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
3π
⎡ ⎛ 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
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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
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