Question

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

• A. $k=-\pi ,K=\pi$
• B. $k=0,K=\frac{\pi }{2}$
• C. $k=\frac{\pi }{4},K=\frac{3\pi }{4}$
• D. $k=0,K=\pi$

Answer 1

Answer: (C)

We have,
${\sin }^{-1}x+{\cos }^{-1}x+{\tan }^{-1}x=\frac{\pi }{2}+{\tan }^{-1}x$
Now ${\sin }^{-1}x$ and ${\cos }^{-1}x$ are defined only if $-1\le x\le 1$
So, $-\frac{\pi }{4}\le {\tan }^{-1}x\le \frac{\pi }{4}\Rightarrow \frac{\pi }{4}\le \frac{\pi }{2}+{\tan }^{-1}x\le \frac{3\pi }{4}$
$\therefore k=\frac{\pi }{4}$ and $k=\frac{3\pi }{4}$