EDEXCEL PURE MATHEMATICS C1 (6663) SPECIMEN PAPER ...

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EDEXCEL PURE MATHEMATICS C1 (6663) SPECIMEN PAPER MARK SCHEME. 42. Question number. Scheme. Marks. 1. a = 7, d = 2. B1. S20 = 2. 1 ´ 20 ´ (2 ...
EDEXCEL PURE MATHEMATICS C1 (6663) SPECIMEN PAPER MARK SCHEME

Question number

Scheme

Marks

a = 7, d = 2

1.

S20 =

1 2

B1

´ 20 ´ (2 ´ 7 + 19 ´ 2) = 520

M1 A1 (3 marks)

2

ó (5 x + 3 Ö x) dx = 5 x + 2 x 32 + C ô 2 õ

2.

M1 A1 A1 B1 (4 marks)

3.

(a)

Ö80 = 4Ö5

(b)

(4 - Ö5) = 16 - 8Ö5 + 5 = 21 - 8Ö5

B1

2

(1)

M1 A1 A1 (3) (4 marks)

Gradient of AB =

4.

Gradient of l = y-4=

4 - (-6) æ 5ö ç= - ÷ 3-7 è 2ø

M1 A1

2 5 2 (x - 3) 5

M1 2x - 5y + 14 = 0

M1 A1

(5)

(5 marks) 5.

(a)

y

O (b)

x

y

O

Position, Shape

B1

(0, 2), (2, 0)

B1 B1

Position, Shape

B1

æ1 ö æ3 ö (0, 1), ç , 2 ÷, ç , 0 ÷ è2 ø è2 ø

B2 (1, 0)

(3)

(3)

x (6 marks)

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EDEXCEL PURE MATHEMATICS C1 (6663) SPECIMEN PAPER MARK SCHEME

Question number 6.

Scheme

(a)

Marks

5 - 2x = 2x2 - 3x - 16

2x2 - x - 21 = 0

(2x - 7)(x + 3) = 0

x = -3, x =

7 2

M1 A1

y = 11, y = -2 (b)

M1 A1

M1 A1ft

Using critical values x = -3,

x=

7 2

M1

x < -3,

x>

7 2

M1 A1ft

(6)

(3)

(9 marks) 7.

(a)

a + (n - 1)d = 250 + (10 ´ 50) = £750

M1 A1

(b)

1 1 n [2a + (n - 1) d ] = ´ 20 ´ (500 + 19 ´ 50), 2 2

(c)

B:

1 ´ 20 ´ (2A + 19 ´ 60) 2

= £14500

[= 10(2A + 1140)], = “14500”

Solve for A: A = 155

(2)

M1 A1, A1 (3) B1, M1 M1 A1

(4)

(9 marks) 8.

(a) (b)

(x + 5)2 - 25 + 36

a = 5, b2 - 4ac = 100 - 144,

< 0, therefore no real

(2)

k =25

M1 A1

(2)

Shape, position

B1 B1

(-5, 0) (0, 25)

B1 B1ft

4k = 100

y

O

B1, M1 A1 (3) M1 A1

roots

(c) Equal roots if b2 - 4ac = 0 (d)

b = 11

(4)

x (11 marks)

43

EDEXCEL PURE MATHEMATICS C1 (6663) SPECIMEN PAPER MARK SCHEME

Question number 9.

(a)

Scheme

Marks

f(x)= x3 - 4x2 + 6x + C

M1 A1

5 = 27 - 36 + 18 + C

C=-4

y = 8 - 16 + 12 - 4 = 0

M1 A1

(4)

M1 A1

(2)

(b)

x = 2:

(c)

f ¢(3) = 27 - 24 + 6 = 9, Parallel therefore equal gradient

B1, M1

3x2 - 8x + 6 = 9

M1

3x2 - 8x - 3 = 0

(3x + 1)(x - 3) = 0

Q: x = -

1 3

M1 A1

(5)

(11 marks) 10.

(a)

dy = 3x2 - 5 - 2x-2 dx

M1 A2(1,0) 2 dy =3´1-5dx 1

At both A and B, (b)

Gradient of normal = y - (-2) =

(c)

(= -4)

1 4

(5)

M1 A1ft

1 (x - 1) 4

4y = x - 9

M1 A1 y= -

Normal at A meets y-axis where x = 0: Similarly for normal at B:

M1 A1

4y = x + 9

Length of PQ =

y=

9 9 9 + = 4 4 2

9 4

9 4

(4)

B1 M1 A1 A1

(4) (13 marks)

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