Question

If $k\le {\sin }^{-1}x+{\cos }^{-1}x+{\tan }^{-1}x\le K$, then -

• A. $k=0,K=\pi .$
• B. $k=0,K=\pi /2.$
• C. $k=\pi /2,K=\pi$
• D. None of these

Answer 1

Answer: (D)

${\sin }^{-1}x+{\cos }^{-1}x+{\tan }^{-1}x=\frac{\pi }{2}+{\tan }^{-1}x$
Since domain of the function $x\in [-1,1]$
$\therefore -\frac{\pi }{4}\le {\tan }^{-1}x\le \frac{\pi }{4}$
Hence, $k=\frac{\pi }{4}K=\frac{3\pi }{4}$