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Question

If sin1(xx2+x34...)+cos1(x2x42+x64...)=π2 for 0<|x|<2, then x equals

A
12
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B
1
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C
12
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D
1
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Solution

The correct option is A 1
sin1(xx2+x24..)+cos1(x2x42+x36+...)

=sin1(x[x21+x22])+cos1(x21+x22)
=sin1(xxx2+2)+cos1(2x2x2+2)
=π2
Therefore
xxx2+2=2x2x2+2
x3+2xx=2x2
x32x2+x=0
x(x22x+1)=0
x((x1)2)=0
x=0 and x=1
But |x|>0
Hence
x=1

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