A) \[k=-\pi ,K=\pi \]
B) \[k=0,K=\frac{\pi }{2}\]
C) \[k=\frac{\pi }{4},K=\frac{3\pi }{4}\]
D) \[k=0,K=\pi \]
Correct Answer: C
Solution :
[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 \[K=\frac{3\pi }{4}\]
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