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SSC Sample Paper Mock Test-3 SSC CGL Tear-II Paper-1

  • question_answer
    If \[\left( x-\frac{1}{x} \right),=\frac{1}{2}\]then the value of \[\left( 6{{x}^{2}}+\frac{6}{{{x}^{2}}} \right)\]is

    A)  0                                

    B)  1

    C)  \[-\,\,1\]                       

    D)  \[\frac{27}{2}\]

    Correct Answer: D

    Solution :

    \[x-\frac{1}{x}=\frac{1}{2}\] On squaring, we get \[{{x}^{2}}+\frac{1}{{{x}^{2}}}-2=\frac{1}{4}\] \[{{x}^{2}}+\frac{1}{{{x}^{2}}}=\frac{1}{4}+2=\frac{9}{4}\] \[\Rightarrow \]   \[6\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)=6\times \frac{9}{4}=\frac{27}{2}\]


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