C4 H Ms

FOR EDEXCEL

GCE Examinations Advanced Subsidiary

Core Mathematics C4 Paper H

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 H – Marking Guide 1.

= 1 + ( 32 )(4x) +

(a)

( 32 )( 12 ) 2

2

(4x)2 +

( 32 )( 12 )( − 12 ) 3×2

(4x)3 + …

3

= 1 + 6x + 6x − 4x + …

2.

u = 1 + sin x ⇒

B1

du = cos x dx

x = 0 ⇒ u = 1, x = I =

2

∫1

A3

1 4

x <

(b)

⇒ u=2

π 2

B1 A1

= [ 14 u4] 12 = 4−

3.

=

M1 15 4

M1 A1

x + 11 ( x + 4)( x − 3)

(a)

≡

A x+4

+

x + 11 ( x + 4)( x − 3) 2

∫0

=

(

≡

2 x−3

−

2 x−3 1 x+4

−

A = −1 B=2

M1 A1 A1

1 x+4

) dx

= [2 lnx − 3 − lnx + 4] 02 = (0 − ln 6) − (2 ln 3 − ln 4) 2 = ln 27

4.

(6)

B x−3

x + 11 ≡ A(x − 3) + B(x + 4) x = −4 ⇒ 7 = −7A ⇒ x=3 ⇒ 14 = 7B ⇒

(b)

(5)

M1

u3 du

1 4

M1

M1 A1 M1 M1 A1

π

= π ∫ π2 (2 sin x + cosec x)2 dx

(8)

M1

6

=π

π 2 π 6 π 2 π 6

∫

= π∫

(4 sin2 x + 4 + cosec2 x) dx

A1

(2 − 2 cos 2x + 4 + cosec2 x) dx

M1

π

= π[6x − sin 2x − cot x] 2π

M1 A2

6

= π{(3π + 0 + 0) − (π − = π(2π +

5.

(a)

3 2

3)=

1 2

2x − 3y − 3x

3 2

−

3 )}

M1

π(4π + 3 3 )

A1

dy dy − 2y =0 dx dx

M1 A2

dy 2x − 3y = 3x + 2 y dx (b)

grad = 5 ∴ y + 2 = 5(x − 2)

M1 A1

[ y = 5x − 12 ]

 Solomon Press C4H MARKS page 2

(8)

M1 M1 A1

(8)

6.

(a)

= =

(b)

1× 6 + 5 × 3 + (−1) × (−6) 27 27 × 81

7.

(a)

(b)

27 9

=

sin (∠AOB) =

area =

(c)

M1 A1

1 + 25 + 1 × 36 + 9 + 36

1 2

= 3 3 = 9

1 3

3

M1 A1

2 3

M1 A1

1 − ( 13 3)2 = 2 3

× 3 3 ×9×

=

= OA × sin (∠AOB) = 3 3 ×

27 2 2 3

2

M1 A1

= 3 2

M1 A1

dx dy = 2t − 1, = 4 × (1 − t ) − 42t × (−1) = 4 2 (1 − t ) (1 − t ) dt dt dy 4 = (2t − 1)(1 − t )2 dx

t = −1, x = 2, y = −2, grad = − 13

M1

3y + 6 = −x + 2 x + 3y + 4 = 0

A1

t(t − 1) + 3 ×

4t 1− t

+4=0

M1

−t(t − 1) + 12t + 4(1 − t) = 0 t3 − 2t2 − 7t − 4 = 0 t = −1 is a solution ∴ (t + 1) is a factor (t + 1)(t2 − 3t − 4) = 0 (t + 1)(t + 1)(t − 4) = 0 t = −1 (at P) or t = 4 ∴ Q (12, − 163 )

(a)

M1 A1

∴ y + 2 = − (x − 2)

2

8.

B1 M1

M1

1 3

(c)

∫

1 P

dP =

∫

k dt

P 300

 = kt,

P 300

A1 M1 M1 A1 M1 A1

= ekt,

A1 M1

P = 300ekt

M1 A1

(b)

t = 1, P = 360 ⇒ 360 = 300ek k = ln 65 = 0.182 (3sf)

M1 A1

(c)

P = 300e0.1823t when t = 2, P = 432; when t = 3, P = 518 model does not seem suitable as data diverges from predictions

B1 B1

(d)

∫

1 P

dP =

∫

(0.4 − 0.25 cos 0.5t) dt

lnP = 0.4t − 0.5 sin 0.5t + c t = 0, P = 300 ⇒ ln 300 = c ln

(e)

P 300

(14)

M1

lnP = kt + c t = 0, P = 300 ⇒ ln 300 = c lnP = kt + ln 300 ln

(10)

 = 0.4t − 0.5 sin 0.5t

M1 A1

[ P = 300e0.4t − 0.5 sin 0.5t ]

second model: t = 1, 2, 3 ⇒ P = 352, 438, 605 the second model seems more suitable as it fits the data better

M1 A1 M1 A1 B1

(16)

Total

(75)

 Solomon Press C4H MARKS page 3

Performance Record – C4 Paper H

Question no.

1

2

3

Topic(s)

binomial series

integration

partial fractions

Marks

5

6

8

4

integration differentiation

8

Student

 Solomon Press C4H MARKS page 4

5

8

6

7

8

vectors

parametric equations

differential equations

10

14

16

Total

75