C4 D Ms

FOR EDEXCEL

GCE Examinations Advanced Subsidiary

Core Mathematics C4 Paper D

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 D – Marking Guide 1.

2.

(a)

= 2−3(1 − =

1 8

=

1 8

3 2

x)−3 =

1 8

(1 −

[1 + (−3)( − 32 x) + +

9 16

x+

27 16

x2 +

3 2

x)−3

( −3)( −4) 2

135 32

B1

( − 32 x)2 +

( −3)( −4)( −5) 3× 2

( − 32 x)3 + …]

x3 + …

A3

2 3

(b)

x <

(a)

2x + 3y + 3x

B1

dy dy =0 − 4y dx dx

3.

(a)

(b)

4.

(a)

grad =

6−6 −8 − 9

M1 A1 =0

M1

∴ normal parallel to y-axis ∴ x = 3

M1 A1

2x3 − 5x2 + 6 ≡ (Ax + B)x(x − 3) + C(x − 3) + Dx x=0 ⇒ 6 = −3C ⇒ C = −2 x=3 ⇒ 15 = 3D ⇒ D=5 coeffs x3 ⇒ A=2 coeffs x2 ⇒ −5 = B − 3A ⇒ B=1

M1

=

2

∫1

(2x + 1 −

2 x

+

5 ) x−3

dx M1 A2 M1 A1

∫

M1

1 2

∫

x2 = k(5t −

k(5 − t) dt 1 2 t) 2

+c

t = 0, x = 0 ⇒ c = 0 t = 2, x = 96 ⇒ 4608 = 8k, t = 1 ⇒ 12 x2 = 576 × 92 , (b)

(10)

M1 A1 k = 576 x = 5184 = 72

3 hours 5 mins ⇒ t = 3.0833, x = 12284 = 110.83 dx dx 576(5 − 3.0833) ∴ = = 9.96, < 10 so she should have left 110.83 dt dt

 Solomon Press C4D MARKS page 2

(8)

A1 B1 M1 A1

= [x2 + x − 2 lnx + 5 lnx − 3] 12 = (4 + 2 − 2 ln 2 + 0) − (1 + 1 + 0 + 5 ln 2) = 4 − 7 ln 2 x dx =

(6)

M1 A2

dy 2x + 3 y = 4 y − 3x dx (b)

M1

B1 M1 A1 M1 A1

M1 A1 M1 A1

(12)

5.

(a) (b)

1 3      4 .  a  5  b     

= 0 ∴ 3 + 4a + 5b = 0

4 + s = −3 + 3t (1) 1 + 4s = 1 + at (2) 1 + 5s = −6 + bt (3) (1) ⇒ s = 3t − 7 sub. (2) ⇒ 1 + 4(3t − 7) = 1 + at 12t − 28 = at, sub. (3) ⇒ 28 12 − a

sub (a) ⇒

 −3    1  −6   

 3

u2 = 1 − x ⇒ x = 1 − u2,

M1 A1

t=

28 15 − b

A1

b=a+3

1  

∴ (3, −3, −4)

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

area =

∫0

x 1 − x dx =

=

∫0

1

(2u2 − 2u4) du 2 5

= ( 23 −

) − (0) =

1

= π∫

0

= π∫

0

2 5

0

∫1

(1 − u ) × u × (−2u) du

4 15

M1 A1

(x2 − x3) dx

1 3

−

, x= 3 2

∴ y−

1 4 1 x ]0 4 1 ) − (0)} 4

M1 A1 1 12

=

π

M1 A1

π 2

,

3π 2

⇒ t=

π 4

,

9 4

M1 A1

3π 4

M1 A1

=−

2 3 3

A1 3 2

, y=

(d)

2

, grad = −

(x −

9 4

2 3 3

B1

)

M1

2

A1 2

2

2

y = sin 2t = 4 sin t cos t = 4(1 − cos t)cos t 2

cos t =

x 3

2

(13)

M1 A1

6 3 y − 9 = −4x + 9 2x + 3 3 y = 9 2

M1

M1

− 23 cot 2t = 0 ⇒ 2t = π 6

(12)

M1 A1

dx dy = 6 cos t × (−sin t), = 2 cos 2t dt dt dy 2 cos 2t 2 cos 2t = = = − 23 cot 2t −6 cos t sin t −3sin 2t dx

t=

M1 A1

A1

∴ ( 32 , 1), ( 32 , −1) (c)

M1 A1

B1 2

x2(1 − x) dx

1

M1

M1

u5] 10

= [ 23 u3 −

= π{(

(b)

 3    −3  ,  −4   

+ 2  −2  =

(a)

(a)

28 12 − a

3 + 4a + 5(a + 3) = 0, a = −2, b = 1

t=2 ∴ r=

= π[ 13 x3 −

7.

t(15 − b) = 28,

12 − a = 15 − b,

,

(c)

(b)

t=

1 + 5(3t − 7) = −6 + bt

28 15 − b

=

B1 M1

t(12 − a) = 28,

15t − 28 = bt,

6.

M1 A1

∴ y = 4(1 −

x 3

x 3

2

) , y =

4 9

x(3 − x)

M2 M1 A1

(14)

Total

(75)

 Solomon Press C4D MARKS page 3

Performance Record – C4 Paper D

Question no.

1

2

3

4

5

6

7

Topic(s)

binomial series

differentiation

partial fractions

differential equation

vectors

integration

parametric equations

Marks

6

8

10

12

12

13

14

Student

 Solomon Press C4D MARKS page 4

Total

75