0 f"(\frac{1}{5}) =\frac{-144}{25 }" >
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UPSEE 2019 Solved Paper
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© examsnet.com
Question : 116
Total: 150
If
f
(
x
)
=
(
x
−
1
)
2
(
x
+
1
)
3
, then the function f has
a local maximum at
x
=
1
5
a local minimum at
x
=
1
5
a local minimum at
x
=
−
1
a local maximum at
x
=
−
1
Validate
Solution:
We have,
f
(
x
)
=
(
x
−
1
)
2
(
x
+
1
)
3
f
′
(
x
)
=
2
(
x
−
1
)
(
x
+
1
)
3
+
3
(
x
−
1
)
2
(
x
+
1
)
2
⇒
f
′
(
x
)
=
(
x
−
1
)
(
x
+
1
)
2
[
2
x
+
2
+
3
x
−
3
]
⇒
f
′
(
x
)
=
(
x
−
1
)
(
x
+
1
)
2
(
5
x
−
1
)
For maxima or minima put
f
′
(
x
)
=
0
∴
(
x
−
1
)
(
x
+
1
)
2
(
5
x
−
1
)
=
0
⇒
x
=
−
1
,
1
5
,
1
f
"
(
x
)
=
(
x
+
1
)
2
(
5
x
−
1
)
+
2
(
x
−
1
)
(
x
+
1
)
(
5
x
−
1
)
+
5
(
x
−
1
)
(
x
+
1
)
2
f
"
(
−
1
)
=
0
f
"
(
1
)
=
16
>
0
f
"
(
1
5
)
=
−
144
25
<
0
∴
f
(
x
)
has local maxima at
x
=
1
5
© examsnet.com
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