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Manipal Entrance Test 2013 Math Paper
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© examsnet.com
Question : 53
Total: 80
The number of solutions of the equation
1
+
sin
x
.
sin
2
x
2
=
0
in
[
−
π
,
π
]
is
zero
1
2
3
Validate
Solution:
1
+
sin
x
.
sin
2
x
2
=
0
⇒
2
+
2
sin
x
.
sin
2
frac
x
2
=
0
⇒
2
+
sin
x
(
1
−
cos
x
)
=
0
⇒
4
+
2
sin
x
(
1
−
cos
x
)
=
0
⇒
4
+
2
sin
x
−
sin
2
x
=
0
⇒
sin
2
x
=
2
sin
x
+
4
Above is not possible for any value of x as LHS has maximum value 1 and RHS has minimum value 2.
Hence, there is no solution.
© examsnet.com
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