A) ½
B) 1
C) -1/2
D) -1
Correct Answer: B
Solution :
[b] \[{{\sin }^{-1}}\left( x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{4}-... \right)+{{\cos }^{-1}}\left( {{x}^{2}}-\frac{{{x}^{2}}}{2}+\frac{{{x}^{6}}}{4}-... \right)=\frac{\pi }{2}\]This is true only when \[x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{4}-...={{x}^{2}}-\frac{{{x}^{4}}}{2}+\frac{{{x}^{6}}}{4}...\] \[\Rightarrow \frac{x}{1+\frac{x}{2}}=\frac{{{x}^{2}}}{1+\frac{{{x}^{2}}}{2}}\] (Common rations are \[-\frac{x}{2}\And -\frac{{{x}^{2}}}{2}\And \] |common ratios|<1, in the given interval) \[\frac{2x}{2+x}=\frac{2{{x}^{2}}}{2+{{x}^{2}}}\Rightarrow x=0orx=1\Rightarrow x=1,\] {x cannot be zero as \[0<\left| x \right|<\sqrt{2}\]}.
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