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Question

Find values of x which satisfy : sin1(1x)2sin1x=π2.

A
x=0
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B
x=1/2
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C
,x=0andx=1/2
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D
None of these
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Solution

The correct option is A x=0
Given, sin1(1x)2sin1x=π/2
Let x=siny
Therefore, the given equation reduces to
sin1(1siny)2sin1siny=π2
sin1(1siny)2y=π2
sin1(1siny)=π2+2y
1siny=sin(π2+2y)
1siny=cos2y
1cos2y=siny
2sin2y=siny
2sin2ysiny=0
siny(2siny1)=0
siny=0 or 1/2
x=0 or 1/2
But, when x=12, it can be observed that
L.H.S. =sin1(112)2sin112
=sin1(12)2sin112
=sin1(12)
=π6π2 R.H.S.
Therefore, x=12 is not the solution of given equation.
Thus, x=0 is only solution.

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