Inverse Trigonometric Functions Question 117
Question: If $ k\le {{\sin }^{-1}}x+{{\cos }^{-1}}x+{{\tan }^{-1}}x\le K, $ then
Options:
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 $
Show Answer
Answer:
Correct 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} $
BETA
