SSC Sample Paper Mock Test-6 SSC CGL Tear-II Paper-1

  • question_answer
    If \[A=\left[ \frac{x-1}{x+1} \right]\]and\[B=\left[ \frac{x+1}{x-1} \right],\]then \[{{(A+B)}^{2}}\]is

    A)  \[\frac{4{{x}^{4}}+8{{x}^{2}}-4}{{{x}^{4}}-2{{x}^{2}}+1}\]

    B)         \[\frac{4{{x}^{4}}+8{{x}^{2}}+4}{{{x}^{4}}-2{{x}^{2}}+1}\]

    C)  \[\frac{4{{x}^{4}}+8{{x}^{2}}+4}{{{x}^{4}}+2x+1}\]

    D)         None of these

    Correct Answer: B

    Solution :

    \[{{(A+B)}^{2}}={{\left[ \left( \frac{x-1}{x+1} \right)+\left( \frac{x+1}{x-1} \right) \right]}^{2}}\]\[={{\left[ \frac{{{(x-1)}^{2}}+{{(x+1)}^{2}}}{(x+1)(x-1)} \right]}^{2}}\] \[={{\left[ \frac{2\,\,({{x}^{2}}+1)}{{{x}^{2}}-1} \right]}^{2}}=\frac{4\,\,({{x}^{4}}+2{{x}^{2}}+1)}{{{x}^{4}}-2{{x}^{2}}+1}\] \[=\frac{4{{x}^{4}}+8{{x}^{2}}+4}{{{x}^{4}}-2{{x}^{2}}+1}\]


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