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Questions  

If  Sin1xx22+x34...+Cos1x2x42+x64...=π2for  0<|x|<2 then x=

a
12
b
1
c
-12
d
-1

detailed solution

Correct option is B

We have sin-1x-x22+x34-.....+cos-1x2-x42+x64-.....=π2 ⇒x-x22+x34-.....=x2-x42+x64-..... ⇒x1+x2=x21+x22 ⇒12+x=x2+x2 ⇒2+x2=2x+x2 ⇒x=1

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