C4 B Ms

FOR EDEXCEL

GCE Examinations Advanced Subsidiary

Core Mathematics C4 Paper B

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 B – Marking Guide 1.

u = x2, u′ = 2x, v′ = sin x, v = −cos x I = −x2 cos x − ∫ −2x cos x dx = −x2 cos x +

∫

2x cos x dx

u = 2x, u′ = 2, v′ = cos x, v = sin x I = −x2 cos x + 2x sin x − ∫ 2 sin x dx

M1 A1

= −x2 cos x + 2x sin x + 2 cos x + c

2.

1

∫

−y−1 =

∫

dy =

y2

2 3

x dx

1 y

y=

=

2 3

x −

1 6

(6)

M1

3

M1 A1

x = 1, y = −2 ⇒ −

A1

x2 + c 3 2

M1 A2

1 2

=

2 3

1 y

,

+ c, 1 6

=

c= 2 3

−

− 16

3 2

M1 A1 3 2

1 6

x =

(1 − 4x )

M1

6

A1

3

(7)

1 − 4x 2

3.

4.

dy dy =0 − 2y dx dx dy dy dy (−1, −3) ⇒ −8 + 6 + 2 +6 = 0, = dx dx dx grad of normal = −4 ∴ y + 3 = −4(x + 1) [ y = −4x − 7 ]

8x − 2y − 2x

(a)

= 1 + (−3)(ax) +

( −3)( −4) 2

2 2

( −3)( −4)( −5) 3× 2

(ax)2 +

M1 A2 1 4

M1 A1 M1 M1 A1

(ax)3 + …

M1 A1

3 3

= 1 − 3ax + 6a x − 10a x + … (b)

6− x (1 + ax)3

A1

= (6 − x)( 1 − 3ax + 6a2x2 + …)

coeff. of x2 = 36a2 + 3a = 3 12a2 + a − 1 = 0 (4a − 1)(3a + 1) = 0 a = − 13 , 14 (c)

a = − 13 ∴

6− x (1 + ax)3

coeff. of x3 = (6 ×

5.

(a)

(b)

5

=

∫1

=

2 3

1 3x + 1

(4 − 2) =

= π∫

5 1

1 3x + 1

M1 A1 M1 A1

= (6 − x)(… +

10 27

) + (−1 ×

2 3

2 3

)=

x2 + 20 9

−

10 27 2 3

x3 + …)

M1

14 9

A1

=

1

dx = [ 23 (3x + 1) 2 ] 15

=

4 3

M1 A1

dx

M1

π(ln 16 − ln 4) =

M1 A1 1 3

π ln 4 =

2 3

π ln 2

[k=

 Solomon Press C4B MARKS page 2

(9)

M1 A1

= π[ 13 ln3x + 1] 15 1 3

(8)

2 3

]

M1 A1

(9)

6.

(a)

(b)

15 − 17x ≡ A(1 − 3x)2 + B(2 + x)(1 − 3x) + C(2 + x) x = −2 ⇒ 49 = 49A ⇒ A=1 1 28 7 x= 3 ⇒ = 3C ⇒ C=4 3

B1 B1

coeffs x2 ⇒

B=3

M1 A1

− 3x)−1] 0−1

M1 A3

=

0

∫ −1

1 2+ x

(

0 = 9A − 3B ⇒ +

3 1 − 3x

+

4 (1 − 3x)2

) dx

4 (1 3 + 13 )

= [ln2 + x − ln1 − 3x + = (ln 2 + 0 +

4 3

) − (0 − ln 4

M1

= 1 + ln 8 7.

(a)

M1 A1

x = 1 ∴ −1 + 4 cos θ = 1, cos θ = y > 0 ∴ sin θ > 0 ∴ θ =

(b)

dx = −4 sin θ, dθ dy ∴ = 2 2 cos θ −4sin θ dx at P, grad = −

grad of normal =

(c)

∴ y−

6 =

y=

6 x,

cos θ = ∴

8.

x +1 , 4

( x + 1)2 16

+

, θ =

π 3

,

5π 3

π 3

3 2

2 3 2

=−

2 2 3

2 2

×

=

M1 6

A1 M1

when x = 0, y = 0 ∴ passes through origin sin θ =

y 2 2

A1 M1

=1

M1 A1

(a)

AB = (7i − j + 12k) − (−3i + 3j + 2k) = (10i − 4j + 10k) ∴ r = (−3i + 3j + 2k) + λ(5i − 2j + 5k)

(b)

OC = [µ i + (5 − 2µ)j + (−7 + 7µ)k]

(c)

M1 M1 A1

6 (x − 1)

y2 8

M1 A1

dy = 2 2 cos θ dθ

2 2 × 12 4×

1 2

(11)

M1 A1

AC = OC − OA = [(3 + µ)i + (2 − 2µ)j + (−9 + 7µ)k]

M1 A1

BC = OC − OB = [(−7 + µ)i + (6 − 2µ)j + (−19 + 7µ)k]

A1

AC . BC = (3 + µ)(−7 + µ)+(2 − 2µ)(6 − 2µ)+(−9 + 7µ)(−19 + 7µ) = 0 µ 2 − 4µ + 3 = 0 (µ − 1)(µ − 3) = 0

M1 A1 M1

µ = 1, 3

A2

∴ OC = (i + 3j) or (3i − j + 14k)

AC =

16 + 0 + 4 = 2 5 , BC =

area =

1 2

36 + 16 + 144 = 14

× 2 5 × 14 = 14 5

(12)

M1 M1 A1

(13)

Total

(75)

 Solomon Press C4B MARKS page 3

Performance Record – C4 Paper B

Question no.

1

Topic(s)

integration

Marks

6

2

3

differential differentiation equation

7

8

4

5

6

7

8

binomial series

integration

partial fractions

parametric equations

vectors

9

9

11

12

13

Student

 Solomon Press C4B MARKS page 4

Total

75