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EAMCET Engineering 2005 Solved Paper
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© examsnet.com
Question : 94
Total: 160
If
x
is real, then the minimum value of
x
2
−
x
+
1
x
2
+
x
+
1
, is
1
3
3
1
2
1
Validate
Solution:
Let
f
(
x
)
=
x
2
−
x
+
1
x
2
+
x
+
1
. . . (i)
On differentiating w.r.t.
x
, we get
f
′
(
x
)
=
(
x
2
+
x
+
1
)
(
2
x
−
1
)
−
(
x
2
−
x
+
1
)
(
2
x
+
1
)
(
x
2
+
x
+
1
)
2
for maximum or minimum, put
f
′
(
x
)
=
0
⇒
(
x
2
+
x
+
1
)
(
2
x
−
1
)
−
(
x
2
−
x
+
1
)
(
2
x
+
1
)
=
0
⇒
x
2
+
x
−
1
−
(
−
x
2
+
x
+
1
)
=
0
⇒
2
x
2
−
2
=
0
⇒
x
=
±
1
Now,
f
′
(
c
)
=
2
x
2
−
2
(
x
2
+
x
+
1
)
2
Again differentiating, we get
(
x
2
+
x
+
1
)
2
(
4
x
)
−
(
2
x
2
−
2
)
f
′
′
(
x
)
=
2
(
x
2
+
x
+
1
)
(
2
x
+
1
)
(
x
2
+
x
+
1
)
4
at
x
=
1
,
f
′
′
(
x
)
>
0
Therefore it is minimum at
x
=
1
Put
x
=
1
in equation (i), we get
f
(
1
)
=
1
−
1
+
1
1
+
1
+
1
=
1
3
∴
The minimum value is
1
3
© examsnet.com
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