Edexcel Gce Core 3 Mathematics C3 6665 Advanced Subsidiary Jan 2006 Mark Scheme

GCE Edexcel GCE

Core Mathematics C3 (6665)

January 2006

Core Mathematics C3 (6665)

Edexcel GCE

Mark Scheme (Results)

January 2006 6665 Core Mathematics C3 Mark Scheme Question Number

1.

Scheme

Marks

(a) y

(2, 7)

O

(b)

Shape unchanged Point

B1 B1

(2)

Shape Point

B1 B1

(2)

Shape (2, 4) ( −2, 4)

B1 B1 B1

x

y

(2, 4)

O

x

(c) ( −2, 4) y

O

(2, 4)

x

(3) [7]

Question Number

Scheme

Marks

x 2 − x − 2 = ( x − 2 )( x +1)

2.

At any stage

x ( 2 x + 3) x 2 x 2 + 3x = = ( 2 x + 3)( x − 2 ) ( 2 x + 3)( x − 2 ) x − 2

B1 B1

x ( x + 1) − 6 2 x 2 + 3x 6 − 2 = ( 2 x + 3)( x − 2 ) x − x − 2 ( x − 2 )( x + 1)

M1

x2 + x − 6 ( x − 2 )( x + 1)

A1

= =

( x + 3)( x − 2 ) ( x − 2 )( x + 1)

M1 A1

=

x+3 x +1

A1

(7) [7]

Alternative method

x 2 − x − 2 = ( x − 2 )( x +1)

At any stage

( 2 x + 3) appearing as a factor of the numerator at any stage 2 x 2 + 3 x ) ( x + 1) − 6 ( 2 x + 3) ( 2 x 2 + 3x 6 − = ( 2 x + 3)( x − 2 )( x + 1) ( 2 x + 3)( x − 2 ) ( x − 2 )( x + 1) =

( x − 2) ( 2x2 + 9x + 9) = ( 2 x + 3)( x − 2 )( x + 1)

or

2 x 3 + 5 x 2 − 9 x − 18 ( 2 x + 3)( x − 2 )( x + 1)

( 2 x + 3) ( x 2 + x − 6 ) ( 2 x + 3)( x − 2 )( x + 1)

or

can be implied

( x + 3) ( 2 x 2 − x − 6 ) ( 2 x + 3)( x − 2 )( x + 1)

Any one linear factor × quadratic ( 2 x + 3)( x − 2 )( x + 3) Complete factors = ( 2 x + 3)( x − 2 )( x + 1) x+3 = x +1

B1 B1 M1 A1

M1

A1 A1

(7)

Question Number

Scheme

Marks

dy 1 = dx x dy 1 At x = 3, = ⇒ m′ = − 3 dx 3

3.

accept

3 3x

Use of mm′ = −1

y − ln1 = −3 ( x − 3) y = −3 x + 9

M1 A1 M1 M1

Accept y = 9 − 3 x

A1

(5) [5]

dy 1 leading to y = −9 x + 27 is a maximum of M1 A0 M1 M1 A0 =3/5 = dx 3 x

4.

(a) (i)

d 3x+2 e ) = 3e3 x + 2 ( or 3e 2 e3 x ) ( dx dy = 3 x 2 e 3 x + 2 + 2 x e3 x + 2 dx

At any stage Or equivalent

B1 M1 A1+A1 (4)

(

)

d cos ( 2 x 3 ) = −6 x 2 sin ( 2 x 3 ) dx

(ii)

At any stage

3 3 3 d y −18 x sin ( 2 x ) − 3cos ( 2 x ) = dx 9 x2 Alternatively using the product rule for second M1 A1 −1 y = ( 3x ) cos ( 2 x3 )

M1 A1 M1 A1 (4)

dy −2 −1 = −3 ( 3 x ) cos ( 2 x 3 ) − 6 x 2 ( 3 x ) sin ( 2 x 3 ) dx Accept equivalent unsimplified forms

(b)

1 = 8cos ( 2 y + 6 )

dy dx

or

dx = 8cos ( 2 y + 6 ) dy

dy 1 = dx 8cos ( 2 y + 6 ) dy = dx

⎛ ⎞ 1 ⎜ = (±) ⎟ ⎛ 2 √ (16 − x 2 ) ⎠⎟ ⎛ x ⎞ ⎞ ⎝⎜ 8cos ⎜ arcsin ⎜ ⎟ ⎟ ⎝ 4 ⎠⎠ ⎝

M1 M1 A1

1

M1 A1 (5) [13]

Question Number

5.

Scheme

Marks

4 =0 x 1 4 x2 = + 2 2x ⎛2 1⎞ x= ⎜ + ⎟ ¿ ⎝ x 2⎠ 2 x2 −1 −

(a)

Dividing equation by x

M1

Obtaining x 2 = …

M1

√

cso

(b) x1 = 1.41, x2 = 1.39, x3 = 1.39 If answers given to more than 2 dp, penalise first time then accept awrt above.

f (1.3915 ) ≈ −3 ×10 , f (1.3925 ) ≈ 7 ×10 −3

Change of sign (and continuity)

−3

M1 Both, awrt

(a)

cso

R = √ (12 2 + 4 2 ) = √ 160

Accept if just written down, awrt 12.6

4 , ⇒ α ≈ 18.43° 12

awrt 18.4°

7 ( ≈ 0.5534 ) their R x + their α = 56.4° awrt 56° 360° − their principal value = … , 303.6° Ignore solutions out of range x = 38.0°, 285.2° If answers given to more than 1 dp, penalise first time then accept awrt above.

(c)(i) (ii)

A1

(3) [9]

R cos α = 12, R sin α = 4

tan α =

(b)

A1

⇒ α ∈ (1.3915, 1.3925) ⇒ α = 1.392 to 3 decimal places ¿

6.

(3)

B1, B1, B1 (3)

Choosing (1.3915, 1.3925 ) or a tighter interval

(c)

A1

cos ( x + their α ) =

minimum value is − √ 160

ft their R

cos ( x + their α ) = −1

x ≈ 161.57°

cao

M1 A1 M1, A1(4)

M1 A1 M1 A1, A1 (5)

B1ft M1 A1

(3) [12]

Question Number

7.

Scheme

Marks

(a) (i) Use of cos 2 x = cos 2 x − sin 2 x in an attempt to prove the identity. cos 2 x cos 2 x − sin 2 x ( cos x − sin x )( cos x + sin x ) = = = cos x − sin x ¿ cos x + sin x cos x + sin x cos x + sin x (ii) Use of cos 2 x = 2 cos 2 x − 1 in an attempt to prove the identity. Use of sin 2 x = 2sin x cos x in an attempt to prove the identity. 1 1 1 ( cos 2 x − sin 2 x ) = ( 2 cos 2 x − 1 − 2sin x cos x ) = cos 2 x − cos x sin x − ¿ 2 2 2 (b)

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

1 2

cso

A1

tan 2θ = 1 π ⎛ 5π 9π 13π ⎞ 2θ = , ⎜ , , ⎟ 4 ⎝ 4 4 4 ⎠

M1 M1 cso

A1

Using (a)(i)

M1

Using (a)(ii)

M1 A1

A1

Obtaining at least 2 solutions in range

M1

The 4 correct solutions If decimals ( 0.393,1.963,3.534,5.105) or degrees ( 22.5°,112.5°, 202.5°, 292.5° ) are

A1

π 5π 9π 13π 8

,

8

,

8

,

8

(3)

(3)

M1 any one correct value of 2θ

θ=

(2)

1 =0 2

1 ( cos 2θ − sin 2θ ) = 0 2 cos 2θ = sin 2θ ¿

(c)

M1

given, but all 4 solutions are found, penalise one A mark only. Ignore solutions out of range.

(4) [12]

Question Number

8.

Scheme

Marks

gf ( x ) = e 2( 2 x + ln 2)

(a)

= e 4 x e 2ln 2 = e 4 x e ln 4 = 4 e4 x ( Hence gf : x a 4 e4 x , x ∈

M1

)

M1 M1 Give mark at this point, cso A1

(4)

Shape and point

B1

(1)

B1

(1)

(b) y

4

O

(c) (d)

Range is

x Accept gf ( x ) > 0, y > 0

+

d ⎡⎣ gf ( x ) ⎤⎦ = 16 e 4 x dx 3 e4 x = 16 3 4 x = ln 16 x ≈ −0.418

M1 A1 M1 A1

(4) [10]