May 2006 - 6665 Core C3

June 2006
6665 Core Mathematics C3
Mark Scheme
Question
number
1.
(a)
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
Expressing
over common denominator
(b)
3x  2
1
x3x  2  1


x 1
xx  1
x x  1
(3 x 2  x  2) x  ( x  1)
]
x( x 2  1)
Multiplying out numerator and attempt to factorise
3x 2  2 x  1  3x  1 x  1
3x  1
Answer:
x
M1
[Or “Otherwise” :


M1
A1
(3)
Total 6 marks
2.
(a)
dy
1
 3 e3x 
dx
x
B1M1A1 (3)
Notes
B1 3e3 x
M1 :
(b)
a
bx
A1: 3 e 3 x 
1
x
5  x 
1
2 2
1
3
2 2


5 x
. 2x
2
B1
= 3x 5  x 
1
2 2
M1 for
M1 A1
(3)
kx(5  x )
2 m
Total 6 marks
Question
Number
3.
(a)
Scheme
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
(3)
(b)
(c)
(0, –1) or equiv.
(1, 0)
(3)
B1
B1
(3)
Total 9 marks
Question
Scheme
Marks
(a)
425 ºC
B1
(b)
300  400 e - 0.05 t  25
Number
4.
 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   )
(4)
M1 A1
A1
(3)
B1
(1)
Total 9 marks
Question
Number
5.
(a)
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
sin 2 x
1
 2(2 x  1)
]
[ 2
cos 2 x
cos 2 2 x
dy
= 0 and multiplying through to eliminate fractions
Setting
dx
[  2 sin 2 x cos 2 x  2(2 x  1) = 0]
Use of “tan 2x =
Completion: producing 4k  sin 4k  2  0 with no wrong
working seen and at least previous line seen. AG
(b)
x1  0.2670, x 2  0.2809, x3  0.2746, x 4  0.2774,
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 x 2 = 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
Marks
M1 A1 A1
M1
M1
A1*
(6)
M1 A1 A1 (3)
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)
Question
Number
6.
(a)
Scheme
Marks
Dividing sin 2   cos 2   1 by sin 2  to give
sin 2  cos 2 
1


2
2
sin 
sin 
sin 2 
Completion: 1  cot 2   cos ec 2  cos ec 2  cot 2   1
AG
(b)
cos ec 4  cot 4   cos ec 2  cot 2  cos ec 2  cot 2  
 cos ec 2  cot 2  
using (a)
AG
Notes:
(i) Using LHS = (1 + cot2  )2 – cot4  , using (a) & elim. cot4 
M1,
conclusion {using (a) again} A1*
(ii) Conversion to sines and cosines: needs
(1  cos 2  )(1  cos 2  )
for M1
sin 4 
cos ec 2  cot 2   2  cot 
(c) Using (b) to form
 1  cot 2   cot 2   2  cot 
Solving:
  135 º
M1
A1*
(2)
M1
A1
 1cot   1  0
1

 cot   
2

(2)
{using (a)}
1 0
2 cot 
A1*
M1
Forming quadratic in cot 
2 cot 2   cot 
M1
or
to cot  =
cot   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)
Question
Number
Scheme
Marks
Log graph: Shape
7.
(a)
Intersection with –ve xaxis
(0, ln k), (1 – k, 0)
Mod graph :V shape,
vertex on +ve x-axis
(b) f(x)  R
,
(d)
dB1
B1
B1
k 
(0, k) and  , 0 
2 
–   f ( x)   , –   y  
B1
(5)
B1
(1)
 k 
f |  |
 2 
M1
2k
k
(c) fg   = ln{k + |  k |} or
4
4
= ln (
B1
3k
)
2
A1
dy
1

dx x  k
Equating (with x = 3) to grad. of line;
(2)
B1
1
2

3 k 9
k = 1½
M1; A1
A1
(4)
(12 marks)
Question
Number
8.
(a)
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:
(b)(i)
3 7
or equivalent exact
8
 f.t.
A1
cos 2x cos
3
(5)
Requires exact value, dependent on 2nd M






− sin 2x sin
+
cos  2 x   + cos  2 x   ≡ cos 2x cos
3
3
3
3



M1

+ sin 2x sin
M1
3
 2 cos 2 x cos

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
just written down, using factor
theorem
Final A1* requires some working after first result.
(b)(ii) dy
 6 sin x cos x  2 sin 2 x
dx
or
(3)
B1 B1




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)

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