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Mark schemeQuestio

nAnswer/Indicative content Marks Part marks and guidance

1 imidpt of oe www

B2

allow unsimplified

B1 for one coordinate

correct

if working shown, should come from

oe

NB B0 for x coord.

(obtained from subtraction instead of addition)

igrad oe

M1

must be obtained

independently of given line;

accept 3 and 5 correctly

shown eg in a sketch,

followed by 3/5

M1 for rise/run = 3/5 etc

M0 for just 3/5 with no

evidence

for those who find eqn of AB first, M0 for just

oe, but M1

for

ignore their going on to find the eqn of AB after

finding grad AB

i using gradient of AB to obtain grad perp bisector M1

for use of m1m2 = − 1 soi or

ft their gradient AB

M0 for just without

AB grad found

this second M1 available for starting with given

and obtaining

grad. of AB from it

i M1

eg M1 for

and

subst of midpt;

ft their gradient of perp

bisector and midpt;

M0 for just rearranging

given equation

no ft for gradient of AB used

i completion to given answer 3y + 5x = 10, showing

at least one interim step

M1 condone a slight slip if they

recover quickly and general

steps are correct (eg

sometimes a slip in working

with the c in -

condone

3y = − 5x + c followed by

substitution and consistent

working)

NB answer given; mark process not answer;

annotate if full marks not earned eg with a tick for

each mark earned

scores such as B2M0M0M1M1 are possible

after B2, allow full marks for complete method of

showing given line has gradient perp to AB (grad

AB must be found independently at some stage)

and passes through midpt of AB

© OCR 2017. You may photocopy this page. Page 1 of 29 Created in ExamBuilder

M0 if clearly ‘fudging’

Examiner's Comments

This part was usually done

well. Most candidates were

confident finding the

gradient of AB, although a

few failed to show their

working. Almost all were

then able to find the

perpendicular gradient. A

minority were unaware that

the perpendicular bisector

would pass through the

midpoint of AB. Most who

realised this were able to

calculate the midpoint

accurately. Once all the

information was combined

into a straight line equation,

a significant minority

struggled to rearrange the

equation correctly because

the arithmetic involved

fractions. Pleasingly almost

all the candidates managed

to work towards the given

equation, rather than trying

to use the given equation to

get back to a common form

with their answer. Some

wasted time finding the

equation of AB first.

ii 3y + 5(4y − 21) = 10 M1

or other valid strategy for

eliminating one variable

attempted eg

condone one error

or eg 20y = 5x + 105 and subtraction of two eqns

attempted

no ft from wrong perp bisector eqn, since given

allow M1 for candidates who reach y = 115/23 and

then make a worse attempt, thinking they have

gone wrong

ii (− 1, 5) or y = 5, x = −1 isw A2 A1 for each value;

if AO allow SC1 for both

values correct but

unsimplified fractions, eg

Examiner's Comments

Some wasted time finding

the equation of CD, which

NB M0A0 in this part for finding E using info from

(iii) that implies E is midpt of CD


was given. Many solved the

simultaneous equations

correctly, but sometimes

using less efficient methods,

giving themselves

complicated fractions to

work with. A few who

eliminated x struggled with

simplifying y = 115/23. A

significant minority used the

implication in part (iii) that E

was the midpoint of CD to

obtain a solution, gaining no

marks for this.

ii

i(x − a)2 + (y − b)2 = r2 seen or used M1

or for (x + 1)2 + (y − 5)2 = k,

or ft their E, where k > 0

ii

i

12 + 42 oe (may be unsimplified), from clear use of

A or BM1

for calculating AE or BE or

their squares, or for subst

coords of A or B into circle

eqn to find r or r2, ft their E;

this M not earned for use of CE or DE or ½ CD

NB some cands finding AB2 = 34 then obtaining 17

erroneously so M0

ii

i(x + 1)2 + (y − 5)2 = 17 A1

for eqn of circle centre E,

through A and B;

allow A1 for r2 = 17 found

after (x + 1)2 + (y − 5)2 = r2

stated and second M1

clearly earned

ii

i

if (x + 1)2 + (y − 5)2 = 17

appears without clear

evidence of using A or B,

allow the first M1 then M0

SC1

SC also earned if circle comes from C or D and E,

but may recover and earn the second M1 later by

using A or B

ii

ishowing midpt of CD = (− 1, 5) M1

ii

i

showing CE or DE = oe or showing one of C

and D on circle

M1 alt M1 for showing CD2 = 68

oe

allow to be earned earlier as

an invalid attempt to find r

showing that both C and D

are on circle and

commenting that E is on CD

is enough for last M1M1;

similarly showing CD2 = 68

and both C and D are on

circle oe earns last M1M1

Examiner's Comments

Most knew the form for the

other methods exist, eg: may find eqn of circle with

centre E and through C or D and then show that A

and B and other of C/D are on this circle – the

marks are then earned in a different order; award

M1 for first fact shown and then final M1 for

completing the argument;

if part-marks earned, annotate with a tick for each

mark earned beside where earned


equation of the circle,

although some used r or

instead of r2. Some

used C or D or the length of

CD to calculate the radius,

instead of using A or B.

Others assumed that AB

was a diameter. Very few

produced enough to show

that CD is a diameter, with

many thinking that showing

that CD is twice the radius

was enough. Some stated

that E was the midpoint of

CD without any working to

support it. This meant that

the full 5 marks on this

question were rarely

awarded, though a

significant number obtained

4 marks.

Total 14

2 i M1

Examiner's Comments

Finding the equation of the

line through A and B was

completed successfully by

the majority of candidates.

The main error seen was a

sign error, either in working

out the gradient (negative

divided by negative given as

negative instead of positive),

or in expanding brackets and

collecting terms.

iy − 7 = their m (x − 4) or

y − 1 = their m (x − 2)M1

or use of y = their gradient x

+ c with coords of A or Ballow step methods used

or eg M1 for 7 = 4m + c and 1 = 2m + c then M1 for

correctly finding one of m and c

i y = 3x − 5 oe A1

accept equivalents if

simplified eg 3x − y = 5

allow B3 for correct eqn

www

allow A1 for c = −5 oe if y = 3x + c oe already seen

B2 for eg y − 1 = 3(x − 2)

ii B1 may be calculation or

showing on diagram

Examiner's Comments

Most tried to use gradients

to show ABC was a right-


angle. Many just stated

‘grad BC = ’ without

showing the calculation. The

m1m2 = -1 rule was well

used on the whole, although

not always explicitly stated,

with some just saying that 3

and were perpendicular

gradients. Those using

Pythagoras to show angle

ABC = 90° were often

successful, but some lost a

mark due to incorrect

notation and / or lack of

convincing steps, with √40 +

√10 = √50 being seen on a

number of scripts. A small

number of candidates

successfully found the

equation of a line

perpendicular to AB that

went through C and then

confirmed that B lay on this

line. Some candidates

worked very hard for their

two marks, unnecessarily

finding equations as they

had not spotted the more

direct methods. These more

long-winded approaches

were variable in terms of

accuracy.

For the area, many correctly

found the lengths needed

but failed to simplify the

surds to obtain 10. The

alternative method of a

rectangle minus three

triangles was seen very

occasionally.

iiand −1/3 × 3 = −1 or grad BC is neg reciprocal of

grad AB, [so 90°]B1

may be earned for

statement / use of m1m2 =

−1 oe, even if first B1 not

earned

for B1 + B1, must be fully

correct, with 3 as gradient in

(i)

eg allow 2nd B1 for statement grad BC = −1/3 with

no working if first B1 not earned

ii or

for finding AC or AC2 independently of AB and BC

or

B1

working needed such as

AC2 = 52 + 52 = 50

condone any confusion between squares and

square roots etc for first B1 and for two M1s eg AC


= 25 + 25

ii for correctly showing AC2 = BC2 + AB2 oe B1

working needed using

correct notation such as BC2

= 32 + 12 = 10; AB2 = 62 + 22

= 40, 40 + 10 = 50 [hence

AC2 = BC2 + AB2]

accept eg 3 and 1 shown on diagram and BC2 = 10

etc

ii

or

finding equation of line through C perpendicular to

AB

or

B1eg B1 for x + 3y = 5

iishowing B is on this line either by substitution or

finding intersection of this line with ABB1

or B1 for finding the equation of the line through B

and C as oe and B1 for using

condition for perp lines and showing true

ii M1

ii M1

both these Ms may be

earned earlier if Pythag

used to show angle ABC =

90°, but are for BC and AB,

not BC2 and AB2

for both M1s accept unsimplified equivs

ii Area = 10 [square units] A1 must be simplified to 10

ii or or

ii area under AC − area under AB − area under BC M1

mark equivalently for other valid methods, eg

trapezium − 2 triangles method, omitting below y =

1:

½ × 7 × 5 − (½ × 3 × 1 + ½ × 2 × 6) = 17.5 − (1.5 +

6)

ii at least two of 22.5, 8 and 4.5 oe M1

ii Area = 10 [square units] A1 must be simplified to 10

ii

i

(1.5, 4.5) oe B1 each coordinate

Examiner's Comments

Most found the mid-point of

AC correctly but failed to

score the final mark, with

some omitting an

explanation. Most

successful explanations

involved showing ABC was

in a semi-circle, but many of

these attempts did not

mention diameter or semi-

circle and were not

sufficiently clear to score the

mark. Some successfully

showed that the right-

angled triangle formed half


of a rectangle with D as the

centre and hence the same

distance from A, B and C.

The most common

explanation was stating that

'D was the midpoint of the

hypotenuse of a right angled

triangle' (or words to that

effect), which did not score.

Weak attempts included the

assumption that ABC was

isosceles or that BD was the

perpendicular bisector of AC

or that A, B and C were

three corners of a square.

ii

i

angle in semicircle oe is a right-angle [so B is on

circle]

and must mention AC as diameter or D as centre

[hence A, B, C all same distance from D]

E1

or ‘[since b = 90°,] ABC are

three vertices of a rectangle.

D is the midpoint of one

diagonal and

so D is the centre of the

rectangle or the diagonals of

a rectangle are equal and

bisect each other, [hence

DA = DB = DC]

or condone showing that

line from D to mid point of

AB is perp to AB, so DBA is

isos [hence DB = DA = DC]

[or equiv using DBC]

E0 for just stating ‘D is midpt of the hypotenuse of a

rt angled triangle ABC so DAB is isos’ without

showing that it is

isw eg wrong calcn of radius

NB some wrongly asserting that ABC is isos

Total 11

3 i f(−3) used M1 Examiner's Comments

Factorising the cubic was

generally done very well. For

the first demand, some did

not use f(3) but divided

successfully, although such

candidates did not always

conclude that finding the

factor (x + 3) meant that x = -

3 was a root. The division,

whether by long division or

inspection, was generally

done well. Candidates seem

well-practised in this

technique. When the

quadratic had been arrived

at correctly, the majority of

candidates successfully

found the two linear factors,

although some using the


formula failed to express the

factors, hence losing the

final two marks. The factor

theorem was occasionally

used to find the remaining

factors, but generally did not

lead to all factors being

found. Some candidates

confused the ideas of roots

and factors.

i −54 −27 + 69 + 12 [= 0] isw A1

or M1 for correct division by

(x + 3) or for the quadratic

factor found by inspection

and A1 for concluding that x

= −3 [is a root] (may be

earned later)

A0 for concluding that x = −3 is a factor

iattempt at division by (x + 3) as far as 2x3 + 6x2 in

workingM1

or inspection with at least

two terms of three–term

quadratic factor correct; or

at least one further root

found using remainder

theorem

i correctly obtaining 2x2 − 9x + 4 A1

or stating further factor,

found from using remainder

theorem again

i factorising the correct quadratic factor M1

for factors giving two terms

of quadratic correct or for

factors ft one error in

quadratic formula or

completing square;

M0 for formula etc without

factors found

allow for (x − 4) and (x − ½) given as factors eg

after using remainder theorem again or quadratic

formula etc

i (2x − 1)(x − 4)[(x + 3)] isw A1

allow 2(x − ½) instead of (2x

− 1), oe condone inclusion

of ‘= 0’

isw (x − ½) as factor and / or roots found, even if

stated as factors

iisketch of cubic right way up, with two turning

pointsB1

0 if stops at x-axis

ignore graph of y = 4x + 12

Examiner's Comments

Sketching the cubic was

well done by most

candidates. A few forgot to

show the y- intercept but

most knew the correct

shape and used their roots

to show intersections.

must not be ruled; no curving back (except

condone between x = 0 and x = 0.5); condone

some ‘flicking out’ at ends but not approaching

more turning points; must continue beyond axes;

allow max on y axis or in 1st or 2nd quadrants

condone some doubling / feathering

ii values of intns on x axis shown, correct (−3, 0.5

and 4) or ft from their factors or roots in (i)

B1 on graph or nearby in this

part

mark intent for intersections

allow if no graph

condone 3 on neg x axis as slip for −3;

condone eg 0.5 roughly halfway between their 0


with both axes and 1 marked on x axis

ii 12 marked on y-axis B1

or x = 0, y = 12 seen in this

part if consistent with graph

drawn

allow if no graph, but eg B0 for graph with intn on

−ve y-axis or nowhere near their indicated 12

ii

i2x3 − 3x2 − 23x + 12 = 4x + 12 oe M1

or ft their factorised f(x)

Examiner's Comments

Most knew they had to

equate the linear and cubic

expressions for y and

usually simplified to the

correct cubic equation.

Many were then unsure how

to solve this. Many lost the x

= 0 root by dividing by x,

although these candidates

often found the other two

roots successfully. Some

candidates started with the

factorised form of f(x) and

divided both sides by x + 3;

many of these lost the x = -3

root. Some tried to use the

quadratic formula on the

cubic equation.

ii

i2x3 − 3x2 − 27x [= 0] A1

after equating, allow A1 for

cancelling (x + 3) factor on

both sides and obtaining 2x2

− 9x [= 0]

condone slip of ‘= y’ instead of ‘= 0’

ii

i[x](2x − 9)(x + 3) [= 0] M1

for linear factors of correct

cubic, giving two terms

correct

or for quadratic formula or

completing square used on

correct quadratic

2x2 − 3x − 27 = 0, condoning

one error in formula etc;

or after cancelling (x + 3) factor allow M1 for x(2x −

9) oe or obtaining x = 0 or 9/2 oe

M0 for eg quadratic formula used on cubic, unless

recovery and all 3 roots given

ii

i[x =] 0, −3 and 9/2 oe A1

need not be all stated

togethereg x = 0 may be earlier

Total 13

4 i B1

Examiner's Comments

The majority of candidates

were able to write down the

centre and radius

successfully. Some radii

were given as 20 instead of


√20 , and some centres as (-

2 , 0) or (0 , 2) instead of (2 ,

0).

i B1

ii subst of x = 0 into circle eqn soi M1

or Pythag used on sketch of

circle: 22 + y2 = 20 oe

Examiner's Comments

A significant minority of

candidates forgot to find the

negative square root when

solving y2 = 16 and so only

found one intersection, but

on the whole this was well

done. Some also found

where the circle cut the x-

axis. Most sketched

correctly, showing the y-

intercepts found and their

centre was correctly placed.

However, a significant

number of candidates took

little care over their

sketches, with many

"circles" drawn poorly.

M0 for just y2 = 20; M1 for y2 = 16 or for y = 4

ignore intns with x-axis also found

ii y = ±4 oe A1

or B2 for just y = ±4 seen

oe;

accept both 4 and −4 shown

on y axis on sketch if both

values not stated

iisketch of circle with centre (2, 0) or ft their centre

from (i)B1

if the centre is not marked, it

should look roughly correct

by eye – coords need not be

given on sketch; condone

intersections with axes not

marked

circle should intersect both +ve and neg x- and y-

axes; must be clear attempt at circle;

ignore any tangents drawn

ii

i(x − 2)2 + (2x + k)2 = 20 M1

for attempt to subst 2x + k

for y

Examiner's Comments

Candidates who substituted

y = 2x + k into the equation

of the circle were generally

very successful, with only a

few minor slips. However,

candidates who decided to

work backwards from the

given result usually

struggled.

allow for attempt to subst k = y − 2x into given eqn


ii

ix2 − 4x + 4 + 4x2 + 4kx + k2 = 20

M1

dep

for correct expansion of at

least one set of brackets,

dependent on first M1

similarly for those working backwards

ii

i5x2 + (4k − 4)x + k2 − 16 = 0 A1

correct completion to given

answer; dependent on both

Ms

condone omission of further interim step if both

sets of brackets expanded correctly, but for cands

working backwards, at least one interim step is

needed;

if cands have made an error and tried to correct it,

corrections must be complete to award this A mark

i

vb2 − 4ac = 0 seen or used M1

need not be substituted into;

may be stated after formula

used

or argument towards

expressing eqn as a perfect

square

Examiner's Comments

Many candidates did not

know where to start, and the

full four marks were rarely

awarded. About a quarter of

the candidates did not

attempt the question and

those that did make an

attempt often substituted x =

2 or x = 0 at the start. Some

successfully used b2 _ 4ac =

0 and found the correct

values of k but many made

errors, particularly taking c

as -16 instead of k2 16.

Some candidates found the

equation of the normal,

although few made further

progress with this approach.

A few candidates offered

solutions using the gradient

of the normal and finding the

intersections with the circle

by using a vector approach

from the centre – a neat

approach which usually

scored full marks.

eg M1 for (4k − 4)2 − 4 × 5 × (k2 − 16) = 0

i

v

4k2 + 32k − 336 [= 0] or

k2 + 8k − 84 [= 0]

M1 expansion and collection of

terms, condoning one error

ft their b2 − 4ac

dep on an attempt at b2 − 4ac with at least two of a,

b and c correct; may be earned with < 0 etc; may

be in formula

i

v

use of factorising or quadratic formula or

completing square

M1 condone one error ft dep on attempt at obtaining required quadratic

equation in k, not for use with any eqn/inequality

they have tried


i

vk = 6 or − 14 A1

i

vor or

i

v

Grad of tgt is 2, and normal passes through centre,

hence finding equation of normal as M1

i

v

finding x values where diameter y = −x/2 + 1

intersects circle as x = 6 or −2 (condone one error

in method)

M1oe for y values; condone

one error in method

or finding intn of tgt and normal as

i

v

finding corresponding y values on circle and subst

into y = 2x + k

or

subst their x values into 5x2 + (4k − 4)x + k2 − 16 =

0

M1

intns are (6, −2) and (−2, 2),

M0 for just (6, 2) and (−2,

−2) used but condone used

as well as correct intns

this last method gives extra

values for k, for the non-

tangent lines y = through (6,

2) and (−2, −2), but allow for

the M mark

or subst their intn of tgt and normal into eqn of

circle:

i

vk = 6 or − 14 A1 and no other values

Total 12

5 iAB2 = (1−(−1))2 + (5 − 1)2

M1

oe, or square root of this;

condone poor notation re

roots; condone (1 + 1)2

instead of (1−(−1))2

allow M1 for vector

, condoning

poor notation, or triangle

with hyp AB and lengths 2

and 4 correctly marked

iBC2 = (3 − (−1))2 + (−1 −1)2

M1

oe, or square root of this;

condone poor notation re

roots; condone (3 + 1)2

instead of (3−(−1))2 oe

allow M1 for vector

, condoning

poor

notation, or triangle with hyp

BC and lengths 4 and 2

correctly marked

i shown equal e.g. A1 or statement that AB and e.g. A0 for AB = 20 etc


AB2 = 22 + 42 [=20] and

BC2 = 42 + 22 [=20] with correct notation for final

comparison

BC are each the hypotenuse

of a right-angled triangle

with sides 2 and 4 so are

equal

SC2 for just AB2 = 22 + 42

and BC2 = 42 + 22 (or roots of

these) with no clearer earlier

working; condone poor

notation

Examiner's Comments

Most candidates showed

good, clear working, but

some used poor notation,

mixing up expressions for

AB2 and AB. Not many used

diagrams, but where these

were used they were

generally good and led to

full marks. Some students

calculated gradients instead

of lengths.

ii M1award at first step shown

even if errors after

ii M1

if one or both of grad AC = −3 and grad BD = 1/3

seen without better working for both gradients,

award one M1 only. For M1M1 it must be clear that

they are obtained independently

ii showing or stating product of gradients = −1 or that

one gradient is the negative reciprocal of the other

oe

B1 e.g. accept m1 × m2 = −1 or

‘one gradient is negative

reciprocal of the other’

B0 for ‘opposite’ used

instead of ‘negative’ or

‘reciprocal’

Examiner's Comments

Showing that the lines are

perpendicular was usually

done well, but some

struggled with arithmetic

involving negative numbers.

Only a few had their

gradients upside down.

Most knew the condition for

perpendicular lines, and

expressed it clearly,

although some just

calculated gradients and

then stated ‘so they are

may be earned independently of correct gradients,

but for all 3 marks to be earned the work must be

fully correct


perpendicular’.

iistate so not E or

show F not on ACA1

ii

imidpoint E of AC = (2, 2) www B1

condone missing brackets

for both B1s0 for ((5 + −1)/2, (1 + 3)/2) = (2, 2)

ii

iM1

accept any correct form isw

or correct ft their gradients

or their midpt F of BD

this mark will often be

gained on the first line of

their working for BD

may be earned using (2, 2) but then must

independently show that B or D or (5, 3) is on this

line to be eligible for A1

ii

ieqn AC is y = −3x + 8 oe M1

accept any correct form isw

or correct ft their gradients

or their midpt E of AC

this mark will often be

gained on the first line of

their working for AC

Alternative Methods

for a mixture of methods,

look for the method which

gives most benefit to

candidate, but take care not

to award the second M1

twice

if equation(s) of lines are seen in part ii, allow the

M1s if seen / used in this part

ii

i

using both lines and obtaining intersection E is (2,

2) (NB must be independently obtained from midpt

of AC)

A1

Alternative Methods

the final A1 is not earned if

there is wrong work leading

to the required statements

ii

i

midpoint F of BD = (5,3) B1 this mark is often earned

earlier

Appendix: alternative

methods for (iii) [details of

equations etc are in main

scheme]

for a mixture of methods,

look for the method which

gives most benefit to

candidate, but take care not

to award the second M1

twice

the final A1 is not earned if

there is wrong work leading

to the required statements

for all methods show annotations M1 B1 etc then

omission mark or A0 if that mark has not been

earned

ignore wrong working which has not been used for

the required statements

for full marks to be earned in this part, there must

be enough to show both the required statements


Examiner's Comments

This question required some

problem-solving skills from

candidates, and most

candidates made a good

attempt. The most common

approach, usually

successful, was to find the

equation of BD, check that

the midpoint of AC lies on

this line, then find the

midpoint of BD and show

that this does not lie on AC.

Most did not realise that

having shown that the

midpoint of AC lies on B,

showing that the midpoint of

BD is not the same as the

midpoint of AC is sufficient

to show that AC does not

bisect BD. Errors in

midpoints or equations of

lines were fairly common.

Some candidates worked

with lengths but these

approaches were often

muddled. Few attempted to

use symmetry arguments,

and those that did usually

did not provide enough

explanation.

ii

i

Alternative Method 1

find midpt E of ACB1

ii

ifind eqn BD M1

ii

ishow E on BD M1

ii

ifind midpt F of BD. B1

ii

istate so not E A1

ii

i

OR


find midpt E of AC


B1

ii

ifind eqn BD M1

ii

ishow E on BD M1

ii

ifind midpt F of BD B1

ii

i

find eqn of AC and correctly show F not on AC (the

correct eqn for AC earns the second M1 as per the

main scheme, if not already earned)

A1

ii

i

OR



ii

ifind eqn BD M1

ii

ishow E on BD M1

ii

ishow BE2 = 10 and DE2 = 90 oe B1

ii

i

showing BE2 = 10 and DE2 = 90 oe earns this A

mark as well as the B1 if there are no errors

elsewhere

A1

ii

i

OR



ii

i

use gradients or vectors to show E is on BD eg

grad and grad

[condone poor vector notation]

M2

ii

ifind midpt F of BD B1

Total 11

6 i (2x + 1)(x + 2)(x − 5) M1

or (x + 1/2)(x + 2)(x − 5);

need not be written as

product

throughout, ignore ‘=0'

icorrect expansion of two linear factors of their

product of three linear factorsM1

for all Ms in this part condone missing brackets if

used correctly

i expansion of their linear and quadratic factors M1 dep on first M1; ft one error


in previous expansion;

condone one error in this

expansion

or for direct expansion of all

three factors, allow M2 for

2x3 − 10x2 + 4x2 + x2 − 20x −

5x + 2x − 10 [or half all

these], or M1 if one or two

errors,

dep on first M1

i [y =] 2x3 − 5x2 − 23x − 10 or a = −5, b = − 23 and c = −

10

A1

for an attempt at setting up

three simultaneous

equations in a, b, and c: M1

for at least two of the three

equations

then M2 for correctly

eliminating any two variables

or M1 for correctly

eliminating one variable to

get two equations in two

unknowns

and then A1 for values.

Examiner's Comments

Most candidates obtained

the first mark for obtaining

the factors from the roots.

Many candidates wrote (x +

½) in place of (2x + 1) as

one of their factors, and

those that did sometimes

omitted to find the equation

of the curve in the required

form and so did not obtain

the last mark. A few

candidates, instead of writing

down the factors as

instructed and then

multiplying out the factors,

attempted to set up

simultaneous equations

using the factor theorem.

One or two marks were

obtained this way but it was

very rare to see the method

taken to a correct

condone poor notation when ‘doubling’ to reach

expression with 2x3…

250 + 25a + 5b + c = 0

−16 + 4a − 2b + c = 0

−¼ + ¼ a − ½ b + c = 0 oe


conclusion.

ii graph of cubic correct way up B1

B0 if stops at x-axis

must not be ruled; no curving back; condone slight

‘flicking out’ at ends; allow min on y axis or in 3rd or

4th quadrants; condone some ‘doubling’ or

‘feathering’ (deleted work still may show in scans)

ii crossing x axis at −2, −1/2 and 5 B1

on graph or nearby in this

part

mark intent for intersections

with both axes

allow if no graph, but marked on x-axis

ii crossing y axis at −10 or ft their cubic in (i) B1

or x = 0, y = −10 or ft in this

part if consistent with graph

drawn;

Examiner's Comments

Most knew the correct

shape for the graph of a

cubic but some were drawn

poorly. A common fault,

leading to a very distorted

graph, was to assume

incorrectly that there was a

minimum at the intersection

with the y-axis. A surprising

number, having obtained the

correct equation in part (i),

thought that the y-

intersection was 5, possibly

because they were starting

again by thinking about the

x-intersections. Some

confused factors with roots

thus reversing the signs.

allow if no graph, but e.g. B0 for graph nowhere

near their indicated −10 or ft

ii

i(0, −18); accept −18 or ft their constant − 8 1

or ft their intn on y-axis −8

Examiner's Comments

Most knew that they needed

to subtract 8 from their y-

intercept, although a few

added 8.


i

vroots at 2.5, 1, 8 M1

or attempt to substitute (x −

3) in

(2x + 1)(x + 2)(x − 5) or in

(x + 1/2)(x + 2)(x − 5) or in

their unfactorised

form of f(x)− attempt need

not be simplified

i

v(2x − 5)(x − 1)(x − 8) A1

accept 2(x − 2.5) oe instead

of (2x − 5)

M0 for use of (x + 3) or roots −3.5, −5,

2 but then allow

SC1 for (2x + 7)(x + 5)(x − 2)

i

v(0, −40); accept −40 B2

M1 for −5 × −1 × −8 or ft or

for f(−3) attempted or g(0)

attempted or for their

answer ft from their

factorised form

Examiner's Comments

The best approach using the

factors was usually only

seen from the better

candidates. Many correctly

found the new roots but

wrote down g(x) using (x –

2.5) rather than (2x – 5).

Some started by substituting

x − 3 into the expanded form

of f(x) and then attempted to

multiply out and simplify –

most of these did not even

attempt to give g(x) in

factorised form as

requested. Many picked up

a final mark by substituting x

= 0 into their g(x). Some

candidates incorrectly

translated to the left by

using (x + 3) and could

obtain 1 or 2 marks.

e.g. M1 for (0, −70) or −70 after

(2x + 7)(x + 5)(x − 2)

after M0, allow SC1 for f(3) = −70

Total 12

7 i(−1, 6) (0,1) (1, −2) (2, −3) (3, −2) (4, 1) (5, 6) seen

plottedB2

or for a curve within 2 mm of

these points; B1 for 3

correct plots or for at least 3

of the pairs of values seen

e.g. in table

use overlay; scroll down to spare copy of graph to

see if used [or click ‘fit height’

also allow B1 for and (2, −3) seen

or plotted and curve not through other correct

points

i smooth curve through all 7 pointsB1

dep

dep on correct points;

tolerance 2 mm;

condone some feathering / doubling (deleted work

still may show in scans); curve should not be flat-

bottomed or go to a point at min. or curve back in at

top;


i(0.3 to 0.5, −0.3 to −0.5) and (2.5 to 2.7, −2.5 to

−2.7) and (4, 1)B2

may be given in form x =…,

y =…

B1 for two intersections

correct or for all the x values

given correctly

Examiner's Comments

A few candidates did not

know where to start and a

significant number confused

the ideas of sketching and

plotting. As a consequence

the full range of integer

values from x = −1 to x = 5

was not used. Some thought

that three or four points

plotted would suffice. Other

candidates found where the

curve would cross the x-axis

and/or determined the

minimum point by

completing the square, and

then relied on a sketch for

the rest of the curve. A

significant minority did not

attempt to find intersections

at all. Of the rest, some

gave only 2 intersections,

and some could not cope

with the scale on the y-axis,

or omitted the negative sign

for the y-coordinates of the

first 2 roots.

ii M1

ii 1 = (x − 3)( x2 − 4x + 1) M1

condone omission of

brackets only if used

correctly afterwards, with at

most one error;

condone omission of ‘=1’ for this M1 only if it

reappears

allow for terms expanded correctly with at most one

error

ii at least one further correct interim step with ‘=1’ or

‘=0’, as appropriate, leading to given answer, which

must be stated correctly

A1 there may also be a

previous step of expansion

of terms without an

equation, e.g. in grid

if M0, allow SC1 for correct

division of given cubic by

quadratic to gain (x − 3) with

remainder −1, or vice-versa

Examiner's Comments

NB mark method not answer - given answer is x3 −

7x2 + 13x − 4 = 0


Deriving the given equation

was usually done well, with

most candidates starting

off with the correct step of

equating to x2 − 4x −

1.

Very few algebraic errors

were seen here. Some

candidates just substituted x

= 4 into the given answer.

Other poor attempts started

with the final expression,

often equating it to ,

so made no progress.

ii

i

quadratic factor is

x2 − 3x + 1B2

found by division or

inspection; allow M1 for

division by x − 4 as far as x3

− 4x2 in the working, or for

inspection with two terms

correct

ii

i

substitution into quadratic formula or for completing

the square used as far asM1 condone one error no ft from a wrong ‘factor’;

ii

iA2

A1 if one error in final

numerical expression, but

only if roots are real

Examiner's Comments

This part was well

attempted. Long division

seemed less successful

than inspection; however

most candidates found the

correct quadratic factor.

Most knew the quadratic

formula and applied it

correctly. Some fully correct

responses were spoilt by

wrong attempts to further

simplify their roots. Some

solved by completing the

square but were usually less

successful in reaching the

correct roots

isw factors

Total 13

8 i f(1) = 1 − 1 + 1 + 9 − 10 [= 0] B1 allow for correct division of

f(x) by (x − 1) showing there

condone 14 – 13 + 12 + 9 – 10


is no remainder,

or for (x − 1) (x3 + x + 10)

found, showing it ‘works’ by

multiplying it out

iattempt at division by (x − 1) as far as x4− x3 in

workingM1

allow equiv for (x + 2) as far

as x4 + 2 x3 in working

or for inspection with at least

two terms of cubic factor

correct

eg for inspection, M1 for two terms right and two

wrong

i correctly obtaining x3 + x + 10 A1

or x3 − 3x2 + 7x − 5

Examiner's Comments

A large number of

candidates successfully

used the factor theorem to

score the first mark and

many went on to find the

correct cubic factor - the

majority of these choosing

to do long division rather

than use the inspection

method. Some did not use

the factor theorem but still

showed that x = 1 was a

root by successful division

with no remainder. Those

who used inspection without

first applying the factor

theorem did not in general

show enough working for a

convincing argument that

there was no remainder and

therefore that x = 1 was a

root. A small number did not

appear to understand what

was meant by ‘express f(x)

as the product of a linear

factor and a quadratic factor’

– some of these gained

partial credit for the correct

division seen in parts (ii) or

(iii).

if M0 and this division / factorising is done in part

(ii) or (iii), allow SC1 if correct cubic obtained there;

attach the relevant part to (i) with a formal chain

link if not already seen in the image zone for (i)

ii [g(− 2) =] − 8 − 2 + 10

or f(− 2) = 16 + 8 + 4 − 18 − 10

M1 [in this scheme g(x) = x3 + x

+ 10]

allow M1 for correct trials

with at least two values of x

eg f(2) = 16 − 8 + 4 + 18 − 10 or 20

f(3) = 81 − 27 + 9 + 27 − 10 or 80

f(0) = − 10

f(− 1) = 1 + 1 + 1 − 9 − 10 or − 16


(other than 1) using g(x) or

f(x) or x3 − 3x2 + 7x − 5

(may allow similar correct

trials using division or

inspection)

No ft from wrong cubic ‘factors’ from (i)

ii x = − 2 isw A1

allow these marks if already

earned in (i)

Examiner's Comments

Many used the correct

method but made careless

errors in calculations

especially when trying

negative values of x. Very

few realised that they could

use the factor theorem on

the cubic they had found to

obtain another root. Many

confused ‘root’ with ‘factor’

and lost a mark.

NB factorising of x3 + x + 10 or x3 − 3x2 + 7x − 5 in

(ii) earns credit for (iii) [annotate with a yellow line in

both parts to alert you – the image zone for (iii)

includes part (ii)]

ii

i

attempted division of x3 + x + 10 by (x + 2) as far

as x3 + 2x2 in workingM1

or x3 − 3x2 + 7x − 5 by (x −

1) as far as x3 − x2 in

working

or inspection with at least

two terms of quadratic factor

correct

alt method: allow M1 for attempted division of

quartic by x2 + x − 2 as far as x4 + x3 − 2 x2 in

working, or inspection etc

ii

icorrectly obtaining x2 − 2x + 5 A1

allow these first 2 marks if

this has been done in (ii),

even if not used here

ii

iuse of b2 − 4ac with x2 − 2x + 5 M1

may be in attempt at formula

(ignore rest of formula)

or completing square form attempted

or attempt at calculus or symmetry to find min pt

NB M0 for use of b2 − 4ac with cubic factor etc

ii

ib2 − 4ac = 4 − 20 [= − 16] A1 may be in formula;

or (x − 1)2 + 4

or min = (1, 4)

ii

i

so only two real roots[ of f(x)] [and hence no more

linear factors]

A1 or no real roots of x2 − 2x +

5 = 0;

allow this last mark if clear

use of x2 − 2x + 5 = 0, even

if error in b2 − 4ac, provided

result negative, but no ft

from wrong factor

or (x − 1)2 + 4 is always positive so no real roots [of

(x − 1)2 + 4 = 0] [ and hence no linear factors]

or similar conclusion from min pt


if last M1 not earned, allow

SC1 for stating that the only

factors of 5 are 1 and 5 and

reasoning eg that (x − 1)(x −

5) and (x + 1)(x + 5) do not

give x2 − 2x + 5 [hence x2 −

2x + 5 does not factorise]

Examiner's Comments

Only about a third of the

candidates found the correct

quadratic factor. Those who

found the quadratic usually

gave sensible arguments

based on the discriminant to

show that only two real roots

existed for the quartic.

Some tried to use b2 – 4ac

on the cubic x3 + x + 10.

Several candidates went

back to square one and

attempted to factorise the

quartic rather than linking

the earlier parts to the

problem. Some candidates

who had not progressed far

in the first two parts

sometimes made no attempt

at this part.

Total 10

9 i translation in the x-direction M1 allow ‘shift’, ‘move’ If just vectors given withhold one ‘A’ mark only

i of π/4 to the right A1 oe (eg using vector)‘Translate is 4 marks; if this is followed by

an additional incorrect transformation, SC

M1M1A1A0

i translation in y–direction M1 allow ‘shift’, ‘move’only is M2A1A0

i of 1 unit up. A1 oe (eg using vector)

Examiner's Comments

We usually insist on the

word ‘translation’ here, but

in this case allowed ‘move’,

‘shift’, etc. A vector on its

own does not in our view

imply a translation.

Occasionally, candidates

clearly knew what the

transformations were, but


wrote the vectors

incorrectly, for example the

wrong way up.

Nevertheless, this topic is

usually well known and

done well.

ii (Can deal with num and denom separately)

ii M1Quotient (or product) rule

consistent with their derivs ; allow one slip, missing brackets

ii A1

Correct expanded

expression (could leave the

‘2’ as a factor)

ii A1 NB AGmust take out 2 as a factor or state

sin2x + cos2x = 1

ii When x = π/4, g′ (π/4) = 2/(1/√2 + 1/√2)2 M1substituting π/4 into correct

deriv

ii = 1 A1

ii f′ (x) = sec2x M1 o.e., e.g. 1/cos2x

ii f′ (0) = sec2 (0) = 1, [so gradient the same here] A1

Examiner's Comments

The quotient rule is generally

well known, and errors here

usually stemmed from faulty

derivatives or poor algebra.

Brackets are not optional in

an expression like this, and

their removal was not always

successfully achieved. We

also needed evidence of the

use of cos2x + sin2x = 1,

either by its direct quotation

or by factoring out the ‘2’ in

the numerator. The

evaluation of g'(x) was

usually correct. With f'(x),

some used a quotient rule on

sin x/cos x rather than

quoting the derivative of tan

x = sec2x; we also got some

occasional ‘translation’

arguments here which

misunderstood the nature of

the verification.


ii

i let u = cos x, du = −sin x dx

when x = 0, u = 1, when x = π/4, u = 1/√2

ii

iM1 substituting to get ∫ −1/u (du)

ignore limits here, condone no du but not dx allow

∫1/u.−du

ii

iA1 NB AG

but for A1 must deal correctly with the -ve sign by

interchanging limits

ii

iM1 [ln u]

ii

i= ln 1 - ln (1/√2)

ii

i= ln √2 = ln 2½ In 2 A1

ln √2, ½ ln 2 or −ln(1/√2)

Examiner's Comments

This was a case where

giving the transformed

integral proved to be of

doubtful value, as many

candidates ‘lost’ the

negative sign in their ∫-1/u

du, and placed the limits the

wrong way round. It appears

that the idea of swapping

limits making the integral

negative was not generally

understood. The evaluation

of the given integral with

respect to u was more

successfully done, though

quite a few candidates

approximated their final

answer.

mark final answer

i

vArea = area in part (iii) translated up 1 unit. M1

soi from π/4 addedor

i

vSo = ½ ln 2 + 1 × π4 = ½ ln 2 + π/4. A1cao

oe (as above)

Examiner's Comments

These marks were gained

by candidates who

managed to spot the

rectangle of area added by

the translation upwards of

the graph of f(x).

Total 17


1

0i 1 − 9a2 = 0 M1 or 1 − 9x2 = 0 √(1 − 9a2) = 1 − 3a is M0

i ⇒ a2 = 1/9 ⇒ a = 1/3 A1

or 0.33 or better √(1/9) is A0

Examiner's Comments

This was usually correct,

though leaving the final

answer as x = 1/3 was quite

common, and some

candidates left their answer

as √(1/9), or gave a = ±1/3.

not a = ± 1/3 nor x = 1/3

ii Range 0 ≤ y ≤ 1 B1

or 0 ≤ f(x) ≤ 1 or 0 ≤ f ≤ 1, not

0 ≤ x ≤ 1

0 ≤ y ≤ √1 is B0

Examiner's Comments

The domain was frequently

given instead of the range.

Other answers scoring zero

included 0 ≤ x ≤ 1, y ≤ 1, 0

to 1 (which does not settle

the inclusion of the

endpoints), and 1.

allow also [0,1], or 0 to 1 inclusive, but not 0 to 1 or

(0,1)

ii

iM1

curve goes from x = −3a to x

= 3a (or −1 to 1)

must have evidence of using s.f. 3 allow also if s.f.3

is stated and stretch is reasonably to scale

ii

iM1 vertex at origin

ii

iA1

curve, ‘centre’ (0, −1), from

(−1, −1) to (1, −1) (y-coords

of −1 can be inferred from

vertex at O and correct

scaling)

allow from (−3a, −1) to (3a, −1) provided a = 1/3 or x

= [±] 1/3 in (i) A0 for badly inconsistent scale(s)

ii

i

Examiner's Comments

Many gained only the

method marks because they

omitted to indicate the

domain or range on their

sketch. Others did not

indicate the x-coordinates of

the endpoints. Some

stretches looked more like

enlargements. To get the

final ‘A’ mark, we required

the axes to have

approximately the same

scale.


Total 6

1

1i (A) (0, 6) and (1, 4) B1B1

Condone P and Q

incorrectly labelled (or

unlabelled)

i (B) (−1, 5) and (0, 4) B1B1

Examiner's Comments

The majority of candidates

obtained full marks; part (A)

was not as well answered as

(B): sometimes marks were

lost through stretching

horizontally rather than

vertically.

ii M1

Quotient or product rule

consistent with their

derivatives, condone

missing brackets

PR: (x2 + 3)(−1)(x + 1)−2 + 2x(x + 1)−1

If formula stated correctly, allow one substitution

error.

ii f′(x) = 0 ⇒ 2x (x + 1) − (x2 + 3) = 0 A1 correct expression condone missing brackets if subsequent

ii ⇒ x2 + 2x – 3 = 0 M1 their derivative = 0 working implies they are intended

ii ⇒ (x − 1)(x + 3) = 0 A1dep

obtaining correct quadratic

equation (soi) dep 1st M1 but

withhold if denominator also

set to zero

Some candidates get x2 + 2x + 3, then realise this

should be x2 + 2x − 3, and correct back, but not for

every occurrence. Treat this sympathetically.

iiWhen x = −3, y = 12/(−2) = −6

so other TP is (−3, −6)

B1B1c

ao

must be from correct work

(but see note re quadratic)

Examiner's Comments

This was all relatively

routine work which good

candidates had little trouble

with; however, several

candidates made slips in

expanding the bracket in the

numerator of the quotient

rule. This was a costly error,

as were errors in the

quotient rule such as uv’−vu’

in the numerator.

Must be supported, but −3 could be verified by

substitution into correct derivative

ii

iM1

substituting x − 1 for both x's

in fallow 1 slip for M1

ii

iA1

ii

i

A1 NB AG

Examiner's Comments


Most candidates were

successful, guided by the

given answer, though a few

found f(x) + 1.

i

vB1

i

vM1

F(b) − F(a) condone missing

brackets

F must show evidence of integration of at least one

term

i

v

= ( b2 − 2b + 4 ln b) − ( a2 − 2a + 4 ln a)

Area is

A1 oe (mark final answer)

i

vSo taking a = 1 and b = 2 M1 or f(x) = x + 1 − 2 + 4/(x + 1)

i

varea = (2 − 4 + 4ln 2) − ( ½ − 2 + 4ln 1)

= ½ − 1 + 4ln 2 = 4ln2 – ½ A1

i

v= 4 ln 2 − ½ A1 cao

must be simplified with ln 1 =

0

Examiner's Comments

Most integrated the function

correctly, though ¼ ln(x)

was seen occasionally. The

question asked candidates

to give the answer in terms

of a and b, but some omitted

this. The final answer

required candidates to

realise that a = 1 and b = 2,

rather than 0 and 1

(notwithstanding the

appearance of ln 0 in the

lower limit), but this was

spotted by only the better

candidates.

Total 18


Documents

Mark scheme - nlcsmaths.com · Web viewWith f'(x), some used a quotient rule on sin x/cos x rather than quoting the derivative of tan x = sec2x; we also got some occasional ‘translation’