C4 J Ms

FOR EDEXCEL

GCE Examinations Advanced Subsidiary

Core Mathematics C4 Paper J

MARKING GUIDE

This guide is intended to be as helpful as possible to teachers by providing concise solutions and indicating how marks could be awarded. There are obviously alternative methods that would also gain full marks. Method marks (M) are awarded for knowing and using a method. Accuracy marks (A) can only be awarded when a correct method has been used. (B) marks are independent of method marks.

Written by Shaun Armstrong

 Solomon Press These sheets may be copied for use solely by the purchaser’s institute.

C4 Paper J – Marking Guide 1.

x(x − 2) = 0, x = 0, 2 ∴ crosses x-axis at (0, 0) and (2, 0) volume = π ∫ = π∫

2 0 2

0

(x2 − 2x)2 dx

M1

(x4 − 4x3 + 4x2) dx

A1

= π[ 15 x5 − x4 +

4 3

= π{( 32 − 16 + 5

2.

x3] 02

32 3

) − (0)} =

∫

1 u

∫

× (2u − 2) du =

16 15

π

dx = −2(1 − u) = 2u − 2 du

1

u = 1 − x 2 ⇒ x = (1 − u)2,

I =

M1 A1

(2 −

2 u

) du

= 2u − 2 lnu + c 1 2

= 2(1 − x ) − 2 ln1 − x  + c

(a)

dy =0 dx

4 cos 2x − sec2 y

4.

(a)

(b)

5.

(a)

grad = 4 × ∴ y−

π 3

1 2

×

=

1 2

y−

π 3

=

1 2

y=

1 2

x+

A1

1 4

(x − x−

(6)

M1 A2 M1 A1

=

1 2

B1

π 6

)

M1

π 12

π 4

A1

dx dy −1 = 12 at 2 , = a(1 − 2t) dt dt dy a(1 − 2t ) = = 2 t (1 − 2t) 1 1 at − 2 dx 2

(8)

M1

M1 A1

y = 0 ⇒ t = 0 (at O) or 1 (at A) t = 1, x = a, y = 0, grad = −2 ∴ y − 0 = −2(x − a) at B, x = 0 ∴ y = 2a area = 12 × a × 2a = a2

B1 M1 A1 M1 M1 A1

(9)

dy = k y dx

∫

y

− 12

dy =

∫

k dx

M1

1 2

(b)

M1 A1

A1

dy = 4 cos 2x cos2 y dx (b)

(6)

M1 A1

1 2

3.

M1 A1

2 y = kx + c (0, 4) ⇒ 4 = c ∴ 2 y = kx + 4

M1 A1 M1 A1

(2, 9) ⇒ 6 = 2k + 4, k = 1 ∴ 2 y = x + 4, y = 12 (x + 4)

M1 A1 M1

y=

1 4

(x + 4)2

A1

 Solomon Press C4J MARKS page 2

(9)

6.

(a)

1 3

V= (b)

πr2h =

2 πh × h =

1 3

3

V = 8 × 120 = 960 =

r h

, r=

h 3

M1

πh3

M1 A1

1 9

πh3 ∴ h =

3

B1 dh = dt

9 × 960 π

360

M1 A1 = 14.011

 −3    6 1  

−

 −4    1  3  

∴ r=

 −4    1  3  

+λ

1    5  −2   

(b)

− 4 + λ = 3 + 2µ (1) 1 + 5λ = −7 − 3µ (2) 3 − 2λ = 9 + µ (3) 2 × (1) + (3): −5 = 15 + 5µ, µ = −4, λ = −1 sub. (2): 1 − 5 = −7 + 12, not true ∴ do not intersect  3 + 2µ     −7 − 3µ   9+µ   

, BC = OC − OB =

 1   6 + 2µ       5  .  −13 − 3µ   −2   8 + µ     

∴

=

1    5  −2   

AB =

OC =

(b)

 6 + 2µ     −13 − 3µ   8+ µ   

M1 A1 M1 A1

 −7     8  4  

M1 A1

M1 A1 A1

coeffs x2 ⇒

A1

=

∫0

(1 +

=

1 4

+ ln

16 9

⇒

3 = 3A

2 1− x

= ( 14 − 2 ln

(c)

B1 M1 A1 M1 A1

x(3x − 7) ≡ A(1 − x)(1 − 3x) + B(1 − 3x) + C(1 − x) x=1 ⇒ − 4 = − 2B ⇒ B=2 x = 13 ⇒ −2 = 23 C ⇒ C = −3 1 4

3 4

3 1 − 3x

−

+ ln

+ ln

1 4

1 4

A=1 1

) dx = [x − 2 ln1 − x + ln1 − 3x] 04

) − (0)

=

1 4

(10)

M1 A1

6 + 2µ − 65 − 15µ − 16 − 2µ = 0

= 0,

µ = −5 ∴ OC =

(a)

M1 A1

(a)

(c)

M1 A1

πh 2

dh = 0.58 cm s−1 (2dp) dt

∴

8.

1 9

=

dV dV = 120, = 13 πh2 dt dh dV dV dh dh = × , 120 = 13 πh2 , dt dh dt dt dh when h = 6, = 3.18 cm s−1 (2dp) dt

(i)

(ii)

7.

1 3

let radius = r, ∴ tan 30° =

+ ln

(13)

M1 A1 M1

4 9

M1 A1

f(x) = 1 + 2(1 − x)−1 − 3(1 − 3x)−1 (1 − x)−1 = 1 + x + x2 + x3 + … (1 − 3x)−1 = 1 + 3x + (3x)2 + (3x)3 + … = 1 + 3x + 9x2 + 27x3 + … ∴ f(x) = 1 + 2(1 + x + x2 + x3 + …) − 3(1 + 3x + 9x2 + 27x3 + …) = −7x − 25x2 − 79x3 + …

B1 M1 A1 M1 A1

(14)

Total

(75)

 Solomon Press C4J MARKS page 3

Performance Record – C4 Paper J

Question no.

1

Topic(s)

integration

Marks

6

2

3

4

integration differentiation parametric equations

6

8

5

6

7

8

differential equation

connected rates

vectors

partial fractions, binomial series

9

10

13

14

9

Student

 Solomon Press C4J MARKS page 4

Total

75