Question

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

• A. 1/2
• B. 1
• C. -1/2
• D. -1

Answer 1

The correct option is B

1

Since,
$si{n}^{-1}x+co{s}^{-1}x=\frac{\pi }{2}$ for 1x1 $\le$ 1
$\therefore x-\frac{x^2}{2}+\frac{x^3}{4}-\cdots =x^2-{\frac{[}{x}}^{4}2+\frac{x^5}{4}-\cdots$
$\Rightarrow \frac{x}{1+\frac{x}{2}}=\frac{x}{1+\frac{x^2}{2}}$ $(\because 0<1x1<\sqrt{2})$
$\Rightarrow \frac{x}{2+x}=\frac{x^2}{2+x^2}\Rightarrow 2x+x^3=2x^2+x^3$
$\Rightarrow x^2=x\Rightarrow x=0,1$
But $x\ne 0,\therefore x=1$