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Question

If sin1(1x)2sin1x=π2, then x is equal to

A
0,12
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B
1,12
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C
12
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D
0
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Solution

The correct option is D 0
sin1(1x)2sin1x=π2

2sin1x=π2sin1(1x)

We know that sin1x+cos1x=π2

Replace x by 1x

sin1(1x)+cos1(1x)=π2

2sin1x=cos1(1x) .....(1)

Let sin1x=a,

Hence our equation becomes

2a=cos1(1x)

cos2a=1x

12sin2a=1x

1x=12sin2a

1x=12x2

2x2+x=0

x(2x1)=0

x=0,x=12

But x=12 does not satisfy the equation

Taking equation sin1(1x)2sin1x=π2

Put x=12 in L.H.S

sin1(112)2sin1(12)

=sin1(12)2sin1(12)

=sin1(12)

=π6

π2

Hence x=12 not possible.

x=0 is the only solution.

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