0 \Rightarrow f(x) is minimum at x=1 Now, f(1)=(1)^{4}-(1)^{2}-2+5=3 ∴ Absolute minimum value is 3" >
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AP EAPCET 23-Aug-2021 Shift 1 Paper
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Section:
Mathematics
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© examsnet.com
Question : 71
Total: 160
The absolute minimum value of
x
4
−
x
2
−
2
x
+
5
is ______
5
3
7
Does not exist
Validate
Solution:
x
4
−
x
2
−
2
x
+
5
=
f
(
x
)
(say)
Then,
f
′
(
x
)
=
4
x
3
−
2
x
−
2
Equate
f
′
(
x
)
=
0
⇒
4
x
3
−
2
x
−
2
=
0
⇒
2
x
3
−
x
−
1
=
0
⇒
2
x
(
x
2
−
1
)
+
(
x
−
1
)
=
0
⇒
2
x
(
x
−
1
)
(
x
+
1
)
+
(
x
−
1
)
=
0
⇒
(
x
−
1
)
[
2
x
(
x
+
1
)
+
1
]
=
0
⇒
(
x
−
1
)
(
2
x
2
+
2
x
+
1
)
=
0
⇒
x
−
1
=
0
and
2
x
2
+
2
x
+
1
=
0
⇒
x
=
1
and
x
=
−
2
±
√
4
−
8
4
(Imaginary)
⇒
x
=
1
and
x
=
(
−
1
±
1
i
)
2
f
"
(
x
)
=
12
x
2
−
2
f
"
(
x
)
x
=
1
=
10
>
0
⇒
f
(
x
)
is minimum at
x
=
1
Now,
f
(
1
)
=
(
1
)
4
−
(
1
)
2
−
2
+
5
=
3
∴ Absolute minimum value is 3
© examsnet.com
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