

= 10dh/dtM1


i

iA1

oe eg 20 − h = 20 − 20(1 − e−t/10) =

20e−t/10

= 10dh/dt (showing sides equivalent)

i

iwhen t = 0, h = 20(1 − 1) = 0 B1 initial conditions

i

i

..............................................

.............OR verifying by integration

.........................................

..................

i

iM1 sep correctly and intending to integrate

i

i ⇒ −ln (20 − h) = 0.1t + c A1

correct result (condone omission of c,

although no further marks are possible)

condone ln (h − 20) as part of the

solution at this stage

i

i

h = 0,t = 0, ⇒ c = −ln 20⇒ ln(20 − h) = −0.1t + ln 20 B1

constant found from expression of

correct form (at any stage) but B0 if say

c = ln (−20) (found using ln (h − 20))

i

i⇒ 20 − h = 20e−0.1t M1

combining logs and anti-logging (correct

rules)

i

i⇒ h = 20(1 − e−0.1t) A1

correct form (do not award if B0 above)

Examiner's Comments

Those who approached the verification

by integration were quite successful. The

common errors were:-

omitting the negative sign

when integrating 1/(20ሢ h)

omitting the constant of

integration

giving ln(hሢ 20) in their

answers (without modulus

signs) despite having usually

given h=20 as a maximum

value in (i)

incorrect anti-logging.

Those who approached from

differentiation usually obtained some

marks, particularly the mark for checking

the initial conditions but many gave

insufficient detail when verifying the given

result.

i

i

i ⇒ 200 = A(20 − h) + B(20 + h)

M1

cover up, substitution or equating coeffs


i

i

i

h = 20 ⇒ 200 = 40 B, B = 5 A1

i

i

i

h = −20 ⇒ 200 = 40A, A = 5

200 dh/dt = 400 − h2

A1

i

i

i

M1separating variables and intending to

integrate (condone sign error)

i

i

i

substituting partial fractions

i

i

i

⇒ 5ln(20 + h) − 5ln(20 − h) = t + c A1ft their A, B, condone absence of c, Do

not allow ln (h-20) for A1.

i

i

i

When t = 0, h = 0 ⇒ 0 = 0 + c ⇒ c = 0 B1 cao need to show this. c can be found at

any stage. NB c = ln (−1) (from ln (h −

20)) or similar scores B0.

i

i

i

i

i

i M1

anti-logging an equation of the correct

form . Allow if c = 0 clearly stated

(provided that c = 0) even if B mark is not

awarded, but do not allow if c omitted.

Can ft their c.

i

i

i

⇒ 20 + h = (20 − h)et/5 = 20 et/5 − h et/5

⇒ h + h et/5 = 20 et/5 − 20⇒ h(et/5 + 1) = 20(et/5 − 1)

DM1making h the subject, dependent on

previous mark

i

i

i

NB method marks can be in either order,

in which case the dependence is the

other way around.(In which case, 20 + h

is divided by 20 − h first to isolate h).

i

i

i

A1 AG must have obtained B1 (for c) in

order to obtain final A1.

Examiner's Comments

There were a few completely correct

solutions to this part. However, many

different errors were seen from the

majority of candidates. There was also a

lot of confused work.

Those who started with the correct partial

fractions, from 200/(20ሢ h)(20+h) or

1/(20ሢ h)(20+h), usually obtained the

first three marks and then integrated


having scored M1A1A1M1 thus far.

Common errors then included omitting

the negative sign when integrating

5/(20ሢ h) (ie giving 5ln(20ሢ h) and

hence A0) or failing to state and then

evaluate a constant. Those who had no

constant were unable to score further

marks. Those who did score the first 5 or

6 marks (dependent upon when the

constant was evaluated) often used the

laws of logarithms correctly and anti-

logged although some fiddled the signs

when subsequently making h the subject.

Some candidates thought that

1/(400ሢ h2)=1/(hሢ 20)(h+20). Marks

were scored for using partial fractions on

1/(hሢ 20)(h+20) but logarithms such as

ln (hሢ 20) for h<20 and constants such

as ln(ሢ 1) could not obtain accuracy

marks although the marks for anti-logging

and making h the subject were still

available.

There were also a number who felt that

1/(400ሢ h2)=1/(200ሢ h)(200+h).

The use of modulus signs was rarely

seen.

i

v

As t → ∞, h → 20. So long-term height is

20m.B1

www

Examiner's Comments

Usually correct.

v 1st model h = 20(1 − e−0.1) = 1.90.. B1 Or 1st model h = 2 gives t = 1.05..

v 2nd model h = 20(e1/5 − 1)/(e1/5 + 1) = 1.99.. B1 2nd model h = 2 gives t = 1.003..

v so 2nd model fits data betterB1

dep

dep previous B1s correct

Examiner's Comments

Most candidates scored all three marks.

Total 19

8 i B1


i M1

i A1 oe

i A1

ag

Examiner's Comments

The differentiation was usually correct

and the use of the chain rule usually lead

to full marks for those that started

correctly.

i

i∫πxdx = ∫kdt M1

separate variables and attempt

integration of both sides

i

i⇒ ½ πx2 = kt + c A1 condone absence of c

i

i

When t = 0, x = 0 ⇒ c = 0⇒ ½ πx2 = ktB1 c = 0 www

i

i

Full when x = 10, t = T⇒ 50π = kTM1

substitute t or T = 50 π/k or x = 10 and

rearranging for the other (dependent on

first M1) oe

i

i⇒ T = 50π/k * A1.

ag, need to have c = 0

Examiner's Comments

The integration here was not difficult.

Most candidates scored either three

marks or five marks depending upon

whether they included a constant of

integration. It was very disappointing to

see how common this error was.

i

i

i

dV/dt = −kx B1 correct

i

i

i

M1 dV/dx.dx/dt = ±kx ft

i

i

i

A1

ag

Examiner's Comments

Those who started correctly with dV/dt =

−kx usually obtained full marks. Some

candidates had given up by this point.

i

v∫ π(20 − x) d x = ∫ −k dt M1

separate variables and intend to

integrate both sides

i π(20x − ½x2) = −kt + c B1 LHS (not dependent on M1)


v

i

vA1 RHS i.e. −kt + c (condone absence of c)

i

v

When t = 0, x = 10⇒ π(200 − 50) = c⇒ c = 150π

A1 evaluation of c cao oe (x = 10, t = 0)

i

v

⇒ π(20x − ½x2) = 150π − kt

x = 0 when 150π − kt = 0M1

substitute x = 0 and rearrange for t -

dependent on first M1 and non-zero c, oe

i

v⇒ t = 150π/k = 3T* A1

ag

Examiner's Comments

The separation of variables and

integration were again generally well

answered by those who attempted them.

As before, the constant was rarely

included or found and as it was non-zero

in this case, some confused attempts at

the final part were seen. As a result,

three marks were usually lost here.

Those who did include the constant were

usually successful in scoring all six

marks.

Total 18

9 e2y = 5 − e−x B1or y = ln√(5 − e−x) o.e (e.g.½ ln(5 − e−x))B1⟹ dy/dx = e−x/[2(5 − e−x)] o.e. B1

B1 = e−x (but must be correct)

M1d

ep

substituting x = 0, y = ln 2 into their dy/dx

dep 1st B1 − allow one slipor substituting x = 0 into their correct dy/dx

A1ca

o

Examiner's Comments

This implicit differentiation was generally

well done. The most common error was

d/dx(e-x) = ex instead of e-x. Some

candidates re-arranged the original

equation correctly to give y = ½ ln(5 –

e−x), though log errors were quite

common here; however, many went on

from here by differentiating this

incorrectly.

Total 4

1

0i y = 2 arc sin ½ = 2 x π/6 M1 y = 2 arcsin ½


i = π/3 A1

must be in terms of π− can isw

approximate answers

Examiner's Comments

This was generally answered

successfully, with only a few failing to

give the exact value π/3.

1.047... implies M1

i

iy = 2 arcsin x x ↔ y

i

i⇒ x = 2 arcsin y

i

i⇒ x/2 = arcsin y M1 or y/2 = arcsin x

i

i⇒ y = sin (x/2) [so g(x) = sin (x/2)] A1

but must interchange x and y at some

stage

i

i⇒ dy/dx = ½ cos(½ x)

A1ca

o

i

iAt Q, x = π/3 M1 substituting their π/3 into their derivative

i

i⇒ dy/dx = ½ cos π/6 = ½ √3/2 = √3/4 A1

must be exact, with their cos(π/6)

evaluated

i

i⇒ gradient at P = 4/√3 B1 ft

o.e. e.g. 4√3/3 but must be exact ft their

√3/4 unless 1

or f′(x) = 2/√(1−x2)

f′(½) = 2/√¾ = 4/√3 cao

i

i

Examiner's Comments

Most candidates successfully found the

inverse function, but ½ sin x was

occasionally seen. Once that hurdle was

crossed, most differentiated sin ½ x

correctly, though cos(½x) and 2 cos(½x)

were seen. The substitution of x = π/3

was usually correct, though a small

number used x = 1. The gradient at P

was usually the reciprocal of that at Q,

with –1/m (instead of 1/m) being the most

common error. A few candidates

differentiated f(x) directly, often with

success.

Total 8

1

1i M1 d/dx(sin 2x) = 2cos 2x soi can be inferred from dy/dx = 2x cos 2x

i A1 cao, mark final answer e.g. dy/dx = tan 2x + 2x is A0

i dy/dx = 0 when sin 2x + 2x cos 2x M1 equating their derivative to zero,


i provided it has two terms

i ⟹ tan 2x + 2x = 0* A1

must show evidence of division by cos

2x

Examiner's Comments

The vast majority differentiated correctly

- though 2xcos2x was seen occasionally

- and equated their derivative to zero.

Most then succeeded in dividing by cos

2x to arrive at the required result. Some

candidates, however, divided before

equating the derivative to zero, and gave

the derivative as 2x + tan 2x.

i

iAt P, x sin 2x = 0 M1 Finding x = π/2 using the given line

i

i⇒ sin 2x = 0, 2x = (0), π ⇒ x = π/2 A1 x = π/2 equation is M0

i

iAt P, dy/dx = sin π + 2(π/2) cos π = −π B1 ft ft their π/2 and their derivative

i

i

Eqn of tangent: y − 0 = −π (x − π/2)

⇒ y = −πx + π2/2M1

substituting 0, their π/2 and their −π into y

− y1 = m(x − x1)

or their −π into y = mx + c, and then evaluating

c: y = (−π)x + c,

i

i⇒ 2πx + 2y = π2 * A1 NB AG

0 = (−π) (π/2) + c M1⇒ c = π2/2

i

iWhen x = 0, y = π2/2, so Q is (0, π2/2)

M1A

1

can isw inexact answers from π2/2

Examiner's Comments

Most candidates solved x sin 2x = 0 to

obtain x = π/2 at P. The derivative was

then required to obtain the gradient of

the tangent and hence its equation, but

some used the given tangent equation

itself to find the gradient. The last part

was successfully completed by nearly all

candidates, with the given tangent

equation being used to obtain the correct

y-coordinate at Q of π2/2.

⇒ y = −πx + π2/2 ⇒ 2πx + 2y = π2 *A1

i

i

i

Area = triangle OPQ − area under curve M1soi (or area under PQ − area under

curve)

area under line may be expressed in integral

form

i

i

i

Triangle OPQ = ½ × π/2 × π2/2 [π3/8]B1ca

oallow art 3.9

or using integral:

i

i

i

Parts: u = x , dv/dx = sin 2x

du/dx = 1, v = −½ cos2xM1 condone v = k cos2x soi v can be inferred from their ‘uv’


i

i

i

A1ft ft their v = −½ cos2x, ignore limits

i

i

i

A1correct at this stage, ignore limits

i

i

i

A1ca

o(so dep previous A1)

i

i

i

So shaded area = π3/8 − π/4 = π(π2 − 2)/8* A1

NB AG must be from fully correct work

Examiner's Comments

Most candidates attempted find the area

of the triangle and the area under the

curve, though a clear statement of

method was not always given. Quite a

few candidates attempted to find the

triangle area by integration, and came

unstuck in the process. The area under

the curve was generally recognised as

integration by parts, but marks were lost

through incorrect v’, or mistakes with

signs. Some tried to combine both

integrals (for line and curve), and got into

a muddle by stock-piling negative signs,

rather than simplifying these on a step-

by-step basis. Nevertheless, good

candidates had little trouble in supplying

a fluent solution.

Total 18

1

2i When t = 2, r = 20(1 − e−0.4) = 6.59 m

M1A

16.6 or art 6.59

idr / dt = −20 × (−0.2e−0.2t)

= 4e−0.2tM1 −0.2e−0.2t soi

i When t = 2, dr / dt = 2.68 A1

2.7 or art 2.68 or 4e−0.4

Examiner's Comments

In part (i), the first two marks for finding

the radius when t = 2 were readily

achieved. Not so the next two, with some

generally rather poor attempts to

differentiate 20(1 − e−0.2t). Quite a few

candidates substituted t = 2 into e−0.2t to

get e−0.4, then differentiated this as

−0.4e−0.4. Some simply divided their value

of r by 2.

mark final answer

i A = πr2 M1 attempt to differentiate πr2 or differentiating 400π(1 − e−0.2t)2 M1


i

i

i⇒ dA / dr = 2πr (= 41.428…) A1 dA / dr = 2πr (not dA / dt, dr / dA etc.) dA / dt = 400π.2(1 − e−0.2t).(−0.2e−0.2t) A1

i

i

dA / dt = (dA / dr) × (dr / dt)

= 41.428… × 2.68M1

(o.e.) chain rule expressed in terms of

their A, r or implied

substitute t = 2 into correct dA / dt M1

(Could use another letter for A)

i

i= 111 m2 / hr A1

110 or art 111

Examiner's Comments

Part (ii) offered some accessible marks

for stating the chain rule, and for dA/dr =

2πr. The final mark depended on getting

dr/dt = 2.68 from part (i).

Total 8

1

3i

x3 + y3 = 3xy⇒ 3x2 + 3y2(dy / dx) = 3x(dy / dx) + 3y

B1B

1

LHS, RHS

Condone 3x dy / dx + y (i.e. with missing

bracket) if recovered thereafter

or equivalent if re-arranged

i⇒ (3y2 − 3x)(dy / dx) = 3y − 3x2

⇒ dy / dx = (3y − 3x2) / (3y2 − 3x)M1 collecting terms in dy / dx and factorising

ft correct algebra on incorrect expressions

with two dy / dx terms

i = (y − x2)/(y2 − x)*A1ca

o

NB AG

Examiner's Comments

Part (i) was very well done – it is

pleasing to see how well implicit

differentiation is understood, and the

algebra to derive the given result was

generally done well.

Ignore starting with ‘dy / dx = …’ unless

pursued

i

iTP when y − x2 = 0

i

i⇒ y = x2 M1 or x = √y

i

i

⇒ x3 + x6 = 3x.x2

⇒ x6 = 2x3M1

substituting for y in implicit eqn (allow

one slip, e.g. x5)or x for y (i.e. y3/2 + y3 = 3y1/2y o.e.)

i

i⇒ x3 = 2 (or x = 0) A1 o.e. (soi) or y3/2 = 2

i

i⇒

A1ca

o

must be exact

Examiner's Comments

In part (ii), many fully correct answers

notwithstanding, some failed to get

beyond the first M1 for y = x2; others who

substituted for y in the implicit function

sometimes erred with (x2)3 = x5.

x = 1.2599… is A0 (but can isw )


Total 8

1

4i When x = 3, y = 3/√(3 − 2) = 3 M1 substituting x = 3 (both x's) or x = x/√(x − 2) M1

i So P is (3, 3) which lies on y = x A1

y = 3 and completion (‘3 = 3’ is enough)

Examiner's Comments

Part (i) was an easy two marks for nearly

all candidates. However, sometimes it

was difficult to tell whether it was made

clear that the point (3, 3) lies on the line y

= x.

⇒ x = 3 A1(by solving or verifying)

i

iM1

Quotient or product rule

PR: −½x(x − 2)−3/2 + (x − 2)−1/2

If correct formula stated, allow one error;

otherwise QR must be on correct u and v,

i

iA1 correct expression

with numerator consistent with their

derivatives and denominator correct initially

i

iM1

× top and bottom by √(x − 2) o.e. e.g.

taking out factor of (x − 2)−3/2

allow ft on correct equivalent algebra from

their incorrect expression

i

iA1 NB AG

i

iWhen x = 3, dy / dx = −½ × 13/2 M1 substituting x = 3

i

i= −½ A1

i

i

This gradient would be −1 if curve were

symmetrical about y = x

A1ca

o

or an equivalent valid argument

Examiner's Comments

In part (ii), both the product and quotient

rules were seen – perhaps the product

rule is slightly easier to sort out in this

case. Although most gained the initial

M1A1 for this, the algebra required to

derive the given answer, either by using

a common denominator or factoring out

(x – 2)− ½ , was poorly done. Most

candidates should have been able to

recover to get the derivative at x = 3, and

4/7 was a common mark for the part. The

final mark, using this result to examine

the symmetry of the function, was the

preserve of more able candidates. Many

thought that the P had to be a turning

point for the graph to be symmetrical

about y = x.

i

i

u = x − 2 ⇒ du / dx = 1 ⇒ du = dx

When x = 3, u = 1 when x = 11, u = 9

B1 or dx / du = 1 No credit for integrating initial integral by

parts. Condone du = 1. Condone missing du's


i in subsequent working.

i

i

i

B1

i

i

i

M1splitting their fraction (correctly) and u/u1/2

= u1/2 (or √u)

or integration by parts:

(must be fully

correct – condone missing bracket

i

i

i

A1 by parts: [2u1/2 (u + 2) − 4u3/2/3]

i

i

i

= (18 + 12) – (2/3 + 4) M1 substituting correct limits

i

i

i

A1ca

oNB AG dep substitution and integration attempted

i

i

i

Area under y = x is ½ (3 + 11) × 8 = 56 B1 o.e. (e.g. 60.5 − 4.5)

i

i

i

Area = (area under y = x) − (area under

curve)M1 soi from working

must be trapezium area:

i

i

i

A1ca

o

30.7 or better

Examiner's Comments

Part (iii) achieved mixed success. It was

pleasing to see that most gained the B1

for du = dx; most got the second B1 for

(u + 2)√u; thereafter, the ‘M’ for splitting

the fraction was often lost – some used

integration by parts here with some

success (a sledgehammer to crack a

nut?). Those who got beyond this hurdle

often gained all 6 marks. The final 3

marks were often omitted, but the best

candidates got all 9 marks; the most

common error here was to use the

triangle with vertices (0, 0), (11, 0) and

(11, 11) rather than the trapezium formed

by removing the triangle with vertices (0,

0), (3, 0) and (3, 3).

Total 18

1

5M1 [k (3x − 2)1/2]

A2 k = 2/3


M1d

epsubstituting limits dep 1st M1

= 2/3* A1 NB AG

OR

M1

A1 × 1/3 (du) × 2/3 w (dw)

A1

M1d

epsubstituting correct limits dep 1st M1 upper − lower, 1 to 4 for u or 1 to 2 for w or

= 2/3* A1

NB AG

Examiner's Comments

This was a straightforward starter

question, for which many candidates

scored full marks. The most popular

strategy was to use the substitution u =

3x – 2, and candidates were generally

adept at replacing dx with 1/3 du,

integrating correctly and substituting

correct limits. In a few cases ln u½ was

obtained after integration. The

substitution u = (3x – 2)½ was less

common and caused greater difficulty.

Relatively few students attempted to

integrate directly without substitution, but

those that did often succeeded, and

gained the 5 marks with ease.

substituting back (correctly) for x and using 1

to 2

Total 5

1

6i

When x = 1, f(1) = ln(2/2) = ln 1 = 0 so P is (1,

0)B1

or ln(2x/1 + x) = 0 ⇒ 2x/(1 + x) = 1⇒ 2x = 1 + x ⇒ x = 1

i f(2) = ln(4/3) B1

Examiner's Comments

Part (i) offered two straightforward marks.

Many approximated for ln (4/3), but we

ignored this in subsequent working.

if approximated, can isw after ln(4/3)

i

iy = ln (2x) ln(1 + x) M1 condone lack of brackets

i

iM1 one term correct 2/2x or −1/(1 + x)

i

i

A1ca

omark final ans


i

iB1 correct quotient or product rule need not be simplified

i

i

M1 chain rule attempted

i

iA1 o.e., but mark final ans need not be simplified

i

iAt P, dy/dx = 1 − ½ = ½

A1ca

o

Examiner's Comments

In part (ii), the hint proved valuable and

was taken by nearly all candidates.

However, many found the derivative of

ln(2x) as 1/(2x) and lost two marks.

Those who avoided this error usually

scored all 4 marks.

i

i

i

x = ln[2y/(1 + y)] or (x↔y here or at end to complete) x = ey/(2 − ey)

i

i

i

⇒ ex = 2y / (1 + y) B1 x(2 − ey) = ey B1

i

i

i

⇒ ex(1 + y) = 2y B1 2x = ey+ xey = ey(1 + x) B1

i

i

i

⇒ ex = 2y − exy = y(2 − ex) B1 2x/(1 + x) = ey B1

i

i

i

⇒ y = ex/(2 − ex) [= g(x)] B1 completion ln[2x/(1 + x)] = y [= f(x)] B1

i

i

i

OR gf(x) = g(2x/(1 + x)) =

eln[2x/(1 + x)]/{2 − eln[2x/(1 + x)]}

M1 forming gf or fg fg(x) = ln{2ex/(2 − ex)/[1 + ex/(2 − ex)]} M1

i

i

i

A1 = ln[2ex/(2 − ex + ex)] A1

i

i

i

M1A

1

= ln(ex) = x M1A1

i

i

i

gradient at R = 1/ ½ = 2 B1 ft 1 / their ans in (ii) unless ±1 or 0

Examiner's Comments

Inverting the function in part (iii) was less

successful than usual. This might have

been caused by candidates using the

2 must follow ½ for (ii) unless g′(x) used


last part: g(x) = ex/(2 − ex) ⇒ g′(x) = [(2 − ex)ex

− ex(−ex)]/(2 − ex)2 = 2ex /(2 − ex)2

or g′(x) = ex(−1) (−ex)]


‘hint’ from the previous part to write x = ln

2y – ln (1 + y), and then getting stuck.

The gradient in the last part as the

reciprocal of that in part (ii) was better

answered than in previous papers.

(2 − ex)−2 + ex(2 − ex)−1

g′(0) = 2.1/12 = 2 B1

i

v

let u = 2 − ex ⇒ du / dx = −ex

x = 0, u = 1, x = ln(4/3), u = 2 − 4/3 = 2/3B1 2−e0 = 1, and 2 − eln(4/3) = 2/3 seen

here or later (i.e. after substituting 0 and

ln(4/3) into ln(2 − ex))

i

v M1 or by inspection [k ln (2 − ex)]

i

vA1 [−ln(u)] (could be [lnu] if limits swapped) k = −1

i

v

A1ca

oNB AG

i

vShaded region = rectangle − integral M1

Allow full marks here for correctly

evaluating

i

v= 2ln(4/3) − ln(3/2) B1 rectangle area = 2ln(4/3)

i

v= ln(16/9 × 2/3)

i

v= ln(32/27)*

A1ca

o

NB AG must show at least one step from

2ln(4/3) − ln(3/2)

Examiner's Comments

Finally, part (iv) was the least well

answered question. The new ‘u’ limits of

1 and 2/3 were usually present, but many

lost the minus sign from du = −e−xdx ,

and few gave fully convincing ‘shows’.

The last result was rarely done, though it

was not possible to gather whether this

was due to difficulty or lack of time.


last part

[x ln 2 + x ln x − x

= 2ln2 + 2ln2 − 2 − 3ln 3 + 2 − (ln2 − 1 − 2ln2

+ 1) = 5ln2 − 3ln3 = ln(32/27)

Total 18

1

7i dV/dt = k√V B1

cao condone different k (allow MR B1 for

= kV2)

i M1

A1

A1

2(1/2 kt + c) × constant multiple of k (or

from multiplying out oe; or implicit

differentiation)

cao www any equivalent form (including

unsimplified)

Allow SCB2 if V = (1/2 kt + c)2 fully

obtained by integration including

convincing change of constant if used

Can score B1 M0 SCB2


Examiner's Comments

Most candidates scored the first mark for

writing down the differential equation.

Those who differentiated often scored full

marks. Common errors included,

incorrectly differentiating the inside of the

bracket- instead of 1/2k, a variety of

errors were seen, including functions of t,

and, for those who did differentiate

correctly, failing to equate this to k√V at

the final stage.

Quite a number omitted this

differentiation. Some others decided to

ignore the instruction given and integrate

instead in order to derive the given result

instead of verifying it. Very few of these

attempts gained any further credit as

they failed to deal with the change in

constant. Those who integrated to reach

2√V=kt+c then, too often, gave √V=

[1/2(kt+c) =] 1/2kt+c when trying to

establish the given result and obtained

no marks unless they explained the

change of constant.

i

iB1

substituting any one from t = 1, V =

10,000 or t = 0, V = 0 or t = 2, V = 40,000

into squared form or rooted form of

equation

(Allow −/±100 or −/±200)

i

i

B1

M1

A1

substituting any other from above

Solving correct equations for both www

(possible solutions are (200,0), (−200,0),

(600, −400), (−600,400) (some from –ve

root))

either form www

SC B2 for V = (100t)2 oe stated without

justification

SCB4 if justification eg showing

substitution

SC those working with (k + c)2 = 30,000

can score a maximum of B1B0 M1A0

(leads to k ≈ 146, c ≈ 26.8)

Examiner's Comments

The majority scored two marks for writing

down two correct equations. Those who

then square rooted say, (1/2k+c)2 =

10,000 to reach 1/2k+c =100, and the


other equation to obtain k+c=200 usually

obtained full marks. Those who did not

square root the equations were

sometimes successful but more often

made errors or abandoned their

attempts.

Some felt that (1/2k+c)2= 1/4k2+c2

Total 8

1

8i dF / dv = −25 v−2 M1 d / dv(v−1) = −v−2 soi

i A1

−25 v−2 o.e mark final ans

Examiner's Comments

This was almost invariably correctly

done. No candidates seemed to be put

off by the rather excessive speed of the

car. Occasionally, the quotient rule was

seen, with errors in differentiating the

‘25’.

i

iWhen v = 50, dF/dv = −25/502 (= −0.01) B1 −25/502

i

iM1 o.e.

i

i= −0.01 × 1.5 = −0.015

A1ca

o

o.e. e.g. −3/200 isw

Examiner's Comments

Again, this was very well answered,

provided part (i) was correct. Almost all

candidates scored an M1 for the chain

rule.

Total 5

1

9Let u = 1 + x ⇒

M1 ∫ (u − 1)u−1/2(du)* condone no du, missing bracket, ignore limits

A1 ∫ (u1/2 − u−1/2)(du)

A1; ignore limits

= (16/3 − 4) − (2/3 − 2)M1d

epupper–lower dep 1st M1 and integration with correct limits e.g. 1, 4 for u or 0, 3 for x

A1ca

oor but must be exact

or using w = (1 + x)1/2 ⇒


OR Let u = x, v′ = (1 + x)−1/2 M1

⇒ u′ = 1, v = 2(1 + x)1/2 A1 upper–lower with correct limits (w = 1,2) M1

⇒ A1 ignore limits, condone no dx 8/3 A1 cao

A1 ignore limits *If du done by parts:

= (2 × 3 × 2 – 4 × 8/3) – (0 – 4/3)2u1/2 (u − 1) −∫2u1/2 du A1

[2u1/2 (u − 1)−4u3/2/3] A1

A1ca

o

or but must be exact

Examiner's Comments

Most candidates used integration by

substitution, though a significant minority

used integration by parts. In general, the

former were more successful, with the

main difficulty being in expanding (u –

1)u-1/2 as u1/2 – u-1/2. Some proceeded

from here using integration by parts, with

mixed success. When parts were used,

the most common error was in deriving v

= 2(1 + x)1/2 from v' = (1 + x)-1/2.

substituting correct limits M1 8/3 A1cao

Total 5


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