1
SUGGESTED SOLUTIONS
3 Unit MATHEMATICS
Question 1
(a)
y
= ln (cot
3
x)
= ln (cot x)
3
= 3 ln (cot x)
dy
= 3 ( cosec
2
x) .
1
dx
cot x
=  3 cosec
2
x
=
−
3
sin
cos
x
x
cot x
(b) Let f(x) = 2x
3
 5x
2
+ 3x  2
f(2) = 16 20 + 6 2
= 0
∴
x  2 is a factor
2x
2
 x + 1_________
x  2
2x
3
5x
2
+3x  2
2x
3
4x
2
 x
2
+3x
 x
2
+2x
x  2
x  2
∴
f(x) = (x  2)(2x
2
x + 1) = 0
when x2 = 0 or 2x
2
 x + 1 = 0
x = 2, the real root
Consider 2x
2
 x + 1 = 0
∆
= b
2
 4ac
= 1  4(2)(1)
=  7 < 0
ie
∆
< 0 so that there are NO real roots
∴
f(x) = 0 has only one real root.
(c)
To find points of intersection
 x
2
= x
2
+ 11x + 15
∴
2x
2
+ 11x + 15 = 0
(x + 3)(2x + 5) = 0
x + 3 = 0 or 2x + 5 = 0
x = 3 or x = 5/2
For y =  x
2
, dy
=  2x = 6 at x = 3
dx
ie the curve y =  x
2
has a gradient
of 6 at x = 3
For y = x
2
+ 11x + 15,
dy
=2x + 11 = 6+11 = 5 at x = 3
dx
ie the curve y = x
2
+ 11x + 15 has a
gradient of 5 at x = 3
If
Π
is the acute angle between the two
curves at x = 3
tan
θ
=
−
+
m
m
m m
1
2
1
2
1
=
6
5
1
6 5
−
+
( )

= 1/31
∴
Π
= 1
!
51’
(d)
(i)
sin 2x
= 2 sin x cox x
1 + cos 2x
1 + (2cos
2
x  1)
= 2 sin x cos x
2 cos
2
x
= sin x
cos x
= tan x
∴
when x=15
!
, tan15
!
=
sin
cos
30
1
30
+
=
1
2
1
3
2
+
=
1
2
2
3
2
+
=
1
2
3
+
(ii)
cot 15
!
= __1
__ = 2 +
3
tan 15
!
∴
tan 15
!
+ cot 15
!
=
1
2
3
2
3
+
+
+
=
2
3
2
3
2
3
2
3
−
+
−
+
+
(
)(
)
=
2
3
4
3
2
3
−
−
+
+
=
2
3
2
3
−
+
+
= 4
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