C4 G Ms

FOR EDEXCEL

GCE Examinations Advanced Subsidiary

Core Mathematics C4 Paper G

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 G – Marking Guide 1.

2x + 2y2 + 2x × 2y

dy dy + =0 dx dx

M2 A2

dy 2x + 2 y2 =− 4 xy + 1 dx 2.

M1 A1

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

∫

u = 2x, u′ = 2, v′ = e−x, v = −e−x I = −x2 e−x − 2x e−x − ∫ −2e−x dx

2x e−x dx

(a)

(1 + ax)n = 1 + nax +

⇒ a=

−4 n

(1 + 8 x)

∴ k=

− 165

x 0 y 2.7183

4.

1 2

(ax)2 + …

, sub. ⇒

=…+

16 n2

×

= 24

n ( n −1) 2

B1

= 24

M1 A1

n = − 12 , a = 8 ( − 12 )( − 32 )( − 52 ) 3× 2

M1 A1

(8x)3 + …

M1

× 512 = −160 0.75 2.0786

1.5 1.0733

A1 2.25 0.5336

3 0.3716

× 1.5 × [2.7183 + 0.3716 + 2(1.0733)] = 3.93 (3sf)

=

(b)

× 0.75 × [2.7183 + 0.3716 + 2(2.0786 + 1.0733 + 0.5336)] = 3.92 (3sf)

(a)

(b)

=

1 2

B1 M1 A1 M1 A1 B2

AB = (4j + k) − (9i − 8j + 2k) = (−9i + 12j − k) ∴ r = (9i − 8j + 2k) + λ(−9i + 12j − k) at C, 2 − λ = −1, λ = 3

M1 A1 M1 A1

∴ OC = (9i − 8j + 2k) + 3(−9i + 12j − k) = (−18i + 28j − k)

A1

AC = 3(−9i + 12j − k), AC = 3 81 + 144 + 1 = 45.10 ∴ distance = 200 × 45.10 = 9020 m = 9.02 km (3sf)

M1 A1 M1 A1

 Solomon Press

(8)

B2

curve must be above top of trapezia in some places and below in others hence position of ordinates determines whether estimate is high or low

C4G MARKS page 2

(7)

B1

(a)

(c)

5.

− 12

n ( n−1) 2

a n ( n−1) 2

8(n − 1) = 24n, (b)

A1

2

∴ an = −4,

M1 A1 A2 M1 A1

= −x2 e−x − 2x e−x − 2e−x + c

3.

(6)

(9)

(9)

6.

(a)

(b)

7.

(a)

1 P

∫

∫

dP =

0.05e−0.05t dt

M1

lnP = −e−0.05t + c t = 0, P = 9000 ⇒ ln 9000 = −1 + c, c = 1 + ln 9000 lnP = 1 + ln 9000 − e−0.05t t = 10 ⇒ lnP = 1 + ln 9000 − e−0.5 = 9.498 P = e9.498 = 13339 = 13300 (3sf)

M1 A1 M1 A1 M1 A1

t → ∞, lnP → 1 + ln 9000 ∴ P → e1 + ln 9000 = 9000e = 24465 = 24500 (3sf)

M1 M1 A1

x = 2 ⇒ t = 1, dx = 3t2 dt

B1

2

∫1

∴ area =

x=9 ⇒ t=2

M1

2 t

2

∫1

× 3t2 dt =

6t dt

A1

= [3t2] 12 = 3(4 − 1) = 9 (b)

= π∫

2

2 t

( )2 × 3t2 dt = π ∫

1

2

M1 A1 12 dt

1

M1

= π[12t] 12 = 12π(2 − 1) = 12π (c)

t=

2 y

2 y

8.

(a)

(b)

8

∴ x = ( )3 + 1 = 8 , x −1

∴ y3 =

y=

3

M1 A1

+1

y3

M1

2 x −1

M1 A1

du = cos x dx

u = sin x ⇒

6 cos x

I =

∫

cos 2 x (2 − sin x)

=

∫

(1 − u 2 )(2 − u )

6

6 (1 + u )(1 − u )(2 − u )

≡

dx =

6 cos x

∫

(1 − sin 2 x)(2 − sin x)

dx

du A 1+ u

(c)

6 2

(1 − u )(2 − u )

≡

1 1+ u

x = 0 ⇒ u = 0, x = I =

1 2

∫0

(

1 1+ u

+

3 1− u

π 6

−

M1 M1 A1

B 1− u

+

C 2−u

+

+

3 1− u

2 2−u

2 2−u

−

⇒ u=

M1 A1 A1 A1

1 2

M1

) du 1

= [ln1 + u − 3 ln1 − u + 2 ln2 − u] 02

M1 A2

= (ln 32 − 3 ln 12 + 2 ln 32 ) − (0 + 0 + 2 ln 2)

M1

= 3 ln

3 2

(11)

B1

6 ≡ A(1 − u)(2 − u) + B(1 + u)(2 − u) + C(1 + u)(1 − u) u = −1 ⇒ 6 = 6A ⇒ A=1 u=1 ⇒ 6 = 2B ⇒ B=3 u=2 ⇒ 6 = −3C ⇒ C = −2 ∴

(10)

+ 3 ln 2 − 2 ln 2

= 3 ln 3 − 3 ln 2 + ln 2 = 3 ln 3 − 2 ln 2

M1 A1

(15)

Total

(75)

 Solomon Press C4G MARKS page 3

Performance Record – C4 Paper G

Question no. Topic(s)

1

2

differentiation integration

Marks

6

7

3

4

5

6

7

8

binomial series

trapezium rule

vectors

differential equation

parametric equations

partial fractions, integration

8

9

9

10

11

15

Student

 Solomon Press C4G MARKS page 4

Total

75