C4 MS Jan 2010

Mark Scheme (Results) January 2010

GCE

GCE Core Mathematics C4 (6666/01)

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January 2010 Publications Code UA022713 All the material in this publication is copyright Š Edexcel Ltd 2010

January 2010 6666 Core Mathematics C4 Mark Scheme

Question Number

Q1

Scheme

(a) (1 − 8 x ) 2 = 1 + ( 12 ) ( −8 x ) + 1

( 12 )( − 12 ) 2

Marks

( −8 x ) + 2

( 12 )( − 12 ) ( − 23 ) 3!

( −8 x )

3

+…

= 1 − 4 x − 8 x 2 ; − 32 x 3 − …

(b)

(1 − 8 x ) = =

A1; A1

8 ⎞ ⎛ ⎜1 − ⎟ ⎝ 100 ⎠ 92 23 23 = = 100 25 5

(4)

M1

¿

cso

(c) 1 − 4 x − 8 x 2 − 32 x3 = 1 − 4 ( 0.01) − 8 ( 0.01) − 32 ( 0.01) = 1 − 0.04 − 0.0008 − 0.000 032 = 0.959 168 2

23 = 5 × 0.959 168 = 4.795 84

GCE Core Mathematics C4 (6666) January 2010

M1 A1

A1

(2)

3

M1 M1

cao

3

A1

(3) [9]

Question Number Q2

Scheme

(a)

1.386, 2.291

Marks

awrt 1.386, 2.291

1 (b) A ≈ × 0.5 ( ... ) 2 = ... ( 0 + 2 ( 0.608 + 1.386 + 2.291 + 3.296 + 4.385 ) + 5.545 )

B1 B1 B1 M1

= 0.25 ( 0 + 2 ( 0.608 + 1.386 + 2.291 + 3.296 + 4.385 ) + 5.545 ) ft their (a) A1ft = 0.25 × 29.477 ... ≈ 7.37 cao A1

(c)(i)

x2 x2 1 ln x − ⌠ ⎮ × dx 2 ⌡ 2 x 2 x x = ln x − ⌠ ⎮ dx ⌡2 2 2 x x2 = ln x − ( +C ) 2 4

∫ x ln x dx =

(2)

(4)

M1 A1

M1 A1

4

⎡ x2 x2 ⎤ ⎛ 1⎞ (ii) ⎢ ln x − ⎥ = ( 8ln 4 − 4 ) − ⎜ − ⎟ 4 ⎦1 ⎝ 4⎠ ⎣2 15 = 8ln 4 − 4 15 = 8 ( 2 ln 2 ) − 4 1 = ( 64 ln 2 − 15 ) 4

GCE Core Mathematics C4 (6666) January 2010

4

M1

ln 4 = 2 ln 2 seen or implied M1 a = 64, b = −15

A1

(7) [13]

Question Number

Q3

Scheme

(a) −2sin 2 x − 3sin 3 y

(b) At x =

π 6

,

dy =0 dx dy 2sin 2 x =− dx 3sin 3 y ⎛ 2π cos ⎜ ⎝ 6

GCE Core Mathematics C4 (6666) January 2010

Accept

2sin 2 x −2sin 2 x , −3sin 3 y 3sin 3 y

π

3

⇒y=

π 9

2sin 2 ( π6 ) 2sin π3 dy 2 =− = − =− π π dx 3sin 3 ( 9 ) 3sin 3 3 2⎛ π⎞ =− ⎜x− ⎟ 9 3⎝ 6⎠ 6 x + 9 y − 2π = 0

y− Leading to

M1 A1

⎞ ⎟ + cos 3 y = 1 ⎠ 1 cos 3 y = 2

3y = ⎛π π ⎞ (c) At ⎜ , ⎟ , ⎝6 9⎠

Marks

π

5

A1

(3)

M1 A1

awrt 0.349 A1

(3)

M1 M1 A1

(3) [9]

Question Number Q4

Scheme

Marks

(a) A: ( −6, 4, − 1)

(b)

Accept vector forms B1

⎛ 4⎞ ⎛ 3⎞ 2 2 ⎜ ⎟⎜ ⎟ 2 2 2 2 ⎜ −1 ⎟ . ⎜ −4 ⎟ = 12 + 4 + 3 = 4 + ( −1) + 3 3 + ( −4 ) + 1 cos θ ⎜ 3⎟ ⎜ 1⎟ ⎝ ⎠⎝ ⎠ 19 awrt 0.73 cos θ = 26

(c) X: (10, 0, 11) ⎛ 10 ⎞ ⎛ −6 ⎞ (d) AX = ⎜⎜ 0 ⎟⎟ − ⎜⎜ 4 ⎟⎟ ⎜ 11 ⎟ ⎜ −1 ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ 16 ⎞ ⎜ ⎟ = ⎜ −4 ⎟ ⎜ 12 ⎟ ⎝ ⎠

(e)

(3)

Accept vector forms

B1

(1)

Either order

M1

cao

A1

2

Do not penalise if consistent incorrect signs in (d)

A1

Use of correct right angled triangle AX = cos θ d 4 26 d = 19 ≈ 27.9 awrt 27.9

M1

(2)

l1

X 4 26

θ d

(2)

M1

= 416 = 16 × 26 = 4 26 ¿

A

M1 A1

A1

AX = 162 + ( −4 ) + 122

(f)

(1)

Y

l2

M1 A1

(3)

26

[12]

GCE Core Mathematics C4 (6666) January 2010

6

Question Number

Q5

Scheme

(a)

Marks

⌠ 9 x + 6 dx = ⌠ ⎛ 9 + 6 ⎞ dx ⎮ ⎮⎜ ⎟ ⌡ x x⎠ ⌡⎝ = 9 x + 6 ln x ( +C )

⌠ 1 ⌠ 9 x + 6 dx ⎮ 13 d y = ⎮ ⌡ x ⌡y −1 ⌠ 9x + 6 ∫ y 3 d y = ⎮⌡ x dx

(b)

M1 A1

(2)

Integral signs not necessary B1

2

y3

= 9 x + 6 ln x ( +C )

± ky 3 = their ( a )

3 23 y = 9 x + 6 ln x ( +C ) 2

ft their ( a )

2 3

y = 8, x =1

3 23 8 = 9 + 61n 1 + C 2 C = −3 2 2 y 3 = ( 9 x + 6 ln x − 3) 3

y 2 = ( 6 x + 4 ln x − 2 )

3

2

M1 A1ft

M1 A1

( = 8 (3x + 2 ln x − 1) ) 3

A1

(6) [8]

GCE Core Mathematics C4 (6666) January 2010

7

Question Number

Q6

Scheme

dA = 1.5 dt dA A = π r2 ⇒ = 2π r dr When A = 2 2 2 = π r2 ⇒ r =

π

Marks

B1 B1

( = 0.797 884 ... )

M1

d A d A dr = × dt dr dt dr 1.5 = 2π r dt

M1

dr 1.5 = ≈ 0.299 dt 2π π2

awrt 0.299

A1 [5]

GCE Core Mathematics C4 (6666) January 2010

8

Question Number Q7

Scheme

(a)

Marks

y = 0 ⇒ t ( 9 − t 2 ) = t ( 3 − t )( 3 + t ) = 0 t = 0, 3, − 3

Any one correct value B1

At t = 0 , x = 5 ( 0 ) − 4 = −4 2

Method for finding one value of x

M1

At A, x = −4 ; at B, x = 41

Both

A1

dx = 10t dt

Seen or implied

B1

At t = 3 , x = 5 ( 3) − 4 = 41 2

( At t = −3,

)

x = 5 ( −3) − 4 = 41 2

(b)

⌠ ⌠ dx 2 ⎮ y dx = ⎮ y dt = ∫ t ( 9 − t )10t dt d t ⌡ ⌡

(3)

M1 A1

= ∫ ( 90t 2 − 10t 4 ) dt =

90t 3 10t 5 − 3 5

( = 30t

( +C )

3

⎡ 90t 3 10t 5 ⎤ 3 5 ⎢ 3 − 5 ⎥ = 30 × 3 − 2 × 3 ⎣ ⎦0

A = 2∫ y dx = 648

GCE Core Mathematics C4 (6666) January 2010

( units ) 2

3

− 2t 5 ( +C ) )

( = 324 )

A1 M1

A1

(6) [9]

9

Question Number

Q8

Scheme

(a)

Marks

dx = −2sin u du

B1

⌠ 1 1 ⌠ dx = ⎮ × −2sin u du ⎮ 2 2 2 2 ⌡ x 4− x ⌡ ( 2 cos u ) 4 − ( 2 cos u ) −2sin u ⌠ du =⎮ ⌡ 4 cos 2 u 4sin 2 u 1 1 =− ⌠ du ⎮ 4 ⌡ cos 2 u 1 = − tan u ( +C ) 4

x = 2 ⇒ 2 = 2 cos u ⇒ u = x = 1 ⇒ 1 = 2 cos u ⇒ u =

π

M1

Use of 1 − cos 2 u = sin 2 u

M1

1 ±k ⌠ du ⎮ ⌡ cos 2 u

M1

± k tan u

M1

π 4 M1

3

π

1⎛ π π⎞ ⎡ 1 ⎤4 − tan u ⎢⎣ 4 ⎥⎦ π = − 4 ⎝⎜ tan 4 − tan 3 ⎠⎟ 3

=−

(b)

⌠ V =π⎮ ⎮ ⌡1

2

(

1 1− 3 4

)

⎛ 4 ⎜ 1 ⎜ x ( 4 − x2 ) 4 ⎝

⎛ 3 −1 ⎞ ⎜⎜ = ⎟ 4 ⎟⎠ ⎝

A1

(7)

2

⎞ ⎟ dx ⎟ ⎠

M1

2

1 ⌠ dx = 16π ⎮ ⌡1 x 2 4 − x 2 ⎛ 3 −1 ⎞ = 16π ⎜⎜ ⎟⎟ ⎝ 4 ⎠

16π × integral in (a) M1 16π × their answer to part (a)

A1ft

(3) [10]

GCE Core Mathematics C4 (6666) January 2010

10

GCE Core Mathematics C4 (6666) January 2010

11

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