3. (m1) 1. a a1 n2 (a1)(a1) a1 n4ibrelics.weebly.com/uploads/2/5/3/9/25397032/paper_1_ms.pdf · 1...

– 8 – M13/5/MATME/SP1/ENG/TZ1/XX/M

1. (a) (i) 4

26

=−

a (A1)

correct expression for 2 +a b A1 N2

eg 52−

, (5, 2)− , 5 2−i j

(ii) correct substitution into length formula (A1)

eg 2 25 2+ , 2 25 2+ −

2 29+ =a b A1 N2 [4 marks] (b) valid approach (M1) eg (2 )= − +c a b , 5 0, 2 0x y+ = − + =

5

2−

=c A1 N2

[2 marks]

Total [6 marks] 2. (a) 1x = , 3x = − ( )accept (1, 0), ( 3, 0)− A1A1 N2 [2 marks] (b) METHOD 1

attempt to find x-coordinate (M1)

eg 1 32+ − ,

2bxa−

= , ( ) 0f x′ =

correct value, 1x = − (may be seen as a coordinate in the answer) A1 attempt to find their y-coordinate (M1)

eg ( 1)f − , 2 2− × , 4D

ya

−=

4y = − A1 vertex ( 1, 4)− − N3 [4 marks] METHOD 2

attempt to complete the square (M1) eg 2 2 1 1 3x x+ + − − attempt to put into vertex form (M1) eg 2( 1) 4x + − , 2( 1) 4x − + vertex ( 1, 4)− − A1A1 N3 [4 marks]

Total [6 marks]


3. (a) evidence of choosing product rule (M1) eg uv vu′ ′+ correct derivatives (must be seen in the product rule) cos x , 2x (A1)(A1) 2( ) cos 2 sinf x x x x x′ = + A1 N4 [4 marks]

(b) substituting 2π into their ( )f x′ (M1)

eg 2

f π′ , 2

cos 2 sin2 2 2 2π π π π

+

correct values for both πsin2

and πcos2

seen in ( )f x′ (A1)

eg 0 2 12π

+ ×

2

f π′ = π A1 N2

[3 marks]

Total [7 marks]


4. (a) attempt to solve for X (M1) eg = −XA C B ,

1−+ =X B CA , 1 ( )− −A C B , 1− −A C B

1( ) −= −X C B A 1 1( )− −= −CA BA A1 N2

[2 marks]

(b) METHOD 1

1 24 2

− =−

C B (seen anywhere) A1

correct substitution into formula for 2 2× inverse A1

eg 1 4 213 14 6

− −=

−−A ,

2 13 12 2

−

−

attempt to multiply ( )−C B and 1−A (in any order) (M1)

eg 2 3 1 1

8 3 4 1− + −+ − − ,

4 6 2 216 6 8 2− − +

− − +, two correct elements

1 011 5

=−

X A2 N3

Note: Award A1 for three correct elements.

[5 marks]

METHOD 2

correct substitution into formula for 2 2× inverse A1

eg 1 4 213 14 6

− −=

−−A ,

2 13 12 2

−

−

attempt to multiply either −1BA or −1CA (in any order) (M1)

eg 0 3 0 114 6 2 22− +−− − +

, 2 3 6 4

13 3 92 22 2 2

− − − +−

+ −, two correct entries

one correct multiplication A1

eg 3 112 02−−−

, 5 12 2

12 5

−

−

1 011 5

=−

X A2 N3

Note: Award A1 for three correct elements. .

Total [7 marks]

[5 marks]


5. (a) METHOD 1

attempt to set up equation (M1) eg 2 5y= − , 2 5x= −

correct working (A1) eg 4 5y= − , 22 5x = +

1 (2) 9f − = A1 N2 [3 marks]

METHOD 2

interchanging x and y (seen anywhere) (M1) eg 5x y= −

correct working (A1) eg 2 5x y= − , 2 5y x= +

1 (2) 9f − = A1 N2 [3 marks]

(b) recognizing 1 (3) 30g − = (M1) eg (30)f

correct working (A1) eg 1( ) (3) 30 5f g − = −D , 25

1( ) (3) 5f g − =D A1 N2

Note: Award A0 for multiple values, eg 5± .

[3 marks]

Total [6 marks]


6. attempt to integrate which involves ln (M1) eg ( )ln 2 5 , 12ln 2 5, ln 2x x x− −

correct expression (accept absence of C)

eg ( ) 112ln 2 52

x C− + , ( )6ln 2 5x − A2

attempt to substitute (4, 0) into their integrated f (M1) eg 0 6ln (2 4 5)= × − , 0 6ln (8 5) C= − +

6ln3C = − (A1)

( ) 6ln (2 5) 6ln 3f x x= − − 2 56ln3

x −= (accept ( ) 66ln 2 5 ln 3x − − ) A1 N5

Note: Exception to the FT rule. Allow full FT on incorrect integration which must involve ln.

Total [6 marks]

7. (a) evidence of correct formula (M1)

eg log log log aa bb

− = , 40log5

, log8 log5 log5+ −

Note: Ignore missing or incorrect base.

correct working (A1) eg 2log 8 , 32 8=

2 2log 40 log 5 3− = A1 N2 [3 marks]

(b) attempt to write 8 as a power of 2 (seen anywhere) (M1) eg 2log 53(2 ) , 32 8= , a2

multiplying powers (M1) eg 23log 52 , 2log 5a

correct working (A1)

eg 2log 1252 , 32log 5 , ( )2

3log 52

2log 58 125= A1 N3 [4 marks]

Total [7 marks]


SECTION B

8. (a) (i) valid approach (M1)

eg 7 14 2

3 1− − −

−, A B− , AB AO OB

→ → →

= +

6AB 2

4

→

= − A1 N2

(ii) any correct equation in the form t= +r a b (accept any parameter for t)

where 12

3= −a and b is a scalar multiple of AB

→

A2 N2

eg 1 62 2

3 4t= − + −r , ( , , ) (1, 2, 3) (3, 1, 2)x y z t= − + − ,

1 62 2

3 4

tt

t

+= − −

+r

Note: Award A1 for t+a b , A1 for 1L t= +a b , A0 for t= +r b a .

[4 marks]

(b) recognizing that scalar product 0= (seen anywhere) R1

correct calculation of scalar product (A1) eg 6(3) 2( 3) 4 p− − + , 18 6 4 p+ +

correct working A1 eg 24 4 0p+ = , 4 24p = −

6p = − AG N0 [3 marks]

continued …


Question 8 continued

(c) setting lines equal (M1)

eg 1 2L L= , 1 6 1 32 2 2 3

3 4 15 6t s

−− + − = + −

−

any two correct equations with different parameters A1A1 eg 1 6 1 3t s+ = − + , 2 2 2 3t s− − = − , 3 4 15 6t s+ = −

attempt to solve their simultaneous equations (M1)

one correct parameter A1

eg 12

t = , 53

s =

attempt to substitute parameter into vector equation (M1)

eg 1 6

12 22

3 4− + − , 11 6

2+ ×

4x = ( )accept (4, 3, 5), ignore incorrect values for andy z− A1 N3

[7 marks]

Total [14 marks]


9. (a) (i) attempt to find P(red) P(red)× (M1)

eg 3 28 7× , 3 3

8 8× , 3 2

8 8×

P(none green) 6 356 28

= = A1 N2

(ii) attempt to find P(red) P(green)× (M1)

eg 5 38 7× , 3 5

8 8× , 15

56

recognizing two ways to get one red, one green (M1)

eg 2P( ) P( )R G× , 5 3 3 58 7 8 7× + × , 3 5 2

8 8× ×

30P(exactly one green)56

= 1528

= A1 N2

[5 marks]

(b) 20P(both green)56

= (seen anywhere) (A1)

correct substitution into formula for E ( )X A1

eg 6 30 200 1 256 56 56

× + × + × , 30 5064 64

+

expected number of green marbles is 7056

54

= A1 N2

[3 marks]

continued …



(c) (i) 4P( jar B)6

= 23

= A1 N1

(ii) ( ) 6P red jar B8

= 34

= A1 N1

[2 marks]

(d) recognizing conditional probability (M1)

eg P ( | )A R , P ( jar A and red )P( red )

, tree diagram

attempt to multiply along either branch (may be seen on diagram) (M1)

eg 1 3P( jar A and red)3 8

= × 18

=

attempt to multiply along other branch (M1)

eg 2 6P( jar B and red)3 8

= × 12

=

adding the probabilities of two mutually exclusive paths (A1)

eg 1 3 2 6P(red)3 8 3 8

= × + × 58

=

correct substitution

eg ( )P jar A|red =

1 33 8

1 3 2 63 8 3 8

×

× + ×,

1858

A1

( ) 1P jar A|red5

= A1 N3

[6 marks]

Total [16 marks]


10. (a) substitute 0 into f (M1) eg ln (0 1)+ , ln1 (0) 0f = A1 N2 [2 marks]

(b) 34

1( ) 41

f x xx

′ = ×+

(seen anywhere) A1A1

Note: Award A1 for 4

11x +

and A1 for 34x . recognizing f increasing where ( ) 0f x′ > (seen anywhere) R1 eg ( ) 0f x′ > , diagram of signs attempt to solve ( ) 0f x′ > (M1) eg 34 0x = ,

3 0x > f increasing for 0x > (accept 0x ≥ ) A1 N1 [5 marks] (c) (i) substituting 1x = into f ′′ (A1)

eg 2

4(3 1)(1 1)

−+

, 4 24×

(1) 2f ′′ = A1 N2 (ii) valid interpretation of point of inflexion (seen anywhere) R1 eg no change of sign in ( )f x′′ , no change in concavity, f ′ increasing both sides of zero attempt to find ( )f x′′ for 0x < (M1)

eg ( 1)f ′′ − , ( )

( )

2 4

24

4( 1) 3 ( 1)

( 1) 1

− − −

− +, diagram of signs

correct working leading to positive value A1 eg ( 1) 2f ′′ − = , discussing signs of numerator and denominator there is no point of inflexion at 0x = AG N0 [5 marks]

continued …


Question 10 continued (d)

A1A1A1 N3 Notes: Award A1 for shape concave up left of POI and concave down right of POI. Only if this A1 is awarded, then award the following: A1 for curve through (0, 0) , A1 for increasing throughout. Sketch need not be drawn to scale. Only essential features need to be clear. [3 marks]

Total [15 marks]

- 9 - M09/5/MATME/SP1/ENG/TZ1/XX/M+

SECTION A QUESTION 1

(a) (i) 4 0

( 4 )0 4§ ·

¨ ¸© ¹

AB I A2 N2

(ii) 1

1 12 11 1 2 4, ,6 5 3 54 4

2 4

�

§ ·�¨ ¸�§ · ¨ ¸¨ ¸� ¨ ¸© ¹ �¨ ¸

© ¹

A B A1 N1

(b) METHOD 1

1xy

�§ · ¨ ¸

© ¹A C (M1)

1 12 1 8 81 2 46 5 4 3 5 44

2 4

§ ·§ ·�¨ ¸¨ ¸�§ ·§ · § ·¨ ¸ ¨ ¸¨ ¸¨ ¸ ¨ ¸� � �¨ ¸¨ ¸© ¹© ¹ © ¹�¨ ¸¨ ¸© ¹© ¹

A1

517

xy

§ · § · ¨ ¸ ¨ ¸�© ¹ © ¹

A1A1 N3

METHOD 2 5 8x y� , 6 2 4x y� � A1 for work towards solving their system (M1)

517

xy

§ · § · ¨ ¸ ¨ ¸�© ¹ © ¹

A1A1 N3

[7 marks] QUESTION 2

(a) 1P( )11

A A1 N1

(b) 2P( | )10

B A A2 N2

(c) recognising that P( ) = P( ) P( )A B A B A� u (M1) correct values (A1)

e.g. 1 2P( )11 10

A B� u

2P( )110

A B� A1 N3

[6 marks]


QUESTION 3 evidence of choosing the product rule (M1) ( ) e ( sin ) cos e (= e cos e sin )x x x xf x x x x xc u � � u � A1A1 substituting S (M1) e.g. ( e cos e sinf S Sc S� S� S , e ( 1 0)S � � , eS� taking negative reciprocal (M1)

e.g. 1(f

�c S�

gradient is 1eS A1 N3

[6 marks] QUESTION 4 (a)

Function Graph displacement A acceleration B

A2A2 N4 (b) 3t A2 N2 [6 marks] QUESTION 5 (a) in any order translated 1 unit to the right A1 N1 stretched vertically by factor 2 A1 N1 (b) METHOD 1 Finding coordinates of image on g (A1)(A1) e.g. 1 1 0, 1 2 2� � u , ( 1, 1) ( 1 1, 2 1)� o � � u , (0, 2) P is (3, 0) A1A1 N4 METHOD 2 2( ) 2( 4) 2h x x � � (A1)(A1) P is (3, 0) A1A1 N4 [6 marks]


QUESTION 6 (a) (i) interchanging x and y (seen anywhere) M1 e.g. 3e yx � correct manipulation A1 e.g. ln 3x y � , ln 3y x � 1( ) ln 3f x x� � AG N0 (ii) 0x ! A1 N1 (b) collecting like terms; using laws of logs (A1)(A1)

e.g. 1ln ln 3xx

§ ·� ¨ ¸© ¹

, ln ln 3x x� ; ln 31x

x

§ ·¨ ¸

¨ ¸¨ ¸© ¹

, 2ln 3x

simplify (A1)

e.g. 3ln2

x , 2 3ex

32ex � �3e A1 N2

[7 marks] QUESTION 7 attempt to substitute into formula 2dV y x S³ (M1) integral expression A1

e.g. � �2

0d

ax xS³ , xS³

correct integration (A1)

e.g. 21d2

x x x ³

correct substitution 212

V aª º S « »¬ ¼ (A1)

equating their expression to 32S M1

e.g. 21 322

aª ºS S« »¬ ¼

2 64a 8a A2 N2 [7 marks]


SECTION B QUESTION 8 (a) (i) 3cosx T A1 N1 (ii) 3siny T A1 N1 [2 marks] (b) finding area (M1)

e.g. 2 2A x y u , 182

A bh u

substituting A1

e.g. 4 3sin 3cosA T T u u , 18 3cos 3sin2

T Tu u u

18(2sin cos )A T T A1 18sin 2A T AG N0 [3 marks]

(c) (i) d 36cos2d

A TT A2 N2

(ii) for setting derivative equal to 0 (M1)

e.g. 36cos2 0T , d 0d

AT

22

T S (A1)

4

T S A1 N2

(iii) valid reason (seen anywhere) R1

e.g. at 2

2

d, 04 d

AT

S� ; maximum when ( ) 0f xcc �

finding second derivative 2

2

d 72sin 2d

A TT

� A1

evidence of substituting 4S M1

e.g. 72sin 2 , 72sin , 724 2S S§ · § ·� u � �¨ ¸ ¨ ¸

© ¹ © ¹

4

T S produces the maximum area AG N0

[8 marks]

Total [13 marks]


QUESTION 9 (a) (i) evidence of approach (M1)

e.g. PQ PO OQo o o

� , Q P�

1

PQ 21

o�§ ·¨ ¸ ¨ ¸¨ ¸© ¹

A1 N2

(ii) 2

PR 24

o§ ·¨ ¸ ¨ ¸¨ ¸© ¹

A1 N1

[3 marks] (b) METHOD 1

choosing correct vectors PQo

and PRo

(A1)(A1)

finding PQ PR, PQ , PRo o o o

(A1) (A1)(A1)

PQ PR 2 4 4( 6)o o

� � �

� �2 2 2PQ ( 1) 2 1 6o

� � � , � �2 2 2PR 2 2 4 24o

� �

substituting into formula for angle between two vectors M1

e.g. 6ˆcosRPQ6 24

u

simplifying to expression clearly leading to 12

A1

e.g. 6

6 2 6u, 6 6,

12144

1ˆcosRPQ2

AG N0

METHOD 2 evidence of choosing cosine rule (seen anywhere) (M1)

3

QR 03

o§ ·¨ ¸ ¨ ¸¨ ¸© ¹

A1

QR 18o

, PQ 6o

and PR 24o

(A1)(A1)(A1)

� � � � � �2 2 2

6 24 18ˆcos RPQ

2 6 24

� �

u A1

6 24 18 12ˆcosRPQ24 24

� � § · ¨ ¸© ¹

A1

1ˆcosRPQ2

AG N0

[7 marks]

continued …



(c) (i) METHOD 1 evidence of appropriate approach (M1) e.g. using 2 2ˆ ˆsin RPQ cos RPQ 1� , diagram

substituting correctly (A1)

e.g. 21ˆsin RPQ 1

2§ · � ¨ ¸© ¹

3 3ˆsin RPQ4 2

§ · ¨ ¸¨ ¸

© ¹ A1 N3

METHOD 2

since 1ˆcosP2

, P̂ 60 (A1)

evidence of approach e.g. drawing a right triangle, finding the missing side (A1)

3ˆsin P2

A1 N3

(ii) evidence of appropriate approach (M1)

e.g. attempt to substitute into 1 sin2

ab C


e.g. area 1 36 242 2

u u A1

area 3 3 A1 N2 [6 marks]

Total [16 marks]


QUESTION 10 (a) METHOD 1 evidence of substituting x� for x (M1)

2

( )( )( ) 1

a xf xx�

� � �

A1

� �2( ) ( )1

axf x f xx�

� ��

AG N0

METHOD 2 ( )y f x � is reflection of ( )y f x in x axis and ( )y f x � is reflection of ( )y f x in y axis (M1) sketch showing these are the same A1

� �2( ) ( )1

axf x f xx�

� ��

AG N0

[2 marks] (b) evidence of appropriate approach (M1) e.g. ( ) 0f xcc to set the numerator equal to 0 (A1) e.g. 2 22 ( 3) 0; ( 3) 0ax x x� �

3 3(0, 0), 3 , , 3 ,4 4

a a§ · § ·� �¨ ¸ ¨ ¸¨ ¸ ¨ ¸

© ¹ © ¹ (accept 0, 0 .)x y etc A1A1A1A1A1 N5

[7 marks]

continued …


Question 10 continued (c) (i) correct expression A2

e.g. 7

2

3

ln ( 1)2a xª º�« »¬ ¼

, ln50 ln102 2a a

� , (ln50 ln10)2a

�

area ln52a

A1A1 N2

(ii) METHOD 1 recognizing the shift that does not change the area (M1)

e.g. 8 7

4 3( 1)d ( )df x x f x x� ³ ³ , ln5

2a

recognizing that the factor of 2 doubles the area (M1)

e.g. � �8 8 7

4 4 32 ( 1)d 2 ( 1)d 2 ( )df x x f x x f x x� � ³ ³ ³

8

42 ( 1)d ln5f x x a� ³ � �. . 2 answer to (c)(i)i e u their A1 N3

METHOD 2 changing variable

let 1w x � , so d 1dwx

222 ( )d ln ( 1)2af w w w c � �³ (M1)

substituting correct limits e.g.

82

4ln[( 1) 1]a xª º� �¬ ¼ ,

72

3ln ( 1)a wª º�¬ ¼ , ln50 ln10a a� (M1)

8

42 ( 1)d ln5f x x a� ³ A1 N3

[7 marks]

Total [16 marks]


SECTION A QUESTION 1 (a) for interchanging x and y (may be done later) (M1) e.g. 2 3x y �

1 3( )2

xg x� � 3 3accept ,

2 2x xy � �§ · ¨ ¸

© ¹ A1 N2

(b) METHOD 1 (4) 5g (A1) evidence of composition of functions (M1) (5) 25f A1 N3 METHOD 2 2( ) (2 3)f g x x � (M1) 2(4) (2 4 3)f g u � (A1) 25 A1 N3 [5 marks] QUESTION 2 finding scalar product and magnitudes (A1)(A1)(A1) scalar product 12 20 15 ( 23) � � �

magnitudes 2 2 2 2 2 23 4 5 , 4 ( 5) ( 3) � � � � � � � �50 , 50

substitution into formula M1

e.g. � � � �2 2 2 2 2 2

12 20 15cos3 4 5 4 ( 5) ( 3)

T � �

� � u � � � �

23cos ( 0.46)50

T � � A2 N4

[6 marks] QUESTION 3 (a) 10n A1 N1 (b) , 2a p b q (or 2 ,a q b p ) A1A1 N1N1

(c) 5 510(2 )

5p q

§ ·¨ ¸© ¹

A1A1A1 N3

[6 marks]


QUESTION 4 (a) 5 A1 N1 (b) METHOD 1

2 2 232log log 32 log 88

xx y

y

§ · �¨ ¸

© ¹ (A1)

2 2log 32 log 8x y � (A1) 2log 8 3 (A1) 5, 3p q � (accept 5 3x y� ) A1 N3 METHOD 2

5

3

32 (2 )8 (2 )

x x

y y (A1)

5

3

22

x

y (A1)

5 32 x y� (A1) 5 3

2log (2 ) 5 3x y x y� � 5, 3p q � (accept 5 3x y� ) A1 N3 [5 marks]


QUESTION 5 (a) METHOD 1 1 1( )� � M M (M1)

2 011 510

§ · ¨ ¸�© ¹

M A1A1 N3

METHOD 2

5 0 1 01 2 0 1

a bc d§ ·§ · § ·

¨ ¸¨ ¸ ¨ ¸© ¹© ¹ © ¹

(M1)

5 1a b� , 2 0b , 5 0c d� , 2 1d (A1)

0.2 00.1 0.5

§ · ¨ ¸�© ¹

M A1 N3

(b) METHOD 1 evidence of appropriate approach (M1) e.g. 1� X M B

5 0 11 2 7

xy

§ · § ·§ · ¨ ¸ ¨ ¸¨ ¸

© ¹ © ¹© ¹ A1

5

15§ ·

¨ ¸© ¹

A1 N2

METHOD 2 evidence of appropriate approach (M1)

e.g. 0.2 0 10.1 0.5 7

xy

§ ·§ · § · ¨ ¸¨ ¸ ¨ ¸�© ¹© ¹ © ¹

0.2 1x , 0.1 0.5 7x y� � A1

5

15xy

§ · § · ¨ ¸ ¨ ¸

© ¹ © ¹ A1 N2

[6 marks]


QUESTION 6 (a) METHOD 1 2( ) 3( 3)f x xcc � A2 N2 METHOD 2 attempt to expand 3( 3)x � (M1) e.g. 3 2( ) 9 27 27f x x x xc � � � 2( ) 3 18 27f x x xcc � � A1 N2 (b) (3) 0f c , (3) 0f cc A1 N1 (c) METHOD 1 f cc does not change sign at P R1 evidence for this R1 N0 METHOD 2 f c changes sign at P so P is a maximum/minimum (i.e. not inflexion) R1 evidence for this R1 N0 METHOD 3

finding 41( ) ( 3)4

f x x c � � and sketching this function R1

indicating minimum at 3x R1 N0 [5 marks]


QUESTION 7 � �2e 3 sin cos 0x x x� (A1)

2e 0x not possible (seen anywhere) (A1) simplifying

e.g. 3 sin cos 0, 3sin cosx x x x� � , sin 1cos 3

xx

� A1

EITHER

1tan3

x � A1

56

x S A2 N4

OR sketch of 30 , 60 , 90 triangle with sides 1, 2, 3 A1

work leading to 5π6

x A1

verifying 5π6

satisfies equation A1 N4

[6 marks]


QUESTION 8 (a) (i) 33e x�� A1 N1

(ii) cos x S§ ·�¨ ¸�© ¹ A1 N1

(b) evidence of choosing product rule (M1) e.g. uv vuc c� correct expression A1

e.g. 3 33e sin e cos3 3

x xx x� �S S§ · § ·� � � �¨ ¸ ¨ ¸© ¹ © ¹

complete correct substitution of π3

x (A1)

e.g. 3 3

3 33e sin e cos3 3 3 3

S S� �S S S S§ · § ·� � � �¨ ¸ ¨ ¸

© ¹ © ¹

eh �SS§ ·c ¨ ¸�© ¹ A1 N3

[6 marks]


SECTION B QUESTION 9 (a)

3, 9 4, 9 5 , 9 3, 10 4, 10 5, 10 3, 10 4, 10 5, 10

A2 N2 [2 marks] (b) 12, 13, 14, 15 (accept 12, 13, 13, 13, 14, 14, 14, 15, 15) A2 N2 [2 marks]

(c) 1 3 3 2P(12) , P(13) , P(14) , P(15)9 9 9 9

A2 N2

[2 marks] (d) correct substitution into formula for E( )X A1

e.g. 1 3 3 2E( ) 12 13 14 159 9 9 9

S u � u � u � u

123E( )9

S A2 N2

[3 marks] (e) METHOD 1 correct expression for expected gain E( )A for 1 game (A1)

e.g. 4 550 309 9u � u

50E( )9

A

amount at end = expected gain for 1 game 36u (M1) 200 (dollars) A1 N2 METHOD 2 attempt to find expected number of wins and losses (M1)

e.g. 4 536 , 369 9u u

attempt to find expected gain E( )G (M1) e.g. 16 50 30 20u � u E( )G 200 (dollars) A1 N2

[3 marks]

Total [12 marks]


QUESTION 10

(a) 1L : 8 01 00 1

t§ · § ·¨ ¸ ¨ ¸ �¨ ¸ ¨ ¸¨ ¸ ¨ ¸© ¹ © ¹

r A2 N2

[2 marks]

(b) evidence of equating r and OAo

(M1)

e.g. 6 2 22 4 19 1 5

s§ · § · § ·¨ ¸ ¨ ¸ ¨ ¸ � �¨ ¸ ¨ ¸ ¨ ¸¨ ¸ ¨ ¸ ¨ ¸�© ¹ © ¹ © ¹

, A r

one correct equation A1 e.g. 6 2 2 , 2 4 , 9 1 5s s s � � � � 2s A1 evidence of confirming for other two equations A1 e.g. 6 2 4, 2 4 2, 9 1 10 � � � � so A lies on 2L AG N0 [4 marks]

continued …


Question 10 continued (c) (i) evidence of approach M1

e.g. 2 2 8 04 1 1 01 5 0 1

s t§ · § · § · § ·¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸� � �¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸�© ¹ © ¹ © ¹ © ¹

1 2L L

one correct equation A1 e.g. 2 2 8, 4 1, 1 5s s s t� � � � attempt to solve (M1) finding 3s A1 substituting M1

e.g. 2 2

OB 4 3 11 5

o§ · § ·¨ ¸ ¨ ¸ � �¨ ¸ ¨ ¸¨ ¸ ¨ ¸�© ¹ © ¹

8

OB 114

o§ ·¨ ¸ ¨ ¸¨ ¸© ¹

AG N0


e.g. AB AO +OBo o o

, AB OB OAo o o

�

2

AB 15

o§ ·¨ ¸ �¨ ¸¨ ¸© ¹

A1 N2

[7 marks] (d) evidence of appropriate approach (M1)

e.g. AB DCo o

correct values A1

e.g. 2 2 2 2 2 2

OD 1 1 , 1 1 , 1 15 4 5 4 5 4

x xy yz z

o�§ · § · § · § · § · § · § ·

¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸� � � � � �¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸� � � �© ¹ © ¹ © ¹ © ¹ © ¹ © ¹ © ¹

0

OD 29

o§ ·¨ ¸ ¨ ¸¨ ¸�© ¹

A1 N2

[3 marks]

Total [16 marks]


QUESTION 11 Note: In this question, do not penalize absence of units. (a) (i) (40 )ds at t �³ (M1)

21402

s t at c � � (A1)(A1)

substituting 100s when 0t ( 100)c (M1)

2140 1002

s t at � � A1 N5

(ii) 21402

s t at � A1 N1

[6 marks] (b) (i) stops at station, so 0v (M1)

40ta

(seconds) A1 N2

(ii) evidence of choosing formula for s from (a) (ii) (M1)

substituting 40ta

(M1)

e.g. 2

2

40 1 40402

aa a

u � u

setting up equation M1

e.g. 500 s , 2

2

40 1 40500 402

aa a

u � u , 1600 800500a a

�

evidence of simplification to an expression which obviously leads to 85

a A1

e.g. 8500 800, 5 , 1000 3200 1600a aa

�

85

a AG N0

[6 marks]

continued …


Question 11 continued (c) METHOD 1 40 4v t � , stops when v = 0 40 4 0t� (A1) 10t A1 substituting into expression for s M1

2140 10 4 102

s u � u u

200s A1 since 200 < 500 (allow FT on their s, if 500s � ) R1 train stops before the station AG N0 METHOD 2

from (b) 40 104

t A2

substituting into expression for s

e.g. 2140 10 4 102

s u � u u M1

200s A1 since 200 < 500, R1 train stops before the station AG N0 METHOD 3 a is deceleration A2

845

! A1

so stops in shorter time (A1) so less distance travelled R1 so stops before station AG N0 [5 marks]

Total [17 marks]

– 9 – M10/5/MATME/SP1/ENG/TZ1/XX/M+

SECTION A QUESTION 1 (a) evidence of setting function to zero (M1) e.g. 0)(xf , 28 2x x evidence of correct working A1

e.g. 0 2 (4 )x x , 8 644

x-intercepts are at 4 and 0 (accept (4, 0) and (0, 0) , or 4x , 0x ) A1A1 N1N1 (b) (i) 2x (must be equation) A1 N1 (ii) substituting 2x into ( )f x (M1)

8y A1 N2 [7 marks] QUESTION 2

(a) 1356

WP A1A1A1 N3

Note: Award A1 for each correct element. (b) Note: The first two steps may be done in any order. subtracting (A1)

e.g. 2612 210

WP

multiplying WP by 2 (A1)

e.g. 261012

022

S A1 N2

[6 marks]


QUESTION 3 (a) evidence of expanding M1 e.g. 4 3 2 2 3 42 4(2 ) 6(2 ) 4(2)x x x x , 2 2(4 4 )(4 4 )x x x x 4 2 3 4(2 ) 16 32 24 8x x x x x A2 N2 (b) finding coefficients 24 and 1 (A1)(A1) term is 225x A1 N3 [6 marks] QUESTION 4

(a) 3tan4

3do not accept4

x A1 N1

(b) (i) 3sin5

, 4cos5

(A1)(A1)

correct substitution A1

e.g. 3 4sin 2 25 5

24sin 225

A1 N3

(ii) correct substitution A1

e.g. 23cos2 1 2

5,

2 24 35 5

7cos225

A1 N1

[7 marks]


QUESTION 5 (a) (i) 0.2p A1 N1 (ii) 0.4q A1 N1 (iii) 0.1r A1 N1

(b) 2P( | )3

A B A2 N2

Note: Award A1 for an unfinished answer such as 0.20.3

. (c) valid reason R1

e.g. 2 0.5, 0.35 0.33

thus, A and B are not independent AG N0 [6 marks] QUESTION 6 attempt to set up integral expression M1

e.g. 2

216 4 dx x , 2 2

02 (16 4 )x ,

2216 4 dx x

16d 16x x , 3

2 44 d3xx x (seen anywhere) A1A1

evidence of substituting limits into the integrand (M1)

e.g. 32 3232 323 3

, 64643

volume 1283

A2 N3

[6 marks]


QUESTION 7 (a) interchanging x and y (seen anywhere) (M1) e.g. logx y (accept any base) evidence of correct manipulation A1

e.g. 3x y , 123y x , 3

1 log2

x y , 32 logy x

1 2( ) 3 xf x AG N0 (b) 0y , 1( ) 0f x A1 N1 (c) METHOD 1 finding 3(2) log 2g (seen anywhere) A1 attempt to substitute (M1) e.g. 32log 21( ) (2) 3f g evidence of using log or index rule (A1)

e.g. 3log 41( ) (2) 3f g , 32log 23

1( )(2) 4f g A1 N1 METHOD 2 attempt to form composite (in any order) (M1) e.g. 32log1( ) ( ) 3 xf g x evidence of using log or index rule (A1) e.g.

23log1( )( ) 3 xf g x , 3

2log3 x

1 2( )( )f g x x A1 1( )(2) 4f g A1 N1 [7 marks]


SECTION B QUESTION 8 (a) 2( ) 2 3f x x x A1A1A1 evidence of solving ( ) 0f x (M1) e.g. 2 2 3 0x x evidence of correct working A1

e.g. ( 1)( 3)x x , 2 162

1x (ignore 3x ) (A1) evidence of substituting their negative x-value into ( )f x (M1)

e.g. 3 21 ( 1) ( 1) 3( 1)3

, 1 1 33

53

y A1

coordinates are 51,3

N3

[8 marks] (b) (i) ( 3, 9) A1 N1 (ii) (1, 4) A1A1 N2 (iii) reflection gives (3, 9) (A1)

stretch gives 3 , 92

A1A1 N3

[6 marks]

Total [14 marks]


QUESTION 9

(a) d sin cosd

x xx

, d cos sind

x xx

(seen anywhere) (A1)(A1)

evidence of using the quotient rule M1 correct substitution A1

e.g. 2

sin ( sin ) cos (cos )sin

x x x xx

, 2 2

2

sin cossinx x

x

2 2

2

(sin cos )( )sinx xf x

x A1

2

1( )sin

f xx

AG N0

[5 marks] (b) METHOD 1 appropriate approach (M1) e.g. 2( ) (sin )f x x

3( ) 2(sin )(cos )f x x x 3

2cossin

xx

A1A1 N3 Note: Award A1 for 32sin x , A1 for cos x . [3 marks] METHOD 2 derivative of 2sin 2sin cosx x x (seen anywhere) A1 evidence of choosing quotient rule (M1)

e.g. 1u , 2sinv x , 2

2 2

sin 0 ( 1) 2sin cossin )(

x x xf

x

2 2

2sin cos( )(sin )

x xf xx

3

2cossin

xx

A1 N3

[3 marks]

(c) evidence of substituting 2

M1

e.g. 2

1

sin2

, 3

2cos2

sin2

1p , 0q A1A1 N1N1 [3 marks] (d) second derivative is zero, second derivative changes sign R1R1 N2 [2 marks]

Total [13 marks]


QUESTION 10 (a) any correct equation in the form tr a b (accept any parameter) A2 N2

e.g. 8 25 1

25 8tr

Note: Award A1 for ta b , A1 for L ta b , A0 for tr b a . [2 marks] (b) recognizing scalar product must be zero (seen anywhere) R1 e.g. 0a b

evidence of choosing direction vectors 2 71 , 28 k

(A1)(A1)

correct calculation of scalar product (A1) e.g. 2( 7) 1( 2) 8k simplification that clearly leads to solution A1 e.g. 16 8k , 16 8 0k 2k AG N0 [5 marks] (c) evidence of equating vectors (M1)

e.g. 1 3L L , 3 2 5 71 1 0 2

25 8 3 2p q

any two correct equations A1A1 e.g. 3 2 5 7p q , 1 2p q , 25 8 3 2p q attempting to solve equations (M1) finding one correct parameter ( 3, 2)p q A1 the coordinates of A are ( 9, 4, 1) A1 N3 [6 marks]

continued …


Question 10 continued (d) (i) evidence of appropriate approach (M1)

e.g. OA AB OB , 8 9

AB 5 425 1

1

AB 126

A1 N2

(ii) finding 7

AC 22

A1

evidence of finding magnitude (M1)

e.g. 2 2 2|AC| 7 2 2

|AC| 57 A1 N3 [5 marks]

Total [18 marks]


SECTION A QUESTION 1 (a) evidence of setting function to zero (M1) e.g. 0)(xf , 28 2x x evidence of correct working A1

e.g. 0 2 (4 )x x , 8 644

x-intercepts are at 4 and 0 (accept (4, 0) and (0, 0) , or 4x , 0x ) A1A1 N1N1 (b) (i) 2x (must be equation) A1 N1 (ii) substituting 2x into ( )f x (M1)

8y A1 N2 [7 marks] QUESTION 2

(a) 1356

WP A1A1A1 N3

Note: Award A1 for each correct element. (b) Note: The first two steps may be done in any order. subtracting (A1)

e.g. 2612 210

WP

multiplying WP by 2 (A1)

e.g. 261012

022

S A1 N2

[6 marks]


QUESTION 3 (a) evidence of expanding M1 e.g. 4 3 2 2 3 42 4(2 ) 6(2 ) 4(2)x x x x , 2 2(4 4 )(4 4 )x x x x 4 2 3 4(2 ) 16 32 24 8x x x x x A2 N2 (b) finding coefficients 24 and 1 (A1)(A1) term is 225x A1 N3 [6 marks] QUESTION 4

(a) 3tan4

3do not accept4

x A1 N1

(b) (i) 3sin5

, 4cos5

(A1)(A1)


e.g. 3 4sin 2 25 5

24sin 225

A1 N3

(ii) correct substitution A1

e.g. 23cos2 1 2

5,

2 24 35 5

7cos225

A1 N1

[7 marks]


QUESTION 5 (a) (i) 0.2p A1 N1 (ii) 0.4q A1 N1 (iii) 0.1r A1 N1

(b) 2P( | )3

A B A2 N2

Note: Award A1 for an unfinished answer such as 0.20.3

. (c) valid reason R1

e.g. 2 0.5, 0.35 0.33

thus, A and B are not independent AG N0 [6 marks] QUESTION 6 attempt to set up integral expression M1

e.g. 2

216 4 dx x , 2 2

02 (16 4 )x ,

2216 4 dx x

16d 16x x , 3

2 44 d3xx x (seen anywhere) A1A1

evidence of substituting limits into the integrand (M1)

e.g. 32 3232 323 3

, 64643

volume 1283

A2 N3

[6 marks]


QUESTION 7 (a) interchanging x and y (seen anywhere) (M1) e.g. logx y (accept any base) evidence of correct manipulation A1

e.g. 3x y , 123y x , 3

1 log2

x y , 32 logy x

1 2( ) 3 xf x AG N0 (b) 0y , 1( ) 0f x A1 N1 (c) METHOD 1 finding 3(2) log 2g (seen anywhere) A1 attempt to substitute (M1) e.g. 32log 21( ) (2) 3f g evidence of using log or index rule (A1)

e.g. 3log 41( ) (2) 3f g , 32log 23

1( )(2) 4f g A1 N1 METHOD 2 attempt to form composite (in any order) (M1) e.g. 32log1( ) ( ) 3 xf g x evidence of using log or index rule (A1) e.g.

23log1( )( ) 3 xf g x , 3

2log3 x

1 2( )( )f g x x A1 1( )(2) 4f g A1 N1 [7 marks]


SECTION B QUESTION 8 (a) 2( ) 2 3f x x x A1A1A1 evidence of solving ( ) 0f x (M1) e.g. 2 2 3 0x x evidence of correct working A1

e.g. ( 1)( 3)x x , 2 162

1x (ignore 3x ) (A1) evidence of substituting their negative x-value into ( )f x (M1)

e.g. 3 21 ( 1) ( 1) 3( 1)3

, 1 1 33

53

y A1

coordinates are 51,3

N3

[8 marks] (b) (i) ( 3, 9) A1 N1 (ii) (1, 4) A1A1 N2 (iii) reflection gives (3, 9) (A1)

stretch gives 3 , 92

A1A1 N3

[6 marks]

Total [14 marks]


QUESTION 9

(a) d sin cosd

x xx

, d cos sind

x xx

(seen anywhere) (A1)(A1)

evidence of using the quotient rule M1 correct substitution A1

e.g. 2

sin ( sin ) cos (cos )sin

x x x xx

, 2 2

2

sin cossinx x

x

2 2

2

(sin cos )( )sinx xf x

x A1

2

1( )sin

f xx

AG N0

[5 marks] (b) METHOD 1 appropriate approach (M1) e.g. 2( ) (sin )f x x

3( ) 2(sin )(cos )f x x x 3

2cossin

xx

A1A1 N3 Note: Award A1 for 32sin x , A1 for cos x . [3 marks] METHOD 2 derivative of 2sin 2sin cosx x x (seen anywhere) A1 evidence of choosing quotient rule (M1)

e.g. 1u , 2sinv x , 2

2 2

sin 0 ( 1) 2sin cossin )(

x x xf

x

2 2

2sin cos( )(sin )

x xf xx

3

2cossin

xx

A1 N3

[3 marks]

(c) evidence of substituting 2

M1

e.g. 2

1

sin2

, 3

2cos2

sin2

1p , 0q A1A1 N1N1 [3 marks] (d) second derivative is zero, second derivative changes sign R1R1 N2 [2 marks]

Total [13 marks]


QUESTION 10 (a) any correct equation in the form tr a b (accept any parameter) A2 N2

e.g. 8 25 1

25 8tr

Note: Award A1 for ta b , A1 for L ta b , A0 for tr b a . [2 marks] (b) recognizing scalar product must be zero (seen anywhere) R1 e.g. 0a b

evidence of choosing direction vectors 2 71 , 28 k

(A1)(A1)

correct calculation of scalar product (A1) e.g. 2( 7) 1( 2) 8k simplification that clearly leads to solution A1 e.g. 16 8k , 16 8 0k 2k AG N0 [5 marks] (c) evidence of equating vectors (M1)

e.g. 1 3L L , 3 2 5 71 1 0 2

25 8 3 2p q

any two correct equations A1A1 e.g. 3 2 5 7p q , 1 2p q , 25 8 3 2p q attempting to solve equations (M1) finding one correct parameter ( 3, 2)p q A1 the coordinates of A are ( 9, 4, 1) A1 N3 [6 marks]

continued …


Question 10 continued (d) (i) evidence of appropriate approach (M1)

e.g. OA AB OB , 8 9

AB 5 425 1

1

AB 126

A1 N2

(ii) finding 7

AC 22

A1

evidence of finding magnitude (M1)

e.g. 2 2 2|AC| 7 2 2

|AC| 57 A1 N3 [5 marks]

Total [18 marks]


SECTION A

QUESTION 1

(a) 2, 4q r= − = or 4, 2q r= = − A1A1 N2

(b) 1x = (must be an equation) A1 N1

(c) substituting (0, 4)− into the equation (M1)

e.g. 4 (0 ( 2))(0 4), 4 ( 4)(2)p p− = − − − − = −

correct working towards solution (A1)

e.g. 4 8p− = −

4 1

8 2p = = A1 N2

[6 marks]

QUESTION 2

(a) evidence of appropriate approach (M1)

e.g. BC BA AC→ → →

= + ,

2 6

3 2

2 3

−

− − −

8

BC 1

1

→−

= −

−

A1 N2

(b) attempt to find the length of AB→

(M1)

( )2 2 2|AB| 6 ( 2) 3 36 4 9 49 7→

= + − + = + + = = (A1)

unit vector is

6

761 2

27 7

33

7

− = − A1 N2

(c) recognizing that the dot product or cosθ being 0 implies perpendicular (M1)

correct substitution in a scalar product formula A1

e.g. (6) ( 2) ( 2) ( 3) (3) (2)× − + − × − + × , 12 6 6

cos7 17

θ− + +

=×

correct calculation A1

e.g. AB AC 0→ →

= , cos 0θ =

therefore, they are perpendicular AG N0 [8 marks]


QUESTION 3

(a) evidence of multiplying (M1)e.g. one correct element

15

5

−=AB A1A1 N3

(b) METHOD 1 evidence of multiplying by A (on left or right) (M1)

e.g. 1− =AA X AB , =X AB

15

5

−=X (accept 15x = − , 5y = ) A1 N2

METHOD 2 attempt to set up a system of equations (M1)

e.g. 4 2

510

x y+= − ,

35

10

x y− +=

15

5

−=X (accept 15x = − , 5y = ) A1 N2

[5 marks]

QUESTION 4

(a) cos2

fπ

= π (A1)

1= − A1 N2

(b) ( ) ( 1)2

g f gπ

= − ( )22( 1) 1= − − (A1)

1= A1 N2

(c) ( ) ( )2 2( )( ) 2 cos(2 ) 1 2cos (2 ) 1g f x x x= − = − A1

evidence of 22cos 1 cos2θ θ− = (seen anywhere) (M1)

( ) ( ) cos4g f x x=

4k = A1 N2

[7 marks]


QUESTION 5

gradient of tangent 8= (seen anywhere) (A1)3( ) 4f x kx′ = (seen anywhere) A1

recognizing the gradient of the tangent is the derivative (M1)

setting the derivative equal to 8 (A1)

e.g. 34 8kx = , 3 2kx =

substituting 1x = (seen anywhere) (M1)

2k = A1 N4

[6 marks]

QUESTION 6

recognizing log log loga b ab+ = (seen anywhere) (A1)

e.g. ( )2log ( 2)x x − , 2 2x x−

recognizing log x

a b x a b= ⇔ = (seen anywhere) (A1)

e.g. 32 8=

correct simplification A1

e.g. 3( 2) 2x x − = , 2 2 8x x− −

evidence of correct approach to solve (M1)

e.g. factorizing, quadratic formula correct working A1

e.g. ( 4)( 2)x x− + , 2 36

2

±

4x = A2 N3

[7 marks]

QUESTION 7

(a) x-intercepts at 3, 0, 2− A2 N2

(b) 3 0x− < < , 2 3x< < A1A1 N2

(c) correct reasoning R2e.g. the graph of f is concave-down (accept convex), the first derivative is decreasing

therefore the second derivative is negative AG

[6 marks]


SECTION B QUESTION 8 (a) substituting into the second derivative M1

e.g. 4

3 13

× − −

45

3f ′′ − =− A1

since the second derivative is negative, B is a maximum R1 N0 [3 marks]

(b) setting ( )f x′ equal to zero (M1)

evidence of substituting 2x = 4

or3

x = − (M1)

e.g. (2)f ′


e.g. 23(2) 2

2p− + ,

23 4 4

2 3 3p− − − +

correct simplification

e.g. 6 2 0p− + = , 8 4

03 3

p+ + = , 4 0p+ = A1

4p = − AG N0

[4 marks]

(c) evidence of integration (M1)

3 21 1( ) 4

2 2f x x x x c= − − + A1A1A1

substituting (2, 4) or 4 358

,3 27

− into their expression (M1)

correct equation A1

e.g. 3 21 12 2 4 2 4

2 2c× − × − × + = ,

1 18 4 4 2 4

2 2c× − × − × + = , 4 2 8 4c− − + =

3 21 1( ) 4 10

2 2f x x x x= − − + A1 N4

[7 marks]

Total [14 marks]


QUESTION 9

(a) (i) 7

24 A1 N1

(ii) evidence of multiplying along the branches (M1)

e.g.2 5 1 7

,3 8 3 8

× ×

adding probabilities of two mutually exclusive paths (M1)

e.g.1 7 2 3 1 1 2 5

,3 8 3 8 3 8 3 8

× + × × + ×

13

P( )24

F = A1 N2

[4 marks]

(b) (i) 1

3×

1

8 (A1)

1

24 A1

(ii) recognizing this is P( )E|F (M1)

e.g.7

24÷

13

24

168 7

312 13= A2 N3

[5 marks]

(c) X (cost in euros) 0 3 6

P( )X1

9

4

9

4

9

A2A1 N3 [3 marks]

(d) correct substitution into E( )X formula (M1)

e.g.1 4 4

0 3 69 9 9

× + × + × , 12 24

9 9+

E( ) 4X = (euros) A1 N2

[2 marks]

Total [14 marks]


QUESTION 10

(a) (i) sin 0x = A1

0x = , x = π A1A1 N2

(ii) sin 1x = − A1

3

2x

π= A1 N1

[5 marks]

(b) 3

2

π A1 N1

[1 mark]

(c) evidence of using anti-differentiation (M1)

e.g.3

2

0(6 6sin )dx x

π

+

correct integral 6 6cosx x− (seen anywhere) A1A1

correct substitution (A1)

e.g. 3 3

6 6cos ( 6cos0)2 2

π π− − − , 9 0 6π − +

9! 6k = + A1A1 N3

[6 marks]

(d) translation of 2

0

π

A1A1 N2

[2 marks] (e) recognizing that the area under g is the same as the shaded region in f (M1)

!

2p = , 0p = A1A1 N3

[3 marks]

Total [17 marks]


SECTION A QUESTION 1

(a) (i) 4 0

( 4 )0 4§ ·

¨ ¸© ¹

AB I A2 N2

(ii) 1

1 12 11 1 2 4, ,6 5 3 54 4

2 4

�

§ ·�¨ ¸�§ · ¨ ¸¨ ¸� ¨ ¸© ¹ �¨ ¸

© ¹

A B A1 N1

(b) METHOD 1

1xy

�§ · ¨ ¸

© ¹A C (M1)

1 12 1 8 81 2 46 5 4 3 5 44

2 4

§ ·§ ·�¨ ¸¨ ¸�§ ·§ · § ·¨ ¸ ¨ ¸¨ ¸¨ ¸ ¨ ¸� � �¨ ¸¨ ¸© ¹© ¹ © ¹�¨ ¸¨ ¸© ¹© ¹

A1

517

xy

§ · § · ¨ ¸ ¨ ¸�© ¹ © ¹

A1A1 N3

METHOD 2 5 8x y� , 6 2 4x y� � A1 for work towards solving their system (M1)

517

xy

§ · § · ¨ ¸ ¨ ¸�© ¹ © ¹

A1A1 N3

[7 marks] QUESTION 2

(a) 1P( )11

A A1 N1

(b) 2P( | )10

B A A2 N2

(c) recognising that P( ) = P( ) P( )A B A B A� u (M1) correct values (A1)

e.g. 1 2P( )11 10

A B� u

2P( )110

A B� A1 N3

[6 marks]


QUESTION 3 evidence of choosing the product rule (M1) ( ) e ( sin ) cos e (= e cos e sin )x x x xf x x x x xc u � � u � A1A1 substituting S (M1) e.g. ( e cos e sinf S Sc S� S� S , e ( 1 0)S � � , eS� taking negative reciprocal (M1)

e.g. 1(f

�c S�

gradient is 1eS A1 N3

[6 marks] QUESTION 4 (a)

Function Graph displacement A acceleration B

A2A2 N4 (b) 3t A2 N2 [6 marks] QUESTION 5 (a) in any order translated 1 unit to the right A1 N1 stretched vertically by factor 2 A1 N1 (b) METHOD 1 Finding coordinates of image on g (A1)(A1) e.g. 1 1 0, 1 2 2� � u , ( 1, 1) ( 1 1, 2 1)� o � � u , (0, 2) P is (3, 0) A1A1 N4 METHOD 2 2( ) 2( 4) 2h x x � � (A1)(A1) P is (3, 0) A1A1 N4 [6 marks]


QUESTION 6 (a) (i) interchanging x and y (seen anywhere) M1 e.g. 3e yx � correct manipulation A1 e.g. ln 3x y � , ln 3y x � 1( ) ln 3f x x� � AG N0 (ii) 0x ! A1 N1 (b) collecting like terms; using laws of logs (A1)(A1)

e.g. 1ln ln 3xx

§ ·� ¨ ¸© ¹

, ln ln 3x x� ; ln 31x

x

§ ·¨ ¸

¨ ¸¨ ¸© ¹

, 2ln 3x

simplify (A1)

e.g. 3ln2

x , 2 3ex

32ex � �3e A1 N2

[7 marks] QUESTION 7 attempt to substitute into formula 2dV y x S³ (M1) integral expression A1

e.g. � �2

0d

ax xS³ , xS³

correct integration (A1)

e.g. 21d2

x x x ³

correct substitution 212

V aª º S « »¬ ¼ (A1)

equating their expression to 32S M1

e.g. 21 322

aª ºS S« »¬ ¼

2 64a 8a A2 N2 [7 marks]


SECTION B QUESTION 8 (a) (i) 3cosx T A1 N1 (ii) 3siny T A1 N1 [2 marks] (b) finding area (M1)

e.g. 2 2A x y u , 182

A bh u

substituting A1

e.g. 4 3sin 3cosA T T u u , 18 3cos 3sin2

T Tu u u

18(2sin cos )A T T A1 18sin 2A T AG N0 [3 marks]

(c) (i) d 36cos2d

A TT A2 N2

(ii) for setting derivative equal to 0 (M1)

e.g. 36cos2 0T , d 0d

AT

22

T S (A1)

4

T S A1 N2

(iii) valid reason (seen anywhere) R1

e.g. at 2

2

d, 04 d

AT

S� ; maximum when ( ) 0f xcc �

finding second derivative 2

2

d 72sin 2d

A TT

� A1

evidence of substituting 4S M1

e.g. 72sin 2 , 72sin , 724 2S S§ · § ·� u � �¨ ¸ ¨ ¸

© ¹ © ¹

4

T S produces the maximum area AG N0

[8 marks]

Total [13 marks]


QUESTION 9 (a) (i) evidence of approach (M1)

e.g. PQ PO OQo o o

� , Q P�

1

PQ 21

o�§ ·¨ ¸ ¨ ¸¨ ¸© ¹

A1 N2

(ii) 2

PR 24

o§ ·¨ ¸ ¨ ¸¨ ¸© ¹

A1 N1

[3 marks] (b) METHOD 1

choosing correct vectors PQo

and PRo

(A1)(A1)

finding PQ PR, PQ , PRo o o o

(A1) (A1)(A1)

PQ PR 2 4 4( 6)o o

� � �

� �2 2 2PQ ( 1) 2 1 6o

� � � , � �2 2 2PR 2 2 4 24o

� �

substituting into formula for angle between two vectors M1

e.g. 6ˆcosRPQ6 24

u

simplifying to expression clearly leading to 12

A1

e.g. 6

6 2 6u, 6 6,

12144

1ˆcosRPQ2

AG N0

METHOD 2 evidence of choosing cosine rule (seen anywhere) (M1)

3

QR 03

o§ ·¨ ¸ ¨ ¸¨ ¸© ¹

A1

QR 18o

, PQ 6o

and PR 24o

(A1)(A1)(A1)

� � � � � �2 2 2

6 24 18ˆcos RPQ

2 6 24

� �

u A1

6 24 18 12ˆcosRPQ24 24

� � § · ¨ ¸© ¹

A1

1ˆcosRPQ2

AG N0

[7 marks]

continued …



(c) (i) METHOD 1 evidence of appropriate approach (M1) e.g. using 2 2ˆ ˆsin RPQ cos RPQ 1� , diagram

substituting correctly (A1)

e.g. 21ˆsin RPQ 1

2§ · � ¨ ¸© ¹

3 3ˆsin RPQ4 2

§ · ¨ ¸¨ ¸

© ¹ A1 N3

METHOD 2

since 1ˆcosP2

, P̂ 60 (A1)

evidence of approach e.g. drawing a right triangle, finding the missing side (A1)

3ˆsin P2

A1 N3


e.g. attempt to substitute into 1 sin2

ab C


e.g. area 1 36 242 2

u u A1

area 3 3 A1 N2 [6 marks]

Total [16 marks]


QUESTION 10 (a) METHOD 1 evidence of substituting x� for x (M1)

2

( )( )( ) 1

a xf xx�

� � �

A1

� �2( ) ( )1

axf x f xx�

� ��

AG N0

METHOD 2 ( )y f x � is reflection of ( )y f x in x axis and ( )y f x � is reflection of ( )y f x in y axis (M1) sketch showing these are the same A1

� �2( ) ( )1

axf x f xx�

� ��

AG N0

[2 marks] (b) evidence of appropriate approach (M1) e.g. ( ) 0f xcc to set the numerator equal to 0 (A1) e.g. 2 22 ( 3) 0; ( 3) 0ax x x� �

3 3(0, 0), 3 , , 3 ,4 4

a a§ · § ·� �¨ ¸ ¨ ¸¨ ¸ ¨ ¸

© ¹ © ¹ (accept 0, 0 .)x y etc A1A1A1A1A1 N5

[7 marks]

continued …


Question 10 continued (c) (i) correct expression A2

e.g. 7

2

3

ln ( 1)2a xª º�« »¬ ¼

, ln50 ln102 2a a

� , (ln50 ln10)2a

�

area ln52a

A1A1 N2

(ii) METHOD 1 recognizing the shift that does not change the area (M1)

e.g. 8 7

4 3( 1)d ( )df x x f x x� ³ ³ , ln5

2a

recognizing that the factor of 2 doubles the area (M1)

e.g. � �8 8 7

4 4 32 ( 1)d 2 ( 1)d 2 ( )df x x f x x f x x� � ³ ³ ³

8

42 ( 1)d ln5f x x a� ³ � �. . 2 answer to (c)(i)i e u their A1 N3

METHOD 2 changing variable

let 1w x � , so d 1dwx

222 ( )d ln ( 1)2af w w w c � �³ (M1)

substituting correct limits e.g.

82

4ln[( 1) 1]a xª º� �¬ ¼ ,

72

3ln ( 1)a wª º�¬ ¼ , ln50 ln10a a� (M1)

8

42 ( 1)d ln5f x x a� ³ A1 N3

[7 marks]

Total [16 marks]

3. (m1) 1. a a1 n2 (a1)(a1) a1 n4ibrelics.weebly.com/uploads/2/5/3/9/25397032/paper_1_ms.pdf · 1...

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