3. (M1) 1. a A1 N2 (A1)(A1) A1 N4ibrelics.weebly.com/uploads/2/5/3/9/25397032/paper_1_ms.pdf · 1 eS A1 N3 [6 marks] e.g. x QUESTION 4 (a) Function Graph displacement A acceleration
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– 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+ = − + =
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]
– 9 – M13/5/MATME/SP1/ENG/TZ1/XX/M
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]
– 10 – M13/5/MATME/SP1/ENG/TZ1/XX/M
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]
– 11 – M13/5/MATME/SP1/ENG/TZ1/XX/M
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]
– 12 – M13/5/MATME/SP1/ENG/TZ1/XX/M
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]
– 13 – M13/5/MATME/SP1/ENG/TZ1/XX/M
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)
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]
– 15 – M13/5/MATME/SP1/ENG/TZ1/XX/M
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 …
– 16 – M13/5/MATME/SP1/ENG/TZ1/XX/M
Question 9 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]
– 17 – M13/5/MATME/SP1/ENG/TZ1/XX/M
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 …
– 18 – M13/5/MATME/SP1/ENG/TZ1/XX/M
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]
(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]
- 10 - M09/5/MATME/SP1/ENG/TZ1/XX/M+
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]
- 11 - M09/5/MATME/SP1/ENG/TZ1/XX/M+
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)
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]
- 13 - M09/5/MATME/SP1/ENG/TZ2/XX/M+
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
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 …
- 17 - M09/5/MATME/SP1/ENG/TZ2/XX/M+
Question 10 continued (c) (i) evidence of approach M1
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 …
- 19 - M09/5/MATME/SP1/ENG/TZ2/XX/M+
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]
– 10 – M10/5/MATME/SP1/ENG/TZ1/XX/M+
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
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]
– 12 – M10/5/MATME/SP1/ENG/TZ1/XX/M+
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]
– 13 – M10/5/MATME/SP1/ENG/TZ1/XX/M+
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)
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]
– 15 – M10/5/MATME/SP1/ENG/TZ1/XX/M+
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 …
– 16 – M10/5/MATME/SP1/ENG/TZ1/XX/M+
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]
– 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]
– 10 – M10/5/MATME/SP1/ENG/TZ1/XX/M+
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
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]
– 12 – M10/5/MATME/SP1/ENG/TZ1/XX/M+
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]
– 13 – M10/5/MATME/SP1/ENG/TZ1/XX/M+
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)
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]
– 15 – M10/5/MATME/SP1/ENG/TZ1/XX/M+
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 …
– 16 – M10/5/MATME/SP1/ENG/TZ1/XX/M+
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]
– 9 – M10/5/MATME/SP1/ENG/TZ2/XX/M+
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
(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]
- 10 - M09/5/MATME/SP1/ENG/TZ1/XX/M+
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]
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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)