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Solution —
2018
Question 2
Differentiation & Applications
|
Integration & Applications
(a)
y
=
3
ln
(
4
x
3
+
2
)
d
y
d
x
=
3
(
12
x
2
4
x
3
+
2
)
=
36
x
2
4
x
3
+
2
=
18
x
2
2
x
3
+
1
∎
(b)
∫
3
x
2
−
2
x
d
x
=
∫
(
3
x
2
x
−
2
x
)
d
x
=
∫
(
3
x
3
2
−
2
x
−
1
2
)
d
x
=
3
(
x
5
2
5
2
(
1
)
)
−
2
(
x
1
2
1
2
(
1
)
)
+
c
=
6
x
5
2
5
−
4
x
1
2
+
c
∎
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