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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 31

  • question_answer
    Let there exist real numbers a and b such that for every positive number x, the identity \[{{\tan }^{-1}}\left( \frac{1}{x}-\frac{x}{8} \right)+{{\tan }^{-1}}(ax)+{{\tan }^{-1}}(bx)=\frac{\pi }{2}\] holds. Then the value of \[{{a}^{2}}+{{b}^{2}}\] is

    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 \[8ab=1\] \[\Rightarrow \,\,\,\,{{a}^{2}}+{{b}^{2}}={{(a+b)}^{2}}-2ab=1-1/4=3//4\]  


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