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)
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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