January 2006
6665 Core Mathematics C3 Mark Scheme
Question Number
1.
Scheme
Marks
(a) y
(2, 7)
O
(b)
Shape unchanged Point
B1 B1
(2)
Shape Point
B1 B1
(2)
Shape (2, 4) (ď€2, 4)
B1 B1 B1
x
y
(2, 4)
O
x
(c)
(ď€2, 4) y
(2, 4)
O
x
1
(3) [7]
January 2006
6665 Core Mathematics C3 Mark Scheme
Question Number
Scheme
Marks
x2 x 2 x 2 x 1
2.
At any stage
x 2 x 3 2 x 3x x 2 x 3 x 2 2 x 3 x 2 x 2
B1
2
B1
x x 1 6 2 x 2 3x 6 2 2 x 3 x 2 x x 2 x 2 x 1
M1
x2 x 6 x 2 x 1
A1
x 3 x 2 x 2 x 1
M1 A1
x3 x 1
A1
(7) [7]
Alternative method
x2 x 2 x 2 x 1
At any stage
2 x 3 appearing as a factor of the numerator at any stage 2 x 2 3x x 1 6 2 x 3 2 x 2 3x 6 2 x 3 x 2 x 2 x 1 2 x 3 x 2 x 1
x 2 2x2 9x 9 2 x 3 x 2 x 1
or
2 x3 5 x 2 9 x 18 2 x 3 x 2 x 1
2 x 3 x 2 x 6 2 x 3 x 2 x 1
or
can be implied
x 3 2 x 2 x 6 2 x 3 x 2 x 1
Any one linear factor × quadratic 2 x 3 x 2 x 3 Complete factors 2 x 3 x 2 x 1
x3 x 1
B1 B1 M1 A1
M1
A1 A1
2
(7)
January 2006
6665 Core Mathematics C3 Mark Scheme
Question Number
Scheme
Marks
dy 1 dx x dy 1 At x = 3, m 3 dx 3
3.
accept
3 3x
Use of mm 1
y ln1 3 x 3 y 3x 9
M1 A1 M1 M1
Accept y 9 3x
A1
(5) [5]
dy 1 leading to y 9 x 27 is a maximum of M1 A0 M1 M1 A0 =3/5 dx 3 x
4.
(a) (i)
d 3x2 e 3e3 x 2 or 3e2e3 x dx dy 3 x 2 e3 x 2 2 x e 3 x 2 dx
At any stage Or equivalent
B1 M1 A1+A1 (4)
d cos 2 x3 6 x 2 sin 2 x3 dx
(ii)
At any stage
3 3 3 d y 18 x sin 2 x 3cos 2 x dx 9x2 Alternatively using the product rule for second M1 A1 1 y 3x cos 2 x3
M1 A1 M1 A1 (4)
dy 2 1 3 3x cos 2 x3 6 x 2 3x sin 2 x3 dx Accept equivalent unsimplified forms
(b)
1 8cos 2 y 6
dy dx
or
dx 8cos 2 y 6 dy
dy 1 dx 8cos 2 y 6 dy dx
1 2 16 x 2 x 8cos arcsin 4
M1 M1 A1
1
M1 A1 (5) [13]
3
January 2006
6665 Core Mathematics C3 Mark Scheme
Question Number
5.
Scheme
Marks
4 0 x 1 4 x2 2 2x 2 1 x x 2 2 x2 1
(a)
Dividing equation by x
M1
Obtaining x 2 …
M1
cso
(b) x1 1.41, x2 1.39, x3 1.39 If answers given to more than 2 dp, penalise first time then accept awrt above.
f 1.3915 3 103 , f 1.3925 7 103 Change of sign (and continuity)
M1 Both, awrt
(a)
cso
R 122 42 160
Accept if just written down, awrt 12.6
4 , 18.43 12
awrt 18.4
7 0.5534 their R x + their 56.4 awrt 56° = … , 303.6 360 their principal value Ignore solutions out of range x 38.0, 285.2 If answers given to more than 1 dp, penalise first time then accept awrt above.
(c)(i) (ii)
A1
(3) [9]
R cos 12, R sin 4
tan
(b)
A1
1.3915,1.3925 1.392 to 3 decimal places
6.
(3)
B1, B1, B1 (3)
Choosing 1.3915,1.3925 or a tighter interval
(c)
A1
cos x their
minimum value is 160
ft their R
cos x their 1 x 161.57
4
cao
M1 A1 M1, A1(4)
M1 A1 M1 A1, A1 (5)
B1ft M1 A1
(3) [12]
January 2006
6665 Core Mathematics C3 Mark Scheme
Question Number
7.
Scheme
Marks
(a) (i) Use of cos 2 x cos2 x sin 2 x in an attempt to prove the identity. cos 2 x cos 2 x sin 2 x cos x sin x cos x sin x cos x sin x cos x sin x cos x sin x cos x sin x (ii) Use of cos 2 x 2cos2 x 1 in an attempt to prove the identity. Use of sin 2 x 2sin x cos x in an attempt to prove the identity. 1 1 1 cos 2 x sin 2 x 2cos2 x 1 2sin x cos x cos2 x cos x sin x 2 2 2 (b)
cos cos sin cos 2 cos sin
1 2
cso
A1
tan 2 1 5 9 13 2 , , , 4 4 4 4
M1 M1 cso
A1
Using (a)(i)
M1
Using (a)(ii)
M1 A1
(3)
A1
Obtaining at least 2 solutions in range
M1
The 4 correct solutions If decimals 0.393,1.963,3.534,5.105 or degrees 22.5,112.5, 202.5, 292.5 are given, but all 4 solutions are found, penalise one A mark only. Ignore solutions out of range.
A1
5 9 13 , , , 8 8 8 8
5
(3)
M1 any one correct value of 2
(2)
1 0 2
1 cos 2 sin 2 0 2 cos 2 sin 2 (c)
M1
(4) [12]
January 2006
6665 Core Mathematics C3 Mark Scheme
Question Number
8.
Scheme
gf x e
(a)
Marks
2 2 x ln 2
e e e4 x eln 4 4e4 x Hence gf : x 4e4 x , x
M1
4 x 2ln 2
M1 M1 Give mark at this point, cso A1
(4)
Shape and point
B1
(1)
B1
(1)
(b) y
4
O
(c) (d)
Range is
x Accept gf x 0, y 0
d gf x 16e4 x dx 3 e4 x 16 3 4 x ln 16 x 0.418
6
M1 A1 M1 A1
(4) [10]