January 2005 6664 Core Mathematics C2 Mark Scheme Question Number 1.
Scheme
Marks
5 5 (3 2 x) 5 (35 ) 3 4 (2 x) 33 (2 x) 2 1 2 243,810 x,1080 x 2
M1 B1, A1, A1 (4) (4 marks)
2.
3.
5 13 1 11 (a) ( , ), (9,5) 2 2
M1, A1 (2)
(b) r 2 (9 5) 2 (5 1) 2 ( 52) or r 2 (13 9)2 (11 5)2 ( 52) (or equiv.) Equation of circle: ( x 9) 2 ( y 5) 2 52 (or equiv.)
M1 M1 A1ft A1 (4) (6 marks) M1
(a) log 3 log 5 log 5 or x log3 = log5 x log 3 1.46 x
A1 A1 cao (3)
(a) log 2 (
2x 1 )2 x
M1
2x 1 2 2 or 4 x 2x 1 4x 1 x or 0.5 2
M1 M1 A1
(4) (7)
4.
(a) 5(1 sin x) 3(1 sin x)
M1
2
5 5 sin 2 x 3 3 sin x 0 5 sin 2 x 3 sin x 2
A1 cso (2)
(b) 0 (5 sin x 2)(sin x 1) 2 sin x , -1 5 2 x 23.6 sin x 5 , 156.4
M1 (both) ( 23.6 or 156.4) (180- )
A1 cso B1 M1 B1
sin x 1 x 270
(ignore extra solutions outside the range)
(5) (7)
Question Number 5.
Scheme
(a)
Marks
f (2) 1 8 2 4 2a b 1 f (1) 28 1 2 a b 28 2 a b 1 solving a 10, b 21 a b 31
(b) f (3) 27 18 3a b 27 18 30 21 0
M1 A1 M1 A1 M1 A1 (6)
( x 3) is a factor
M1 A1 c.s.o (2) (8)
6.
5.832 ( 0.81) 7.2 r 0.9
(a) ar 7.2, ar 3 5.832 r 2
(b) a
7.2 , 8 (a)
(c) s50
M1 A1 (2) M1, A1 (2)
8(1 (0.9) 50 ) 1 0.9
M1
79.588 (3dp)
A1 c.a.o (2) 8 ( 80) 1 0.9 s s50 80 (c) 0.412
(d) s
7.
M1 A1
(2) (8)
(a) r 8 0.7, 5.6(cm)
M1, A1
(b) BC 8 11 2 8 11 cos 0.7 BC 7.098 Perimeter = (a) (11 8) BC , 15.7(cm)
(2) M1 A1 M1, A1cao (4)
2
2
2
1 1 (c) ab sin c 11 8 sin 0.7, AWRT 28.3 2 2 1 2 1 2 Sector r , 8 0.7 2 2 Area of R 28.345..... 22.4 5.9455 5.95(cm 2 )
M1, A1 M1, A1 A1
(5) (11)
Question Number 8. (a) x 2 6 x 10 3x 20
Scheme
x 3x 10 0 ( x 5)( x 2) 0 so x ,5 or 2 sub a value for x to obtain a value for y in y 3x 20, y 5 or 26
Marks
M1
2
(b) line – curve =, 10 3x x 2 3 2 x3 2 ( 10 3 x x ) dx 10 x x 2 3
M1, A1 M1, A1 (5) M1, A1 M1 A2/1/0
2
3 2 x3 3 8 3 125 10 x x (20 4 ) (50 25 ) 2 3 5 2 3 2 3 1 5 1 11 45 57 3 6 6 ALT (b)
3
x (-1 each incorrect term) 3x 2 10 x 3 125 use of limits ( 8 12 20) ( 75 50) (51 1 3 ) 3 3 1 Area of Trapezium (5 26)(2 5) (108 1 2) or 46 - - 62.5 (from integration) 2 343 Shaded area = Trapezium - 108 1 2 51 1 3 57 1 6 or or 57.16 6
(x
2
6 x 10)dx
M1 A1 (7) (12) M1 A2/1/0
M1 B1 M1 A1 (7)
Question Number 9. (a) Perimeter 2 x 2 y x 80
Scheme
Marks
B1
1 A 2 xy x 2 2 80 2 x x and sub in to A y 2 1 A 80 x 2 x 2 x 2 x 2 2
Area
i.e. A 80 x (2
B1 M1
A1 c.s.o (4)
)x 2
2 dA (b) 80 2(2 ) x or 80 4x x (or equiv.) dx 2 40 dA 80 so x , or or Awrt 11.2 0 40 (2 ) x dx 2 4 2 2 2 d A (c) 4 dx 2 0 A is Max
(d) Max Area 80(b) (2 = 448(m 2 )
2
)(b) 2
M1, A1 M1, A1 (4)
M1 A1
(2)
M1 (448 only for A1)
A1 cao (2)
(12)