Edexcel Gce Core 3 Mathematics C3 6665 Advanced Subsidiary Jan 2007 Mark Scheme

Mark Scheme (Results) January 2007

GCE

GCE Mathematics Core Mathematics C3 (6665)

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

January 2007 6665 Core Mathematics C3 Mark Scheme Question Number 1.

Scheme

Marks

(a) sin 3θ = sin ( 2θ + θ ) = sin 2θ cos θ + cos 2θ sin θ

B1

= 2sin θ cos 2 θ + (1 − 2sin 2 θ ) sin θ

= 2sin θ − 2sin θ + sin θ − 2sin θ = 3sin θ − 4sin 3 θ ¿ 3

3

B1 B1

M1 cso A1

(5)

3

⎛ 3⎞ 3 3 3 3 9 3 3 (b) sin 3θ = 3 × √ − 4 ⎜ √ ⎟ = √ − √ = √ 4 4 16 16 ⎝ 4 ⎠ equivalent

or exact

M1 A1

(2) [7]

2.

(a) f ( x ) =

( x + 2) =

2

, −3 ( x + 2 ) + 3

( x + 2)

x 2 + 4 x + 4 − 3x − 6 + 3

( x + 2)

2

M1 A1, A1

2

=

x2 + x + 1

( x + 2)

2

¿

cso A1

(4)

2

1⎞ 3 ⎛ (b) x + x + 1 = ⎜ x + ⎟ + , > 0 for all values of x. 2⎠ 4 ⎝ 2

M1 A1, A1 (3)

2

1⎞ 3 ⎛ ⎜x+ ⎟ + 2⎠ 4 (c) f ( x ) = ⎝ 2 ( x + 2)

Numerator is positive from (b) 2 x ≠ −2 ⇒ ( x + 2 ) > 0 (Denominator is positive) Hence f ( x ) > 0

B1

Alternative to (b) d 2 1 3 x + x + 1) = 2 x + 1 = 0 ⇒ x = − ⇒ x 2 + x + 1 = ( dx 2 4

A parabola with positive coefficient of x 2 has a minimum ⇒ x 2 + x + 1 > 0 Accept equivalent arguments

1

(1) [8]

M1 A1 A1

(3)

Question Number

3.

Scheme

π

π

1 = √ 2 ⇒ P∈C 4 4 √2 Accept equivalent (reversed) arguments. In any method it must be clear π 1 or exact equivalent is used. that sin = 4 √2

(a) y =

⇒ x = 2sin

= 2×

dx dy = 2 cos y or 1 = 2 cos y dy dx dy 1 = May be awarded after substitution dx 2 cos y π dy 1 y= ⇒ = ¿ cso 4 dx √ 2

(b)

m′ = − √ 2

(c)

y−

π

4

Marks

B1

(1)

M1 A1 M1 A1

(4)

B1

(

= −√ 2 x −√ 2

y = − √ 2x + 2 +

)

π

M1 A1 A1

4

(4) [9]

4.

(i)

2 ⎛ ⎞ dy (9 + x ) − x ( 2x ) ⎜ 9 − x2 ⎟ = = 2 ⎜ ( 9 + x 2 )2 ⎟ dx (9 + x2 ) ⎝ ⎠ dy = 0 ⇒ 9 − x 2 = 0 ⇒ x = ±3 dx

⎛ 1⎞ ⎜ 3, ⎟ , ⎝ 6⎠

1⎞ ⎛ ⎜ −3, − ⎟ 6⎠ ⎝

x=

M1 A1

Final two A marks depend on second M only A1, A1

1 dy 3 = (1 + e 2 x ) 2 × 2 e 2 x dx 2

(ii)

M1 A1

1 1 1 dy 3 ln 3 ⇒ = (1 + eln 3 ) 2 × 2 eln 3 = 3 × 4 2 × 3 = 18 2 dx 2

(6)

M1 A1 A1 M1 A1

(5) [11]

2

Question Number

5.

Scheme

(a)

( )

2

R 2 = √ 3 + 12 ⇒ R = 2

tan α = √ 3 ⇒ α =

(b)

Marks

sin ( x + their α ) = x + their α =

x=

π 11π 2

,

M1 A1

π

accept awrt 1.05

3

M1 A1

1 2

M1

⎜ , ⎟ 6 ⎝ 6 6 ⎠

A1

π ⎛ 5π 13π ⎞ accept awrt 1.57, 5.76

6

M1 A1

(4)

(4) [8]

The use of degrees loses only one mark in this question. Penalise the first time it occurs in an answer and then ignore.

3

Question Number

6.

Scheme

Marks

(a) y = ln ( 4 − 2 x ) 1 e y = 4 − 2 x leading to x = 2 − e y Changing subject and removing ln 2 1 x 1 y = 2 − e ⇒ f −1 a 2 − e x ¿ cso 2 2 Domain of f −1 is (b) Range of f −1 is f −1 ( x ) < 2 (and f -1 (x ) ∈ )

M1 A1 A1 B1

(4)

B1

(1)

(c) Shape B1 1.5 B1 ln 4 B1 y=2

ln 4

B1

(d) x1 ≈ −0.3704, x2 ≈ −0.3452 cao If more than 4 dp given in this part a maximum on one mark is lost. Penalise on the first occasion. (e) x3 = −0.354 030 19 … x4 = −0.350 926 88 … x5 = −0.352 017 61 … x6 = −0.351 633 86 … k ≈ −0.352

B1, B1 (2)

Calculating to at least x6 to at least four dp M1 cao A1

Alternative to (e) k ≈ −0.352

(4)

(2) [13]

Found in any way

1 Let g ( x ) = x + e x 2 g ( −0.3515 ) ≈ +0.0003, g ( −0.3525 ) ≈ −0.001 Change of sign (and continuity) ⇒ k ∈ ( −0.3525, − 0.3515 ⇒ k = −0.352 (to 3 dp)

4

M1

) A1

(2)

Question Number

7.

Scheme

Marks

(a) f ( −2 ) = 16 + 8 − 8 ( = 16 ) > 0

B1

f ( −1) = 1 + 4 − 8 ( = −3) < 0

B1

Change of sign (and continuity) ⇒ root in interval ( −2, − 1) ft their calculation as long as there is a sign change (b)

B1ft

dy = 4 x3 − 4 = 0 ⇒ x = 1 dx

M1 A1

Turning point is (1, − 11)

A1

(c) a = 2, b = 4, c = 4

(3)

(3)

B1 B1 B1 (3)

(d) Shape ft their turning point in correct quadrant only 2 and −8

B1 B1 ft B1

(3)

B1

(1) [13]

(e)

Shape

5

Question Number

8.

Scheme

Marks

sec2 x − cosec2 x = (1 + tan 2 x ) − (1 + cot 2 x )

(i)

M1 A1

= tan 2 x − cot 2 x ¿ (ii)(a)

cso

A1

(3)

y = arccos x ⇒ x = cos y

B1

π ⎛π ⎞ x = sin ⎜ − y ⎟ ⇒ arcsin x = − y 2 ⎝2 ⎠

B1

(2)

B1

(1)

Accept

arcsin x = arcsin cos y (b) arccos x + arcsin x = y +

π 2

−y=

π 2

[6]

Alternatives for (i)

Rearranging cso

sec 2 x − tan 2 x = 1 = cosec 2 x − cot 2 x sec2 x − cosec 2 x = tan 2 x − cot 2 x ¿

M1 A1 A1

(3)

⎛ 1 1 sin 2 x − cos 2 x ⎞ − = ⎜ LHS = ⎟ cos 2 x sin 2 x cos 2 x sin 2 x ⎠ ⎝ 2 2 2 2 sin 2 x cos 2 x sin 4 x − cos 4 x ( sin x − cos x )( sin x + cos x ) − = = RHS = cos 2 x sin 2 x cos 2 x sin 2 x cos 2 x sin 2 x

sin 2 x − cos 2 x cos 2 x sin 2 x = LHS ¿ =

6

M1 A1

or equivalent

A1

(3)