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

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GCE

Edexcel GCE Mathematics Core Mathematics C3 (6665)

June 2006

Mathematics

Edexcel GCE

Mark Scheme (Results)

EDEXCEL CORE MATHEMATICS C3 (6665) – JUNE 2006 Question Number 1.

(a)

MARK SCHEME

Scheme

(3x + 2)(x − 1) , (x + 1)(x − 1)

=

Marks

3x + 2 x +1

M1B1, A1 (3)

Notes M1 attempt to factorise numerator, usual rules B1 factorising denominator seen anywhere in (a), A1 given answer If factorisation of denom. not seen, correct answer implies B1 (b) Expressing over common denominator x(3 x + 2 ) − 1 3x + 2 1 − = x +1 x( x + 1) x(x + 1)

M1

(3 x 2 − x − 2) x − ( x − 1) ] [Or “Otherwise” : x( x 2 − 1) Multiplying out numerator and attempt to factorise

[3x

2

M1

]

+ 2 x − 1 ≡ (3 x − 1)( x + 1) Answer:

3x − 1 x

A1

(3) (6 marks)

2.

(a)

dy 1 = 3 e 3x + x dx

B1M1A1(3)

Notes B1 3e3 x M1 : (b)

a bx

A1: 3 e 3 x +

1 x

(5 + x )

1 2 2

(

3 5 + x2 2

B1

)

1 2

. 2x

= 3x (5 + x 2 )2 1

M1 for kx(5 + x 2 ) m

M1 A1

(3)

(6 marks)

2

EDEXCEL CORE MATHEMATICS C3 (6665) – JUNE 2006 Question Number 3.

MARK SCHEME

Scheme

(a)

(b)

(c)

Marks

Mod graph, reflect for y < 0

M1

(0, 2), (3, 0) or marked on axes

A1

Correct shape, including cusp

A1

Attempt at reflection in y = x

M1

Curvature correct

A1

(–2, 0), (0, 3) or equiv.

B1

Attempt at ‘stretches’

M1

(0, –1) or equiv.

B1

(1, 0)

B1

(3)

(3)

(3) (9 marks)

4.

(a)

425 ºC

B1

(b)

300 = 400 e - 0.05 t + 25

⇒ 400 e - 0.05 t = 275

sub. T = 300 and attempt to rearrange to e–0.05t = a, where a ∈ Q e −0.05t =

(c)

275 400

M1 A1

M1 correct application of logs

M1

t = 7.49

A1

dT = − 20 e - 0.05 dt

( M1 for k e - 0.05 t )

t

At t = 50, rate of decrease = ( ± ) 1.64 ºC/min (d) T > 25,

(1)

(since e - 0.05

t

→ 0 as t → ∞ )

M1 A1 A1

(3)

B1

(1)

(9 marks)

3

(4)

EDEXCEL CORE MATHEMATICS C3 (6665) – JUNE 2006 Question Number 5.

(a)

MARK SCHEME

Scheme Using product rule:

dy = 2 tan 2 x + 2( 2 x − 1) sec 2 2 x dx

sin 2 x 1 " and “ sec 2 x = " cos 2 x cos 2 x 1 sin 2 x ] [= 2 + 2( 2 x − 1) cos 2 2 x cos 2 x dy Setting = 0 and multiplying through to eliminate fractions dx [ ⇒ 2 sin 2 x cos 2 x + 2( 2 x − 1) = 0]

Use of “tan 2x =

(b)

Marks M1 A1 A1

M1

M1

Completion: producing 4k + sin 4k − 2 = 0 with no wrong working seen and at least previous line seen. AG

A1*

x1 = 0.2670, x 2 = 0.2809, x3 = 0.2746, x 4 = 0.2774,

M1 A1 A1 (3)

Note: M1 for first correct application, first A1 for two correct, second A1 for all four correct Max –1 deduction, if ALL correct to > 4 d.p. M1 A0 A1 SC: degree mode: M1 x1 = 0.4948 , A1 for x2 = 0 .4914, then A0; max 2 (c) Choose suitable interval for k: e.g. [0.2765, 0.2775] and evaluate f(x) at these values Show that 4k + sin 4k − 2 changes sign and deduction

(6)

M1 A1

(2)

[f(0.2765) = –0.000087.., f(0.2775) = +0.0057] Note: Continued iteration: (no marks in degree mode) Some evidence of further iterations leading to 0.2765 or better M1; Deduction A1 (11 marks)

4

EDEXCEL CORE MATHEMATICS C3 (6665) – JUNE 2006 Question Number 6.

(a)

Scheme

Marks

Dividing sin 2 θ + cos 2 θ ≡ 1 by sin 2 θ to give

Completion: (b)

MARK SCHEME

sin 2 θ cos 2 θ 1 + ≡ 2 2 sin θ sin θ sin 2 θ 1 + cot 2 θ ≡ cos ec 2θ ⇒

(

M1

cos ec 2θ − cot 2 θ ≡ 1

)(

cos ec 4θ − cot 4 θ ≡ cos ec 2θ − cot 2 θ cos ec 2θ + cot 2 θ

(

≡ cos ec 2θ + cot 2 θ

)

using (a)

AG

)

AG

⇒ 1 + cot 2 θ + cot 2 θ ≡ 2 − cot θ

1⎞ ⎛ ⎜ cot θ = ⎟ 2⎠ ⎝

θ = 135 º

M1

{using (a)}

−1= 0

(2 cot θ

(2)

M1

Forming quadratic in cot θ

Solving:

(2)

M1 A1*

Notes: (i) Using LHS = (1 + cot2 θ )2 – cot4 θ , using (a) & elim. cot4 θ M1, conclusion {using (a) again} A1* (1− cos 2 θ )(1+ cos 2 θ ) (ii) Conversion to sines and cosines: needs for M1 sin 4 θ cos ec 2θ + cot 2 θ ≡ 2 − cot θ (c) Using (b) to form

2 cot 2 θ + cot θ

A1*

A1

− 1)(cot θ + 1) = 0 or

to cot θ =

cot θ = − 1

(or correct value(s) for candidate dep. on 3Ms)

M1 A1 A1√

(6)

Note: Ignore solutions outside range Extra “solutions” in range loses A1√, but candidate may possibly have more than one “correct” solution. (10 marks)

5

EDEXCEL CORE MATHEMATICS C3 (6665) – JUNE 2006 Question Number 7.

Scheme

(a)

(b) f(x) ∈ R (c)

,

Marks

Log graph: Shape

B1

Intersection with –ve x-axis

dB1

(0, ln k), (1 – k, 0) Mod graph :V shape, vertex on +ve x-axis

B1 B1

⎛k ⎞ (0, k) and ⎜ , 0 ⎟ ⎝2 ⎠ – ∞ < f ( x) < ∞ , – ∞ < y < ∞

B1

(5)

B1

(1)

⎛ k ⎞ f ⎜| − |⎟ ⎝ 2 ⎠

M1

2k ⎛k⎞ fg ⎜ ⎟ = ln{k + | − k |} or 4 ⎝4⎠

= ln ( (d)

MARK SCHEME

3k ) 2

A1

dy 1 = dx x + k

(2)

B1 1 2 = 3+ k 9

Equating (with x = 3) to grad. of line;

k = 1½

M1; A1 A1√

(4)

(12 marks)

6

EDEXCEL CORE MATHEMATICS C3 (6665) – JUNE 2006 Question Number 8.

(a)

MARK SCHEME

Scheme

Marks

Method for finding sin A

M1

7 4

sin A = –

A1 A1

7 , exact. 4 Second A1 for sign (even if dec. answer given) Use of sin 2 A ≡ 2 sin A cos A

Note:

First A1 for

sin 2 A = −

Note:

M1

3 7 or equivalent exact 8

± f.t.

A1√

(5)

Requires exact value, dependent on 2nd M

(b)(i) π⎞ π⎞ π π ⎛ ⎛ cos⎜ 2 x + ⎟ + cos⎜ 2 x − ⎟ ≡ cos 2 x cos − sin 2 x sin 3⎠ 3⎠ 3 3 ⎝ ⎝

≡ 2 cos 2 x cos

+ cos 2 x cos

π 3

+ sin 2 x sin

π 3

π M1 3

A1

[This can be just written down (using factor formulae) for M1A1] ≡ cos 2 x

AG

Note: M1A1 earned, if ≡ 2 cos 2 x cos

A1*

(3)

π

just written down, using factor theorem 3 requires some working after first result.

Final A1* (b)(ii) dy = 6 sin x cos x − 2 sin 2 x dx or

B1 B1

π⎞ π⎞ ⎛ ⎛ 6 sin x cos x − 2 sin ⎜ 2 x + ⎟ − 2 sin ⎜ 2 x − ⎟ 3⎠ 3⎠ ⎝ ⎝ = 3 sin 2 x − 2 sin 2 x = sin 2x

AG

M1 A1*

(4)

Note: First B1 for 6 sin x cos x ; second B1 for remaining term(s) (12 marks)

7

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