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AMU Engineering 2019 solved paper Solved Paper
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© examsnet.com
Question : 148
Total: 150
If
y
=
1
√
a
2
−
b
2
cos
−
1
[
a
cos
(
x
−
a
)
+
b
θ
]
where
θ
=
a
+
b
cos
(
x
−
α
)
, then
d
y
d
x
is equal to
1
∕
θ
2
∕
θ
1
∕
θ
2
2
∕
θ
2
Validate
Solution:
Given that
y
=
1
√
a
2
−
b
2
cos
−
1
[
a
cos
(
x
−
a
)
+
b
θ
]
⇒
cos
√
a
2
−
b
2
y
=
a
cos
(
x
−
α
)
+
b
θ
........(i)
and
θ
=
a
+
b
cos
(
x
−
α
)
.......(ii)
Differentiating Eq. (ii) w.r.t.
x
, we get
d
θ
d
x
=
−
b
sin
(
x
−
α
)
.........(iii)
Differentiating Eq. (i) w.r.t.
x
, we get
−
sin
(
√
a
2
−
b
2
y
)
√
a
2
−
b
2
d
y
d
x
=
θ
[
−
a
sin
(
x
−
α
)
]
−
(
a
cos
(
x
−
α
)
+
b
)
d
θ
d
x
θ
2
=
θ
(
−
a
sin
(
x
−
α
)
)
+
(
a
cos
(
x
−
α
)
+
b
)
(
−
b
sin
(
x
−
α
)
)
θ
2
=
sin
(
x
−
α
)
[
−
a
2
−
a
b
cos
(
x
−
α
)
+
a
b
cos
(
x
−
α
)
+
b
2
]
θ
2
=
sin
(
x
−
α
)
(
b
2
−
a
2
)
θ
2
⇒
d
y
d
x
=
(
a
2
−
b
2
)
sin
(
x
−
α
)
√
a
2
−
b
2
sin
(
√
a
2
−
b
2
)
y
(
1
θ
2
)
=
√
a
2
−
b
2
sin
(
x
−
α
)
θ
2
√
(
1
−
cos
2
√
a
2
−
b
2
y
)
=
√
a
2
−
b
2
sin
(
x
−
α
)
θ
2
√
θ
2
−
(
a
cos
(
x
−
α
)
+
b
)
2
θ
2
√
a
2
−
b
2
sin
(
x
−
α
)
θ
a
2
+
(
b
2
cos
2
(
x
−
α
)
+
2
a
b
cos
(
x
−
α
)
−
(
a
2
cos
2
(
x
−
α
)
+
b
2
+
2
a
b
cos
(
x
−
α
)
)
=
√
a
2
−
b
2
sin
(
x
−
α
)
θ
√
a
2
−
b
2
sin
(
x
−
α
)
⇒
d
y
d
x
=
1
θ
© examsnet.com
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