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IfSin1xx22+x34.....+cos1x2x42+x64....=π2for0<x<2 thenx=

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-.... ⇒x=x2 ⇒xx-1=0 ⇒x=0,1 Since 0

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