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12th Class Mathematics Sample Paper Mathematics Olympiad - Sample Paper-5

  • question_answer
    If \[ta{{n}^{-1}}\left( \frac{x+1}{x-1} \right)+ta{{n}^{-1}}\left( \frac{x-1}{x} \right)=ta{{n}^{-1}}(-7)\], then \[\mathbf{x}=\]

    A)  1                                

    B)  2             

    C)  \[-2\]                           

    D)  \[-1\]

    Correct Answer: B

    Solution :

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


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