6665 Core C3
Mark Scheme (Post standardisation)
Question
Number
1
(b)
(a)
Scheme
Marks
sin 2  cos 2 
1


2
2
cos  cos 
cos 2 
2
2
Completion : 1 + tan   sec  (no errors seen)
M1
Use of 1  tan 2   sec 2  :
M1
Dividing by cos 2  :
2 (sec 2   1)  sec   1
A1
(2)
[ 2 sec 2   sec   3  0 ]
Factorising or solving: (2 sec   3)(sec  1)  0
3
[ sec   –
cos   –
or sec   1 ]
2
 0
M1
B1
2
3
;
M1 A1
1  131.8 °
 2  228.2 °
A1√
(6)
[A1ft for  2  360° – 1 ]
[8]
1
6665 Core C3
Mark Scheme (Post standardisation)
Question
Number
2
Scheme
Marks
(a) (i) 6 sin x cos x  2 sec 2 x tan 2 x
or 3 sin 2 x
(ii)
M1A1A1
(3)
 2sec 2 x tan 2 x
[M1 for 6 sin x]
B1M1A1
(3)
1
3( x  ln 2 x) 2 (1  )
x
[ B1 for 3( x  ln 2 x) 2 ]
(b)
Differentiating numerator to obtain 10x – 10
Differentiating denominator to obtain 2(x-1)
Using quotient rule formula correctly:
To obtain
dy ( x  1) 2 (10 x  10)  (5 x 2  10 x  9)2( x  1)

dx
( x  1) 4
2( x  1)[5( x  1) 2  (5 x 2  10 x  9)
( x  1) 4
Simplifying to form
=
3
(a)

8
( x  1) 3
*
(c.s.o.)
5x  1
3

( x  2)( x  1) x  2
5 x  1  3( x  1)
=
( x  2)( x  1)
M1 for combining fractions even if the denominator is not lowest common
=
2x  4
=
( x  2)( x  1)
2( x  2)
2
=
( x  2)( x  1)
x 1
B1
M1
M1 A1 cso
(4)
*
M1 must have linear numerator
(b)
2
 xy  y  2  xy = 2 + y
x 1
2x
f –1(x) =
o.e.
x
2
2
(attempt)
[
]
fg(x) = 2
" g"  1
x 4
1
2
=
and finding x2 = …;
x = 2
Setting 2
4
x 4
y=
M1A1
A1
(3)
M1
M1; A1
(3)
[10]
2
6665 Core C3
Mark Scheme (Post standardisation)
Question
Number
4
(a)
Scheme
f  (x) = 3 ex –
3e x

(c)
M1A1A1
1
2x
(3)
M1
1
=0
2x
 6 e   1
Marks
 
1
6
e

A1 cso
(*)
x1  0.0613..., x 2  0.1568.., x3  0.1425..., x 4  0.1445....
(2)
M1 A1 (2)
[M1 at least x1 correct , A1 all correct to 4 d.p.]
(d)
1
with suitable interval
2x
e.g. f  (0.14425) = – 0.0007
Using f  (x) = 3 ex –
M1
f  (0.14435) = + 0.002(1)
Accuracy (change of sign and correct values)
A1
(2)
[9]
3
6665 Core C3
Mark Scheme (Post standardisation)
Question
Number
5
Scheme
Marks
(a) cos 2A = cos2 A – sin2 A ( + use of cos2 A + sin2 A  1)
= (1 – sin2 A); – sin2 A = 1 – 2 sin2 A
(*)
A1
2sin 2  3cos 2  3sin   3  4sin  cos ; 3(1  2sin 2  )  3sin   3
(b)
(2)
B1; M1
M1
 4 sin  cos   6 sin 2   3 sin 
(*)
 sin  (4 cos   6 sin   3)
(c)
M1
A1
(4)
4 cos   6 sin   R sin  cos   R cos  sin 
Complete method for R (may be implied by correct answer)
[ R 2  4 2  6 2 , R sin   4 , R cos   6 ]
R=
A1
52 or 7.21
Complete method for  ;
(d)
  0.588 (allow 33.7º)
sin  ( 4 cos  + 6 sin   3 ) = 0
3
M1 A1 (4)
M1
 0
sin(   0.588) =
M1
B1
 0.4160..
(24.6º)
M1
52
  0.588 = (0.4291), 2.7125 [or   33.7 ° = (24.6º), 155.4°]
dM1
  2.12
A1
cao
(5)
[15]
4
6665 Core C3
Mark Scheme (Post standardisation)
Question
Number
6.
Scheme
(a)
Marks
Translation  by 1
M1
Intercepts correct
A1
x  0, correct “shape”
B1
y
0
–2
x
2
(2)
a
(b)
y
[provided not just original]
–3
0
3
x
b
(c)
Reflection in y-axis
B1
Intercepts correct
B1
(3)
B1 B1
(2)
a = – 2, b = – 1
(d) Intersection of y = 5x with y = – x – 1
Solving to give x = –
M1A1
M1A1 (4)
1
6
[11]
[Notes:
(i) If both values found for 5x = – x – 1 and 5x = x – 3, or solved
algebraically, can score 3 out of 4 for x = –
1
6
and x = – ¾;
required to eliminate x = – ¾ for final mark.
(ii) Squaring approach: M1 correct method,
24x2 + 22x + 3 = 0 ( correct 3 term quadratic, any form) A1
Solving M1, Final correct answer A1.]
5
6665 Core C3
Mark Scheme (Post standardisation)
Question
Number
7
(a)
Scheme
(300 = 2500a);
(b)
2800a
1 a
a = 0.12 (c.s.o) *
Setting p = 300 at t = 0  300 =
Marks
M1
dM1A1 (3)
2800(0.12)e 0.2 t
;
e 0.2 t  16.2...
0.2 t
1  0.12e
Correctly taking logs to 0.2 t = ln k
M1A1
t = 14
A1
(4)
B1
(1)
1850 =
M1
(13.9..)
(c) Correct derivation:
(Showing division of num. and den. by e 0.2 t ; using a)
M1
(d) Using t   , e  0.2t  0,
p
A1
336
 2800
0.12
(2)
[10]
6