C4 A Ms

FOR EDEXCEL

GCE Examinations Advanced Subsidiary

Core Mathematics C4 Paper A

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 A – Marking Guide 1.

2x(2 + y) + x2

dy = dx

2.

M2 A2

2 x (2 + y )

M1 A1

2 y − x2

3

(a)

f( 101 ) =

(b)

= 3(1 − x) =3+

3 2

− 12

x+

3

=

1 1− 10

3

=

9 10

(

3 ) 10

10

=

= 3[1 + ( − 12 )(−x) + 9 8

x2 +

10 − 3.1621875 10

3 20

+

9 800

( − 12 )( − 32 ) 2

(−x)2 +

( − 12 )( − 32 )( − 52 ) 3× 2

(−x)3 + …] M1 A2

+

15 16000

= 3.1621875 (8sf)

× 100% = 0.003% (1sf)

=

(a)

1 + 3 λ = −5 ∴ λ = −2 p−λ=9 ∴ p=7 −5 + 2λ = 11 ∴ q = 2

M1 A1 A1

(b)

1 + 3λ = 25 ∴ λ = 8 when λ = 8, r = (i + 7j − 5k) + 8(3i − j + 2k) = (25i − j + 11k) ∴ (25, −1, 11) lies on l

M1

OC = [(1 + 3λ)i + (7 − λ)j + (−5 + 2λ)k] ∴ [(1 + 3λ)i + (7 − λ)j + (−5 + 2λ)k].(3i − j + 2k) = 0 3 + 9λ − 7 + λ − 10 + 4λ = 0

λ = 1 ∴ OC = (4i + 6j − 3k), C (4, 6, −3)

4.

B1

(d)

(c)

(d)

A : λ = −2, B : λ = 8, C : λ = 1

(a)

∫

1 ( x − 6)( x − 3)

1 ( x − 6)( x − 3)

≡

∫

dx = A x−6

+

∴ AC : CB = 3 : 7

2 dt

∫

(

1 x−6

−

1 ) x−3

ln x − 6

2( x − 3)

= 6t,

dx =

t=

∫

A1 M1 A1 M1 A1 M1 A1

(11)

1 6

M1 A2

2 dt M1 A1 M1 A1 M1 A1

ln x − 6

2( x − 3)

as x → 3, t → ∞ ∴ cannot make 3 g

 Solomon Press C4A MARKS page 2

(8)

B x−3

lnx − 6 − lnx − 3 = 6t + c t = 0, x = 0 ∴ ln 6 − ln 3 = c, c = ln 2 x = 2 ⇒ ln 4 − 0 = 6t + ln 2 t = 16 ln 2 = 0.1155 hrs = 0.1155 × 60 mins = 6.93 mins ≈ 7 mins (b)

M1 A1

M1

1 ≡ A(x − 3) + B(x − 6) x = 6 ⇒ A = 13 , x = 3 ⇒ B = − 13 1 3

(6)

M1 A1

x3 + …

15 16

10 = f( 101 ) ≈ 3 +

(c)

3.

dy dy =0 − 2y dx dx

B2

(12)

5.

(a)

(b)

x 0 y 0 area ≈

0.5 1 1.5 2 1.716 1.472 1.093 1.083 × 0.5 × [0 + 1.083 + 2(1.716 + 1.472 + 1.093)] = 2.41 (3sf)

1 2

volume = π ∫

2

16x e−2x dx

0

M1

u = 16x, u′ = 16, v′ = e−2x, v = − 12 e−2x I = −8 x e

−2x

∫

−

−2x

−8e

−2x

M1

dx

A2

−2x

= −8x e − 4e + c volume = π[−8x e−2x − 4e−2x] 02 = π{(−16e−4 − 4e−4) − (0 − 4)} = 4π(1 − 5e−4) 6.

(a)

=

∫

(b)

1 5

∫1

2

(u − 1) u

= ( 64 − 5

(b)

cos 2t =

1 2

2

× 2u du =

2

∫1

B1

(2u4 − 4u2 + 2) du

u3 + 2u] 12

4 3

32 3

, 2t =

π 3

, t=

π 6

M1 A1 M1 A1

+ 4) − ( 25 −

4 3

+ 2) = 5 151

M1 A1

π 6

(12)

M1 A1 M1 M1 A1

, y = 2, grad = 2

∴ y − 2 = 2(x − y = 2x + 1 x=0 ⇒ t=

∴ area =

∫

1 2

)

M1 A1

π 4

π 6 π 4

B1 cosec t × (−2 sin 2t) dt

π

= − ∫ π6 cosec t × 4 sin t cos t dt = 4

(d)

M1

dx dy = −2 sin 2t, = −cosec t cot t dt dt dy −cosec t cot t = −2sin 2t dx t=

(c)

M1 A1

dx = 2u du x = 0 ⇒ u = 1, x = 3 ⇒ u = 2 2

(12)

M1 A1

sin 5x + c

= [ 25 u5 −

(a)

M1 A1

u2 = x + 1 ⇒ x = u2 − 1,

I =

7.

A1

(cos x − cos 5x) dx

= sin x −

B2 B1 M1 A1

M1 π 4

∫ π6

4 cos t dt

π

= [4 sin t] 4π = 2 2 − 2 = 2( 2 − 1) 6

M1 A1 M2 A1

(14)

Total

(75)

 Solomon Press C4A MARKS page 3

Performance Record – C4 Paper A

Question no.

1

2

3

4

5

6

7

Topic(s)

differentiation

binomial series

vectors

differential equation, partial fractions

trapezium rule, integration

integration

parametric equations

Marks

6

8

11

12

12

12

14

Student

 Solomon Press C4A MARKS page 4

Total

75