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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
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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
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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
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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
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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
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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
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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
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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
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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
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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)
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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
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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,
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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’
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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
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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 )
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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
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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
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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
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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
Additional notes and solutions
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)]
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‘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.
Additional notes and solutions
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
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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
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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 ⇒
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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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