0 f''(\frac{\pi}{6})=-2\sin\frac{\pi}{6}-4\cos(\frac{2\pi}{6}) =-1-2=-3" >
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NIMCET 2015 Question Paper with solutions
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© examsnet.com
Question : 43
Total: 120
A function f : (0, π) → R defined by f(x) = 2 sin x + cos 2x has
a local minimum but no local maximum
a local maximum but no local minimum
both local minimum and local maximum
neither a local minimum nor a local maximum
Validate
Solution:
We have,
f
(
x
)
=
2
sin
x
+
cos
2
x
f
′
(
x
)
=
2
cos
x
−
2
sin
2
x
f
"
(
x
)
=
−
2
sin
x
−
4
cos
2
x
For maxima or minima,
f
′
(
x
)
=
0
⇒
2
cos
x
−
2
sin
2
x
=
0
⇒
cos
x
−
2
sin
x
cos
x
=
0
⇒
cos
x
(
1
−
2
sin
x
)
=
0
cos
x
=
0
or
1
−
2
sin
x
=
0
Now,
f
′
′
(
π
2
)
=
−
2
sin
π
2
−
4
cos
(
2
π
2
)
=
−
2
+
4
=
2
>
0
f
′
′
(
π
6
)
=
−
2
sin
π
6
−
4
cos
(
2
π
6
)
=
−
1
−
2
=
−
3
<
0
f
′
′
(
5
π
6
)
=
−
2
sin
5
π
6
−
4
cos
(
10
π
6
)
=
−
1
−
2
=
−
3
<
0
Hence, fix) is minimum at
π
2
and maximum at
π
6
∴ f(x) has both local minimum and local maximum.
© examsnet.com
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