Mark Scheme (Results) Summer 2008

GCE

GCE Mathematics (6665/01)

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

June 2008 6665 Core Mathematics C3 Mark Scheme Question Number

1.

Scheme

e2 x+1 = 2 2 x + 1 = ln 2 1 x = ( ln 2 − 1) 2

(a)

dy = 8e 2 x +1 dx

(b) x=

1 ( ln 2 − 1) ⇒ 2

Marks

M1

A1

(2)

B1 dy = 16 dx

1 ⎛ ⎞ y − 8 = 16 ⎜ x − ( ln 2 − 1) ⎟ 2 ⎝ ⎠ y = 16 x + 16 − 8ln 2

B1

M1 A1

(4) [6]

Question Number

2.

Scheme

R 2 = 52 + 122 R = 13 12 tan α = 5 α ≈ 1.176

(a)

(b)

cos ( x − α ) =

Marks

M1 A1 M1 cao A1

6 13

M1

6 = 1.091 … 13 x = 1.091 … + 1.176 … ≈ 2.267 …

x − α = arccos

A1 awrt 2.3

A1

x − α = −1.091 … accept … = 5.19 … for M M1 x = −1.091 … + 1.176 … ≈ 0.0849 … awrt 0.084 or 0.085 A1

(c)(i)

(4)

Rmax = 13

ft their R

B1 ft

awrt 1.2, ft their α

M1 A1ft

(ii) At the maximum, cos ( x − α ) = 1 or x − α = 0 x = α = 1.176 …

(5)

(3) [12]

Question Number

3.

Scheme

Marks

y

(a)

shape Vertices correctly placed O

(2)

shape B1 Vertex and intersections with axes correctly placed B1

(2)

x

y

(b)

x

O

(c)

(d)

B1 B1

P : ( −1, 2 )

B1

Q : ( 0, 1)

B1

R : (1, 0 )

B1

1 x 2 2 Leading to x = 3 1 x < −1 ; 2 + x +1 = x 2 Leading to x = −6

x > −1 ;

2 − x −1 =

(3)

M1 A1 A1 M1 A1

(5) [12]

Question Number

4.

Scheme

Marks

x 2 − 2 x − 3 = ( x − 3)( x + 1)

(a)

⎛ 2 ( x − 1) x +1 − ⎜⎜ or ⎝ ( x − 3)( x + 1) ( x − 3)( x + 1) x−3 1 ¿ = = ( x − 3)( x + 1) x + 1

f ( x) =

2 ( x − 1) − ( x + 1) ( x − 3)( x + 1)

B1

⎛ 1⎞ ⎜ 0, ⎟ ⎝ 4⎠

(b)

(c)

Let y = f ( x )

M1 A1 cso

1 1 Accept 0 < y < , 0 < f ( x ) < etc. 4 4

1 x +1 1 x= y +1 yx + x = 1 1− x y= x 1 − x f −1( x ) = x

A1

(4)

B1 B1

(2)

y=

⎛ 1⎞ Domain of f −1 is ⎜ 0, ⎟ ⎝ 4⎠ (d)

⎞ ⎟⎟ ⎠

1 2x − 3 +1 1 1 = 2 2x − 2 8 x2 = 5 x = ±√5 fg ( x ) =

or

1 −1 x

M1 A1

ft their part (b) B1 ft

(3)

2

M1 both

A1 A1

(3) [12]

Question Number

5.

Scheme

Marks

sin 2 θ + cos 2 θ = 1 sin 2 θ cos 2 θ 1 + = 2 2 sin θ sin θ sin 2 θ 1 + cot 2 θ = cosec2 θ ¿

(a) ÷ sin 2 θ

M1 cso

A1

(2)

Alternative for (a) 1 + cot 2 θ = 1 +

cos 2 θ sin 2 θ + cos 2 θ 1 = = 2 2 sin θ sin θ sin 2 θ = cosec 2 θ ¿

2 ( cosec 2 θ − 1) − 9 cosec θ = 3

(b)

2 cosec θ − 9 cosec θ − 5 = 0 ( 2 cosec θ + 1)( cosec θ − 5 ) = 0 2

or or

cosec θ = 5 or

5sin θ + 9sin θ − 2 = 0 ( 5sin θ − 1)( sin θ + 2 ) = 0

θ = 11.5°, 168.5°

cso

A1

M1

2

sin θ =

M1

1 5

M1 M1 A1 A1 A1

(6) [8]

Question Number

6.

Scheme

(a)(i)

Marks

d 3x e ( sin x + 2 cos x ) ) = 3e3 x ( sin x + 2 cos x ) + e3 x ( cos x − 2sin x ) ( dx

M1 A1 A1 (3)

( = e ( sin x + 7 cos x ) ) 3x

d 3 5 x3 2 (ii) ( x ln ( 5x + 2 ) ) = 3x ln ( 5x + 2 ) + 5x + 2 dx

M1 A1 A1 (3)

2 d y ( x + 1) ( 6 x + 6 ) − 2 ( x + 1) ( 3 x + 6 x − 7 ) = 4 dx ( x + 1) 2

(b)

M1

( x + 1) ( 6 x 2 + 12 x + 6 − 6 x 2 − 12 x + 14 ) = 4 ( x + 1)

=

(c)

20

( x + 1)

3

M1 cso

¿

A1

d2y 60 15 =− =− 4 2 dx 4 ( x + 1)

M1

( x + 1)

M1

4

= 16

x = 1, − 3

A1 A1

both

A1

Note: The simplification in part (b) can be carried out as follows 2 ( x + 1) ( 6 x + 6 ) − 2 ( x + 1) ( 3x 2 + 6 x − 7 )

(6x =

( x + 1)

3

4

+ 18 x 2 + 18 x + 6 ) − ( 6 x 3 + 18 x 2 − 2 x − 14 )

( x + 1) 20 x + 20 20 ( x + 1) 20 = = = 4 4 3 ( x + 1) ( x + 1) ( x + 1) 4

M1 A1

(5)

(3) [14]

Question Number

7.

Scheme

(a)

Marks

f (1.4 ) = −0.568 … < 0 f (1.45 ) = 0.245 … > 0

M1

Change of sign (and continuity) ⇒ α ∈ (1.4, 1.45 ) (b)

3x3 = 2 x + 6 2x +2 x3 = 3 2 2 x2 = + 3 x ⎛ 2 2⎞ x= ⎜ + ⎟ ¿ ⎝ x 3⎠

√

(c)

A1

(2)

M1 A1 cso

x1 = 1.4371 x2 = 1.4347 x3 = 1.4355

(d) Choosing the interval (1.4345, 1.4355 ) or appropriate tighter interval.

A1

(3)

B1 B1 B1

(3)

M1

f (1.4345 ) = −0.01 ... f (1.4355 ) = 0.003 …

M1

Change of sign (and continuity) ⇒ α ∈ (1.4345, 1.4355 ) ⇒ α = 1.435 , correct to 3 decimal places

Note: α = 1.435 304 553 …

¿ cso

A1

(3) [11]