A) \[\frac{1}{\sqrt{2}}\]
B) \[-\frac{1}{\sqrt{2}}\]
C) \[\pm \sqrt{\frac{5}{2}}\]
D) \[\pm \frac{1}{2}\]
Correct Answer: C
Solution :
We have \[{{\tan }^{-1}}\,\frac{x-1}{x+2}+{{\tan }^{-1}}\frac{x+1}{x+2}=\frac{\pi }{4}\] \[\Rightarrow \,\,{{\tan }^{-1}}\,\left[ \frac{\frac{x-1}{x+2}+\frac{x+1}{x+2}}{1-\left( \frac{x-1}{x+2} \right)\,\left( \frac{x+1}{x+2} \right)} \right]=\frac{\pi }{4}\] \[\Rightarrow \,\,\left[ \frac{2x\,(x+2)}{{{x}^{2}}+4+4x-{{x}^{2}}+1} \right]=\tan \frac{\pi }{4}\] Þ \[\,\,\frac{2x\,(x+2)}{4x+5}=\tan \frac{\pi }{4}=1\] \[\Rightarrow \,\,2{{x}^{2}}+4x=4x+5\]\[\Rightarrow \,\,x=\pm \sqrt{\frac{5}{2}}\].You need to login to perform this action.
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