C4 F Ms

FOR EDEXCEL

GCE Examinations Advanced Subsidiary

Core Mathematics C4 Paper F

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 F – Marking Guide 1.

M1 A2

dy = 0 ∴ 4x + y = 0, y = −4x dx 2x2 − 4x2 − 16x2 + 18 = 0 x2 = 1, x = ± 1 ∴ (−1, 4), (1, −4)

SP: sub.

2.

dy dy =0 − 2y dx dx

4x + y + x

∫0

=

∫0

4 tan 2 u

π 4

π 4

× 2 sec2 u du =

4sec2 u

B1 π 4

∫0

2 tan2 u du

A1

(2 sec2 u − 2) du

M1

π

= [2 tan u − 2u] 04 = (2 −

3.

π 2

1 2

(4 − π)

25 = ( 24 )

=

4 ×6 25

(b)

= 1 + ( − 12 )( 12 x) +

( − 12 )( − 32 ) 2

1 4

(c)

x=

x=

1 12 1 12

∴ (a)

(b)

=

x+

24 25

M1 A1

(a)

=1−

4.

M1 A1

) − (0) = − 12

3 2 x 32

⇒ (1 + ⇒ (1 +

6 ≈

5 2

− 1 2 1 2

5 128

=

2 5

[k=

6

( 12 x)2 +

2 5

( − 12 )( − 32 )( − 52 ) 3× 2

]

M1 A1

( 12 x)3 + …

M1

x3 + …

= cos−1

x)

− 12

1 = (1 24 )

x)

− 12

≈1−

− 12

1 4

=

2 5

( 121 ) +

6 3 32

( 121 )2 −

5 128

= 0.97979510 × 0.97979510 = 2.44949 (5dp)

4 × (−3) + (−3) × 0 + 1× 2 = 57.0° (1dp)

M1 M1 A1

(9)

M1 B1 M1 A1 A1

M1 A1

 Solomon Press C4F MARKS page 2

( 121 )3

M1 A1

16 + 9 + 1 × 9 + 0 + 4 −10 26 × 13

(8)

A3

4s = −7 − 3t (1) 7 − 3s = 1 (2) −4 + s = 8 + 2 t (3) (2) ⇒ s = 2, sub. (1) ⇒ t = −5 check (3) −4 + 2 = 8 − 10, true ∴ intersect intersect at (7j − 4k) + 2(4i − 3j + k) = (8i + j − 2k) = cos−1

(8)

M1

x = 0 ⇒ u = 0, x = 2 ⇒ u = I =

M1 A2

dx = 2 sec2 u du

x = 2 tan u ⇒

π 4

M1 A1

(9)

5.

(a)

(b)

dx dy = 1× (2 − t ) − t2× (−1) = 2 2 , = −(1 + t)−2 (2 − t ) (2 − t ) dt dt 2 2 dy 2−t  = − 1 2 ÷ 2 2 = − (2 − t ) 2 = − 12   + 1 t   (1 + t ) (2 − t ) 2(1 + t ) dx 1 2

t = 1, x = 1, y =

15 2

[ y = 8x −

]

M1 A1 M1

=

2x 1+ x

A1

1 1+ x 1+ x ∴y= = 2 x (1 ) 2 + x + x 1 + 3x 1 + 1+ x

y=

(a)

B1

x(2 − t) = t 2x = t(1 + x), t =

6.

M1 A1

, grad = − 18

grad of normal = 8 ∴ y − 12 = 8(x − 1)

(c)

M1 B1

∫

M1 A1

(sec2 x − 1) dx

M1

= tan x − x + c

(b)

=

∫

sin x cos x

=

∫

1 u

M1 A1

dx,

du = −sin x dx

let u = cos x,

× (−1) du = − ∫

1 u

M1

du

A1

= −lnu + c = lnu−1 + c = lnsec x + c

(c)

volume = π ∫

π 3

0

M1 A1

x tan2 x dx

M1

u = x, u′ = 1, v′ = tan2 x, v = tan x − x I = x(tan x − x) − ∫ (tan x − x) dx = x tan x − x2 − lnsec x + 1 2

volume = π[x tan x −

(a)

(b)

1 2

x2 + c

A1

x2 − lnsec x] 0

π2 − ln 2) − (0)} =

1 18

π2( 6 3 − π) − π ln 2

dV = −kV, dt dV dV = × dt dh

dV = 10πh − πh2 dh dh dh ∴ −kV = (10πh − πh2) dt dt d h − 13 kπh2(15 − h) = πh(10 − h) dt dh dh kh(15 − h) −kh(15 − h) = 3(10 − h) ∴ =− 3(10 − h) dt dt 3(10 − h) h(15 − h)

≡

A h

+

B 15 − h

,

∫

3(10 − h) h(15 − h)

dh =

∫

∫

−k dt,

(

2 h

−

3(10 − h) h(15 − h) 1 15 − h

250

(d)

h2 (15 − h) 250

= e−kt,

(13)

M1

M1 A1 M1

2 h

−

1 15 − h

A2

) dh =

∫

−k dt

M1

≡

2 lnh + ln15 − h = −kt + c t = 0, h = 5 ⇒ 2 ln 5 + ln 10 = c, c = ln 250 2 lnh + ln15 − h − ln 250 = −kt 2 ln h (15 − h) = −kt,

M1 A1 B2

3(10 − h) ≡ A(15 − h) + Bh

h = 0 ⇒ A = 2, h = 15 ⇒ B = −1 ∴ (c)

M1 A1

π 3

1 18

= π{( 13 3 π − 7.

(11)

h2(15 − h) = 250e−kt

t = 2, h = 4 ⇒ 176 = 250e−2k k = − 12 ln 176 = 0.175 (3sf) 250

M1 A1 M1 M1 A1 M1 M1 A1

(17)

Total

(75)

 Solomon Press C4F MARKS page 3

Performance Record – C4 Paper F

Question no. Topic(s)

Marks

1

2

differentiation integration

8

8

3

4

5

6

7

binomial series

vectors

parametric equations

integration

differential equation, partial fractions

9

9

11

13

17

Student

 Solomon Press C4F MARKS page 4

Total

75