Question

If the range of $2{\sin }^{-1}x+{\cos }^{-1}x$ is $[\alpha ,\beta ]$, then

• A. $\alpha =\pi$
• B. $\alpha =0$
• C. $\beta =\pi$
• D. $\beta =2\pi$

Answer 1

Answer: (B), (C)

$\beta =\pi$

We know,
${\cos }^{-1}x+{\sin }^{-1}x=\frac{\pi }{2}$
Now,
$\frac{-\pi }{2}\le {\sin }^{-1}x\le \frac{\pi }{2}\Rightarrow \frac{-\pi }{2}+\frac{\pi }{2}\le {\sin }^{-1}x+\frac{\pi }{2}\le \frac{\pi }{2}+\frac{\pi }{2}\Rightarrow 0\le {\sin }^{-1}x+({\cos }^{-1}x+{\sin }^{-1}x)\le \pi \Rightarrow 0\le 2{\sin }^{-1}x+{\cos }^{-1}x\le \pi \therefore \alpha =0,\beta =\pi$