





Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
The mark scheme for questions 1 to 9 of the GCE Mathematics June 2009 exam. It includes the question number, the correct answer scheme, and the marks awarded for each question. provided by PhysicsAndMathsTutor.com.
Typology: Schemes and Mind Maps
1 / 9
This page cannot be seen from the preview
Don't miss anything!
Question
Number
Scheme Marks
Q ( ) ( )
( )
f 4 4
x x x
−
√
M
( ) ( )
(^12) ... 4 1 ...
− = + (^) ( )
B
( )
( )( ) ( )( )( )
1 3 2 1 3 5 3 1 2 2 2 2 2 ... 1 2 ... 4 2 4 3! 4
(^) x (^) − − (^) x (^) − − − x = (^) + − (^) + (^) + (^) + (^)
M1 A1ft
ft their 4
^ x
(^1 1 3 2 ) , ... 2 16 256 2048
= − x + x − x + A1, A1 (6)
[6]
Alternative
( ) ( )
( )
f 4 4
x x x
−
√
M
( )
(^12 32) ( )( ) 52 ( )( )( ) (^72) 1 3 1 3 5 1 2 2 2 2 2 2 3 4 2 4 4 4 ... 1.2 1.2.
x x x
= − x + x − x + A1, A1^ (6)
Question
Number
Scheme Marks
Q3 (a) (^ )^ ( )( )( )
f 2 1 1 3 2 1 1 3
x A B C x x x x x x x
4 − 2 x = A x ( + (^1) ) (^) ( x + (^3) ) + B (^) ( 2 x + (^1) ) (^) ( x + (^3) ) + C (^) ( 2 x + (^1) ) (^) ( x + (^1) ) M
A method for evaluating one constant M
1 x → − 2 , (^) ( )( ) 1 5 5 = A (^) 2 2 ⇒ A = 4 any one correct constant A
x → − 1 , 6 = B (^) ( − (^1) ) (^) ( 2 ) ⇒ B = − 3
x → − 3 , 10 = C (^) ( − (^5) )( − (^2) )⇒ C = 1 all three constants correct A1 (4)
(b) (i)
d 2 1 1 3
x x x x
( ) ( ) ( )
ln 2 1 3ln 1 ln 3 2
= x + − x + + x + + C A1 two ln terms correct M1 A1ft
All three ln terms correct and “+ C ” ; ft constants A1ft^ (3)
(ii) ( ) ( ) ( )
2 0
2 ln 2 x + 1 − 3ln x + 1 + ln x + 3
= (^) ( 2 ln 5 − 3ln 3 + ln 5) − (^) ( 2 ln1 − 3ln1 + ln 3) M
= 3ln 5 −4 ln 3 3
4
ln 3
M
ln 81
A1 (3)
[10]
Question
Number
Scheme Marks
Q4 (a)
2 d^2 d e 2 e 2 2 d d
x y^ x y y y x x
− − − = + A1 correct RHS (^) M1 A
( )
d 2 2 d 2 e e 2 e d d
x x y x y y x x
− − − = − (^) B
( )
2 d 2 e 2 2 2 e d
x y x y y x
− − − = + (^) M
2
2
d 2 2 e
d e 2
x
x
y y
x y
−
−
A1 (5)
(b) At P ,
0
0
d 2 2 e 4 d e 2
y
x
M
Using (^) mm ′ (^) = − 1
1
4
m ′ = M
( )
y − = x − M
x − 4 y + 4 = 0 or any integer multiple A1 (4)
[9]
Alternative for (a) differentiating implicitly with respect to y.
2 2 d^ d e 2 e 2 2 d d
x x x^ x y y y y
− − − = + A1 correct RHS (^) M1 A
( )
d 2 2 2 d e e 2 e d d
x x x x y y y y
− − − = − (^) B
( )
2 d 2 2 2 e e 2 d
x x x y y y
− −
2
2
d e 2
d 2 2 e
x
x
x y
y y
−
−
2
2
d 2 2 e
d e 2
x
x
y y
x y
−
−
A1 (5)
Question
Number
Scheme Marks
Q6 (a) (^) ( ) ( )
( ) ( )
(^32) 1 2 3 2
5 d 5 d
x x x x x C
∫ ∫
M1 A1 (2)
( )
x C
(b) (^) (i) (^) ( ) ( ) ( )( ) ( )
3 3 1 5 d 2 1 5 2 2 5 2 d 3 3
x − (^) √ − x x = − x − − x + ⌠ − x x ∫ (^) ⌡
M1 A1ft
( ) ( )
(^52)
5 2
x C
M
( )( ) ( ) ( )
= − x − − x − − x + C A1^ (4)
(ii) ( )( ) ( )
(^32 )
5
1
x x x
( )
awrt 8.53 (^) M1 A1 (2)
[8]
Alternatives for ( b) and (c)
(b) 2
d 5 2 1 d
u u x u x
d 2 d
x u u
( ) ( ) (^) ( ) ( ) ( )
2 d 2 1 5 d 4 d 4 2 d d
x x x x u u u u u u u u
∫ −^ √ −^ =^ −^ =^ ∫ −^ − ∫
M1 A
( ) (^ )
2 8 d 5 3
= (^) ∫ u − u u = u − u + C M
( ) ( ) ( )
= − x − − x + C A
(c) x = 1 ⇒ u = 2 , x = 5 ⇒ u = 0
( )
0 5 3
2
u u
M
awrt 8.53 A1 (2)
Question
Number
Scheme Marks
Q7 (a)
or
M
r or
r accept equivalents M1 A1ft (3)
(b)
or
( (^ )) (^ )^ (^ )
2 2 2 CB = (^) √ 1 + 5 + − 10 = (^) √ 126 = (^3) √ 14 ≈11.2 awrt 11.2 M1 A1 (2)
(c) CB AB. = CB AB cos θ
( ±^ )( 2 +^5 +^20 )=^ √^126 √9 cos^ θ M1 A
3 cos 36. 14
√
= ⇒ ≈ ° awrt (^) 36.7° A1 (3)
(d)
sin 126
d
√
= M1 A1ft
d = 3 √ (^5) ( ≈ 6.7) awrt 6.7 A1 (3)
(e) (^) BX^2 = BC^2 − d^2 = 126 − 45 = 81 M
( )
CBX BX d
√ ! = × × = × × √ = ≈ awrt 30.1 or 30.2^ M1 A1^ (3)
[14]
Alternative for (e)
1 sin 2
! CBX = × d × BC ∠ XCB M
( )
3 5 126sin 90 36. 2
= × √ × √ − ° sine of correct angle M
√ , awrt 30.1 or 30.2 A1 (3)
θ
l
X
B
C
d √^126