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Manipal Entrance Test 2020 Math Paper
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© examsnet.com
Question : 15
Total: 70
Let
f
(
x
)
=
|
sin
3
x
1
2
(
cos
3
x
2
+
sin
3
x
2
)
2
cos
3
x
−
1
2
(
cos
2
3
x
2
−
sin
2
3
x
2
)
tan
3
x
4
1
+
2
tan
3
x
|
Then, the value of
f
′
(
x
)
at
x
=
(
2
n
+
1
)
π
,
n
∈
I
(the set of integers) is equal to
(
−
1
)
n
(
−
1
)
n
+
1
3
9
Validate
Solution:
Given,
f
(
x
)
=
|
sin
3
x
1
2
(
cos
3
x
2
+
sin
3
x
2
)
2
cos
3
x
−
1
2
(
cos
2
3
x
2
−
sin
2
3
x
2
)
tan
3
x
4
1
+
2
tan
3
x
|
On differentiating w.r.t.
x
, we get
f
′
(
x
)
=
|
d
d
x
(
sin
3
x
)
1
2
(
cos
3
x
2
+
sin
3
x
2
)
2
d
d
x
(
cos
3
x
)
−
1
2
(
cos
2
3
x
2
−
sin
2
3
x
2
)
d
d
x
(
tan
3
x
)
4
1
+
2
tan
3
x
|
+
|
sin
3
x
d
d
x
(
1
)
2
(
cos
3
x
2
+
sin
3
x
2
)
2
cos
3
x
d
d
x
(
−
1
)
2
(
cos
2
3
x
2
−
sin
2
3
x
2
)
tan
3
x
d
d
x
(
4
)
1
+
2
tan
3
x
|
+
|
sin
3
x
1
2
d
d
x
(
cos
3
x
2
+
sin
3
x
2
)
2
cos
3
x
−
1
2
d
d
x
(
cos
2
3
x
2
−
sin
2
3
x
2
)
tan
3
x
4
d
d
x
(
1
+
2
tan
3
x
)
|
=
|
3
cos
3
x
1
2
(
cos
3
x
2
+
sin
3
x
2
)
2
3
sin
3
x
−
1
2
(
cos
2
3
x
2
−
sin
2
3
x
2
)
3
sec
2
3
x
4
1
+
2
tan
3
x
|
+
|
sin
3
x
0
2
(
cos
3
x
2
+
sin
3
x
2
)
2
cos
3
x
0
2
(
cos
2
3
x
2
−
sin
2
3
x
2
)
tan
3
x
0
1
+
2
tan
3
x
|
+
|
sin
3
x
1
2
×
2
(
cos
3
x
2
+
sin
3
x
2
)
×
(
−
3
2
sin
3
x
2
+
3
2
cos
3
x
2
)
cos
3
x
−
1
2
(
−
2
cos
3
x
2
×
3
2
sin
3
x
2
−
2
sin
3
x
2
×
3
2
cos
3
x
2
)
tan
3
x
4
(
0
+
2
×
3
sec
2
3
x
)
|
At
x
=
(
2
n
+
1
)
π
f
′
(
x
)
=
|
3
(
−
1
)
1
2
(
1
)
0
−
1
2
(
−
1
)
3
4
1
+
0
|
+
0
+
|
0
1
4
(
0
−
1
)
×
[
−
3
2
(
−
1
)
+
3
2
×
0
]
−
1
−
1
0
0
4
0
|
=
[
−
3
(
−
1
+
8
)
−
1
(
0
+
6
)
+
2
(
0
+
3
)
]
+
[
0
−
1
(
0
−
0
)
−
6
(
−
4
)
]
=
−
21
+
24
=
3
© examsnet.com
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