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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2002

  • question_answer
    If \[{{\tan }^{-1}}\frac{x+1}{x-1}+{{\tan }^{-1}}\frac{x-1}{x}={{\tan }^{-1}}(-7)\] then the value of \[x\] is:

    A)  zero                                     

    B) \[~-2\]

    C)  1                                            

    D)  2

    Correct Answer: D

    Solution :

    \[{{\tan }^{-1}}\left( \frac{x+1}{x-1} \right)+{{\tan }^{-1}}\left( \frac{x-1}{x} \right)={{\tan }^{-1}}(-7)\] \[\frac{\frac{x+1}{x-1}+\frac{x-1}{x}}{1-\frac{(x+1)(x-1)}{x(x-1)}}=(-7)\] \[\frac{x(x+1)+{{(x-1)}^{2}}}{x(x-1)-({{x}^{2}}-1)}=-7\] \[\frac{{{x}^{2}}+x+{{x}^{2}}+1-2x}{{{x}^{2}}-x-{{x}^{2}}+1}=-7\] \[\frac{2{{x}^{2}}-x+1}{1-x}=\frac{-7}{1}\] \[-7+7x=2{{x}^{2}}-x+1\] \[2{{x}^{2}}-8x+8=0\] \[{{x}^{2}}-4x+4x=0\] \[{{(x-2)}^{2}}=0\Rightarrow x=2\]


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