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KEAM 2013 Math Paper
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© examsnet.com
Question : 30
Total: 120
The value of
x
satisfying the equation
|
c
o
s
2
x
s
i
n
2
x
s
i
n
2
x
s
i
n
2
x
c
o
s
2
x
s
i
n
2
x
s
i
n
2
x
s
i
n
2
x
c
o
s
2
x
|
=
0
and
x
∊
[
0
,
π
4
]
is
π
4
π
2
π
16
π
3
π
8
Validate
Solution:
Given,
|
cos
2
x
sin
2
x
sin
2
x
sin
2
x
cos
2
x
sin
2
x
sin
2
x
sin
2
x
cos
2
x
|
=
0
;
x
∈
[
0
,
π
4
]
Applying operation:
R
1
→
R
1
+
R
2
+
R
3
,
we get
|
cos
2
x
+
2
sin
2
x
2
sin
2
x
+
cos
2
x
2
sin
2
x
+
cos
2
x
sin
2
x
cos
2
x
sin
2
x
sin
2
x
sin
2
x
cos
2
x
|
⇒
(
2
sin
2
x
+
cos
2
x
)
|
1
1
1
sin
2
x
cos
2
x
sin
2
x
sin
2
x
sin
2
x
cos
2
x
|
=
0
Now, apply operation
C
2
→
C
2
−
C
1
and
C
3
→
C
3
−
C
1
,
we get
(
2
sin
2
x
+
cos
2
x
)
|
1
0
0
sin
2
x
cos
2
x
−
sin
2
x
0
sin
2
x
0
cos
2
x
−
sin
2
x
|
=
0
Expanding along
R
1
, we get
(
2
sin
2
x
+
cos
2
x
)
(
cos
2
x
−
sin
2
x
)
2
=
0
⇒
2
sin
2
x
+
cos
2
x
=
0
or
cos
2
x
−
sin
2
x
=
0
⇒
tan
2
x
=
−
1
2
or
tan
2
x
=
1
=
tan
π
4
∵
x
≠
1
2
tan
−
1
(
−
1
2
)
∴
x
=
π
8
∈
[
0
,
π
4
]
© examsnet.com
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