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AP EAPCET 06-Jul-2022 Shift 2 Paper
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© examsnet.com
Question : 25
Total: 160
Let
θ
∈
R
such that
3
sinh
(
2
θ
)
=
13
−
3
e
2
θ
, then
θ
=
1
2
log
3
1
3
log
3
log
3
1
2
log
5
Validate
Solution:
👈: Video Solution
"
3
s
i
n
h
(
2
θ
)
=
13
−
3
e
2
θ
3
(
e
2
θ
−
e
−
2
θ
2
)
=
13
−
3
e
2
θ
Let
x
=
e
2
θ
3
2
(
x
−
1
x
)
=
13
−
3
x
⇒
3
x
2
−
3
=
2
x
(
13
−
3
x
)
⇒
3
x
2
−
3
=
26
x
−
6
x
2
⇒
9
x
2
−
26
x
−
3
=
0
⇒
9
x
2
−
27
x
+
x
−
3
=
0
⇒
9
x
(
x
−
3
)
+
1
(
x
−
3
)
=
0
⇒
(
9
x
+
1
)
(
x
−
3
)
=
0
⇒
x
=
3
[
∵
x
≠
−
1
9
]
∴
e
2
θ
=
3
⇒
2
θ
=
log
3
⇒
θ
=
1
2
log
3
© examsnet.com
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