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GCE Mathematics June 2009 Exam Mark Scheme for Questions 1 to 9, Schemes and Mind Maps of Physics

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

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Mark Scheme (Results)
Summer 2009
GCE
GCE Mathematics (6666/01)
PhysicsAndMathsTutor.com
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Download GCE Mathematics June 2009 Exam Mark Scheme for Questions 1 to 9 and more Schemes and Mind Maps Physics in PDF only on Docsity!

Mark Scheme (Results)

Summer 2009

GCE

GCE Mathematics (6666/01)

June 2009

6666 Core Mathematics C

Mark Scheme

Question

Number

Scheme Marks

Q ( ) ( )

( )

f 4 4

x x x

M

( ) ( )

(^12) ... 4 1 ...

− = + (^) ( )

  • or ( )

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 + xx + 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

− − −^ −^ − −^ −^ − −

= + − + + + B1 M1 A

= − x + xx + 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

− −

  • = − (^) M

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

= (^) ∫ uu u = uu + 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)

AB OB OA

uuur uuur uuur

or

BA

uuur

M

r or

r accept equivalents M1 A1ft (3)

(b)

CB OB OC

 −^     − 

uuur uuur uuur

or

BC

uuur

( (^ )) (^ )^ (^ )

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

uuur uuur uuur uuur

( ±^ )( 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 × BCXCB 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