Question

If $\alpha \le 2{\sin }^{-1}x+{\cos }^{-1}x\le \beta$, then

• A. $\alpha =\frac{\pi }{2},\beta =\frac{\pi }{2}$
• B. $\alpha =\frac{\pi }{2},\beta =\frac{3\pi }{2}$
• C. $\alpha =0,\beta =\pi$
• D. $\alpha =0,\beta =2\pi$

Answer 1

The correct option is C $\alpha =0,\beta =\pi$

We know that, $-\frac{\pi }{2}\le {\sin }^{-1}x\le \frac{\pi }{2}$

Now add $\frac{\pi }{2}$ each sides

$\frac{\pi }{2}-\frac{\pi }{2}\le {\sin }^{-1}x+\frac{\pi }{2}\le \frac{\pi }{2}+\frac{\pi }{2}$

$\Rightarrow 0\le {\sin }^{-1}+\frac{\pi }{2}\le \pi$

$\Rightarrow 0\le {\sin }^{-1}+({\sin }^{-1}x+{\cos }^{-1}x)\le \pi$, use the identity ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$

$\therefore 0\le 2{\sin }^{-1}x+{\cos }^{-1}x\le \pi$

Hence $\alpha =0,\beta =\pi$