Mathematics: analysis and approaches Higher level and Standard

Mathematics: analysis and

approaches

Higher level and Standard level

Specimen papers 1, 2 and 3

First examinations in 2021

CONTENTS

Mathematics: analysis and approaches higher level

paper 1 specimen paper

M

athematics: analysis and approaches higher level

paper 1 markscheme

M

athematics: analysis and approaches higher level

paper 2 specimen paper

M

athematics: analysis and approaches higher level

paper 2 markscheme

M

athematics: analysis and approaches higher level

paper 3 specimen paper

Mathematics: analysis and approaches higher level

paper 3 markscheme

M

athematics: analysis and approaches standard level

paper 1 specimen paper

M

athematics: analysis and approaches standard level

paper 1 markscheme

M

athematics: analysis and approaches standard level

paper 2 specimen paper

M

athematics: analysis and approaches standard level

paper 2 markscheme

Candidate session number

Mathematics: analysis and approaches

Higher level

Paper 1

13 pages

Specimen paper

2 hours

16EP01

Instructions to candidates

• Write your session number in the boxes above.

• Do not open this examination paper until instructed to do so.

• You are not permitted access to any calculator for this paper.

• Section A: answer all questions. Answers must be written within the answer boxes provided.

• Section B: answer all questions in the answer booklet provided. Fill in your session number

on the front of the answer booklet, and attach it to this examination paper and your

cover sheet using the tag provided.

• Unless otherwise stated in the question, all numerical answers should be given exactly or

correct to three signicant gures.

• A clean copy of the mathematics: analysis and approaches formula booklet is required for

this paper.

• The maximum mark for this examination paper is [110 marks].

SPEC/5/MATAA/HP1/ENG/TZ0/XX

Full marks are not necessarily awarded for a correct answer with no working. Answers must be

supported by working and/or explanations. Where an answer is incorrect, some marks may be given

for a correct method, provided this is shown by written working. You are therefore advised to show all

working.

Section A

Answer all questions. Answers must be written within the answer boxes provided. Working may be

continued below the lines, if necessary.

1. [Maximum mark: 5]

Let A and B be events such that P (A) = 0.5 , P (B) = 0.4 and P (A ∪ B) = 0.6 .

Find P (A | B) .

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

16EP02

SPEC/5/MATAA/HP1/ENG/TZ0/XX– 2 –

2. [Maximum mark: 5]

(a) Show that (2n - 1)

2

+ (2n + 1)

2

= 8n

2

+ 2 , where n ∈  . [2]

(b) Hence, or otherwise, prove that the sum of the squares of any two consecutive odd

integers is even. [3]

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

Turn over

16EP03

SPEC/5/MATAA/HP1/ENG/TZ0/XX– 3 –

3. [Maximum mark: 5]

Let

′

=

+

fx

x

()

8

21

2

. Given that f (0) = 5 , nd f (x) .

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

16EP04

SPEC/5/MATAA/HP1/ENG/TZ0/XX– 4 –

4. [Maximum mark: 5]

The following diagram shows the graph of y = f (x) . The graph has a horizontal asymptote

at y = -1 . The graph crosses the x-axis at x = -1 and x = 1 , and the y-axis at y = 2 .

y

x

1

−1

−2

−2−3

−

4 2

2

3 4

3

4

y = f (x)

On the following set of axes, sketch the graph of y =

[

f (x)

]

2

+ 1 , clearly showing any

asymptotes with their equations and the coordinates of any local maxima or minima.

y

x

1

−1 0

−1

−2

−2−3

−

4 2

2

3 4

3

4

5

6

Turn over

16EP05

SPEC/5/MATAA/HP1/ENG/TZ0/XX– 5 –

5. [Maximum mark: 5]

The functions f and g are dened such that

fx

x

()=

+

3

4

and g (x) = 8x + 5 .

(a) Show that ( g  f )(x) = 2x + 11 . [2]

(b) Given that ( g  f )

-1

(a) = 4 , nd the value of a . [3]

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

16EP06

SPEC/5/MATAA/HP1/ENG/TZ0/XX– 6 –

6. [Maximum mark: 8]

(a) Show that

log(cos)logcos

93

22

22xx

+= +

. [3]

(b) Hence or otherwise solve log

3

(2 sin x) = log

9

(cos 2x + 2) for

0

2

<<

x

�

. [5]

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

Turn over

16EP07

SPEC/5/MATAA/HP1/ENG/TZ0/XX– 7 –

7. [Maximum mark: 7]

A continuous random variable X has the probability density function f given by

fx

xx

x

()

sin,

,

=













≤≤











� �

36 6

06

0

otherwise

.

Find P (0 ≤ X ≤ 3).

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

16EP08

SPEC/5/MATAA/HP1/ENG/TZ0/XX– 8 –

8. [Maximum mark: 7]

The plane

П

has the Cartesian equation 2x + y + 2z = 3 .

The line L has the vector equation

r =−













+−













3

5

1

2

µ

p

,

µ

, p ∈  . The acute angle between

the line L and the plane

П

is 30



.

Find the possible values of p .

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

Turn over

16EP09

SPEC/5/MATAA/HP1/ENG/TZ0/XX– 9 –

9. [Maximum mark: 8]

The function f is dened by f (x) = e

2x

- 6e

x

+ 5 , x ∈  , x ≤ a . The graph of y = f (x) is

shown in the following diagram.

x

y

2

0

−2

−2 2

−4

4

5

(a) Find the largest value of a such that f has an inverse function. [3]

(b) For this value of a , nd an expression for f

-1

(x) , stating its domain. [5]

(This question continues on the following page)

16EP10

SPEC/5/MATAA/HP1/ENG/TZ0/XX– 10 –

(Question 9 continued)

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

Turn over

16EP11

SPEC/5/MATAA/HP1/ENG/TZ0/XX– 11 –

Do not write solutions on this page.

Section B

Answer all questions in the answer booklet provided. Please start each question on a new page.

10. [Maximum mark: 16]

Let

fx

x

kx

()

ln

=

5

where x > 0 , k ∈ 

+

.

(a) Show that

′

=

−

fx

x

kx

()

ln

15

2

. [3]

The graph of f has exactly one maximum point P .

(b) Find the x-coordinate of P . [3]

The second derivative of f is given by

′′

=

−

fx

x

kx

()

ln25 3

3

. The graph of f has exactly one

point of inexion Q .

(c) Show that the x-coordinate of Q is

1

5

3

2

e

. [3]

The region R is enclosed by the graph of f , the x-axis, and the vertical lines through the

maximum point P and the point of inexion Q .

y

x

R

P

Q

(d) Given that the area of R is 3 , nd the value of k . [7]

16EP12

SPEC/5/MATAA/HP1/ENG/TZ0/XX– 12 –

Do not write solutions on this page.

11. [Maximum mark: 18]

(a) Express

−+

33

i

in the form re

iθ

, where r > 0 and -π < θ ≤ π . [5]

Let the roots of the equation

z

3

33

=− + i

be u , v and w .

(b) Find u , v and w expressing your answers in the form re

iθ

, where r > 0 and -π < θ ≤ π . [5]

On an Argand diagram, u , v and w are represented by the points U , V and W respectively.

(c) Find the area of triangle UVW . [4]

(d) By considering the sum of the roots u , v and w , show that

co

scos cos

5

18

7

18

17

18

0

� � �

++ =

. [4]

12. [Maximum mark: 21]

The function f is dened by f (x) = e

sin x

.

(a) Find the rst two derivatives of f (x) and hence nd the Maclaurin series for f (x) up to

and including the x

2

term. [8]

(b) Show that the coecient of x

3

in the Maclaurin series for f (x) is zero. [4]

(c) Using the Maclaurin series for arctan x and e

3x

- 1 , nd the Maclaurin series

for arctan

(

e

3x

- 1

)

up to and including the x

3

term. [6]

(d) Hence, or otherwise, nd

lim

()

arctan

x

fx

→

−

()

0

3

1

1e

. [3]

16EP13

SPEC/5/MATAA/HP1/ENG/TZ0/XX– 13 –

Please do not write on this page.

Answers written on this page

will not be marked.

16EP14

Please do not write on this page.

Answers written on this page

will not be marked.

16EP15

Please do not write on this page.

Answers written on this page

will not be marked.

16EP16

SPEC/5/MATAA/HP1/ENG/TZ0/XX/M

16 pages

Markscheme

Specimen paper

Mathematics:

analysis and approaches

Higher level

Paper 1

– 2 – SPEC/5/MATAA/HP1/ENG/TZ0/XX/M

Instructions to Examiners

Abbreviations

M Marks awarded for attempting to use a correct Method.

A Marks awarded for an Answer or for Accuracy; often dependent on preceding M marks.

R Marks awarded for clear Reasoning.

AG Answer given in the question and so no marks are awarded.

Using the markscheme

1 General

Award marks using the annotations as noted in the markscheme eg M1, A2.

2 Method and Answer/Accuracy marks

 Do not automatically award full marks for a correct answer; all working must be checked, and

marks awarded according to the markscheme.

 It is generally not possible to award M0 followed by A1, as A mark(s) depend on the preceding

M mark(s), if any.

 Where M and A marks are noted on the same line, e.g. M1A1, this usually means M1 for an

attempt to use an appropriate method (e.g. substitution into a formula) and A1 for using the

correct values.

 Where there are two or more A marks on the same line, they may be awarded independently;

so if the first value is incorrect, but the next two are correct, award A0A1A1.

 Where the markscheme specifies M2, A2

, etc., do not split the marks, unless there is a note.

 Once a correct answer to a question or part-question is seen, ignore further correct working.

However, if further working indicates a lack of mathematical understanding do not award the final

A1. An exception to this may be in numerical answers, where a correct exact value is followed by

an incorrect decimal. However, if the incorrect decimal is carried through to a subsequent part,

and correct working shown, award FT marks as appropriate but do not award the final A1 in that

part.

Examples

Correct answer seen Further working seen Action

1.

82

5.65685...

(incorrect decimal value)

Award the final A1

(ignore the further working)

2.

1

sin 4

4

x

sin

x

Do not award the final A1

3.

log logab

log ( )ab

Do not award the final A1

– 3 – SPEC/5/MATAA/HP1/ENG/TZ0/XX/M

3 Implied marks

Implied marks appear in brackets e.g. (M1), and can only be awarded if correct work is seen or if

implied in subsequent working.

 Normally the correct work is seen or implied in the next line.

 Marks without brackets can only be awarded for work that is seen.

4 Follow through marks (only applied after an error is made)

Follow through (FT) marks are awarded where an incorrect answer from one part of a question is

used correctly in subsequent part(s) or subpart(s). Usually, to award FT marks, there must be

working present and not just a final answer based on an incorrect answer to a previous part.

However, if the only marks awarded in a subpart are for the answer (i.e. there is no working

expected), then FT marks should be awarded if appropriate.

 Within a question part, once an error is made, no further A marks can be awarded for work

which uses the error, but M marks may be awarded if appropriate.

 If the question becomes much simpler because of an error then use discretion to award fewer

FT marks.

 If the error leads to an inappropriate value (e.g. probability greater than 1, use of

1r 

for the

sum of an infinite GP,

sin 1.5





, non integer value where integer required), do not award the

mark(s) for the final answer(s).

 The markscheme may use the word “their” in a description, to indicate that candidates may be

using an incorrect value.

 Exceptions to this rule will be explicitly noted on the markscheme.

 If a candidate makes an error in one part, but gets the correct answer(s) to subsequent part(s),

award marks as appropriate, unless the question says hence. It is often possible to use a

different approach in subsequent parts that does not depend on the answer to previous parts.

5 Mis-read

If a candidate incorrectly copies information from the question, this is a mis-read (MR). Apply a MR

penalty of 1 mark to that question

 If the question becomes much simpler because of the MR, then use discretion to award

fewer marks.

 If the MR leads to an inappropriate value (e.g. probability greater than

1,

sin 1.5





, non-integer

value where integer required), do not award the mark(s) for the final answer(s).

 Miscopying of candidates’ own work does not constitute a misread, it is an error.

 The MR penalty can only be applied when work is seen. For calculator questions with no

working and incorrect answers, examiners should not infer that values were read incorrectly.

– 4 – SPEC/5/MATAA/HP1/ENG/TZ0/XX/M

6 Alternative methods

 Alternative methods for complete questions are indicated by METHOD 1,

METHOD 2, etc.

 Alternative solutions for part-questions are indicated by EITHER . . . OR.

7 Alternative forms

Unless the question specifies otherwise, accept equivalent forms.

 As this is an international examination, accept all alternative forms of notation.

 In the markscheme, equivalent numerical and algebraic forms will generally be written in

brackets immediately following the answer.

 In the markscheme, simplified answers, (which candidates often do not write in examinations),

will generally appear in brackets. Marks should be awarded for either the form preceding the

bracket or the form in brackets (if it is seen).

8 Accuracy of Answers

If the level of accuracy is specified in the question, a mark will be linked to giving the answer to the

required accuracy. There are two types of accuracy errors, and the final answer mark should not be

awarded if these errors occur.

 Rounding errors: only applies to final answers not to intermediate steps.

 Level of accuracy: when this is not specified in the question the general rule applies to final

answers: unless otherwise stated in the question all numerical answers must be given exactly or

correct to three significant figures.

9 Calculators

No calculator is allowed. The use of any calculator on paper 1 is malpractice, and will result in no

grade awarded. If you see work that suggests a candidate has used any calculator, please follow

the procedures for malpractice. Examples: finding an angle, given a trig ratio of 0.4235.

Candidates will sometimes use methods other than those in the markscheme. Unless the question

specifies a method, other correct methods should be marked in line with the markscheme

– 5 – SPEC/5/MATAA/HP1/ENG/TZ0/XX/M

Section A

1. attempt to substitute into



PPPP

A

BABAB   

(M1)

Note: Accept use of Venn diagram or other valid method.

0.6 0.5 0.4 P( )

A

B 

(A1)



P0.3AB

(seen anywhere) A1

attempt to substitute into









P

P|

P

A

B

AB

B





(M1)

0.3

0.4







P|

A

B

3

0.75

4









A1

Total [5 marks]

2. (a) attempting to expand the LHS (M1)



22

LHS 4 4 1 4 4 1nn nn A1





2

82RHSn AG

[2 marks]

(b) METHOD 1

recognition that

21n 

and

21n 

represent two consecutive odd

integers (for all odd integers

n

) R1



22

82241nn  A1

valid reason eg divisible by

2 (2 is a factor) R1

so the sum of the squares of any two consecutive odd integers is even AG

[3 marks]

METHOD 2

recognition, eg that

n

and

2n 

represent two consecutive odd integers

(for

n

) R1



2

22

22 22nn n n  

A1

valid reason eg divisible by

2 (2 is a factor) R1

so the sum of the squares of any two consecutive odd integers

is even AG

[3 marks]

Total [5 marks]

– 6 – SPEC/5/MATAA/HP1/ENG/TZ0/XX/M

3. attempt to integrate (M1)

2

d

21 4

d

u

ux x

x



2

82

dd

21

x

u

x







(A1)

EITHER



4 uC

A1

OR



2

42 1

x

C A1

THEN

correct substitution into their integrated function (must have

C

) (M1)

54 1CC  



2

42 11fx x A1

Total [5 marks]

– 7 – SPEC/5/MATAA/HP1/ENG/TZ0/XX/M

4.

no

y values below 1 A1

horizontal asymptote at

2y 

with curve approaching from below as

x



A1



1,1

local minima A1



0,5 local maximum A1

smooth curve and smooth stationary points A1

Total [5 marks]

5. (a) attempt to form composition M1

correct substitution

33

85

44

xx

g











A1



211gf x x

AG

[2 marks]

(b) attempt to substitute 4 (seen anywhere) (M1)

correct equation

2411a 

(A1)

19a 

A1

[3 marks]

Total [5 marks]

– 8 – SPEC/5/MATAA/HP1/ENG/TZ0/XX/M

6. (a) attempting to use the change of base rule M1

3

9

3

log (cos 2 2)

log 9

x





A1

3

1

log (cos 2 2)

2

x A1

3

log cos 2 2x AG

[3 marks]

(b)

33

log (2sin ) log cos 2 2xx

2sin cos2 2xx

M1

2

4sin cos2 2xx (or equivalent) A1

use of

2

cos 2 1 2sin

x

x (M1)

2

6sin 3x 



1

sin

2

x 

A1

π

4

x  A1

Note: Award A0 if solutions other than

π

4

x  are included.

[5 marks]

Total [8 marks]

– 9 – SPEC/5/MATAA/HP1/ENG/TZ0/XX/M

7. attempting integration by parts, eg

ππ π 6π

,d d ,d sin d , cos

36 36 6 π 6

x

xx

uuxv xv

 

  

 

 

(M1)



3

0

π6 π 6 π

P0 3 cos cos d

36 π 6 π 6

xx x

X

x





 

  



 





 







(or equivalent) A1A1

Note: Award A1 for a correct

uv

and A1 for a correct dvu



.

attempting to substitute limits M1

3

0

π6 π

cos 0

36 π 6

xx













(A1)

so



3

0

1π

P0 3 sin

π6

x

X















(or equivalent) A1

1

π



A1

Total [7 marks]

8. recognition that the angle between the normal and the line is

60

(seen anywhere) R1

attempt to use the formula for the scalar product M1

2

21

12

2

cos 60

914

p

 

 



 

 







A1

2

1

2

35

p





A1

2

35 4

p

p

attempt to square both sides M1





222

95 16 7 45ppp  

5

3

7

p 

(or equivalent) A1A1

Total [7 marks]

– 10 – SPEC/5/MATAA/HP1/ENG/TZ0/XX/M

9. (a) attempt to differentiate and set equal to zero M1

2

( ) 2e 6e 2e (e 3) 0

xx xx

fx



 

A1

minimum at

ln 3x 

ln 3a 

A1

[3 marks]

(b)

Note: Interchanging

x

and

y

can be done at any stage.



2

e3 4

x

y 

(M1)

e3 4

x

y 

A1

as

ln 3x 

,



ln 3 4xy

R1

so



1

ln 3 4fx x





A1

domain of

1

f



is

,4 5xx

A1

[5 marks]

Total [8 marks]

– 11 – SPEC/5/MATAA/HP1/ENG/TZ0/XX/M

Section B

10. (a) attempt to use quotient rule (M1)

correct substitution into quotient rule



2

1

5ln5

5

kx k x

x

fx

kx













(or equivalent) A1



22

ln 5

,

kk x

k

kx







 A1

2

1ln5

x

kx





AG

[3 marks]

(b)



0fx





M1

2

1ln5

0

x

kx





ln 5 1

x



(A1)

e

5

x 

A1

[3 marks]

(c)



0fx





M1

3

2ln5 3

0

x

kx





3

ln 5

2

x 

A1

3

2

5ex 

A1

so the point of inflexion occurs at

3

2

1

e

5

x 

AG

[3 marks]

continued…

– 12 – SPEC/5/MATAA/HP1/ENG/TZ0/XX/M

Question 10 continued

(d) attempt to integrate (M1)

d1

ln 5

d

u

ux

x



ln 5 1

d d

x

uu

kx k





(A1)

EITHER

2

u

k



A1

so

3

2

1

d

2

u

uu

kk











A1

OR



2

ln 5

2

x

k



A1

so



3

2

1

e

2

5

e

5

ln 5

d

2

x

kx k











A1

THEN

19

1

24k









5

8k



A1

setting their expression for area equal to

3 M1

5

3

8k



5

24

k 

A1

[7 marks]

Total [16 marks]

– 13 – SPEC/5/MATAA/HP1/ENG/TZ0/XX/M

11. (a) attempt to find modulus (M1)





23 12r 

A1

attempt to find argument in the correct quadrant (M1)

3

πarctan

3





 





A1

5π

6



A1

5πi 5πi

66

33i12e 23e



  





[5 marks]

(b) attempt to find a root using de Moivre’s theorem M1

15πi

618

12 e

A1

attempt to find further two roots by adding and subtracting

2π

3

to

the argument M1

17πi

618

12 e



A1

117πi

618

12 e A1

Note: Ignore labels for

,uv

and

w

at this stage.

[5 marks]

continued…

– 14 – SPEC/5/MATAA/HP1/ENG/TZ0/XX/M

Question 11 continued

(c) METHOD 1

attempting to find the total area of (congruent) triangles

UOV, VOW

and

UOW

M1

11

66

12π

Area 3 12 12 sin

23















A1A1

Note: Award A1 for

11

66

12 12







and A1 for

2π

sin

3

.

1

3

33

12

4









(or equivalent) A1

[4 marks]

METHOD 2

22

11 11

2

66 66

2π

UV 12 12 2 12 12 cos

3

 



 

 

(or equivalent) A1

1

6

UV 3 12









(or equivalent) A1

attempting to find the area of

UVW

using

1

Area UV VW sin

2



  

for example M1

11

66

1π

Area 3 12 3 12 sin

23



 





1

3

33

12

4









(or equivalent) A1

[4 marks]

(d)

0uvw 

R1

1

6

7π 7π 5π 5π 17π 17π

12 cos isin cos i sin cos isin 0

18 18 18 18 18 18



 

     

 



 



A1

consideration of real parts M1

1

6

7π 5π 17π

12 cos cos cos 0

18 18 18



  





7π 7π

cos cos

18 18









explicitly stated A1

5π 7π 17π

cos cos cos 0

18 18 18

 

AG

[4 marks]

Total [18 marks]

– 15 – SPEC/5/MATAA/HP1/ENG/TZ0/XX/M

12. (a) attempting to use the chain rule to find the first derivative M1









sin

cos e

x

fx x



 A1

attempting to use the product rule to find the second derivative M1



sin 2

ecos sin

x

f

xxx





(or equivalent) A1

attempting to find



0f

,



0f



and





0f



M1





01f 

;

  

sin 0

0cos0e 1f





;



sin 0 2

0e cos0sin01f





A1

substitution into the Maclaurin formula

2

( ) (0) (0) (0) ...

2!

x

fx f xf f



  

M1

so the Maclaurin series for



f

x

up to and including the

2

x

term is

2

1

2

x

x

A1

[8 marks]

(b) METHOD 1

attempting to differentiate

()

f

x



M1

  



 

sin 2 sin

cos e cos sin cos e 2sin 1

xx

fx x x x x x





(or equivalent) A2

substituting

0x 

into their



f

x



M1

    

01101010f





so the coefficient of

3

x

in the Maclaurin series for



f

x is zero AG

METHOD 2

substituting

sin

x

into the Maclaurin series for e

x

(M1)

23

sin

sin sin

e 1 sin ...

2! 3!

x

xx

x   

substituting Maclaurin series for

sin

x

M1

23

33

3

sin

... ...

3! 3!

e 1 ... ...

3! 2! 3!

x

xx

x



 







     





A1

coefficient of

3

x

is

11

0

3! 3!

 A1

so the coefficient of

3

x

in the Maclaurin series for



f

x

is zero AG

[4 marks]

continued…

– 16 – SPEC/5/MATAA/HP1/ENG/TZ0/XX/M

Question 12 continued

(c) substituting

3

x

into the Maclaurin series for e

x

M1

 

23

3

33

e 1 3 ...

2! 3!

x

xx

x   

A1

substituting





3

e1

x

 into the Maclaurin series for

arctan

x

M1





35

33

e1 e1

arctan e 1 e 1 ...

35

xx



   

 

3

23

33

3

2! 3!

33

3 ...

2! 3! 3

xx

x

xx

x











   





A1

selecting correct terms from above M1

  

23 3

33 3

3

2! 3! 3

x

xx

x



  





23

99

3

22

x

x 

A1

[6 marks]

(d) METHOD 1

substitution of their series M1

2

0

...

2

lim

9

3 ...

2

x





A1

0

1...

2

lim

9

3...

2

x









1

3



A1

METHOD 2

use of l’Hôpital’s rule M1







sin

3

0

2

3

cos e

lim

3e

1e 1

x





(or equivalent) A1

1

3



A1

[3 marks]

Total [21 marks]

12EP01

Candidate session number

Mathematics: analysis and approaches

Higher level

Paper 2

12 pages

Specimen

2 hours

Instructions to candidates

• Write your session number in the boxes above.

• Do not open this examination paper until instructed to do so.

• A graphic display calculator is required for this paper.

• Section A: answer all questions. Answers must be written within the answer boxes provided.

• Section B: answer all questions in the answer booklet provided. Fill in your session number

on the front of the answer booklet, and attach it to this examination paper and your

cover sheet using the tag provided.

• Unless otherwise stated in the question, all numerical answers should be given exactly or

correct to three signicant gures.

• A clean copy of the mathematics: analysis and approaches formula booklet is required for

this paper.

• The maximum mark for this examination paper is [110 marks].

SPEC/5/MATAA/HP2/ENG/TZ0/XX

Full marks are not necessarily awarded for a correct answer with no working. Answers must be

supported by working and/or explanations. Solutions found from a graphic display calculator should be

supported by suitable working. For example, if graphs are used to nd a solution, you should sketch

these as part of your answer. Where an answer is incorrect, some marks may be given for a correct

method, provided this is shown by written working. You are therefore advised to show all working.

Section A

Answer all questions. Answers must be written within the answer boxes provided. Working may be

continued below the lines, if necessary.

1. [Maximum mark: 6]

The following diagram shows part of a circle with centre O and radius 4 cm .

O

B

A

θ

4 cm

5 cm

Chord AB has a length of 5 cm and AÔB = θ .

(a) Find the value of θ , giving your answer in radians. [3]

(b) Find the area of the shaded region. [3]

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

12EP02

SPEC/5/MATAA/HP2/ENG/TZ0/XX– 2 –

2. [Maximum mark: 6]

On 1st January 2020, Laurie invests $P in an account that pays a nominal annual interest

rate of 5.5 % , compounded quarterly.

The amount of money in Laurie’s account at the end of each year follows a geometric

sequence with common ratio, r .

(a) Find the value of r , giving your answer to four signicant gures. [3]

Laurie makes no further deposits to or withdrawals from the account.

(b) Find the year in which the amount of money in Laurie’s account will become double the

amount she invested. [3]

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

12EP03

Turn over

SPEC/5/MATAA/HP2/ENG/TZ0/XX– 3 –

3. [Maximum mark: 6]

A six-sided biased die is weighted in such a way that the probability of obtaining a “six” is

7

10

.

The die is tossed ve times. Find the probability of obtaining

(a) at most three “sixes”. [3]

(b) the third “six” on the fth toss. [3]

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

12EP04

SPEC/5/MATAA/HP2/ENG/TZ0/XX– 4 –

4. [Maximum mark: 7]

The following table below shows the marks scored by seven students on two dierent

mathematics tests.

Test 1 (x)

15 23 25 30 34 34 40

Test 2 ( y)

20 26 27 32 35 37 35

Let L

1

be the regression line of x on y . The equation of the line L

1

can be written in the

form x =ay + b .

(a) Find the value of a and the value of b . [2]

Let L

2

be the regression line of y on x . The lines L

1

and L

2

pass through the same point

with coordinates ( p , q) .

(b) Find the value of p and the value of q . [3]

(c) Jennifer was absent for the rst test but scored 29 marks on the second test. Use an

appropriate regression equation to estimate Jennifer’s mark on the rst test. [2]

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

12EP05

Turn over

SPEC/5/MATAA/HP2/ENG/TZ0/XX– 5 –

5. [Maximum mark: 7]

The displacement, in centimetres, of a particle from an origin, O, at time t seconds, is given

by s (t) = t

2

cos t+2t sin t , 0 ≤ t ≤ 5 .

(a) Find the maximum distance of the particle from O. [3]

(b) Find the acceleration of the particle at the instant it rst changes direction. [4]

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

12EP06

SPEC/5/MATAA/HP2/ENG/TZ0/XX– 6 –

6. [Maximum mark: 6]

In a city, the number of passengers, X , who ride in a taxi has the following probability distribution.

x

1 2 3 4 5

P (X = x)

0.60 0.30 0.03 0.05 0.02

After the opening of a new highway that charges a toll, a taxi company introduces a charge

for passengers who use the highway. The charge is $ 2.40 per taxi plus $ 1.20 per passenger.

Let T represent the amount, in dollars, that is charged by the taxi company per ride.

(a) Find E (T ) . [4]

(b) Given that Var (X ) = 0.8419 , nd Var (T ) . [2]

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

12EP07

Turn over

SPEC/5/MATAA/HP2/ENG/TZ0/XX– 7 –

7. [Maximum mark: 5]

Two ships, A and B , are observed from an origin O . Relative to O , their position vectors at

time t hours after midday are given by

r

A

=













+













4

3

5

8

t

r

B

=

−













+













7

3

0

12

t

where distances are measured in kilometres.

Find the minimum distance between the two ships.

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

12EP08

SPEC/5/MATAA/HP2/ENG/TZ0/XX– 8 –

8. [Maximum mark: 7]

The complex numbers w and z satisfy the equations

w

z

= 2i

z

∗

- 3w = 5 + 5i .

Find w and z in the form a + bi where a , b ∈  .

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

12EP09

Turn over

SPEC/5/MATAA/HP2/ENG/TZ0/XX– 9 –

9. [Maximum mark: 5]

Consider the graphs of

y

x

=

−

2

3

and y = m (x + 3) , m ∈  .

Find the set of values for m such that the two graphs have no intersection points.

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

12EP10

SPEC/5/MATAA/HP2/ENG/TZ0/XX– 10 –

Do not write solutions on this page.

Section B

Answer all questions in the answer booklet provided. Please start each question on a new page.

10. [Maximum mark: 15]

The length, X mm , of a certain species of seashell is normally distributed with mean 25 and

variance, σ

2

.

The probability that X is less than 24.15 is 0.1446 .

(a) Find P (24.15 < X < 25) . [2]

(b) (i) Find σ , the standard deviation of X .

(ii) Hence, nd the probability that a seashell selected at random has a length

greater than 26 mm . [5]

A random sample of 10 seashells is collected on a beach. Let Y represent the number of

seashells with lengths greater than 26 mm .

(c) Find E (Y ) . [3]

(d) Find the probability that exactly three of these seashells have a length greater

than 26 mm . [2]

A seashell selected at random has a length less than 26 mm .

(e) Find the probability that its length is between 24.15 mm and 25 mm . [3]

12EP11

Turn over

SPEC/5/MATAA/HP2/ENG/TZ0/XX– 11 –

Do not write solutions on this page.

11. [Maximum mark: 21]

A large tank initially contains pure water. Water containing salt begins to ow into the tank

The solution is kept uniform by stirring and leaves the tank through an outlet at its base.

Let x grams represent the amount of salt in the tank and let t minutes represent the time

since the salt water began owing into the tank.

The rate of change of the amount of salt in the tank,

d

x

t

, is described by the dierential

equation

d

e

x

t

x

t

=−

+

−

10

1

4

.

(a) Show that t + 1 is an integrating factor for this dierential equation. [2]

(b) Hence, by solving this dierential equation, show that

xt

t

()

=

−+

+

−

200 40 5

1

4

e

. [8]

(c) Sketch the graph of x versus t for 0 ≤ t ≤ 60 and hence nd the maximum amount of

salt in the tank and the value of t at which this occurs. [5]

(d) Find the value of t at which the amount of salt in the tank is decreasing most rapidly. [2]

The rate of change of the amount of salt leaving the tank is equal to

x

t + 1

.

(e) Find the amount of salt that left the tank during the rst 60 minutes. [4]

12. [Maximum mark: 19]

(a) Show that

cot

tan

2

1

2

θ

=

−

. [1]

(b) Verify that x = tan θ and x =- cot θ satisfy the equation x

2

+ (2 cot 2θ) x - 1= 0 . [7]

(c) Hence, or otherwise, show that the exact value of

ta

n

�

12

23

= −

. [5]

(d) Using the results from parts (b) and (c) nd the exact value of

tanc

ot

� �

24 24

−

.

Give your answer in the form

ab+ 3

where a , b ∈  . [6]

12EP12

SPEC/5/MATAA/HP2/ENG/TZ0/XX– 12 –

SPEC/5/MATAA/HP2/ENG/TZ0/XX/M

16 pages

Markscheme

Specimen paper

Mathematics:

analysis and approaches

Higher level

Paper 2

– 2 – SPEC/5/MATAA/HP2/ENG/TZ0/XX/M

Instructions to Examiners

Abbreviations

M Marks awarded for attempting to use a correct Method.

A Marks awarded for an Answer or for Accuracy; often dependent on preceding M marks.

R Marks awarded for clear Reasoning.

AG Answer given in the question and so no marks are awarded.

Using the markscheme

1 General

Award marks using the annotations as noted in the markscheme eg M1, A2.

2 Method and Answer/Accuracy marks

 Do not automatically award full marks for a correct answer; all working must be checked, and

marks awarded according to the markscheme.

 It is generally not possible to award M0 followed by A1, as A mark(s) depend on the preceding

M mark(s), if any.

 Where M and A marks are noted on the same line, e.g. M1A1, this usually means M1 for an

attempt to use an appropriate method (e.g. substitution into a formula) and A1 for using the

correct values.

 Where there are two or more A marks on the same line, they may be awarded independently;

so if the first value is incorrect, but the next two are correct, award A0A1A1.

 Where the markscheme specifies M2, A2

, etc., do not split the marks, unless there is a note.

 Once a correct answer to a question or part-question is seen, ignore further correct working.

However, if further working indicates a lack of mathematical understanding do not award the final

A1. An exception to this may be in numerical answers, where a correct exact value is followed by

an incorrect decimal. However, if the incorrect decimal is carried through to a subsequent part,

and correct working shown, award FT marks as appropriate but do not award the final A1 in that

part.

Examples

Correct answer seen Further working seen Action

1.

82

5.65685...

(incorrect decimal value)

Award the final A1

(ignore the further working)

2.

1

sin 4

4

x

sin

x

Do not award the final A1

3.

log logab

log ( )ab

Do not award the final A1

– 3 – SPEC/5/MATAA/HP2/ENG/TZ0/XX/M

3 Implied marks

Implied marks appear in brackets e.g. (M1), and can only be awarded if correct work is seen or if

implied in subsequent working.

 Normally the correct work is seen or implied in the next line.

 Marks without brackets can only be awarded for work that is seen.

4 Follow through marks (only applied after an error is made)

Follow through (FT) marks are awarded where an incorrect answer from one part of a question is

used correctly in subsequent part(s) or subpart(s). Usually, to award FT marks, there must be

working present and not just a final answer based on an incorrect answer to a previous part.

However, if the only marks awarded in a subpart are for the answer (i.e. there is no working

expected), then FT marks should be awarded if appropriate.

 Within a question part, once an error is made, no further A marks can be awarded for work

which uses the error, but M marks may be awarded if appropriate.

 If the question becomes much simpler because of an error then use discretion to award fewer

FT marks.

 If the error leads to an inappropriate value (e.g. probability greater than 1, use of

1r  for the

sum of an infinite GP,

sin 1.5





, non integer value where integer required), do not award the

mark(s) for the final answer(s).

 The markscheme may use the word “their” in a description, to indicate that candidates may be

using an incorrect value.

 Exceptions to this rule will be explicitly noted on the markscheme.

 If a candidate makes an error in one part, but gets the correct answer(s) to subsequent part(s),

award marks as appropriate, unless the question says hence. It is often possible to use a

different approach in subsequent parts that does not depend on the answer to previous parts.

5 Mis-read

If a candidate incorrectly copies information from the question, this is a mis-read (MR). Apply a MR

penalty of 1 mark to that question

 If the question becomes much simpler because of the MR, then use discretion to award

fewer marks.

 If the MR leads to an inappropriate value (e.g. probability greater than

1,

sin 1.5





, non-integer

value where integer required), do not award the mark(s) for the final answer(s).

 Miscopying of candidates’ own work does not constitute a misread, it is an error.

 The MR penalty can only be applied when work is seen. For calculator questions with no

working and incorrect answers, examiners should not infer that values were read incorrectly.

– 4 – SPEC/5/MATAA/HP2/ENG/TZ0/XX/M

6 Alternative methods

 Alternative methods for complete questions are indicated by METHOD 1,

METHOD 2, etc.

 Alternative solutions for part-questions are indicated by EITHER . . . OR.

7 Alternative forms

Unless the question specifies otherwise, accept equivalent forms.

 As this is an international examination, accept all alternative forms of notation.

 In the markscheme, equivalent numerical and algebraic forms will generally be written in

brackets immediately following the answer.

 In the markscheme, simplified answers, (which candidates often do not write in examinations),

will generally appear in brackets. Marks should be awarded for either the form preceding the

bracket or the form in brackets (if it is seen).

8 Accuracy of Answers

If the level of accuracy is specified in the question, a mark will be linked to giving the answer to the

required accuracy. There are two types of accuracy errors, and the final answer mark should not be

awarded if these errors occur.

 Rounding errors: only applies to final answers not to intermediate steps.

 Level of accuracy: when this is not specified in the question the general rule applies to final

answers: unless otherwise stated in the question all numerical answers must be given exactly or

correct to three significant figures.

9 Calculators

A GDC is required for paper 2, but calculators with symbolic manipulation features/ CAS functionality

are not allowed.

Calculator notation

The subject guide says:

Students must always use correct mathematical notation, not calculator notation.

Do not accept final answers written using calculator notation. However, do not penalize the use of

calculator notation in the working.

Candidates will sometimes use methods other than those in the markscheme. Unless the question

specifies a method, other correct methods should be marked in line with the markscheme

– 5 – SPEC/5/MATAA/HP2/ENG/TZ0/XX/M

Section A

1. (a) METHOD 1

attempt to use the cosine rule (M1)

222

445

cos

244









(or equivalent) A1

1.35





A1

[3 marks]

METHOD 2

attempt to split triangle

AOB into two congruent right triangles (M1)

2.5

sin

24











A1

1.35





A1

[3 marks]

(b) attempt to find the area of the shaded region (M1)

1

44(2

2

    

A1

39.5

(cm

2

) A1

[3 marks]

Total [6 marks]

2. (a)

4

5.5

1

4100











(M1)(A1)

1.056

A1

[3 marks]

continued…

– 6 – SPEC/5/MATAA/HP2/ENG/TZ0/XX/M

Question 2 continued

(b) EITHER

4

5.5

21

100 4

n

PP











OR



2their()

m

P

Pa

(M1)(A1)

Note: Award (M1) for substitution into loan payment formula. Award (A1) for correct substitution.

OR

PV 1

FV 2 

I% 5.5

P/Y  4

C/Y  4

50.756n 

(M1)(A1)

OR

PV 1

FV 2 

I% 100 (their ( ) 1)a

P/Y  1

C/Y  1 (M1)(A1)

THEN



12.7 years

Laurie will have double the amount she invested during

2032 A1

[3 marks]

Total [6 marks]

3. (a) recognition of binomial (M1)

~B(5,0.7)X

attempt to find



P3X 

M1



0.472 0.47178

A1

[3 marks]

(b) recognition of

2 sixes in 4 tosses (M1)



22

4

P 3rd six on the 5th toss 0.7 0.3 0.7( 0.2646 0.7)

2





   









A1



0.185 0.18522

A1

[3 marks]

Total [6 marks]

– 7 – SPEC/5/MATAA/HP2/ENG/TZ0/XX/M

4. (a)

1.29a 

and

10.4b 

A1A1

[2 marks]

(b) recognising both lines pass through the mean point (M1)

28.7, 30.3pq A2

[3 marks]

(c) substitution into their

x on y equation (M1)

1.29082(29) 10.3793x 

27.1x 

A1

Note: Accept

27.

[2 marks]

Total [7 marks]

5. (a) use of a graph to find the coordinates of the local minimum (M1)

16.513...s 

(A1)

maximum distance is

16.5 cm (to the left of O) A1

[3 marks]

(b) attempt to find time when particle changes direction eg considering the

first maximum on the graph of

s

or the first t – intercept on the graph of

.s



(M1)

1.51986...t 

(A1)

attempt to find the gradient of

s



for their value of t,



1.51986...s



(M1)

8.92

(cm/s

2

) A1

[4 marks]

Total [7 marks]

– 8 – SPEC/5/MATAA/HP2/ENG/TZ0/XX/M

6. (a) METHOD 1

attempting to use the expected value formula (M1)

     

E 1 0.60 2 0.30 3 0.03 4 0.05 5 0.02X     

 

E1.59$X 

(A1)

use of



E 1.20 2.40 1.20E 2.40XX 

(M1)









E 1.20 1.59 2.40T 



4.31 $

A1

METHOD 2

attempting to find the probability distribution for

T (M1)

t

3.60 4.80 6.00 7.20 8.40



P Tt

0.60 0.30 0.03 0.05 0.02

(A1)

attempting to use the expected value formula (M1)

























E 3.60 0.60 4.80 0.30 6.00 0.03 7.20 0.05 8.40 0.02T 



4.31 $

A1

[4 marks]

(b) METHOD 1

using



2

Var 1.20 2.40 1.20 Var

X

X

with



Var 0.8419X 

(M1)



Var 1.21T 

A1

METHOD 2

finding the standard deviation for their probability distribution found

in part (a) (M1)

  

2

Var 1.101...T 

1.21

A1

Note: Award M1A1 for

  

2

Var 1.093... 1.20T 

.

[2 marks]

Total [6 marks]

– 9 – SPEC/5/MATAA/HP2/ENG/TZ0/XX/M

7. attempting to find

BA

rr

for example (M1)

BA

35

64

t





 







rr

attempting to find

BA

rr

M1

distance







22

2

( ) 3 5 4 6 41 78 45dt t t t t  

A1

using a graph to find the

d 

coordinate of the local minimum M1

the minimum distance between the ships is

 

11 41

2.81 km km

41









A1

Total [5 marks]

8. substituting

2iwz

into

355izw





M1

6i 5 5izz





A1

let

izxy

comparing real and imaginary parts of



i6i i55ixy xy 

M1

to obtain

65xy

and

65xy

A1

attempting to solve for

x

and

y

M1

1x 

and

1y 

and so

1iz  

A1

hence

22iw  

A1

– 10 – SPEC/5/MATAA/HP2/ENG/TZ0/XX/M

9. METHOD 1

sketching the graph of

2

3

x

y

x





(

9

3

yx

x





) M1

the (oblique) asymptote has a gradient equal to

1

and so the maximum value of

m

is 1 R1

consideration of a straight line steeper than the horizontal line joining



3, 0

and



0, 0

M1

so

0m 

R1

hence

01m

A1

METHOD 2

attempting to eliminate y to form a quadratic equation in

x

M1



22

9xmx



2

190mxm 

A1

EITHER

attempting to solve



4190mm

for

m

M1

OR

attempting to solve

2

0x  ie



9

01

1

m





for

m

M1

THEN

01m 

A1

a valid reason to explain why

1m 

gives no solutions eg if

1m 

,



2

19090mxm

and so

01m

R1

Total [5 marks]

– 11 – SPEC/5/MATAA/HP2/ENG/TZ0/XX/M

Section B

10. (a) attempt to use the symmetry of the normal curve (M1)

eg diagram,

0.5  0.1446



P 24.15 25 0.3554X 

A1

[2 marks]

(b) (i) use of inverse normal to find

z score (M1)

1.0598z 

correct substitution

24.15 25

1.0598







(A1)

0.802





A1

(ii)



P 26 0.106X 

(M1)A1

[5 marks]

(c) recognizing binomial probability (M1)

E( ) 10 0.10621Y 

(A1)

 1.06

A1

[3 marks]

(d)





P3Y 

(M1)

0.0655

A1

[2 marks]

(e) recognizing conditional probability (M1)

correct substitution A1

0.3554

1 0.10621

 0.398

A1

[3 marks]

Total [15 marks]

– 12 – SPEC/5/MATAA/HP2/ENG/TZ0/XX/M

11. (a) METHOD 1

using



d

e

P

tt

It





M1

1

d

1

e

t

t 





ln 1

e

t 



A1

1t

AG

METHOD 2

attempting product rule differentiation on



d

1

d

xt

t



M1



dd

11

dd

x

ttx

tt

 



d

1

d1

x

t

tt



 







A1

so

1t 

is an integrating factor for this differential equation AG

[2 marks]

continued…

– 13 – SPEC/5/MATAA/HP2/ENG/TZ0/XX/M

Question 11 continued

(b) attempting to multiply through by





1t 

and rearrange to give (M1)

 

4

d

1101e

d

t

x

txt

t





A1



4

d

1101e

d

t

xt t

t



 

 

4

1101e d

t

x

ttt



 



A1

attempting to integrate the RHS by parts M1



44

40 1 e 40 e d

tt



  





44

40 1 e 160e

tt

tC



   

A1

Note: Condone the absence of

C

.

EITHER

substituting 0, 0tx

200C

M1



44

40 1 e 160e 200

1

tt

t

x

t



  





A1

using

4

40e

t



as the highest common factor of



4

40 1 e

t





and

4

160e

t



M1

OR

using

4

40e

t



as the highest common factor of



4

40 1 e

t





and

4

160e

t



giving

 

4

140e 5

t

x

ttC



 

(or equivalent) M1A1

substituting

0, 0tx

200C

M1

THEN



4

200 40e 5

1

t

xt

t









AG

[8 marks]

continued…

– 14 – SPEC/5/MATAA/HP2/ENG/TZ0/XX/M

Question 11 continued

(c)

graph starts at the origin and has a local maximum (coordinates not required) A1

sketched for

060t

A1

correct concavity for

060t

A1

maximum amount of salt is

14.6 (grams) at

6.60t 

(minutes) A1A1

[5 marks]

(d) using an appropriate graph or equation (first or second derivative) M1

amount of salt is decreasing most rapidly at

12.9t 

(minutes) A1

[2 marks]

(e) EITHER

attempting to form an integral representing the amount of salt that left

the tank M1

60

0

()

d

1

x

t

t 





60

4

2

0

200 40e 5

d

1

t









A1

OR

attempting to form an integral representing the amount of salt that entered the

tank minus the amount of salt in the tank at

60t 

(minutes) M1

amount of salt that left the tank is



60

4

0

10e d 60

t

tx





A1

THEN

36.7

(grams) A2

[4 marks]

Total [21 marks]

– 15 – SPEC/5/MATAA/HP2/ENG/TZ0/XX/M

12. (a) stating the relationship between cot and tan and stating the identity

for

tan 2



M1

1

cot 2

tan 2





and

2

2tan

tan 2

1tan







2

1tan

cot 2

2tan







AG

[1 mark]

(b) METHOD 1

attempting to substitute

tan



for x and using the result from (a) M1

2

1tan

LHS tan 2 tan 1

2tan











 





A1



22

tan 1 tan 1 0 RHS



  

A1

so

tanx





satisfies the equation AG

attempting to substitute

cot





for x and using the result from (a) M1

2

1tan

LHS cot 2 cot 1

2tan











 





A1

2

22

11tan

1

tan tan









 





A1



22

11

11 0 RHS

tan tan





A1

so

cotx





satisfies the equation AG

METHOD 2

let

tan







and

cot







attempting to find the sum of roots M1

1

tan

 



 

2

tan 1

tan







A1

2cot2





(from part (a)) A1

attempting to find the product of roots M1



tan cot



 



A1

1

A1

the coefficient of

x and the constant term in the quadratic are

2cot2



and

1 respectively R1

hence the two roots are

tan







and

cot







AG

[7 marks]

continued…

– 16 – SPEC/5/MATAA/HP2/ENG/TZ0/XX/M

Question 12 continued

(c) METHOD 1

π

tan

12

x 

and

π

cot

12

x 

are roots of

2

π

2cot 1 0

6

xx









R1

Note: Award R1 if only

π

tan

12

x 

is stated as a root of

2

π

2cot 1 0

6

xx









.

2

23 1 0xx A1

attempting to solve their quadratic equation M1

32x   A1

π

tan 0

12



(

π

cot 0

12



) R1

so

π

tan 2 3

12



AG

METHOD 2

attempting to substitute

π

12





into the identity for

tan 2



M1

2

π

2tan

π

12

tan

π

6

1tan

12





2

ππ

tan 2 3 tan 1 0

12 12



A1

attempting to solve their quadratic equation M1

π

tan 3 2

12

 

A1

π

tan 0

12



R1

so

π

tan 2 3

12



AG

[5 marks]

(d)

ππ

tan cot

24 24



is the sum of the roots of

2

π

2cot 1 0

12

xx









R1

ππ π

tan cot 2 cot

24 24 12



A1

2

23







A1

attempting to rationalise their denominator (M1)

423  A1A1

[6 marks]

Total [19 marks]

Mathematics: analysis and approaches

Higher level

Paper 3

5 pages

Specimen

1 hour

Instructions to candidates

• Do not open this examination paper until instructed to do so.

• A graphic display calculator is required for this paper.

• Answer all the questions in the answer booklet provided.

• Unless otherwise stated in the question, all numerical answers should be given exactly or

correct to three signicant gures.

• A clean copy of the mathematics: analysis and approaches formula booklet is required for

this paper.

• The maximum mark for this examination paper is [55 marks].

SPEC/5/MATAA/HP3/ENG/TZ0/XX

Answer all questions in the answer booklet provided. Please start each question on a new page. Full

marks are not necessarily awarded for a correct answer with no working. Answers must be supported

by working and/or explanations. Solutions found from a graphic display calculator should be supported

by suitable working. For example, if graphs are used to nd a solution, you should sketch these as part

of your answer. Where an answer is incorrect, some marks may be given for a correct method, provided

this is shown by written working. You are therefore advised to show all working.

1. [Maximum mark: 30]

This question asks you to investigate regular n-sided polygons inscribed and circumscribed

in a circle, and the perimeter of these as n tends to innity, to make an approximation for π .

(a) Consider an equilateral triangle ABC of side length, x units, inscribed in a circle of

radius 1 unit and centre O as shown in the following diagram.

O

BC

A

x

The equilateral triangle ABC can be divided into three smaller isosceles triangles, each

subtending an angle of

2

3

�

at O, as shown in the following diagram.

2π

3

11

O

BC x

Using right-angled trigonometry or otherwise, show that the perimeter of the equilateral

triangle ABC is equal to

33

units. [3]

(b) Consider a square of side length, x units, inscribed in a circle of radius 1 unit.

By dividing the inscribed square into four isosceles triangles, nd the exact perimeter of

the inscribed square. [3]

(This question continues on the following page)

SPEC/5/MATAA/HP3/ENG/TZ0/XX– 2 –

(Question 1 continued)

(c) Find the perimeter of a regular hexagon, of side length, x units, inscribed in a circle of

radius 1 unit. [2]

Let P

i

(n) represent the perimeter of any n-sided regular polygon inscribed in a circle of

radius 1 unit.

(d) Show that

Pn

n

i

() sin=













2

�

. [3]

(e) Use an appropriate Maclaurin series expansion to nd

lim()

n

i

Pn

→∞

and interpret this

result geometrically. [5]

Consider an equilateral triangle ABC of side length, x units, circumscribed about a circle of

radius 1 unit and centre O as shown in the following diagram.

x x

A

BC

O

x

1

Let P

c

(n) represent the perimeter of any n-sided regular polygon circumscribed about a

circle of radius 1 unit.

(f) Show that

Pn n

n

c

() tan=













2

�

. [4]

Consider the function

Px x

x

() tan=













2

�

where x ∈  , x > 2 .

(g) (i) By writing P (x) in the form

2

1

tan

�

x













, nd

lim()

x

Px

→∞

.

(ii) Hence state the value of

li

m(

)

n

c

Pn

→∞

for integers n > 2 . [5]

(This question continues on the following page)

Turn over

SPEC/5/MATAA/HP3/ENG/TZ0/XX– 3 –

(Question 1 continued)

(h) Use the results from part (d) and part (f) to determine an inequality for the value of π in

terms of n . [2]

The inequality found in part (h) can be used to determine lower and upper bound

approximations for the value of π .

(i) Determine the least value for n such that the lower bound and upper bound

approximations are both within 0.005 of π . [3]

SPEC/5/MATAA/HP3/ENG/TZ0/XX– 4 –

2. [Maximum mark: 25]

This question asks you to investigate some properties of the sequence of functions of the

form f

n

(x) = cos

(

n arccos x

)

, -1 ≤ x ≤ 1 and n ∈ 

+

.

Important: When sketching graphs in this question, you are not required to nd the

coordinates of any axes intercepts or the coordinates of any stationary points unless requested.

(a) On the same set of axes, sketch the graphs of y = f

1

(x) and y = f

3

(x) for -1 ≤ x ≤ 1 . [2]

(b) For odd values of n > 2 , use your graphic display calculator to systematically vary the

value of n . Hence suggest an expression for odd values of n describing, in terms of n ,

the number of

(i) local maximum points;

(ii) local minimum points. [4]

(c) On a new set of axes, sketch the graphs of y = f

2

(x) and y = f

4

(x) for -1 ≤ x ≤ 1 . [2]

(d) For even values of n > 2 , use your graphic display calculator to systematically vary

the value of n . Hence suggest an expression for even values of n describing, in terms

of n , the number of

(i) local maximum points;

(ii) local minimum points. [4]

(e) Solve the equation f

n

' (x) = 0 and hence show that the stationary points on the graph

of y = f

n

(x) occur at

x

k

n

= cos

�

where k ∈ 

+

and 0 < k < n . [4]

The sequence of functions, f

n

(x) , dened above can be expressed as a sequence of

polynomials of degree n .

(f) Use an appropriate trigonometric identity to show that f

2

(x) = 2x

2

- 1 . [2]

Consider f

n + 1

(x) = cos

(

(n + 1) arccos x

)

.

(g) Use an appropriate trigonometric identity to show that

f

n + 1

(x) = cos (n arccos x)

cos (arccos x) - sin (n arccos x) sin (arccos x) . [2]

(h) Hence

(i) show that f

n + 1

(x) + f

n - 1

(x) = 2xf

n

(x) , n ∈ 

+

;

(ii) express f

3

(x) as a cubic polynomial. [5]

SPEC/5/MATAA/HP3/ENG/TZ0/XX– 5 –

SPEC/5/MATAA/HP3/ENG/TZ0/XX/M

9 pages

Markscheme

Specimen paper

Mathematics:

analysis and approaches

Higher level

Paper 3

– 2 – SPEC/5/MATAA/HP3/ENG/TZ0/XX/M

Instructions to Examiners

Abbreviations

M Marks awarded for attempting to use a correct Method.

A Marks awarded for an Answer or for Accuracy; often dependent on preceding M marks.

R Marks awarded for clear Reasoning.

AG Answer given in the question and so no marks are awarded.

Using the markscheme

1 General

Award marks using the annotations as noted in the markscheme eg M1, A2.

2 Method and Answer/Accuracy marks

 Do not automatically award full marks for a correct answer; all working must be checked, and

marks awarded according to the markscheme.

 It is generally not possible to award M0 followed by A1, as A mark(s) depend on the preceding

M mark(s), if any.

 Where M and A marks are noted on the same line, e.g. M1A1, this usually means M1 for an

attempt to use an appropriate method (e.g. substitution into a formula) and A1 for using the

correct values.

 Where there are two or more A marks on the same line, they may be awarded independently;

so if the first value is incorrect, but the next two are correct, award A0A1A1.

 Where the markscheme specifies M2, A2

, etc., do not split the marks, unless there is a note.

 Once a correct answer to a question or part-question is seen, ignore further correct working.

However, if further working indicates a lack of mathematical understanding do not award the final

A1. An exception to this may be in numerical answers, where a correct exact value is followed by

an incorrect decimal. However, if the incorrect decimal is carried through to a subsequent part,

and correct working shown, award FT marks as appropriate but do not award the final A1 in that

part.

Examples

Correct answer seen Further working seen Action

1.

82

5.65685...

(incorrect decimal value)

Award the final A1

(ignore the further working)

2.

1

sin 4

4

x

sin

x

Do not award the final A1

3.

log logab

log ( )ab

Do not award the final A1

– 3 – SPEC/5/MATAA/HP3/ENG/TZ0/XX/M

3 Implied marks

Implied marks appear in brackets e.g. (M1), and can only be awarded if correct work is seen or if

implied in subsequent working.

 Normally the correct work is seen or implied in the next line.

 Marks without brackets can only be awarded for work that is seen.

4 Follow through marks (only applied after an error is made)

Follow through (FT) marks are awarded where an incorrect answer from one part of a question is

used correctly in subsequent part(s) or subpart(s). Usually, to award FT marks, there must be

working present and not just a final answer based on an incorrect answer to a previous part.

However, if the only marks awarded in a subpart are for the answer (i.e. there is no working

expected), then FT marks should be awarded if appropriate.

 Within a question part, once an error is made, no further A marks can be awarded for work

which uses the error, but M marks may be awarded if appropriate.

 If the question becomes much simpler because of an error then use discretion to award fewer

FT marks.

 If the error leads to an inappropriate value (e.g. probability greater than 1, use of

1r 

for the

sum of an infinite GP,

sin 1.5





, non integer value where integer required), do not award the

mark(s) for the final answer(s).

 The markscheme may use the word “their” in a description, to indicate that candidates may be

using an incorrect value.

 Exceptions to this rule will be explicitly noted on the markscheme.

 If a candidate makes an error in one part, but gets the correct answer(s) to subsequent part(s),

award marks as appropriate, unless the question says hence. It is often possible to use a

different approach in subsequent parts that does not depend on the answer to previous parts.

5 Mis-read

If a candidate incorrectly copies information from the question, this is a mis-read (MR). Apply a MR

penalty of 1 mark to that question

 If the question becomes much simpler because of the MR, then use discretion to award

fewer marks.

 If the MR leads to an inappropriate value (e.g. probability greater than

1,

sin 1.5





, non-integer

value where integer required), do not award the mark(s) for the final answer(s).

 Miscopying of candidates’ own work does not constitute a misread, it is an error.

 The MR penalty can only be applied when work is seen. For calculator questions with no

working and incorrect answers, examiners should not infer that values were read incorrectly.

– 4 – SPEC/5/MATAA/HP3/ENG/TZ0/XX/M

6 Alternative methods

 Alternative methods for complete questions are indicated by METHOD 1,

METHOD 2, etc.

 Alternative solutions for part-questions are indicated by EITHER . . . OR.

7 Alternative forms

Unless the question specifies otherwise, accept equivalent forms.

 As this is an international examination, accept all alternative forms of notation.

 In the markscheme, equivalent numerical and algebraic forms will generally be written in

brackets immediately following the answer.

 In the markscheme, simplified answers, (which candidates often do not write in examinations),

will generally appear in brackets. Marks should be awarded for either the form preceding the

bracket or the form in brackets (if it is seen).

8 Accuracy of Answers

If the level of accuracy is specified in the question, a mark will be linked to giving the answer to the

required accuracy. There are two types of accuracy errors, and the final answer mark should not be

awarded if these errors occur.

 Rounding errors: only applies to final answers not to intermediate steps.

 Level of accuracy: when this is not specified in the question the general rule applies to final

answers: unless otherwise stated in the question all numerical answers must be given exactly or

correct to three significant figures.

9 Calculators

A GDC is required for paper 3, but calculators with symbolic manipulation features/ CAS functionality

are not allowed.

Calculator notation

The subject guide says:

Students must always use correct mathematical notation, not calculator notation.

Do not accept final answers written using calculator notation. However, do not penalize the use of

calculator notation in the working.

Candidates will sometimes use methods other than those in the markscheme. Unless the question

specifies a method, other correct methods should be marked in line with the markscheme

– 5 – SPEC/5/MATAA/HP3/ENG/TZ0/XX/M

1. (a) METHOD 1

consider right-angled triangle

OCX where

CX

2

x



π

2

sin

31

x



M1A1

3

22

x

x 

A1

333

i

Px

AG

METHOD 2

eg use of the cosine rule



222

2π

11211cos

3

x 

M1A1

3x 

A1

333

i

Px

AG

Note: Accept use of sine rule.

[3 marks]

(b)

π1

sin

4

x



where

x

 side of square M1

2x 

A1

42

i

P 

A1

[3 marks]

(c) 6 equilateral triangles

1

x



A1

6

i

P 

A1

[2 marks]

(d) in right-angled triangle

π

2

sin

1

x

n









M1

π

2sinx

n









A1

i

Pnx

π

2sin

i

Pn

n









M1

π

2sin

i

Pn

n









AG

[3 marks]

continued…

– 6 – SPEC/5/MATAA/HP3/ENG/TZ0/XX/M

Question 1 continued

(e) consider

π

lim 2 sin

n









use of

35

sin

3! 5!

xx

xx  

M1

35

ππππ

2sin 2

6120

nn

nnnn









(A1)

35

24

ππ

2π

6120nn



 





A1

π

lim 2 sin 2π

n











A1

as

n 

polygon becomes a circle of radius 1 and

2π

i

P 

R1

[5 marks]

(f) consider an

n-sided polygon of side length

x

2n right-angled triangles with angle

2π π

2nn



at centre M1A1

opposite side

ππ

tan 2tan

2

x

nn

 



 

 

M1A1

Perimeter

π

2tan

c

Pn

n









AG

[4 marks]

(g) (i)



π

2tan

0

lim lim

1

0

xx

x

Px

x

 









R1

attempt to use L’Hôpital’s rule M1



2

2π π

sec

lim lim

1

xx

x

Px

x

 













A1



lim 2π

x

Px





A1

(ii)



lim 2π

c

n

Pn





A1

[5 marks]

continued…

– 7 – SPEC/5/MATAA/HP3/ENG/TZ0/XX/M

Question 1 continued

(h)

2π

ic

PP

ππ

2sin 2π 2tannn

nn

 



 

 

M1

ππ

sin π tannn

nn

 



 

 

A1

[2 marks]

(i) attempt to find the lower bound and upper bound approximations within

0.005 of

π

(M1)

46n 

A2

[3 marks]

Total [30 marks]

2. (a) correct graph of

1

()

y

fx

A1

correct graph of

3

()

y

fx

A1

[2 marks]

(b) (i) graphical or tabular evidence that

n

has been systematically varied M1

eg

3n 

,

1

local maximum point and

1

local minimum point

5n 

,

2

local maximum points and

2

local minimum points

7n 

,

3

local maximum points and

3

local minimum points (A1)

1

2

n 

local maximum points A1

(ii)

1

2

n 

local minimum points A1

Note: Allow follow through from an incorrect local maximum formula expression.

[4 marks]

continued…

– 8 – SPEC/5/MATAA/HP3/ENG/TZ0/XX/M

Question 2 continued

(c) correct graph of

2

()

y

fx

A1

correct graph of

4

()

y

fx

A1

[2 marks]

(d) (i) graphical or tabular evidence that

n

has been systematically varied M1

eg

2n 

,

0

local maximum point and

1

local minimum point

4n 

,

1

local maximum points and

2

local minimum points

6n 

,

2

local maximum points and

3

local minimum points (A1)

2

n 

local maximum points A1

(ii)

2

n

local minimum points A1

[4 marks]

(e)













cos arccos

n

f

xnx



2

sin arccos

1

n

nn x

fx

x







M1A1

Note: Award M1 for attempting to use the chain rule.

 





0 sin arccos 0

n

fx n n x



 

M1





arccos πnxk

(

k



 

) A1

leading to

π

cos

k

x

n



(

k



 

and

0 kn

) AG

[4 marks]

continued…

– 9 – SPEC/5/MATAA/HP3/ENG/TZ0/XX/M

Question 2 continued

(f)

  

2

cos 2arccos

f

xx



2

2 cos arccos 1x

M1

stating that



cos arccos

x

x

A1

so





2

21fx x

AG

[2 marks]

(g)

  



1

cos 1 arccos

n

f

xn x







cos arccos arccosnx x

A1

use of





cos cos cos sin sin

A

BABAB 

leading to M1

















cos arccos cos arccos sin arccos sin arccosnx x nx x

AG

[2 marks]

(h) (i)

  



1

cos 1 arccos

n

f

xn x





A1



cos arccos cos arccos sin arccos sin arccosnx x nx x

M1

















11

2cos arccos cos arccos

nn

f

xfx n x x





A1



2

n

x

fx

AG

(ii)

  

321

2

f

xxfxfx

(M1)



2

22 1

x

xx

3

43

x

x

A1

[5 marks]

Total [25 marks]

Candidate session number

Mathematics: analysis and approaches

Standard level

Paper 1

10 pages

Specimen

1 hour 30 minutes

Instructions to candidates

• Write your session number in the boxes above.

• Do not open this examination paper until instructed to do so.

• You are not permitted access to any calculator for this paper.

• Section A: answer all questions. Answers must be written within the answer boxes provided.

• Section B: answer all questions in the answer booklet provided. Fill in your session number

on the front of the answer booklet, and attach it to this examination paper and your

cover sheet using the tag provided.

• Unless otherwise stated in the question, all numerical answers should be given exactly or

correct to three signicant gures.

• A clean copy of the mathematics: analysis and approaches formula booklet is required for

this paper.

• The maximum mark for this examination paper is [80 marks].

12EP01

SPEC/5/MATAA/SP1/ENG/TZ0/XX

Full marks are not necessarily awarded for a correct answer with no working. Answers must be

supported by working and/or explanations. Where an answer is incorrect, some marks may be given

for a correct method, provided this is shown by written working. You are therefore advised to show all

working.

Section A

Answer all questions. Answers must be written within the answer boxes provided. Working may be

continued below the lines, if necessary.

1. [Maximum mark: 5]

The following diagram shows triangle ABC, with AB = 6 and AC = 8 .

diagram not to scale

B

CA

8

6

(a) Given that

co

s A =

5

6

, nd the value of sin

cos A =

5

6

. [3]

(b) Find the area of triangle ABC . [2]

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

12EP02

SPEC/5/MATAA/SP1/ENG/TZ0/XX– 2 –

2. [Maximum mark: 5]

Let A and B be events such that P (A) = 0.5 , P (B) = 0.4 and P (A ∪ B) = 0.6 .

Find P (A | B) .

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

Turn over

12EP03

SPEC/5/MATAA/SP1/ENG/TZ0/XX– 3 –

3. [Maximum mark: 5]

(a) Show that (2n - 1)

2

+ (2n + 1)

2

= 8n

2

+ 2 , where n ∈  . [2]

(b) Hence, or otherwise, prove that the sum of the squares of any two consecutive odd

integers is even. [3]

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

12EP04

SPEC/5/MATAA/SP1/ENG/TZ0/XX– 4 –

4. [Maximum mark: 5]

Let

′

=

+

fx

x

()

8

21

2

. Given that f (0) = 5 , nd f (x) .

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

Turn over

12EP05

SPEC/5/MATAA/SP1/ENG/TZ0/XX– 5 –

5. [Maximum mark: 5]

The functions f and g are dened such that

fx

x

()=

+

3

4

and g (x) = 8x + 5 .

(a) Show that ( g  f )(x) = 2x + 11 . [2]

(b) Given that ( g  f )

-1

(a) = 4 , nd the value of a . [3]

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

12EP06

SPEC/5/MATAA/SP1/ENG/TZ0/XX– 6 –

6. [Maximum mark: 8]

(a) Show that

log(cos)logcos

93

22

22xx

+= +

. [3]

(b) Hence or otherwise solve log

3

(2 sin x) = log

9

(cos 2x + 2) for

0

2

<<

x

�

. [5]

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

Turn over

12EP07

SPEC/5/MATAA/SP1/ENG/TZ0/XX– 7 –

Do not write solutions on this page.

Section B

Answer all questions in the answer booklet provided. Please start each question on a new page.

7. [Maximum mark: 15]

A large company surveyed 160 of its employees to nd out how much time they spend

traveling to work on a given day. The results of the survey are shown in the following

cumulative frequency diagram.

10 20 40 50 60

0

7030

20

40

60

80

100

120

140

160

travelling time (minutes)

cumulative frequency

(This question continues on the following page)

12EP08

SPEC/5/MATAA/SP1/ENG/TZ0/XX– 8 –

Do not write solutions on this page.

(Question 7 continued)

(a) Find the median number of minutes spent traveling to work. [2]

(b) Find the number of employees whose travelling time is within 15 minutes of the median. [3]

Only 10% of the employees spent more than k minutes traveling to work.

(c) Find the value of k . [3]

The results of the survey can also be displayed on the following box-and-whisker diagram.

a

b

47

5

travelling times (minutes)

(d) Write down the value of b . [1]

(e) (i) Find the value of a .

(ii) Hence, nd the interquartile range. [4]

Travelling times of less than p minutes are considered outliers.

(f) Find the value of p . [2]

8. [Maximum mark: 16]

Let

fx xx x()=+

−+

1

3

15 17

32

.

(a) Find f ' (x) . [2]

The graph of f has horizontal tangents at the points where x = a and x = b , a < b .

(b) Find the value of a and the value of b . [3]

(c) (i) Sketch the graph of y = f ' (x) .

(ii) Hence explain why the graph of f has a local maximum point at x = a . [2]

(d) (i) Find f " (b) .

(ii) Hence, use your answer to part (d)(i) to show that the graph of f has a local

minimum point at x = b . [4]

The normal to the graph of f at x = a and the tangent to the graph of f at x = b intersect at

the point ( p , q) .

(e) Find the value of p and the value of q . [5]

Turn over

12EP09

SPEC/5/MATAA/SP1/ENG/TZ0/XX– 9 –

Do not write solutions on this page.

9. [Maximum mark: 16]

Let

fx

x

kx

()

ln

=

5

where x > 0 , k ∈ 

+

.

(a) Show that

′

=

−

fx

x

kx

()

ln

15

2

. [3]

The graph of f has exactly one maximum point P .

(b) Find the x-coordinate of P . [3]

The second derivative of f is given by

′′

=

−

fx

x

kx

()

ln25 3

3

. The graph of f has exactly one

point of inexion Q .

(c) Show that the x-coordinate of Q is

1

5

3

2

e

. [3]

The region R is enclosed by the graph of f , the x-axis, and the vertical lines through the

maximum point P and the point of inexion Q .

y

x

R

P

Q

(d) Given that the area of R is 3, nd the value of k . [7]

12EP10

SPEC/5/MATAA/SP1/ENG/TZ0/XX– 10 –

Please do not write on this page.

Answers written on this page

will not be marked.

12EP11

Please do not write on this page.

Answers written on this page

will not be marked.

12EP12

SPEC/5/MATAA/SP1/ENG/TZ0/XX/M

11 pages

Markscheme

Specimen paper

Mathematics:

analysis and approaches

Standard level

Paper 1

– 2 – SPEC/5/MATAA/SP1/ENG/TZ0/XX/M

Instructions to Examiners

Abbreviations

M Marks awarded for attempting to use a correct Method.

A Marks awarded for an Answer or for Accuracy; often dependent on preceding M marks.

R Marks awarded for clear Reasoning.

AG Answer given in the question and so no marks are awarded.

Using the markscheme

1 General

Award marks using the annotations as noted in the markscheme eg M1, A2.

2 Method and Answer/Accuracy marks

 Do not automatically award full marks for a correct answer; all working must be checked, and

marks awarded according to the markscheme.

 It is generally not possible to award M0 followed by A1, as A mark(s) depend on the preceding

M mark(s), if any.

 Where M and A marks are noted on the same line, e.g. M1A1, this usually means M1 for an

attempt to use an appropriate method (e.g. substitution into a formula) and A1 for using the

correct values.

 Where there are two or more A marks on the same line, they may be awarded independently;

so if the first value is incorrect, but the next two are correct, award A0A1A1.

 Where the markscheme specifies M2, A2

, etc., do not split the marks, unless there is a note.

 Once a correct answer to a question or part-question is seen, ignore further correct working.

However, if further working indicates a lack of mathematical understanding do not award the final

A1. An exception to this may be in numerical answers, where a correct exact value is followed by

an incorrect decimal. However, if the incorrect decimal is carried through to a subsequent part,

and correct working shown, award FT marks as appropriate but do not award the final A1 in that

part.

Examples

Correct answer seen Further working seen Action

1.

82

5.65685...

(incorrect decimal value)

Award the final A1

(ignore the further working)

2.

1

sin 4

4

x

sin

x

Do not award the final A1

3.

log logab

log ( )ab

Do not award the final A1

– 3 – SPEC/5/MATAA/SP1/ENG/TZ0/XX/M

3 Implied marks

Implied marks appear in brackets e.g. (M1), and can only be awarded if correct work is seen or if

implied in subsequent working.

 Normally the correct work is seen or implied in the next line.

 Marks without brackets can only be awarded for work that is seen.

4 Follow through marks (only applied after an error is made)

Follow through (FT) marks are awarded where an incorrect answer from one part of a question is

used correctly in subsequent part(s) or subpart(s). Usually, to award FT marks, there must be

working present and not just a final answer based on an incorrect answer to a previous part.

However, if the only marks awarded in a subpart are for the answer (i.e. there is no working

expected), then FT marks should be awarded if appropriate.

 Within a question part, once an error is made, no further A marks can be awarded for work

which uses the error, but M marks may be awarded if appropriate.

 If the question becomes much simpler because of an error then use discretion to award fewer

FT marks.

 If the error leads to an inappropriate value (e.g. probability greater than 1, use of

1r 

for the

sum of an infinite GP,

sin 1.5





, non integer value where integer required), do not award the

mark(s) for the final answer(s).

 The markscheme may use the word “their” in a description, to indicate that candidates may be

using an incorrect value.

 Exceptions to this rule will be explicitly noted on the markscheme.

 If a candidate makes an error in one part, but gets the correct answer(s) to subsequent part(s),

award marks as appropriate, unless the question says hence. It is often possible to use a

different approach in subsequent parts that does not depend on the answer to previous parts.

5 Mis-read

If a candidate incorrectly copies information from the question, this is a mis-read (MR). Apply a MR

penalty of 1 mark to that question

 If the question becomes much simpler because of the MR, then use discretion to award

fewer marks.

 If the MR leads to an inappropriate value (e.g. probability greater than

1,

sin 1.5





, non-integer

value where integer required), do not award the mark(s) for the final answer(s).

 Miscopying of candidates’ own work does not constitute a misread, it is an error.

 The MR penalty can only be applied when work is seen. For calculator questions with no

working and incorrect answers, examiners should not infer that values were read incorrectly.

– 4 – SPEC/5/MATAA/SP1/ENG/TZ0/XX/M

6 Alternative methods

 Alternative methods for complete questions are indicated by METHOD 1,

METHOD 2, etc.

 Alternative solutions for part-questions are indicated by EITHER . . . OR.

7 Alternative forms

Unless the question specifies otherwise, accept equivalent forms.

 As this is an international examination, accept all alternative forms of notation.

 In the markscheme, equivalent numerical and algebraic forms will generally be written in

brackets immediately following the answer.

 In the markscheme, simplified answers, (which candidates often do not write in examinations),

will generally appear in brackets. Marks should be awarded for either the form preceding the

bracket or the form in brackets (if it is seen).

8 Accuracy of Answers

If the level of accuracy is specified in the question, a mark will be linked to giving the answer to the

required accuracy. There are two types of accuracy errors, and the final answer mark should not be

awarded if these errors occur.

 Rounding errors: only applies to final answers not to intermediate steps.

 Level of accuracy: when this is not specified in the question the general rule applies to final

answers: unless otherwise stated in the question all numerical answers must be given exactly or

correct to three significant figures.

9 Calculators

No calculator is allowed. The use of any calculator on paper 1 is malpractice, and will result in no

grade awarded. If you see work that suggests a candidate has used any calculator, please follow

the procedures for malpractice. Examples: finding an angle, given a trig ratio of 0.4235.

Candidates will sometimes use methods other than those in the markscheme. Unless the question

specifies a method, other correct methods should be marked in line with the markscheme

– 5 – SPEC/5/MATAA/SP1/ENG/TZ0/XX/M

Section A

1. (a) valid approach using Pythagorean identity (M1)

2

5

sin + 1

6

A









(or equivalent) (A1)

11

sin =

6

A

A1

[3 marks]

(b)

111

86

26



(or equivalent) (A1)

area = 4 11

A1

[2 marks]

Total [5 marks]

2. attempt to substitute into

















PPPP

A

BABAB   

(M1)

Note: Accept use of Venn diagram or other valid method.

0.6 0.5 0.4 P( )

A

B 

(A1)



P0.3AB

(seen anywhere) A1

attempt to substitute into



P

P|

P

A

B

AB

B





(M1)

0.3

0.4







P|

A

B

3

0.75

4









A1

Total [5 marks]

– 6 – SPEC/5/MATAA/SP1/ENG/TZ0/XX/M

3. (a) attempting to expand the LHS (M1)



22

LHS 4 4 1 4 4 1nn nn

A1





2

82RHSn

AG

[2 marks]

(b) METHOD 1

recognition that

21n

and

21n 

represent two consecutive odd

integers (for

n

) R1



22

82241nn 

A1

valid reason eg divisible by

2 (2 is a factor) R1

so the sum of the squares of any two consecutive odd integers is even AG

[3 marks]

METHOD 2

recognition, eg that

n

and

2n 

represent two consecutive odd integers

(for

n

) R1



2

22

22 22nn n n   

A1

valid reason eg divisible by

2 (2 is a factor) R1

so the sum of the squares of any two consecutive odd integers is even AG

[3 marks]

Total [5 marks]

4. attempt to integrate (M1)

2

d

21 4

d

u

ux x

x



2

82

dd

21

x

u

x







(A1)

EITHER





4 uC

A1

OR



2

42 1

x

C

A1

THEN

correct substitution into their integrated function (must have

C

) (M1)

54 1CC  



2

42 11fx x

A1

Total [5 marks]

– 7 – SPEC/5/MATAA/SP1/ENG/TZ0/XX/M

5. (a) attempt to form composition M1

correct substitution

33

85

44

xx

g











A1



211gf x x

AG

[2 marks]

(b) attempt to substitute 4 (seen anywhere) (M1)

correct equation

2411a 

(A1)

19a 

A1

[3 marks]

Total [5 marks]

6. (a) attempting to use the change of base rule M1

3

9

3

log (cos 2 2)

log 9

x





A1

3

1

log (cos 2 2)

2

x

A1

3

log cos 2 2x

AG

[3 marks]

(b)

33

log (2sin ) log cos 2 2xx

2sin cos2 2xx

M1

2

4sin cos2 2xx

(or equivalent) A1

use of

2

cos 2 1 2sin

x

x

(M1)

2

6sin 3x 



1

sin

2

x 

A1

π

4

x 

A1

Note: Award A0 if solutions other than

π

4

x 

are included.

[5 marks]

Total [8 marks]

– 8 – SPEC/5/MATAA/SP1/ENG/TZ0/XX/M

Section B

7. (a) evidence of median position (M1)

80th employee

40 minutes A1

[2 marks]

(b) valid attempt to find interval (

2555) (M1)

18 (employees), 142 (employees) A1

124 A1

[3 marks]

(c) recognising that there are

16 employees in the top 10% (M1)

144 employees travelled more than k minutes (A1)

A1

[3 marks]

(d)

70b 

A1

[1 mark]

(e) (i) recognizing

a is first quartile value (M1)

40 employees

A1

(ii)

47 33

(M1)

IQR 14

A1

[4 marks]

(f) attempt to find

1.5 IQR their

(M1)

33 21

12

(A1)

[2 marks]

[Total 15 marks]

8. (a)

2

() 2 15fx x x





(M1)A1

[2 marks]

(b) correct reasoning that

() 0fx





(seen anywhere) (M1)

2

2150xx

valid approach to solve quadratic M1

(3)(5)xx

, quadratic formula

correct values for

x

3, 5

correct values for

a and b

5a 

and

3b 

A1

[3 marks]

continued…

56k 

33a 

– 9 – SPEC/5/MATAA/SP1/ENG/TZ0/XX/M

Question 8 continued

(c) (i)

A1

(ii) first derivative changes from positive to negative at

x

a

A1

so local maximum at

x

a

AG

[2 marks]

(d) (i)

() 2 2fx x





A1

substituting their

b into their second derivative (M1)

(3) 2 3 2f





() 8fb





(A1)

(ii)

()

f

b



is positive so graph is concave up R1

so local minimum at

x

 b

AG

[4 marks]

(e) normal to

f at

x

a

is

5x 

(seen anywhere) (A1)

attempt to find

y-coordinate at their value of b (M1)

(3) 10f 

(A1)

tangent at

x

b

has equation

10y 

(seen anywhere) A1

intersection at



5, 10

5p 

and

10q 

A1

[5 marks]

[Total 16 marks]

– 10 – SPEC/5/MATAA/SP1/ENG/TZ0/XX/M

9. (a) attempt to use quotient rule (M1)

correct substitution into quotient rule



2

1

5ln5

5

kx k x

x

fx

kx













(or equivalent) A1



22

ln 5

,

kk x

k

kx







A1

2

1ln5

x

kx





AG

[3 marks]

(b)



0fx





M1

2

1ln5

0

x

kx





ln 5 1

x



(A1)

e

5

x 

A1

[3 marks]

(c)



0fx





M1

3

2ln5 3

0

x

kx





3

ln 5

2

x 

A1

3

2

5ex 

A1

so the point of inflexion occurs at

3

2

1

e

5

x 

AG

[3 marks]

continued…

– 11 – SPEC/5/MATAA/SP1/ENG/TZ0/XX/M

Question 9 continued

(d) attempt to integrate (M1)

d1

ln 5

d

u

ux

x



ln 5 1

d d

x

uu

kx k





(A1)

EITHER

2

u

k



A1

so

3

2

1

d

2

u

uu

kk











A1

OR



2

ln 5

2

x

k

 A1

so



3

2

1

e

2

5

e

5

ln 5

d

2

x

kx k











A1

THEN

19

1

24k









5

8k



A1

setting their expression for area equal to

3 M1

5

3

8k



5

24

k 

A1

[7 marks]

Total [16 marks]

12EP01

Candidate session number

Mathematics: analysis and approaches

Standard level

Paper 2

10 pages

Specimen

1 hour 30 minutes

Instructions to candidates

• Write your session number in the boxes above.

• Do not open this examination paper until instructed to do so.

• A graphic display calculator is required for this paper.

• Section A: answer all questions. Answers must be written within the answer boxes provided.

• Section B: answer all questions in the answer booklet provided. Fill in your session number

on the front of the answer booklet, and attach it to this examination paper and your

cover sheet using the tag provided.

• Unless otherwise stated in the question, all numerical answers should be given exactly or

correct to three signicant gures.

• A clean copy of the mathematics: analysis and approaches formula booklet is required for

this paper.

• The maximum mark for this examination paper is [80 marks].

SPEC/5/MATAA/SP2/ENG/TZ0/XX

Full marks are not necessarily awarded for a correct answer with no working. Answers must be

supported by working and/or explanations. Solutions found from a graphic display calculator should be

supported by suitable working. For example, if graphs are used to nd a solution, you should sketch

these as part of your answer. Where an answer is incorrect, some marks may be given for a correct

method, provided this is shown by written working. You are therefore advised to show all working.

Section A

Answer all questions. Answers must be written within the answer boxes provided. Working may be

continued below the lines, if necessary.

1. [Maximum mark: 6]

A metal sphere has a radius 12.7 cm.

(a) Find the volume of the sphere expressing your answer in the form a × 10

k

, 1 ≤ a < 10

and k ∈  . [3]

The sphere is to be melted down and remoulded into the shape of a cone with a height of 14.8 cm.

(b) Find the radius of the base of the cone, correct to 2 signicant gures. [3]

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

12EP02

SPEC/5/MATAA/SP2/ENG/TZ0/XX– 2 –

2. [Maximum mark: 6]

The following diagram shows part of a circle with centre O and radius 4 cm .

O

B

A

θ

4 cm

5 cm

Chord AB has a length of 5 cm and AÔB = θ .

(a) Find the value of θ , giving your answer in radians. [3]

(b) Find the area of the shaded region. [3]

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

12EP03

Turn over

SPEC/5/MATAA/SP2/ENG/TZ0/XX– 3 –

3. [Maximum mark: 6]

On 1st January 2020, Laurie invests $P in an account that pays a nominal annual interest

rate of 5.5 % , compounded quarterly.

The amount of money in Laurie’s account at the end of each year follows a geometric

sequence with common ratio, r .

(a) Find the value of r , giving your answer to four signicant gures. [3]

Laurie makes no further deposits to or withdrawals from the account.

(b) Find the year in which the amount of money in Laurie’s account will become double the

amount she invested. [3]

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

12EP04

SPEC/5/MATAA/SP2/ENG/TZ0/XX– 4 –

4. [Maximum mark: 6]

A six-sided biased die is weighted in such a way that the probability of obtaining a “six” is

7

10

.

The die is tossed ve times. Find the probability of obtaining

(a) at most three “sixes”. [3]

(b) the third “six” on the fth toss. [3]

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

12EP05

Turn over

SPEC/5/MATAA/SP2/ENG/TZ0/XX– 5 –

5. [Maximum mark: 5]

The following table below shows the marks scored by seven students on two dierent

mathematics tests.

Test 1 (x)

15 23 25 30 34 34 40

Test 2 ( y)

20 26 27 32 35 37 35

Let L

1

be the regression line of x on y . The equation of the line L

1

can be written in the

form x = ay + b .

(a) Find the value of a and the value of b . [2]

Let L

2

be the regression line of y on x . The lines L

1

and L

2

pass through the same point

with coordinates ( p , q) .

(b) Find the value of p and the value of q . [3]

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

12EP06

SPEC/5/MATAA/SP2/ENG/TZ0/XX– 6 –

6. [Maximum mark: 7]

The displacement, in centimetres, of a particle from an origin, O, at time t seconds, is given

by s (t) = t

2

cos t + 2t sin t , 0 ≤ t ≤ 5 .

(a) Find the maximum distance of the particle from O. [3]

(b) Find the acceleration of the particle at the instant it rst changes direction. [4]

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

12EP07

Turn over

SPEC/5/MATAA/SP2/ENG/TZ0/XX– 7 –

Do not write solutions on this page.

Section B

Answer all questions in the answer booklet provided. Please start each question on a new page.

7. [Maximum mark: 16]

Adam sets out for a hike from his camp at point A . He hikes at an average speed of 4.2 km/h

for 45 minutes, on a bearing of 035



from the camp, until he stops for a break at point B .

(a) Find the distance from point A to point B . [2]

35



A

B

N

Adam leaves point B on a bearing of 114



and continues to hike for a distance of 4.6 km until

he reaches point C .

35



A

B

C

N

(b) (i) Show that AB

̂

C is 101



.

(ii) Find the distance from the camp to point C . [5]

(c) Find BA . [3]

Adam’s friend Jacob wants to hike directly from the camp to meet Adam at point C .

(d) Find the bearing that Jacob must take to point C . [3]

Jacob hikes at an average speed of 3.9 km/h.

(e) Find, to the nearest minute, the time it takes for Jacob to reach point C . [3]

12EP08

SPEC/5/MATAA/SP2/ENG/TZ0/XX– 8 –

Do not write solutions on this page.

8. [Maximum mark: 15]

The length, X mm , of a certain species of seashell is normally distributed with mean 25 and

variance, σ

2

.

The probability that X is less than 24.15 is 0.1446 .

(a) Find P (24.15 < X < 25) . [2]

(b) (i) Find σ , the standard deviation of X .

(ii) Hence, nd the probability that a seashell selected at random has a length

greater than 26 mm . [5]

A random sample of 10 seashells is collected on a beach. Let Y represent the number of

seashells with lengths greater than 26 mm .

(c) Find E (Y ) . [3]

(d) Find the probability that exactly three of these seashells have a length greater

than 26 mm . [2]

A seashell selected at random has a length less than 26 mm .

(e) Find the probability that its length is between 24.15 mm and 25 mm . [3]

12EP09

Turn over

SPEC/5/MATAA/SP2/ENG/TZ0/XX– 9 –

Do not write solutions on this page.

9. [Maximum mark: 13]

Consider a function f , such that

fx xb

xb

() .sin ()

,,

= +













+≤

≤∈

58

6

1010

�



.

(a) Find the period of f . [2]

The function f has a local maximum at the point (2 , 21.8) , and a local minimum at (8 , 10.2) .

(b) (i) Find the value of b .

(ii) Hence, nd the value of f (6) . [4]

A second function g is given by

gx qx pq() sin( .)

,;

,= −













+≤

≤∈

2

9

375010

�

px

.

The function g passes through the points (3 , 2.5) and (6 , 15.1) .

(c) Find the value of p and the value of q . [5]

(d) Find the value of x for which the functions have the greatest dierence. [2]

12EP10

SPEC/5/MATAA/SP2/ENG/TZ0/XX– 10 –

12EP11

Please do not write on this page.

Answers written on this page

will not be marked.

12EP12

Please do not write on this page.

Answers written on this page

will not be marked.

SPEC/5/MATAA/SP2/ENG/TZ0/XX/M

10 pages

Markscheme

Specimen paper

Mathematics:

analysis and approaches

Standard level

Paper 2

– 2 – SPEC/5/MATAA/SP2/ENG/TZ0/XX/M

Instructions to Examiners

Abbreviations

M Marks awarded for attempting to use a correct Method.

A Marks awarded for an Answer or for Accuracy; often dependent on preceding M marks.

R Marks awarded for clear Reasoning.

AG Answer given in the question and so no marks are awarded.

Using the markscheme

1 General

Award marks using the annotations as noted in the markscheme eg M1, A2.

2 Method and Answer/Accuracy marks

 Do not automatically award full marks for a correct answer; all working must be checked, and

marks awarded according to the markscheme.

 It is generally not possible to award M0 followed by A1, as A mark(s) depend on the preceding

M mark(s), if any.

 Where M and A marks are noted on the same line, e.g. M1A1, this usually means M1 for an

attempt to use an appropriate method (e.g. substitution into a formula) and A1 for using the

correct values.

 Where there are two or more A marks on the same line, they may be awarded independently;

so if the first value is incorrect, but the next two are correct, award A0A1A1.

 Where the markscheme specifies M2, A2

, etc., do not split the marks, unless there is a note.

 Once a correct answer to a question or part-question is seen, ignore further correct working.

However, if further working indicates a lack of mathematical understanding do not award the final

A1. An exception to this may be in numerical answers, where a correct exact value is followed by

an incorrect decimal. However, if the incorrect decimal is carried through to a subsequent part,

and correct working shown, award FT marks as appropriate but do not award the final A1 in that

part.

Examples

Correct answer seen Further working seen Action

1.

82

5.65685...

(incorrect decimal value)

Award the final A1

(ignore the further working)

2.

1

sin 4

4

x

sin

x

Do not award the final A1

3.

log logab

log ( )ab

Do not award the final A1

– 3 – SPEC/5/MATAA/SP2/ENG/TZ0/XX/M

3 Implied marks

Implied marks appear in brackets e.g. (M1), and can only be awarded if correct work is seen or if

implied in subsequent working.

 Normally the correct work is seen or implied in the next line.

 Marks without brackets can only be awarded for work that is seen.

4 Follow through marks (only applied after an error is made)

Follow through (FT) marks are awarded where an incorrect answer from one part of a question is

used correctly in subsequent part(s) or subpart(s). Usually, to award FT marks, there must be

working present and not just a final answer based on an incorrect answer to a previous part.

However, if the only marks awarded in a subpart are for the answer (i.e. there is no working

expected), then FT marks should be awarded if appropriate.

 Within a question part, once an error is made, no further A marks can be awarded for work

which uses the error, but M marks may be awarded if appropriate.

 If the question becomes much simpler because of an error then use discretion to award fewer

FT marks.

 If the error leads to an inappropriate value (e.g. probability greater than 1, use of

1r  for the

sum of an infinite GP,

sin 1.5





, non integer value where integer required), do not award the

mark(s) for the final answer(s).

 The markscheme may use the word “their” in a description, to indicate that candidates may be

using an incorrect value.

 Exceptions to this rule will be explicitly noted on the markscheme.

 If a candidate makes an error in one part, but gets the correct answer(s) to subsequent part(s),

award marks as appropriate, unless the question says hence. It is often possible to use a

different approach in subsequent parts that does not depend on the answer to previous parts.

5 Mis-read

If a candidate incorrectly copies information from the question, this is a mis-read (MR). Apply a MR

penalty of 1 mark to that question

 If the question becomes much simpler because of the MR, then use discretion to award

fewer marks.

 If the MR leads to an inappropriate value (e.g. probability greater than

1,

sin 1.5





, non-integer

value where integer required), do not award the mark(s) for the final answer(s).

 Miscopying of candidates’ own work does not constitute a misread, it is an error.

 The MR penalty can only be applied when work is seen. For calculator questions with no

working and incorrect answers, examiners should not infer that values were read incorrectly.

– 4 – SPEC/5/MATAA/SP2/ENG/TZ0/XX/M

6 Alternative methods

 Alternative methods for complete questions are indicated by METHOD 1,

METHOD 2, etc.

 Alternative solutions for part-questions are indicated by EITHER . . . OR.

7 Alternative forms

Unless the question specifies otherwise, accept equivalent forms.

 As this is an international examination, accept all alternative forms of notation.

 In the markscheme, equivalent numerical and algebraic forms will generally be written in

brackets immediately following the answer.

 In the markscheme, simplified answers, (which candidates often do not write in examinations),

will generally appear in brackets. Marks should be awarded for either the form preceding the

bracket or the form in brackets (if it is seen).

8 Accuracy of Answers

If the level of accuracy is specified in the question, a mark will be linked to giving the answer to the

required accuracy. There are two types of accuracy errors, and the final answer mark should not be

awarded if these errors occur.

 Rounding errors: only applies to final answers not to intermediate steps.

 Level of accuracy: when this is not specified in the question the general rule applies to final

answers: unless otherwise stated in the question all numerical answers must be given exactly or

correct to three significant figures.

9 Calculators

A GDC is required for paper 2, but calculators with symbolic manipulation features/ CAS functionality

are not allowed.

Calculator notation

The subject guide says:

Students must always use correct mathematical notation, not calculator notation.

Do not accept final answers written using calculator notation. However, do not penalize the use of

calculator notation in the working.

Candidates will sometimes use methods other than those in the markscheme. Unless the question

specifies a method, other correct methods should be marked in line with the markscheme

– 5 – SPEC/5/MATAA/SP2/ENG/TZ0/XX/M

Section A

1. (a)



3

4

π 12.7

3

(or equivalent) A1

8580.24 (A1)

V

 8.58 10

3

A1

[3 marks]

(b) recognising volume of the cone is same as volume of their sphere (M1)



2

1

π 14.8 8580.24

3

r 

(or equivalent) A1

r

 23.529

24r 

(cm) correct to 2 significant figures A1

[3 marks]

Total [6 marks]

2. (a) METHOD 1

attempt to use the cosine rule (M1)

222

445

cos

244









(or equivalent) A1

1.35





A1

[3 marks]

METHOD 2

attempt to split triangle

AOB into two congruent right triangles (M1)

2.5

sin

24











A1

1.35





A1

[3 marks]

(b) attempt to find the area of the shaded region (M1)

1

44(2

2

    

A1

39.5

(cm

2

) A1

[3 marks]

Total [6 marks]

3. (a)

4

5.5

1

4100











(M1)(A1)

1.056

A1

[3 marks]

continued…

– 6 – SPEC/5/MATAA/SP2/ENG/TZ0/XX/M

Question 3 continued

(b) EITHER

4

5.5

21

100 4

n

PP











OR



2their()

m

PP a

(M1)(A1)

Note: Award (M1) for substitution into loan payment formula. Award (A1) for correct substitution.

OR

PV 1

FV 2 

I% 5.5

P/Y  4

C/Y

 4

50.756n 

(M1)(A1)

OR

PV 1

FV 2

 

I% 100 (their ( ) 1)a

P/Y

 1

C/Y

 1 (M1)(A1)

THEN



12.7 years

Laurie will have double the amount she invested during

2032 A1

[3 marks]

Total [6 marks]

4. (a) recognition of binomial (M1)

~B(5,0.7)X

attempt to find

P( 3)X 

M1

0.472( 0.47178)

A1

[3 marks]

(b) recognition of

2 sixes in 4 tosses (M1)



22

4

P 3rd six on the 5th toss 0.7 0.3 0.7( 0.2646 0.7)

2





   









A1



0.185 0.18522 A1

[3 marks]

Total [6 marks]

– 7 – SPEC/5/MATAA/SP2/ENG/TZ0/XX/M

5. (a)

1.29a 

and

10.4b 

A1A1

[2 marks]

(b) recognising both lines pass through the mean point (M1)

28.7, 30.3pq

A2

[3 marks]

Total [5 marks]

6. (a) use of a graph to find the coordinates of the local minimum (M1)

16.513...s 

(A1)

maximum distance is

16.5 cm (to the left of O) A1

[3 marks]

(b) attempt to find time when particle changes direction eg considering the

first maximum on the graph of

s

or the first t – intercept on the graph of

.s



(M1)

1.51986...t 

(A1)

attempt to find the gradient of

s



for their value of t,



1.51986...s



(M1)

8.92

(cm/s

2

) A1

[4 marks]

Total [7 marks]

– 8 – SPEC/5/MATAA/SP2/ENG/TZ0/XX/M

Section B

7. (a)

4.2

60

 45

A1

AB  3.15

(km) A1

[2 marks]

(b) (i)

or

180 114



A1

35  66

A1

ˆ

ABC 101



AG

(ii) attempt to use cosine rule (M1)

222

4.6 cos 1AC 3.15 4.6 2 3.15 01 

(or equivalent) A1

AC 6.05

(km) A1

[5 marks]

(c) valid approach to find angle

BCA

(M1)

eg sine rule

correct substitution into sine rule A1

eg

sin B

ˆ

CA



3.15



sin101

6.0507...

B

ˆ

CA  30.7

A1

[3 marks]

(d)

B

ˆ

AC  48.267

(seen anywhere) A1

valid approach to find correct bearing (M1)

eg

48.267  35

bearing

 83.3

(accept

083

) A1

[3 marks]

(e) attempt to use time

distance

speed



M1

6.0507

3.9

or

0.065768

km/min (A1)

t  93

(minutes) A1

[3 marks]

Total [16 marks]

– 9 – SPEC/5/MATAA/SP2/ENG/TZ0/XX/M

8. (a) attempt to use the symmetry of the normal curve (M1)

eg diagram,

0.5  0.1446



P 24.15 25 0.3554X  A1

[2 marks]

(b) (i) use of inverse normal to find

z score (M1)

1.0598z 

correct substitution

24.15 25

1.0598







(A1)

0.802





A1

(ii)



P 26 0.106X  (M1)A1

[5 marks]

(c) recognizing binomial probability (M1)

E( ) 10 0.10621Y 

(A1)

 1.06

A1

[3 marks]

(d)





P3Y 

(M1)

0.0655

A1

[2 marks]

(e) recognizing conditional probability (M1)

correct substitution A1

0.3554

1 0.10621

 0.398

A1

[3 marks]

Total [15 marks]

9. (a) correct approach A1

eg

2

6

p

eriod





(or equivalent)

period 12

A1

[2 marks]

(b) (i) valid approach (M1)

eg

max + min

2

max amplitudeb 

21.8 10.2

2



, or equivalent

16b 

A1

continued…

– 10 – SPEC/5/MATAA/SP2/ENG/TZ0/XX/M

Question 9 continued

(ii) attempt to substitute into their function (M1)



π

5.8sin 6 1 16

6









(6) 13.1f 

A1

[4 marks]

(c) valid attempt to set up a system of equations (M1)

two correct equations A1



2π

sin 3 3.75 2.5

9

pq









,



2π

sin 6 3.75 15.1

9

pq









valid attempt to solve system (M1)

8.4; 6.7pq

A1A1

[5 marks]

(d) attempt to use

() ()

f

xgx to find maximum difference (M1)

1.64x 

A1

[2 marks]

Total [13 marks]