Q60 / 125 · BITSAT 2012 · Mathematics
If $k \leq \sin ^{-1} x+\cos ^{-1} x+\tan ^{-1} x \leq K,$ then $-$
- A$\mathrm{k}=0, \mathrm{~K}=\pi$
- B$\mathrm{k}=0, \mathrm{~K}=\pi / 2$
- C$\mathrm{k}=\pi / 2, \mathrm{~K}=\pi$
- Correct answerDNone of these
Solution
$\sin ^{-1} x+\cos ^{-1} x+\tan ^{-1} x=\frac{\pi}{2}+\tan ^{-1} x$
Since domain of the function $\mathrm{x} \in[-1,1]$ $\therefore-\frac{\pi}{4} \leq \tan ^{-1} x \leq \frac{\pi}{4}$
Hence, $\mathrm{k}=\frac{\pi}{4}$ and $\mathrm{K}=\frac{3 \pi}{4}$
Hence, the correct answer is option: 'D'.