Question
If $$2{\tan ^{ - 1}}x + {\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}}$$ is independent of $$x$$ then
A.
$$x \in \left[ {1, + \infty } \right)$$
B.
$$x \in \left[ { - 1,1} \right]$$
C.
$$x \in \left( { - \infty , - 1} \right]$$
D.
None of these
Answer :
$$x \in \left[ {1, + \infty } \right)$$
Solution :
Let $$x = \tan \theta .$$ Then $${\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}} = {\sin ^{ - 1}}\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }} = {\sin ^{ - 1}}\left( {\sin 2\theta } \right).$$
$$\eqalign{
& \therefore \,\,2{\tan ^{ - 1}}x + {\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}} = 2\theta + {\sin ^{ - 1}}\left( {\sin 2\theta } \right). \cr
& {\text{If }} - \frac{\pi }{2} \leqslant 2\theta \leqslant \frac{\pi }{2},2{\tan ^{ - 1}}x + {\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}} = 2\theta + 2\theta = 4{\tan ^{ - 1}}x \ne {\text{independent of }}x. \cr
& {\text{If }} - \frac{\pi }{2} \leqslant \pi - 2\theta \leqslant \frac{\pi }{2},2{\tan ^{ - 1}}x + {\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}} = 2\theta + {\sin ^{ - 1}}\left\{ {\sin \left( {\pi - 2\theta } \right)} \right\} = 2\theta + \pi - 2\theta = \pi = {\text{independent of }}x. \cr} $$
$$\therefore \,\,\theta \notin \left[ { - \frac{\pi }{4},\frac{\pi }{4}} \right]\,{\text{but }}\theta \in \left[ {\frac{\pi }{4},\frac{{3\pi }}{4}} \right]$$ and from the principal value of $${\tan ^1}x,\theta \in \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right).\,{\text{Hence, }}\theta \in \left( {\frac{\pi }{4},\frac{\pi }{2}} \right).$$
$$\eqalign{
& \therefore \,\,\theta \in \left( {\frac{\pi }{4},\frac{\pi }{2}} \right) \cr
& \Rightarrow \,\,{\tan ^{ - 1}}x + {\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}} = \pi . \cr
& {\text{Also at, }}\theta = \frac{\pi }{4},2{\tan ^{ - 1}}x + {\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}} = 2 \cdot \frac{\pi }{4} + {\sin ^{ - 1}}1 = \frac{\pi }{2} + \frac{\pi }{2} = \pi . \cr} $$
∴ the given function $$ = \pi = $$ constant if $$\theta \in \left[ {\frac{\pi }{4},\frac{\pi }{2}} \right),\,{\text{i}}{\text{.e}}{\text{., }}x \in \left[ {1, + \infty } \right).$$