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June 2009 6666 Core Mathematics C4 Mark Scheme Question Number Q1

Scheme

f ( x) =

Marks

1 −1 = (4 + x) 2 √ (4 + x)

= ( 4)

− 12

(1 +

...

)

M1

1 (1 + ... 2

...

)

...

or

1 2 √ (1 + ...)

2 3 ⎛ ⎞ − 12 ) ( − 32 ) ⎛ x ⎞ ( − 12 ) ( − 32 )( − 52 ) ⎛ x ⎞ ( 1 ⎛ x⎞ ... = ... ⎜1 + ( − 2 ) ⎜ ⎟ + + + ⎟⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 2 3! ⎝4⎠ ⎝4⎠ ⎝ 4⎠ ⎝ ⎠ ⎛ x⎞ ft their ⎜ ⎟ ⎝4⎠ 1 1 3 2 5 3 = − x, + x − x + ... 2 16 256 2048

B1 M1 A1ft

A1, A1

(6) [6]

Alternative f ( x) =

1 −1 = (4 + x) 2 √ (4 + x)

= 4 2 + ( − 12 ) 4 2 x + −1

−3

( − 12 ) ( − 32 ) 4−

M1 5 2

x2 +

1.2 1 1 3 2 5 3 = − x, + x − x + ... 2 16 256 2048

8371_9374 GCE Mathematics January 2009

( − 12 ) ( − 32 )( − 52 ) 4− 1.2.3

7 2

x3 + ...

B1 M1 A1 A1, A1

39

(6)


Question Number Q2

(a) (b)

Scheme

1.14805

awrt 1.14805

1 3π A ≈ × ( ... ) 2 8 = ... ( 3 + 2 ( 2.77164 + 2.12132 + 1.14805 ) + 0 ) 3π ( 3 + 2 ( 2.77164 + 2.12132 + 1.14805) ) 16 3π = × 15.08202 ... = 8.884 16

=

(c)

Marks B1

(1)

B1

0 can be implied ft their (a) cao

⎛ x⎞ 3sin ⎜ ⎟ ⌠ ⎛ x⎞ ⎝3⎠ ⎮ 3cos ⎜ ⎟ dx = 1 ⌡ ⎝3⎠ 3 ⎛x⎞ = 9sin ⎜ ⎟ ⎝3⎠

M1 A1ft A1

(4)

M1 A1

⎡ ⎛ x ⎞⎤ 2 A = ⎢9sin ⎜ ⎟ ⎥ = 9 − 0 = 9 ⎝ 3 ⎠⎦ 0 ⎣

cao

A1

(3)

[8]

8371_9374 GCE Mathematics January 2009

40


Question Number

Q3

(a)

Scheme

f ( x) =

Marks

4 − 2x A B C = + + ( 2 x + 1)( x + 1)( x + 3) 2 x + 1 x + 1 x + 3

4 − 2 x = A ( x + 1)( x + 3) + B ( 2 x + 1)( x + 3) + C ( 2 x + 1)( x + 1)

M1

A method for evaluating one constant

M1

any one correct constant

A1

all three constants correct

A1

x → − 12 , 5 = A ( 12 ) ( 52 ) ⇒ A = 4 x → −1 , 6 = B ( −1)( 2 ) ⇒ B = −3

x → −3 , 10 = C ( −5 )( −2 ) ⇒ C = 1 (b)

3 1 ⎞ ⌠⎛ 4 − + (i) ⎮ ⎜ ⎟ dx ⌡ ⎝ 2x +1 x +1 x + 3 ⎠ 4 = ln ( 2 x + 1) − 3ln ( x + 1) + ln ( x + 3) + C A1 two ln terms correct 2 All three ln terms correct and “+C” ; ft constants (ii)

(4)

M1 A1ft A1ft

(3)

⎡⎣ 2 ln ( 2 x + 1) − 3ln ( x + 1) + ln ( x + 3) ⎤⎦ 0

2

= ( 2 ln 5 − 3ln 3 + ln 5 ) − ( 2 ln1 − 3ln1 + ln 3) = 3ln 5 − 4 ln 3 ⎛ 53 ⎞ = ln ⎜ 4 ⎟ ⎝3 ⎠

M1

M1

⎛ 125 ⎞ = ln ⎜ ⎟ ⎝ 81 ⎠

A1

(3) [10]

8371_9374 GCE Mathematics January 2009

41


Question Number

Q4

Scheme

dy dy − 2 y e−2 x = 2 + 2 y dx dx d dy y e −2 x ) = e −2 x − 2 y e−2 x ( dx dx ( e−2 x − 2 y ) ddxy = 2 + 2 y e−2 x

e −2 x

(a)

(b)

At P ,

Marks

A1 correct RHS

M1 A1 B1 M1

d y 2 + 2 y e −2 x = −2 x dx e − 2y

A1

d y 2 + 2 e0 = 0 = −4 dx e −2

M1

(5)

Using mm′ = −1 1 4 1 y −1 = ( x − 0) 4 x − 4y + 4 = 0

m′ =

M1 M1

or any integer multiple

A1

(4) [9]

Alternative for (a) differentiating implicitly with respect to y. dx dx e −2 x − 2 y e −2 x = 2 + 2y A1 correct RHS dy dy d dx y e −2 x ) = e−2 x − 2 y e−2 x ( dy dy ( 2 + 2 y e−2 x ) ddxy = e−2 x − 2 y

M1 A1 B1 M1

dx e −2 x − 2 y = d y 2 + 2 y e −2 x d y 2 + 2 y e −2 x = −2 x dx e − 2y

8371_9374 GCE Mathematics January 2009

42

A1

(5)


Question Number

Q5

Scheme

Marks

dx dy = −4sin 2t , = 6 cos t dt dt

(a)

At t = (b)

π 3

,

Use of

dy 6 cos t ⎛ 3 ⎞ =− ⎜= − ⎟ dx 4sin 2t ⎝ 4sin t ⎠ 3 √3 m=− =− accept equivalents, awrt −0.87 √3 4× 2 2

cos 2t = 1 − 2sin 2 t x y cos 2t = , sin t = 2 6 2 x ⎛ y⎞ = 1− 2⎜ ⎟ 2 ⎝6⎠ y = √ (18 − 9 x )

Leading to

(c)

B1, B1 M1 A1

(4)

M1

M1

( = 3 √ ( 2 − x ))

cao

A1

−2 ≤ x ≤ 2

k =2

B1

0 ≤ f ( x) ≤ 6

either 0 ≤ f ( x ) or f ( x ) ≤ 6

B1

Fully correct. Accept 0 ≤ y ≤ 6 , [ 0, 6]

B1

(4)

(2) [10]

Alternatives to (a) where the parameter is eliminated 1

y = (18 − 9 x ) 2 1

1 dy 1 − = (18 − 9 x ) 2 × ( −9 ) dx 2 π 2π At t = , x = cos = −1 3 3 dy 1 1 √3 = × × −9 = − dx 2 √ ( 27 ) 2

2

y 2 = 18 − 9 x dy 2y = −9 dx At t =

π 3

, y = 6sin

B1 M1 A1

(4)

B1

π

3 dy 9 √3 =− =− dx 2× 3√ 3 2

8371_9374 GCE Mathematics January 2009

B1

= 3√ 3

B1 M1 A1

43

(4)


Question Number

Q6

Scheme

∫ √ ( 5 − x ) dx = ∫ ( 5 − x )

(a)

1 2

(5 − x ) dx =

3 2

− 32

Marks

( +C )

M1 A1

3 2⌠ + ⎮ 5 − x ) 2 dx ( 3⌡

M1 A1ft

(2)

3 2 ⎛ ⎞ 2 = − − x +C⎟ 5 ( ) ⎜ 3 ⎝ ⎠

(i)

(b)

2 3

∫ ( x − 1) √ ( 5 − x ) dx = − ( x − 1)( 5 − x )

3 2

2 (5 − x )2 + × − 52 3 5 4 − (5 − x ) 2 15 5

=

=−

3 2 ( x − 1)( 5 − x ) 2 3

( +C )

M1

( +C )

A1

(4)

M1 A1

(2)

5

3 5 5 ⎞ 4 4 ⎛ ⎡ 2 ⎤ (ii) ⎢ − ( x − 1)( 5 − x ) 2 − ( 5 − x ) 2 ⎥ = ( 0 − 0 ) − ⎜ 0 − × 4 2 ⎟ 15 ⎝ 15 ⎠ ⎣ 3 ⎦1 128 ⎛ 8 ⎞ = awrt 8.53 ⎜ = 8 ≈ 8.53 ⎟ 15 ⎝ 15 ⎠

[8]

Alternatives for ( b) and (c) ⎛ ⎞ du dx = −1 ⎜ ⇒ (b) u 2 = 5 − x ⇒ 2u = −2u ⎟ dx du ⎝ ⎠ dx 2 2 ∫ ( x − 1) √ ( 5 − x ) dx = ( 4 − u ) u du du = ∫ ( 4 − u ) u ( −2u ) du 2 8 = ∫ ( 2u 4 − 8u 2 ) du = u 5 − u 3 ( +C ) 5 3 5 3 2 8 = ( 5 − x ) 2 − ( 5 − x ) 2 ( +C ) 5 3

(c)

M1 A1 M1 A1

x =1 ⇒ u = 2, x = 5 ⇒ u = 0 0

⎡2 5 8 3⎤ ⎛ 64 64 ⎞ ⎢⎣ 5 u − 3 u ⎥⎦ = ( 0 − 0 ) − ⎝⎜ 5 − 3 ⎠⎟ 2 128 ⎛ 8 ⎞ = ⎜ = 8 ≈ 8.53 ⎟ 15 ⎝ 15 ⎠

8371_9374 GCE Mathematics January 2009

44

M1

awrt 8.53

A1

(2)


Question Number

Q7

(a)

Scheme

⎛ −2 ⎞ uuur ⎜ ⎟ or BA = ⎜ −1 ⎟ ⎜ 2⎟ ⎝ ⎠

⎛ 10 ⎞ ⎛ 8 ⎞ ⎛ 2 ⎞ uuur uuur uuur ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ AB = OB − OA = ⎜ 14 ⎟ − ⎜ 13 ⎟ = ⎜ 1 ⎟ ⎜ −4 ⎟ ⎜ −2 ⎟ ⎜ −2 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛8⎞ ⎛ 2⎞ ⎛ 10 ⎞ ⎛ 2⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ r = ⎜ 13 ⎟ + λ ⎜ 1 ⎟ or r = ⎜ 14 ⎟ + λ ⎜ 1 ⎟ ⎜ −2 ⎟ ⎜ −2 ⎟ ⎜ −4 ⎟ ⎜ −2 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

(b)

Marks

accept equivalents

(

2

) = √ (126)

M1 A1ft (3)

⎛ −1 ⎞ uuur ⎜ ⎟ or BC = ⎜ −5 ⎟ ⎜ 10 ⎟ ⎝ ⎠

⎛ 10 ⎞ ⎛ 9 ⎞ ⎛ 1 ⎞ uuur uuur uuur ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ CB = OB − OC = ⎜ 14 ⎟ − ⎜ 9 ⎟ = ⎜ 5 ⎟ ⎜ −4 ⎟ ⎜ 6 ⎟ ⎜ −10 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ CB = √ 12 + 52 + ( −10 )

M1

( = 3 √ 14 ≈ 11.2 )

awrt 11.2

M1 A1

(2)

uuur uuur uuur uuur CB. AB = CB AB cos θ

(c)

( ± )( 2 + 5 + 20 ) = √ 126 √ 9 cos θ cos θ =

M1 A1

3 ⇒ θ ≈ 36.7° √ 14

awrt 36.7°

A1

(3)

B

(d) θ

X

l

√ 126

d = sin θ √ 126 d = 3 √ 5 ( ≈ 6.7 )

M1 A1ft

awrt 6.7

A1

BX = BC − d 2 = 126 − 45 = 81 1 1 27 √ 5 ! CBX = × BX × d = × 9 × 3 √ 5 = ( ≈ 30.2 ) awrt 30.1 or 30.2 2 2 2

M1

d

(3)

C

(e)

2

2

M1 A1

(3) [14]

Alternative for (e) 1 ! CBX = × d × BC sin ∠XCB 2 1 = × 3 √ 5 × √ 126sin ( 90 − 36.7 ) ° 2

M1

sine of correct angle 27 √ 5 , awrt 30.1 or 30.2 2

≈ 30.2

8371_9374 GCE Mathematics January 2009

45

M1 A1

(3)


Question Number Q8

(a)

(b)

Scheme

Marks

1 1 1 ∫ sin θ dθ = 2 ∫ (1 − cos 2θ ) dθ = 2 θ − 4 sin 2θ ( +C ) 2

M1 A1

dx = sec 2 θ dθ dx 2 π ∫ y 2 dx = π y 2 dθ = π ∫ ( 2sin 2θ ) sec 2θ dθ dθ

(2)

x = tan θ ⇒

M1 A1

( 2 × 2sin θ cos θ ) dθ =π 2 2

M1

cos θ 2 = 16π ∫ sin θ dθ

x = 0 ⇒ tan θ = 0 ⇒ θ = 0 , x = ⎛ ⎜ V = 16π ⎜ ⎝

π 6 0

k = 16π

1 √3

⇒ tan θ =

1 √3

⇒ θ=

π 6

A1 B1

(5)

⎞ sin 2 θ dθ ⎟ ⎟ ⎠ π

(c)

sin 2θ ⎤ 6 ⎡1 V = 16π ⎢ θ − 4 ⎥⎦ 0 ⎣2

M1

⎡⎛ π 1 π⎞ ⎤ = 16π ⎢⎜ − sin ⎟ − ( 0 − 0 ) ⎥ 3⎠ ⎣⎝ 12 4 ⎦ ⎛ π √3⎞ 4 = 16π ⎜ − ⎟ = π 2 − 2π √ 3 ⎝ 12 8 ⎠ 3

Use of correct limits p=

4 , q = −2 3

M1 A1

(3) [10]

8371_9374 GCE Mathematics January 2009

46


C4 MS Jun 2009