A) 0
B) - 1
C) - 2
D) - 3
Correct Answer: B
Solution :
\[{{({{\tan }^{-1}}x)}^{2}}+{{\left( \frac{\pi }{2}-{{\tan }^{-1}} \right)}^{2}}=\frac{5{{\pi }^{2}}}{8}\] \[\Rightarrow \] \[2\,{{({{\tan }^{-1}}x)}^{2}}-\pi \,({{\tan }^{-1}}x)-\frac{3{{\pi }^{2}}}{8}=0\] \[\therefore \] \[{{\tan }^{-1}}x=\frac{\pi \pm \,\sqrt{4{{\pi }^{2}}}}{4}\]\[=\frac{\pi \pm 2\pi }{4}\] \[\Rightarrow \] \[{{\tan }^{-1}}x=\frac{3\pi }{4}\] or \[-\frac{\pi }{4}\] \[\therefore \] \[x=-1\] for \[-1\] \[\Rightarrow \] \[x=-1\]
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