Question

If $si{n}^{-1}\left(x-\frac{x^2}{2}+\frac{x^3}{4}-...\right)+co{s}^{-1}\left(x^2-\frac{x^4}{2}+\frac{x^6}{4}-...\right)=\frac{\pi }{2}$ for $0<\left|x\right|<\sqrt{2}$, then $x$ equals

• A. $1/2$
• B. $1$
• C. $-1/2$
• D. $-1$

Answer 1

Answer: (B)

$si{n}^{-1}\left(x-\frac{x^2}{2}+\frac{x^3}{4}-...\right)+co{s}^{-1}\left(x^2-\frac{x^4}{2}+\frac{x^6}{4}-...\right)=\frac{\pi }{2}$
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 ratios are $-\frac{x}{2}\&-\frac{x^2}{2}\&$ |common ratios $|<1$, in the given interval)
$\frac{2x}{2+x}=\frac{2x^2}{2+x^2}\Rightarrow x=0$ or $x=1\Rightarrow x=1,$
$\left\{xcannotbezeroas0<|x|<\sqrt{2}\right\}.$