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MHTCET 14 May 2023 Shift 1 Solved Paper
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© examsnet.com
Question : 72
Total: 150
If
∫
cos
θ
5
+
7
sin
θ
−
2
cos
2
θ
d
θ
=
A
log
e
|
f
(
θ
)
|
+
c
(where
c
is a constant of integration), then
f
(
θ
)
A
can be
2
sin
θ
+
1
sin
θ
+
3
2
sin
θ
+
1
5
(
sin
θ
+
3
)
5
(
sin
θ
+
3
)
2
sin
θ
+
1
5
(
2
sin
θ
+
1
)
sin
θ
+
3
Validate
Solution:
Let
I
=
∫
cos
θ
5
+
7
sin
θ
−
2
cos
2
θ
d
θ
=
∫
cos
θ
5
+
7
s
i
n
θ
−
2
(
1
−
s
i
n
2
θ
)
d
θ
=
∫
cos
θ
2
s
i
n
2
θ
+
7
s
i
n
θ
+
3
d
θ
=
∫
cos
θ
(
s
i
n
θ
+
3
)
(
2
s
i
n
θ
+
1
)
d
θ
Put
s
i
n
θ
=
t
⇒
cos
θ
d
θ
=
d
t
∴
I
=
∫
d
t
(
t
+
3
)
(
2
t
+
1
)
=
∫
[
2
5
(
2
t
+
1
)
−
1
5
(
t
+
3
)
]
dt
=
2
5
⋅
log
∣
2
t
+
1
2
−
1
5
log
|
t
+
3
|
+
c
=
1
5
log
|
2
s
i
n
θ
+
1
|
−
1
5
log
|
s
i
n
θ
+
3
|
+
c
=
1
5
log
|
2
s
i
n
θ
+
1
s
i
n
θ
+
3
|
+
c
∴
A
=
1
5
,
f
(
θ
)
=
2
s
i
n
θ
+
1
s
i
n
θ
+
3
∴
f
(
θ
)
A
=
5
(
2
s
i
n
θ
+
1
)
s
i
n
θ
+
3
© examsnet.com
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