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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)


C3 MS Jan 2007