A) \[1/4\]
B) \[3/4\]
C) \[5/4\]
D) \[7/4\]
Correct Answer: B
Solution :
[b] We have \[{{\tan }^{-1}}(ax)+{{\tan }^{-1}}(bx)=\frac{\pi }{2}-{{\tan }^{=1}}\left( \frac{1}{x}-\frac{x}{8} \right)\] \[\Rightarrow \,\,\,\,\tan ({{\tan }^{-1}}(ax)+{{\sin }^{-1}}(bx))=\cot \left( {{\tan }^{-1}}\left( \frac{1}{x}-\frac{x}{8} \right) \right)\]\[\Rightarrow \,\,\,\frac{ax+bx}{1-ab{{x}^{2}}}=\frac{8x}{8-{{x}^{2}}}\] \[\Rightarrow \,\,\,(a+b)\,(8-{{x}^{2}})=8(1-ab{{x}^{2}})\] \[\Rightarrow \,\,\,a+b=1\] and \[8ab=1\] \[\Rightarrow \,\,\,\,{{a}^{2}}+{{b}^{2}}={{(a+b)}^{2}}-2ab=1-1/4=3//4\]You need to login to perform this action.
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