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
${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}\forall x\in R\cdots \left(1\right)$
$\text{Also,}-\frac{\pi }{2}\le {\sin }^{-1}x\le \frac{\pi }{2}\cdots \left(2\right)$
From $\left(1\right)\&\left(2\right),$
$0\le 2{\sin }^{-1}x+{\cos }^{-1}x\le \pi$