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AIEEE 2004 Math Solved Paper
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© examsnet.com
Question : 1
Total: 68
If
u
=
√
a
2
cos
2
θ
+
b
2
s
i
n
2
θ
+
√
a
2
s
i
n
2
θ
+
b
2
cos
2
θ
then the difference between the maximum and minimum values of
u
2
is given by
(
a
−
b
)
2
2
√
a
2
+
b
2
(
a
+
b
)
2
2
(
a
2
+
b
2
)
Validate
Solution:
Given
u
=
√
a
2
cos
2
θ
+
b
2
s
i
n
2
θ
+
√
a
2
s
i
n
2
θ
+
b
2
cos
2
θ
∴
u
2
=
a
2
cos
2
θ
+
b
2
s
i
n
2
θ
+
a
2
s
i
n
2
θ
+
b
2
cos
2
θ
+
2
√
(
a
2
cos
2
θ
+
b
2
s
i
n
2
θ
)
×
√
(
a
2
s
i
n
2
θ
+
b
2
cos
2
θ
)
⇒
u
2
=
a
2
+
b
2
+
2
√
(
a
4
+
b
4
)
cos
2
θ
s
i
n
2
θ
+
a
2
b
2
(
cos
4
θ
+
s
i
n
4
θ
)
⇒
u
2
=
a
2
+
b
2
+
2
√
(
a
4
+
b
4
)
cos
2
θ
s
i
n
2
θ
+
a
2
b
2
(
1
−
2
cos
2
θ
s
i
n
2
θ
)
⇒
u
2
=
a
2
+
b
2
+
2
√
(
a
4
+
b
4
−
2
a
2
b
2
)
cos
2
θ
s
i
n
2
θ
+
a
2
b
2
⇒
u
2
=
a
2
+
b
2
+
2
√
(
a
2
−
b
2
)
2
(
2
cos
θ
s
i
n
θ
)
2
4
+
a
2
b
2
⇒
u
2
=
a
2
+
b
2
+
2
√
(
a
2
−
b
2
)
2
s
i
n
2
2
θ
4
+
a
2
b
2
We know
0
≤
s
i
n
2
2
θ
≤
1
∴
0
≤
(
a
2
−
b
2
)
2
s
i
n
2
2
θ
4
≤
(
a
2
−
b
2
)
2
4
⇒
a
2
b
2
≤
(
a
2
−
b
2
)
2
s
i
n
2
2
θ
4
+
a
2
b
2
≤
(
a
2
−
b
2
)
2
4
+
a
2
b
2
∴
Min value of
u
2
=
a
2
+
b
2
+
2
√
a
2
b
2
=
(
a
+
b
)
2
and Max value of
u
2
=
a
2
+
b
2
+
2
√
(
a
2
−
b
2
)
2
4
+
a
2
b
2
=
a
2
+
b
2
+
2
√
(
a
2
−
b
2
)
2
+
4
a
2
b
2
4
=
a
2
+
b
2
+
2
√
(
a
2
+
b
2
)
2
4
=
a
2
+
b
2
+
a
2
+
b
2
=
2
(
a
2
+
b
2
)
Max of
u
2
−
Min
of
u
2
=
2
(
a
2
+
b
2
)
−
(
a
+
b
)
2
=
2
(
a
2
+
b
2
)
−
(
a
2
+
b
2
+
2
a
b
)
=
√
−
2
(
a
2
+
b
2
−
2
a
b
)
4
=
(
a
−
b
)
2
© examsnet.com
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