1.3 HL

1a. [2 marks]

Markscheme

     A1A1

Note: Award A1 for −0.414, 2.41 and A1 for correct inequalities.

[2 marks]

Examiners report

[N/A]

1b. [7 marks]

Markscheme

check for ,

16 > 9 so true when          A1

assume true for 

       M1

Note: Award M0 for statements such as “let ”.

Note: Subsequent marks after this M1 are independent of this mark and can be awarded.

prove true for

               M1

               (M1)

        (from part (a))        A1

1

      which is true for  ≥ 3        R1

Note: Only award the A1 or the R1 if it is clear why. Alternate methods are possible.

hence if true for true for , true for so true for all  ≥ 3        R1

Note: Only award the final R1 provided at least three of the previous marks are awarded.

[7 marks]

Examiners report

[N/A]

2. [7 marks]

Markscheme

, so true for                R1

assume true for  , ie.                M1

Note: Award M0 for statements such as “let ”.

Note: Subsequent marks after the first M1 are independent of this mark and can be awarded.

              M1

         A1

         A1

         A1

true for and if true for then true for , the statement is true for any positive

integer (or equivalent).        R1

2

Note: Award the final R1 mark provided at least four of the previous marks are gained.

[7 marks]

Examiners report

[N/A]

3. [6 marks]

Markscheme

consider  .   and    therefore true for        R1

Note: There must be evidence that has been substituted into both expressions, or an

expression such LHS=RHS=1 is used. “therefore true for ” or an equivalent statement must be

seen.

 

assume true for  , (so that  )       M1

Note: Assumption of truth must be present.

 

consider 

      (M1)

      A1

       M1

Note: M1 is for factorising 

 

3

so if true for  , then also true for  , and as true for   then true for all       

R1

Note: Only award final R1 if all three method marks have been awarded.

Award R0 if the proof is developed from both LHS and RHS.

 

[6 marks]

Examiners report

[N/A]

4. [7 marks]

Markscheme

if 

     M1

hence true for 

assume true for       M1

Note: Assumption of truth must be present. Following marks are not dependent on the first two M1

marks.

so 

if

      M1A1

finding a common denominator for the two fractions      M1

4

     A1

hence if true for  then also true for  , as true for  , so true (for all  )     R1

Note: Award the final R1 only if the first four marks have been awarded.

[7 marks]

Examiners report

[N/A]

5. [7 marks]

Markscheme

Let   be the statement:  for some  where   consider

the case       M1

because  . Therefore  is true     R1

assume  is true for some 

     M1

Note: Assumption of truth must be present. Following marks are not dependent on this M1.

EITHER

consider       M1

      A1

is true (as )     R1

OR

multiply both sides by  (which is positive)      M1

     A1

5

 is true (as  )     R1

THEN

is true  is true  is true so  true for all   (or equivalent)      R1

Note: Only award the last R1 if at least four of the previous marks are gained including the A1.

[7 marks]

Examiners report

[N/A]

6a. [2 marks]

Markscheme

even function     A1

since and is a product of even functions     R1

OR

even function     A1

since     R1

 

Note:     Do not award A0R1.

 

[2 marks]

Examiners report

[N/A]

6b. [8 marks]

Markscheme

consider the case

6

    M1

hence true for     R1

assume true for , ie,     M1

 

Note:     Do not award M1 for “let ” or “assume ” or equivalent.

 

consider :

    (M1)

    A1

    A1

    A1

so true and true true. Hence true for all     R1

 

Note:     To obtain the final R1, all the previous M marks must have been awarded.

 

[8 marks]

Examiners report

[N/A]

6c. [3 marks]

Markscheme

attempt to use (or correct product rule)     M1

7

    A1A1

 

Note:     Award A1 for correct numerator and A1 for correct denominator.

 

[3 marks]

Examiners report

[N/A]

6d. [8 marks]

Markscheme

    (M1)(A1)

    (A1)

    A1

    A1

    A1

 

Note:     This A mark is independent from the previous marks.

 

    M1A1

    AG

[8 marks]8

Examiners report

[N/A]

7. [6 marks]

Markscheme

let be the proposition that is divisible by 9

showing true for      A1

ie for

which is divisible by 9, therefore is true

assume is true so      M1

 

Note:     Only award M1 if “truth assumed” or equivalent.

 

consider

     M1

     A1

which is divisible by 9     R1

 

Note:     Award R1 for either the expression or the statement above.

 

since is true and true implies is true, therefore (by the principle of mathematical

induction) is true for      R1

 

9

Note:     Only award the final R1 if the 2 M1s have been awarded.

 

[6 marks]

Examiners report

[N/A]

8. [9 marks]

Markscheme

show true for      (M1)

     A1

hence true for

assume true for      M1

consider for      (M1)

     A1

or any correct expression with a visible common

factor     (A1)

or any correct expression with a common denominator     (A1)

 

10

Note:     At least one of the above three lines or equivalent must be seen.

 

or equivalent     A1

Result is true for . If result is true for it is true for . Hence result is true for all .

Hence proved by induction.     R1

 

Note:     In order to award the R1 at least [5 marks] must have been awarded.

 

[9 marks]

Examiners report

[N/A]

9a. [2 marks]

Markscheme

   (M1)A1

 

Note: Award M1 for 5 equal terms with \) + \) or signs.

 

[2 marks]

Examiners report

[N/A]

9b. [2 marks]

Markscheme11

   M1

   A1

   AG

[2 marks]

Examiners report

[N/A]

9c. [9 marks]

Markscheme

let 

if 

which is true (as proved in part (b))     R1

assume  true,      M1

 

Notes: Only award M1 if the words “assume” and “true” appear. Do not award M1 for “let ” only.

Subsequent marks are independent of this M1.

 

consider :

   M1

   A1

12

   M1

   M1

   A1

   A1

so if true for , then also true for 

as true for then true for all      R1

 

Note: Accept answers using transformation formula for product of sines if steps are shown clearly.

 

Note: Award R1 only if candidate is awarded at least 5 marks in the previous steps.

 

[9 marks]

Examiners report

[N/A]

9d. [6 marks]

Markscheme

EITHER

   M1

   A1

   M1

13

   M1

 or      A1

and 

OR

   M1A1

   M1A1

of     A1

and 

THEN

 and      A1

 

Note: Do not award the final A1 if extra solutions are seen.

 

[6 marks]

Examiners report

[N/A]

10a. [2 marks]

Markscheme

   M1

   A1

[2 marks]

14

Examiners report

This was well done by most candidates who correctly applied de Moivre’s theorem.

10b. [6 marks]

Markscheme

show the expression is true for      R1

assume true for      M1

Note:     Do not accept “let ” or “assume ”, assumption of truth must be present.

   M1

   A1

Note:     Award A1 for any correct expansion.

   A1

therefore if true for true for , true for , so true for all      R1

Note:     To award the final R mark the first 4 marks must be awarded.

[6 marks]

Examiners report

This question was poorly done, which was surprising as it is very similar to the proof of de Moivre’s

theorem which is stated as being required in the course guide. Many candidates spotted that they

needed to use trigonometric identities but fell down through not being able to set out the proof in a

logical form.

10c. [2 marks]

Markscheme

   (M1)A1

15

[2 marks]

Examiners report

This was well done by the majority of candidates.

10d. [5 marks]

Markscheme

(i)     

   A1

   AG

Note:     Allow justification starting with .

(ii)          A1

(iii)          A1

   M1A1

  AG

Note:     M1 is for using , this might be seen in d(ii).

[5 marks]

Examiners report

(d) parts (i) and (ii) were well done by the candidates, who were able to successfully use

trigonometrical identities and the binomial theorem.

(d)(iii) This is a familiar technique that has appeared in several recent past papers and was successfully

completed by many of the better candidates. Some candidates though neglected the instruction ‘hence’

and tried to derive the expression using trigonometric identities.

10e. [6 marks]

Markscheme

16

   A1A1

Note:     A1 for and A1 for .

   A1

or      M1

   A1A1

Note:     Do not accept solutions via factor theorem or other methods that do not follow “hence”.

[6 marks]

Examiners report

Again some candidates ignored ‘hence’ and tried to form a polynomial equation. Many candidates

obtained the solution and hence the solution . Few were able to find the

other solutions which can be obtained from consideration of the unit circle or similar methods.

11. [8 marks]

Markscheme

let  be the proposition that  is divisible by 6 for 

consider P(1):

when  and so P(1) is true     R1

assume  is true ie, where      M1

Note:     Do not award M1 for statements such as “let ”.

consider :

   M1

17

   (A1)

   A1

   A1

is even hence all three terms are divisible by 6     R1

is true whenever is true and P(1) is true, so  is true for      R1

Note:     To obtain the final R1, four of the previous marks must have been awarded.

[8 marks]

Examiners report

This proved to be a good discriminator. The average candidate seemed able to work towards

, and a number made some further progress.

Unfortunately, even otherwise good candidates are still writing down incorrect or incomplete induction

statements, such as ‘Let ’ rather than ‘Suppose true for ’ (or equivalent).

It was also noted than an increasing number of candidates this session assumed ‘ to be true’

before going to consider . Showing a lack of understanding of the induction argument, these

approaches scored very few marks.

12a. [1 mark]

Markscheme

    M1

    AG

 

Note:     Accept a transformation/graphical based approach.

[1 mark]

Examiners report18

[N/A]

12b. [7 marks]

Markscheme

consider     M1

since then the proposition is true for     R1

assume that the proposition is true for so     M1

    M1

(using part (a))     A1

    A1

given that the proposition is true for then we have shown that the proposition is true for

. Since we have shown that the proposition is true for then the proposition is true for

all     R1

 

Note:     Award final R1 only if all prior M and R marks have been awarded.

[7 marks]

Total [8 marks]

Examiners report

[N/A]

13a. [5 marks]

Markscheme

if then     A1

so true for 19

assume true for     M1

so

consider

    M1

    A1

hence if true for then also true for . As true for , so true for all .     R1

 

Note:     Do not award the R1 if the two M marks have not been awarded.

[5 marks]

Examiners report

[N/A]

13b. [6 marks]

Markscheme

consider the series , where     R1

 

Note:     Award the R1 for starting at

 

compare to the series where     M1

is an infinite Geometric Series with and hence converges     A1

20

 

Note:     Award the A1 even if series starts at .

 

as so for all     M1R1

as converges and so must converge

 

Note:     Award the A1 even if series starts at .

 

as is finite, so must converge     R1

 

Note:     If the limit comparison test is used award marks to a maximum of R1M1A1M0A0R1.

[6 marks]

Total [11 marks]

Examiners report

[N/A]

14a. [2 marks]

Markscheme

    M1A1

[2 marks]

Examiners report

Well done.

14b. [7 marks]

21

Markscheme

let be the statement

prove for     M1

of is which is and  is     R1

as is true

assume is true and attempt to prove is true     M1

assuming

    (M1)

    A1

(as required)     A1

 

Note:     Can award the A marks independent of the M marks

 

since is true and is true is true

then (by ), is true     R1

 

Note: To gain last R1 at least four of the above marks must have been gained.

[7 marks]

Examiners report

The logic of an induction proof was not known well enough. Many candidates used what they had to

prove rather than differentiating what they had assumed. They did not have enough experience in

doing Induction proofs.22

14c. [5 marks]

Markscheme

    M1A1

point is     A1

EITHER

when therefore the point is a minimum     M1A1

OR

nature table shows point is a minimum     M1A1

[5 marks]

Examiners report

Good, some forgot to test for min/max, some forgot to give the value.

14d. [5 marks]

Markscheme

    A1

    M1A1

point is     A1

23

since the curvature does change (concave down to concave up) it is a point of inflection     R1

 

Note:     Allow derivative is not zero at

[5 marks]

Examiners report

Again quite good, some forgot to check for change in curvature and some forgot the value.

15a. [2 marks]

Markscheme

    M1

    A1

    AG

[2 marks]

Examiners report

[N/A]

15b. [2 marks]

Markscheme

    A2

    AG

24

[2 marks]

Examiners report

[N/A]

15c. [9 marks]

Markscheme

consider the case : required to prove that     M1

from part (b)

hence is true for     A1

now assume true for     M1

attempt to prove true for     (M1)

from assumption, we have that     M1

so attempt to show that     (M1)

EITHER

    A1

, (from part a), which is true     A1

OR

    A1

25

    A1

THEN

so true for and true true. Hence true for all     R1

 

Note:     Award R1 only if all previous M marks have been awarded.

[9 marks]

Total [13 marks]

Examiners report

[N/A]

16. [7 marks]

Markscheme

let be the proposition that

consider :

and so is true     R1

assume is true ie     M1

 

Note:     Do not award M1 for statements such as “let ”.

 

consider :

    M1

    A1

26

    R1

    A1

is true whenever is true and is true, so is true for     R1

 

Note:     To obtain the final R1, four of the previous marks must have been awarded.

 

[7 marks]

Examiners report

An easy question, but many candidates exhibited discomfort and poor reasoning abilities. The difficulty

for most was that the proposition was expressed in terms of an inequality. Hopefully, as most

publishers of IB textbooks have realised, inequalities in such questions are within the syllabus.

17. [8 marks]

Markscheme

if

which is divisible by 5, hence true for     A1

 

Note:     Award A0 for using but do not penalize further in question.

 

assume true for     M1

 

Note:     Only award the M1 if truth is assumed.

 

so     A1

if

27

    M1

    M1

    A1

    A1

hence if true for , then also true for . Since true for , then true for all     R1

 

Note:     Only award the R1 if the first two M1s have been awarded.

 

[8 marks]

Examiners report

[N/A]

18a. [7 marks]

Markscheme

let be the proposition 

let

is true     R1

assume true for     M1

 

Note:     Only award the M1 if truth is assumed.

28

 

now show true implies also true

    M1

    A1

    A1

is true     A1

true implies true and is true, therefore by mathematical induction statement is

true for     R1

 

Note:     Only award the final R1 if the first 4 marks have been awarded.

 

[7 marks]

Examiners report

[N/A]

18b. [4 marks]

Markscheme

(i)         A1

    A1

 

Notes:     Accept 3 sf answers only. Accept equivalent forms.

     Accept and .

29

 

(ii)    

    (M1)

    A1

 

Notes:     Award (M1) for an attempt to find and .

     Accept equivalent forms.

 

[4 marks]

Examiners report

[N/A]

18c. [1 mark]

Markscheme

     A1

 

30

Note:     Award A1 if A or and B or are in their correct quadrants, are aligned vertically

and it is clear that .

 

[1 mark]

Examiners report

[N/A]

18d. [3 marks]

Markscheme

Area     M1A1

    A1

 

Notes:     Award M1A0A0 for using .

 

[3 marks]

Examiners report

[N/A]

18e. [5 marks]

Markscheme

    M1A1

 

Note:     Award M1 for recognition that a complex conjugate is also a root.

 

31

    A1

    M1A1

 

Note:     Award M1 for an attempt to expand two quadratics.

 

[5 marks]

Examiners report

[N/A]

19. [7 marks]

Markscheme

or a multiple of 3     A1

assume the proposition is true for       M1

 

Note:     Do not award M1 for statements with “Let ”.

 

consider       M1

    A1

    M1

    A1

 

Note:     Accept or statement that is a

multiple of 3.

32

 

true for , and

hence true for all     R1

 

Note:     Only award the final R1 if at least 4 of the previous marks have been achieved.

 

[7 marks]

Examiners report

It was pleasing to see a great many clear and comprehensive answers for this relatively straightforward

induction question. The inductive step only seemed to pose problems for the very weakest candidates.

As in previous sessions, marks were mainly lost by candidates writing variations on ‘Let ’, rather

than ‘Assume true for ’. The final reasoning step still needs attention, with variations on ‘

’ evident, suggesting that mathematical induction as a technique is not

clearly understood.

20. [7 marks]

Markscheme

is divisible by 576 for

for

Zero is divisible by 576, (as every non-zero number divides zero), and so P(1) is true.     R1

Note: Award R0 for P(1) = 0 shown and zero is divisible by 576 not specified.

 

Note: Ignore P(2) = 576 if P(1) = 0 is shown and zero is divisible by 576 is specified.

 

Assume is true for some .     M1

Note: Do not award M1 for statements such as “let n = k”.

33

 

consider     M1

    A1

EITHER

    A1

which is a multiple of 576     A1

OR

    A1

(or equivalent) which is a multiple of 576     A1

THEN

P(1) is true and true true, so is true for all     R1

Note: Award R1 only if at least four prior marks have been awarded.

 

[7 marks]

Examiners report

This proof by mathematical induction challenged most candidates. While most candidates were able to

show that P(1) = 0, a significant number did not state that zero is divisible by 576. A few candidates

started their proof by looking at P(2). It was pleasing to see that the inductive step was reasonably well

done by most candidates. However many candidates committed simple algebraic errors. The most

common error was to state that . The concluding statement often omitted the

required implication statement and also often omitted that P(1) was found to be true.

21a. [3 marks]

Markscheme

    M1A1

34

    A1

[3 marks]

Examiners report

Part a) proved to be an easy 3 marks for most candidates. 

 

21b. [8 marks]

Markscheme

    A1A1

assume that P(k) is true, i.e.,     M1

consider

EITHER

    (M1)

    A1

    A1

OR

35

    (M1)

    A1

    A1

THEN

    A1

P(k) true implies P(k + 1) true, P(1) true so P(n) true for all     R1

[8 marks]

Examiners report

Part b) was often answered well, and candidates were well prepared in this session for this type of

question. Candidates still need to take care when showing explicitly that P(1) is true, and some are still

writing ‘Let n = k’ which gains no marks. The inductive step was often well argued, and given in clear

detail, though the final inductive reasoning step was incorrect, or appeared rushed, even from the

better candidates. ‘True for n =1, n = k and n = k + 1’ is still disappointingly seen, as were some even

more unconvincing variations.

21c. [6 marks]

Markscheme

METHOD 1

    M1A1

    A1

    A1

    M1

36

    A1

    AG

METHOD 2

attempt     M1

    A1A1

    A1A1

Note: Award A1 marks for numerators and denominators.

 

    A1AG

METHOD 3

attempt     M1

    A1A1

    A1A1

Note: Award A1 marks for numerators and denominators.

 

    A1AG

[6 marks]

Examiners report

Part c) was again very well answered by the majority. A few weaker candidates attempted to find an

inverse for the individual case n = 1 , but gained no credit for this.

37

21d. [6 marks]

Markscheme

(i)         A1

 

(ii)     METHOD 1

    (M1)

    A1

so     R1

    AG

METHOD 2

    (M1)

    A1

true in the interval     R1

 

(iii)         (M1)A1

[6 marks]

Examiners report

Part d) was not at all well understood, with virtually no candidates able to tie together the hints given

by connecting the different parts of the question. Rash, and often thoughtless attempts were made at

each part, though by this stage some seemed to be struggling through lack of time. The inequality part

of the question tended to be ‘fudged’, with arguments seen by examiners being largely unconvincing

and lacking clarity. A tiny number of candidates provided the correct answer to the final part, though a

surprising number persisted with what should have been recognised as fruitless working – usually in

the form of long-winded integration attempts.

38

22a. [4 marks]

Markscheme

let and using the result

    M1A1

    A1

    A1

    AG

[4 marks]

Examiners report

Even though the definition of the derivative was given in the question, solutions to (a) were often

disappointing with algebraic errors fairly common, usually due to brackets being omitted or

manipulated incorrectly. Solutions to the proof by induction in (b) were often poor. Many candidates

fail to understand that they have to assume that the result is true for and then show that this

leads to it being true for . Many candidates just write ‘Let ’ which is of course

meaningless. The conclusion is often of the form ‘True for therefore

true by induction’. Credit is only given for a conclusion which includes a statement such as ‘True for

true for ’.

22b. [9 marks]

Markscheme

let

we want to prove that

39

let     M1

which is the same result as part (a)

hence the result is true for     R1

assume the result is true for     M1

    M1

    (A1)

    A1

    (A1)

    A1

hence if the result is true for , it is true for

since the result is true for , the result is proved by mathematical induction     R1 

Note: Only award final R1 if all the M marks have been gained.

 

[9 marks]

Examiners report

Even though the definition of the derivative was given in the question, solutions to (a) were often

disappointing with algebraic errors fairly common, usually due to brackets being omitted or

manipulated incorrectly. Solutions to the proof by induction in (b) were often poor. Many candidates

fail to understand that they have to assume that the result is true for and then show that this

leads to it being true for . Many candidates just write ‘Let ’ which is of course

40

meaningless. The conclusion is often of the form ‘True for therefore

true by induction’. Credit is only given for a conclusion which includes a statement such as ‘True for

true for ’.

23. [7 marks]

Markscheme

proposition is true for n = 1 since     M1

    A1

Note: Must see the 1! for the A1.

 

assume true for n = k, , i.e.     M1

consider     (M1)

    A1

    A1

hence, is true whenever is true, and is true, and therefore the proposition is true for all

positive integers     R1

Note: The final R1 is only available if at least 4 of the previous marks have been awarded.

 

[7 marks]

Examiners report

Most candidates were awarded good marks for this question. A disappointing minority thought that the

th derivative was the th derivative multiplied by the first derivative. Providing an acceptable

final statement remains a perennial issue.

41

24a. [8 marks]

Markscheme

prove that

for n = 1

so true for n = 1     R1

assume true for n = k     M1

so

now for n = k +1

LHS:     A1

    M1A1

(or equivalent)     A1

(accept )     A1

Therefore if it is true for n = k it is true for n = k + 1. It has been shown to be true for n = 1 so it is true

for all .     R1

Note: To obtain the final R mark, a reasonable attempt at induction must have been made.

 

[8 marks]

Examiners report

Part A: Given that this question is at the easier end of the ‘proof by induction’ spectrum, it was

disappointing that so many candidates failed to score full marks. The n = 1 case was generally well

done. The whole point of the method is that it involves logic, so ‘let n = k’ or ‘put n = k’, instead of

42

‘assume ... to be true for n = k’, gains no marks. The algebraic steps need to be more convincing than

some candidates were able to show. It is astonishing that the R1 mark for the final statement was so

often not awarded.

24b. [17 marks]

Markscheme

(a)

METHOD 1

    M1A1A1

    A1A1

    M1

    AG 

METHOD 2

    M1A1A1

    A1A1

    M1

    AG

[6 marks]

 

(b)

    M1A1

    A1

43

when     M1

    A1

[5 marks]

 

(c)

(i)    

   

A1

P is (1.16, 0)     A1

Note: Award A1 for 1.16 seen anywhere, A1 for complete sketch.

44

 

Note: Allow FT on their answer from (b)

 

(ii)         M1A1

    A2

Note: Allow FT on their answers from (b) and (c)(i).

 

[6 marks]

Examiners report

Part B: Part (a) was often well done, although some faltered after the first integration. Part (b) was also

generally well done, although there were some errors with the constant of integration. In (c) the graph

was often attempted, but errors in (b) usually led to manifestly incorrect plots. Many attempted the

volume of integration and some obtained the correct value.

25. [20 marks]

Markscheme

(a)         M1A1

    AG

[2 marks]

 

(b)     if n = 1     M1

    M1

so LHS = RHS and the statement is true for n = 1     R1

assume true for n = k     M1

45

Note: Only award M1 if the word true appears.

   Do not award M1 for ‘let n = k’ only.

   Subsequent marks are independent of this M1.

 

so

if n = k + 1 then

    M1

    A1

    M1

    M1

    A1

    M1

    A1

so if true for n = k, then also true for n = k + 1

as true for n = 1 then true for all     R1

Note: Final R1 is independent of previous work.

[12 marks]

 

(c)         M1A1

but this is impossible

46

    A1

    A1

    A1

for not including any answers outside the domain     R1

Note: Award the first M1A1 for correctly obtaining or equivalent and

subsequent marks as appropriate including the answers .

[6 marks]

 

Total [20 marks]

Examiners report

This question showed the weaknesses of many candidates in dealing with formal proofs and showing

their reasoning in a logical manner. In part (a) just a few candidates clearly showed the result and part

(b) showed that most candidates struggle with the formality of a proof by induction. The logic of many

solutions was poor, though sometimes contained correct trigonometric work. Very few candidates

were successful in answering part (c) using the unit circle. Most candidates attempted to manipulate

the equation to obtain a cubic equation but made little progress. A few candidates guessed as a

solution but were not able to determine the other solutions.

26. [10 marks]

Markscheme

(a)     (i)         R1

 

(ii)     LHS = 40; RHS = 40     A1

[2 marks]

 

47

(b)     the sequence of values are:

5, 7, 11, 19, 35 … or an example     A1

35 is not prime, so Bill’s conjecture is false     R1AG

[2 marks]

 

(c)     is divisible by 6

is divisible by true     A1

assume is true     M1

Note: Do not award M1 for statement starting ‘let n = k’.

Subsequent marks are independent of this M1.

 

consider     M1

    (A1)

is true     A1

P(1) true and true true, so by MI is true for all     R1

Note: Only award R1 if there is consideration of P(1), and in the final statement.

Only award R1 if at least one of the two preceding A marks has been awarded.

 

[6 marks]

Total [10 marks]

Examiners report

Although there were a good number of wholly correct solutions to this question, it was clear that a

number of students had not been prepared for questions on conjectures. The proof by induction was

relatively well done, but candidates often showed a lack of rigour in the proof. It was fairly common to 48

see students who did not appreciate the idea that is assumed not given and this was penalised.

Also it appeared that a number of students had been taught to write down the final reasoning for a

proof by induction, even if no attempt of a proof had taken place. In these cases, the final reasoning

mark was not awarded.

27a. [14 marks]

Markscheme

(a)         M1A1

    M1A1

solving simultaneously, a = 6 , d = 3     A1A1

[6 marks]

 

(b)         A1

    A1

obtaining     M1

(since all terms are positive)     A1

    A1

[5 marks]

 

49

(c)         A1

    A1

    M1AG

[3 marks]

 

Total [14 marks]

Examiners report

Parts (a), (b) and (c) were answered successfully by a large number of candidates. Some, however, had

difficulty with the arithmetic.

27b. [7 marks]

Markscheme

prove:

show true for n = 1 , i.e.

    A1

assume true for n = k , i.e.     M1

consider n = k +1

    M1A1

    A1

    A1

50

hence true for n = k + 1

is true whenever is true, and is true, therefore is true     R1

for

[7 marks]

Examiners report

In part (d) many candidates showed little understanding of sigma notation and proof by induction.

There were cases of circular reasoning and using n, k and r randomly. A concluding sentence almost

always appeared, even if the proof was done incorrectly, or not done at all.

28. [8 marks]

Markscheme

let

LHS

RHS

hence true for     R1

assume true for

    M1

    M1A1

    A1

    A1

hence if true for , true for     R1

51

since the result is true for and the result is proved by mathematical

induction     R1

[8 marks]

Examiners report

This question was done poorly on a number of levels. Many students knew the structure of induction

but did not show that they understood what they were doing. The general notation was poor for both

the induction itself and the sigma notation.

In noting the case for too many stated the equation rather than using the LHS and RHS

separately and concluding with a statement. There were also too many who did not state the conclusion

for this case.

Many did not state the assumption for as an assumption.

Most stated the equation for and worked with the equation. Also common was the lack of

sigma and inappropriate use of n and k in the statement. There were some very nice solutions however.

The final conclusion was often not complete or not considered which would lead to the conclusion that

the student did not really understand what induction is about.

29a. [3 marks]

Markscheme

    (A1)

    M1

    A1     N1

[3 marks]

Examiners report

Part (a) was correctly answered by the majority of candidates, although a few found r = –3.

29b. [7 marks]

Markscheme52

Attempting to show that the result is true for n = 1     M1

LHS = a and     A1

Hence the result is true for n = 1

Assume it is true for n = k

    M1

Consider n = k + 1:

    M1

    A1

Note: Award A1 for an equivalent correct intermediate step.

 

    A1

Note: Illogical attempted proofs that use the result to be proved would gain M1A0A0 for the last three

above marks.

 

The result is true for it is true for and as it is true for , the result is proved

by mathematical induction.     R1     N0

Note: To obtain the final R1 mark a reasonable attempt must have been made to prove the k + 1 step.

 

[7 marks]

Examiners report

53

Part (b) was often started off well, but a number of candidates failed to initiate the n = k + 1 step in a

satisfactory way. A number of candidates omitted the ‘P(1) is true’ part of the concluding statement.

30. [4 marks]

Markscheme

the number of ways of allocating presents to the first child is     (A1)

multiplying by     (M1)(A1)

 

Note:     Award M1 for multiplication of combinations.

 

    A1

[4 marks]

Examiners report

[N/A]

31a. [4 marks]

Markscheme

    M1A1A1

    A1

[4 marks]

Examiners report

[N/A]

31b. [4 marks]

54

Markscheme

METHOD 1

    A1A1A1

probability is     A1

METHOD 2

P (5 eaten) =P (M eats 1) P (N eats 4) + P (M eats 0) P (N eats 5)     (M1)

    (A1)(A1)

    A1

[4 marks]

Examiners report

[N/A]

32. [6 marks]

Markscheme

expanding     A1

expanding gives

    (M1)A1A1

 

Note:     Award (M1) for an attempt at expanding using binomial.

     Award A1 for .

     Award A1 for .

55

 

    (M1)

 

Note:     Award (M1) only if both terms are considered.

 

therefore coefficient is     A1

 

Note:     Accept

 

Note:     Award full marks if working with the required terms only without giving the entire expansion.

 

[6 marks]

Examiners report

[N/A]

33a. [2 marks]

Markscheme

    M1

    A1

    AG

[2 marks]

Examiners report

Part a) has appeared several times before, though with it again being a ‘show that’ question, some

candidates still need to be more aware of the need to show every step in their working, including the

result that .

56

33b. [1 mark]

Markscheme

(b)         A1

 

Note:     Accept .

 

[1 mark]

Examiners report

Part b) was usually answered correctly.

33c. [4 marks]

Markscheme

METHOD 1

    M1

    A1A1

 

Note:     Award A1 for RHS, A1 for LHS, independent of the M1.

 

    A1

METHOD 2

    M1

57

    A1

    A1

    A1

[4 marks]

Examiners report

Part c) was again often answered correctly, though some candidates often less successfully utilised a

trig-only approach rather than taking note of part b).

33d. [3 marks]

Markscheme

    M1

    A1A1

 

Note:     Award A1 for RHS, A1 for LHS, independent of the M1.

 

    AG

 

Note:     Accept a purely trigonometric solution as for (c).

 

[3 marks]

58

Examiners report

Part d) was a good source of marks for those who kept with the spirit of using complex numbers for

this type of question. Some limited attempts at trig-only solutions were seen, and correct solutions

using this approach were extremely rare.

33e. [3 marks]

Markscheme

    M1A1

    A1

[3 marks]

Examiners report

Part e) was well answered, though numerical slips were often common. A small number integrated

as .

A large number of candidates did not realise the help that part e) inevitably provided for part f). Some

correctly expressed the volume as and thus gained the first 2 marks but

were able to progress no further. Only a small number of able candidates were able to obtain the

correct answer of .

33f. [4 marks]

Markscheme

    M1

    M1

    A1

59

    A1

 

Note:     Follow through from an incorrect r in (c) provided the final answer is positive.

Examiners report

[N/A]

33g. [3 marks]

Markscheme

(i)     constant term =     A1

(ii)         A1

              A1

[3 marks]

Examiners report

Part g) proved to be a challenge for the vast majority, though it was pleasing to see some of the highest

scoring candidates gain all 3 marks.

34a. [3 marks]

Markscheme

    (M1)

    A1

    A1

60

    AG

[3 marks]

Examiners report

This question linked the binomial distribution with binomial expansion and coefficients and was

generally well done.

(a) Candidates need to be aware how to work out binomial coefficients without a calculator

34b. [4 marks]

Markscheme

(i)     2 outcomes for each of the 6 games or     R1

 

(ii)        

A1

Note: Accept notation or

 

setting x = 1 in both sides of the expression     R1

Note: Do not award R1 if the right hand side is not in the correct form.

 

    AG

 

(iii)     the total number of outcomes = number of ways Alfred can win no games, plus the number of

ways he can win one game etc.     R1 

[4 marks]

Examiners report61

This question linked the binomial distribution with binomial expansion and coefficients and was

generally well done.

(b) (ii) A surprising number of candidates chose to work out the values of all the binomial coefficients

(or use Pascal’s triangle) to make a total of 64 rather than simply putting 1 into the left hand side of the

expression.

34c. [9 marks]

Markscheme

(i)     Let be the probability that Alfred wins x games on the first day and y on the second.

    M1A1

or     A1

r = 2 or 4, s = t = 6

 

(ii)     P(Total = 6) =

P(0, 6) + P(1, 5) + P(2, 4) + P(3, 3) + P(4, 2) + P(5, 1) + P(6, 0)     (M1)

    A2

Note: Accept any valid sum of 7 probabilities.

 

(iii)     use of     (M1)

(can be used either here or in (c)(ii))

62

P(wins 6 out of 12)     A1

    A1

therefore     AG

[9 marks]

Examiners report

This question linked the binomial distribution with binomial expansion and coefficients and was

generally well done.

34d. [6 marks]

Markscheme

(i)    

(a = 2, b = 3)     M1A1

Note: M0A0 for a = 2, b = 3 without any method.

 

(ii)         A1A1

(sigma notation not necessary)

(if sigma notation used also allow lower limit to be r = 0)

let x = 2     M1

multiply by 2 and divide by     (M1)

63

    AG

[6 marks]

Examiners report

This question linked the binomial distribution with binomial expansion and coefficients and was

generally well done.

(d) This was poorly done. Candidates were not able to manipulate expressions given using sigma

notation.

35. [4 marks]

Markscheme

clear attempt at binomial expansion for exponent 5     M1

    (A1)

Note: Only award M1 if binomial coefficients are seen.

 

    A2

Note: Award A1 for correct moduli of coefficients and powers. A1 for correct signs.

 

Total [4 marks]

Examiners report

Generally well done. The majority of candidates obtained a quintic with correct alternating signs. A few

candidates made arithmetic errors. A small number of candidates multiplied out the linear expression,

often correctly.

36a. [3 marks]

64

Markscheme

the three girls can sit together in 3! = 6 ways     (A1)

this leaves 4 ‘objects’ to arrange so the number of ways this can be done is 4!     (M1)

so the number of arrangements is     A1

[3 marks]

Examiners report

Some good solutions to part (a) and certainly fewer completely correct answers to part (b). Many

candidates were able to access at least partial credit, if they were showing their reasoning.

36b. [4 marks]

Markscheme

Finding more than one position that the girls can sit     (M1)

Counting exactly four positions     (A1)

number of ways     M1A1     N2

[4 marks]

Examiners report

Some good solutions to part (a) and certainly fewer completely correct answers to part (b). Many

candidates were able to access at least partial credit, if they were showing their reasoning.

37a. [4 marks]

Markscheme

(i)     (or equivalent)     A1

Note: Award A0 for or equivalent.

 

(ii)     EITHER

65

    M1A1

OR

    M1A1

OR

    M1A1

THEN

    AG

 

(iii)         A1

[4 marks]

Examiners report

In part (a) (i), a large number of candidates were unable to correctly use sigma notation to express the

sum of the first n positive odd integers. Common errors included summing from 1 to n and

specifying sums with incorrect limits. Parts (a) (ii) and (iii) were generally well done.

37b. [7 marks]

Markscheme

(i)     EITHER

a pentagon and five diagonals     A1

OR

five diagonals (circle optional)     A1

 

(ii)     Each point joins to n – 3 other points.     A1

a correct argument for     R1

66

a correct argument for     R1

 

(iii)     attempting to solve for n.     (M1)

    (A1)

    A1

[7 marks]

Examiners report

Parts (b) (i) and (iii) were generally well done. In part (b) (iii), many candidates unnecessarily

simplified their quadratic when direct GDC use could have been employed. A few candidates gave

as their final answer. While some candidates displayed sound reasoning in part (b) (ii),

many candidates unfortunately adopted a ‘proof by example’ approach.

37c. [8 marks]

Markscheme

(i)     np = 4 and npq = 3     (A1)

attempting to solve for n and p     (M1)

and     A1

 

(ii)         (A1)

    (A1)

    (A1)

    (M1)

= 0.261     A1

67

[8 marks]

Examiners report

Part (c) was generally well done. In part (c) (ii), some candidates multiplied the two probabilities

rather than adding the two probabilities.

38. [4 marks]

Markscheme

    (M1)(A1)

Note: Award M1 for attempt to expand and A1 for correct unsimplified expansion.

 

    A1A1

Note: Award A1 for powers, A1 for coefficients and signs.

 

Note: Final two A marks are independent of first A mark.

 

[4 marks]

Examiners report

This was generally very well answered. Those who failed to gain full marks often made minor sign slips.

A surprising number obtained the correct simplified expression, but continued to rearrange their

expressions, often doing so incorrectly. Fortunately, there were no penalties for doing so.

39a. [3 marks]

Markscheme

    (A2)

 Note: Award (A1) for 3 or 4 correct terms. 

68

 

Note: Accept combinatorial expressions, e.g. for 6.

 

    A1

[3 marks]

Examiners report

It was disappointing to see many candidates expanding by first expanding and

then either squaring the result or multiplying twice by , processes which often resulted in

arithmetic errors being made. Candidates at this level are expected to be sufficiently familiar with

Pascal’s Triangle to use it in this kind of problem. In (b), some candidates appeared not to understand

the phrase ‘constant term’.

39b. [2 marks]

Markscheme

constant term from expansion of     A2

Note: Award A1 for –64 or 24 seen.

[2 marks]

Examiners report

It was disappointing to see many candidates expanding by first expanding and

then either squaring the result or multiplying twice by , processes which often resulted in

arithmetic errors being made. Candidates at this level are expected to be sufficiently familiar with

Pascal's Triangle to use it in this kind of problem. In (b), some candidates appeared not to understand

the phrase "constant term".

40. [7 marks]

69

Markscheme

    (M1)(A1)

    (M1)(A1)

 Note: Accept unsimplified or uncalculated coefficients in the constant term

 

    (M1)(A1)

    A1

[7 marks]

Examiners report

Many correct answers were seen, although most candidates used rather inefficient methods (e.g.

expanding the brackets in multiple steps). In a very few cases candidates used the binomial theorem to

obtain the answer quickly.

41a. [3 marks]

Markscheme

number of arrangements of boys is and number of arrangements of girls is     (A1)

total number of arrangements is     M1A1 

Note: If 2 is omitted, award (A1)M1A0.

 

[3 marks]

Examiners report

A good number of correct answers were seen to this question, but a significant number of candidates

forgot to multiply by 2 in part (a) and in part (b) the most common error was to add the combinations

rather than multiply them.

70

41b. [3 marks]

Markscheme

number of ways of choosing two boys is and the number of ways of choosing three girls is

    (A1)

number of ways of choosing two boys and three girls is     M1A1

[3 marks]

Examiners report

A good number of correct answers were seen to this question, but a significant number of candidates

forgot to multiply by 2 in part (a) and in part (b) the most common error was to add the combinations

rather than multiply them.

42a. [2 marks]

Markscheme

for     M1A1

    AG

[2 marks]

Examiners report

Part (a) of this question was found challenging by the majority of candidates, a fairly common ‘solution’

being that the result is true for n = 1, 2, 3 and therefore true for all n. Some candidates attempted to use

induction which is a valid method but no completely correct solution using this method was seen.

Candidates found part (b) more accessible and many correct solutions were seen. The most common

problem was candidates using an incorrect comparison test, failing to realise that what was required

was a comparison between and .

42b. [3 marks]

71

Markscheme

    A1

is a positive converging geometric series     R1

hence converges by the comparison test     R1

[3 marks]

Examiners report

Part (a) of this question was found challenging by the majority of candidates, a fairly common ‘solution’

being that the result is true for n = 1, 2, 3 and therefore true for all n. Some candidates attempted to use

induction which is a valid method but no completely correct solution using this method was seen.

Candidates found part (b) more accessible and many correct solutions were seen. The most common

problem was candidates using an incorrect comparison test, failing to realise that what was required

was a comparison between and .

43a. [1 mark]

Markscheme

    A1

Note: Accept .

 

Note: Either of these may be seen in (b) and if so A1 should be awarded.

 

[1 mark]

Examiners report

Many candidates struggled to find an efficient approach to this problem by applying the Binomial

Theorem. A disappointing number of candidates attempted the whole expansion which was clearly an

72

unrealistic approach when it is noted that the expansion is to the 8 power. The fact that some

candidates wrote down Pascal’s Triangle suggested that they had not studied the Binomial Theorem in

enough depth or in a sufficient variety of contexts.

43b. [4 marks]

Markscheme

EITHER

     M1

     (A1)

coefficient of      M1

= −17 496     A1

Note: Under ft, final A1 can only be achieved for an integer answer.

 

OR

     M1

     (A1)

coefficient of      M1

= −17 496     A1

Note: Under ft, final A1 can only be achieved for an integer answer.

 

[4 marks]

Examiners report

Many candidates struggled to find an efficient approach to this problem by applying the Binomial

Theorem. A disappointing number of candidates attempted the whole expansion which was clearly an

unrealistic approach when it is noted that the expansion is to the 8 power. The fact that some 73

candidates wrote down Pascal’s Triangle suggested that they had not studied the Binomial Theorem in

enough depth or in a sufficient variety of contexts.

44. [4 marks]

Markscheme

    (M1)

    A3

Note: Deduct one A mark for each incorrect or omitted term.

 

[4 marks]

Examiners report

Most candidates solved this question correctly with most candidates who explained how they obtained

their coefficients using Pascal’s triangle rather than the combination formula.

45. [7 marks]

Markscheme

(a)     There are 3! ways of arranging the Mathematics books, 5! ways of arranging the English books

and 4! ways of arranging the Science books.     (A1)

Then we have 4 types of books which can be arranged in 4! ways.     (A1)

    (M1)A1

 

(b)     There are 3! ways of arranging the subject books, and for each of these there are 2 ways of putting

the dictionary next to the Mathematics books.     (M1)(A1)

    A1

[7 marks]

Examiners report

74

Many students added instead of multiplying. There were, however, quite a few good answers to this

question.

46. [6 marks]

Markscheme

(a)     coefficient of is      M1(A1)

    (A1)

     (M1)

    A1

 

(b)          A1

 

[6 marks]

Examiners report

Most candidates were able to answer this question well.

47. [5 marks]

Markscheme

EITHER

with no restrictions six people can be seated in ways     A1

we now count the number of ways in which the two restricted people will be sitting next to each other

call the two restricted people and

they sit next to each other in two ways     A1

the remaining people can then be seated in ways     A1

75

the six may be seated and next to each other) in ways     M1

with and ( not next to each other the number of ways      A1     N3

[5 marks]

OR

person seated at table in way    A1

then sits in any of seats (not next to )     M1A1

the remaining people can then be seated in ways     A1

number ways with not next to ways      A1        N3

Note: If candidate starts with instead of , potentially leading to an answer of , do not penalise.

[5 marks]

Examiners report

Very few candidates provided evidence of a clear strategy for solving such a question. The problem

which was set in a circular scenario was no more difficult than an analogous linear one.

48. [5 marks]

Markscheme

METHOD 1

constant term:     A1

term in x:     (M1)A1

term in :     M1A1     N3

[5 marks]

METHOD 2

76

    M1M1

    A1A1A1     N3

[5 marks]

Examiners report

Although the majority of the candidates understood the question and attempted it, excessive time was

spent on actually expanding the expression without consideration of the binomial theorem. A fair

amount of students confused “ascending order”, giving the last three instead of the first three terms.

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