Inverse Trigonometric Functions Question 214
Question: If $ k\le {{\sin }^{-1}}x+{{\cos }^{-1}}x+{{\tan }^{-1}}x\le K, $ then
Options:
A) $ k=0,K=\pi $
B) $ k=0,K=\frac{\pi }{2} $
C) $ k=\frac{\pi }{2},K=\pi $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
We have $ {{\sin }^{-1}}x+{{\cos }^{-1}}x+{{\tan }^{-1}}x=\frac{\pi }{2}+{{\tan }^{-1}}x $
Since $ \frac{-\pi }{2}\le {{\tan }^{-1}}x\le \frac{\pi }{2}\Rightarrow 0\le \frac{\pi }{2}+{{\tan }^{-1}}x\le \pi $ \ $ K=\pi ,k=0 $ .
BETA
