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Mark Scheme (Results) January 2008

GCE

GCE Mathematics (6665/01)

Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WC1V 7BH


January 2008 6665 Core Mathematics C3 Mark Scheme Question Number

Scheme

Marks

1.

x2 −1

2 x4 2 x4

−1 2x 2 2 − 3x + x + 1 − 2x2 − x2 + x + 1 − x2 +1 x

a = 2 stated or implied c = −1 stated or implied

M1 A1 A1

a = 2, b = 0, c = −1, d = 1, e = 0 d = 1 and b = 0, e = 0 stated or implied

A1

2 x2 − 1 +

x x −1 2

[4] 2.

(a) dy = 2 e 2 x tan x + e 2 x sec 2 x dx

M1 A1+A1

dy = 0 ⇒ 2 e 2 x tan x + e 2 x sec 2 x = 0 dx

M1

2 tan x + 1 + tan 2 x = 0 2 ( tan x + 1) = 0 tan x = −1 ¿

(b)

⎛ dy ⎞ ⎜ ⎟ =1 ⎝ dx ⎠ 0

Equation of tangent at ( 0, 0 ) is y = x

A1 cso

A1

(6)

M1 A1

(2) [8]


Question Number

3.

Scheme

(a)

Marks

f ( 2 ) = 0.38 … f ( 3) = −0.39 …

Change of sign (and continuity) ⇒ root in (b)

(c)

M1

( 2, 3)

cso

¿

x1 = ln 4.5 + 1 ≈ 2.504 08 x2 ≈ 2.50498 x3 ≈ 2.50518

A1

(2)

M1 A1 A1

(3)

Selecting [ 2.5045, 2.5055] , or appropriate tighter range, and evaluating at both ends. f ( 2.5045 ) ≈ 6 × 10−4

M1

f ( 2.5055 ) ≈ −2 × 10−4

Change of sign (and continuity) ⇒ root ∈ ( 2.5045, 2.5055

)

⇒ root = 2.505 to 3 dp ¿ Note: The root, correct to 5 dp, is 2.50524

cso

A1

(2) [7]


Question Number

4.

Scheme

Marks

(a)

( −5, 4 )

y

( 5, 4 )

x

O

Shape ( 5, 4 )

B1 B1

( −5, 4 )

B1

(b) For the purpose of marking this paper, the graph is identical to (a) Shape ( 5, 4 )

B1 B1

( −5, 4 )

B1

(3)

(3)

(c)

y

O

( 4, 8)

x

( −6, − 8) General shape – unchanged B1 Translation to left B1 ( 4, 8 ) B1

( −6, − 8 ) In all parts of this question ignore any drawing outside the domains shown in the diagrams above.

B1

(4) [10]


Question Number

5.

Scheme

Marks

(a) 1000 (b)

(c)

B1

1000 e −5730 c = 500 1 e−5730 c = 2 1 −5730c = ln 2 c = 0.000121

cao

R = 1000 e−22920 c = 62.5

Accept 62-63

(1)

M1

A1 M1 A1

(4)

M1 A1

(2)

(d) R 1000

Shape 1000 O

t

B1 B1

(2) [9]


Question Number

6.

Scheme

Marks

(a) cos ( 2 x + x ) = cos 2 x cos x − sin 2 x sin x

M1

= ( 2 cos 2 x − 1) cos x − ( 2sin x cos x ) sin x

M1

= ( 2 cos 2 x − 1) cos x − 2(1 − cos 2 x) cos x any correct expression

A1

= 4 cos x − 3cos x

A1

3

2 cos x 1 + sin x cos x + (1 + sin x ) (b)(i) + = 1 + sin x cos x (1 + sin x ) cos x

=

(c)

2 (1 + sin x ) (1 + sin x ) cos x

=

2 = 2sec x cos x

sec x = 2 or cos x =

x=

π 5π 3

,

2

M1

cos 2 x + 1 + 2sin x + sin 2 x (1 + sin x ) cos x

=

3

(4)

A1

M1 ¿

cso A1

1 2

(4)

M1 accept awrt 1.05, 5.24

A1, A1

(3) [11]

7.

(a)

dy = 6 cos 2 x − 8sin 2 x dx ⎛ dy ⎞ ⎜ ⎟ =6 ⎝ dx ⎠ 0 1 y−4= − x 6

M1 A1 B1 or equivalent

(b) R = √ ( 32 + 42 ) = 5 4 tan α = , α ≈ 0.927 3

M1 A1

(5)

M1 A1 awrt 0.927

M1 A1

(4)

(c) sin ( 2 x + their α ) = 0

M1

x = −2.03, − 0.46, 1.11, 2.68 First A1 any correct solution; second A1 a second correct solution; third A1 all four correct and to the specified accuracy or better. Ignore the y-coordinate.

A1 A1 A1 (4) [13]


Question Number

8.

Scheme

(a) x = 1 − 2 y

⎛ 1− x ⎞ f :xa⎜ ⎟ ⎝ 2 ⎠ −1

(b) gf ( x ) = =

⎛ 1− x ⎞ y=⎜ ⎟ ⎝ 2 ⎠

3

1

1

3

or

3

1− x 2

3

3 −4 1 − 2 x3 3 − 4 (1 − 2 x3 )

8 x3 − 1 = 0 1 x= 2

(2)

M1 A1 M1 cso

A1

(4)

Ignore domain Attempting solution of numerator = 0 M1 Correct answer and no additional answers A1

3 2 3 2 d y (1 − 2 x ) × 24 x + ( 8 x − 1) × 6 x (d) = 2 dx (1 − 2 x3 )

=

M1 A1 Ignore domain

1 − 2 x3 8 x3 − 1 = ¿ 1 − 2 x3 8 x3 − 1 gf : x a 1 − 2 x3 (c)

Marks

18 x 2

(1 − 2 x )

3 2

Solving their numerator = 0 and substituting to find y.

x = 0, y = −1

(2)

M1 A1

A1 M1 A1

(5) [13]


C3 MS Jan 2008