Question
If $$k \leqslant {\sin ^{ - 1}}x + {\cos ^{ - 1}}x + {\tan ^{ - 1}}x \leqslant K,$$ 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 :
$$k = \frac{\pi }{4},K = \frac{{3\pi }}{4}$$
Solution :
We have,
$$\eqalign{
& {\sin ^{ - 1}}x + {\cos ^{ - 1}}x + {\tan ^{ - 1}}x = \frac{\pi }{2} + {\tan ^{ - 1}}x \cr
& {\text{Now, }}{\sin ^{ - 1}}x\,\,{\text{and }}\,{\cos ^{ - 1}}x\,\,{\text{defined only if}} - 1 \leqslant x \leqslant 1 \cr
& {\text{So, }} - \frac{\pi }{4} \leqslant {\tan ^{ - 1}}x \leqslant \frac{\pi }{4} \cr
& \Rightarrow \frac{\pi }{4} \leqslant \frac{\pi }{2} + {\tan ^{ - 1}}x \leqslant \frac{{3\pi }}{4} \cr
& \therefore k = \frac{\pi }{4}{\text{ and }}\,K = \frac{{3\pi }}{4} \cr} $$