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SSC Sample Paper SSC CHSL (10+2) Sample Test Paper-1

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

    A) \[1\]                               

    B) \[-1\]

    C) \[2\]                 

    D)        \[-2\]

    Correct Answer: A

    Solution :

     Given,\[x+\frac{2}{x}=1\] Expression             \[=\frac{{{x}^{2}}+x+2}{{{x}^{2}}(1-x)}=\frac{x+1+\frac{2}{x}}{x(1-x)}\]            (Dividing numerator ad denominator b\[x\])             \[=\frac{x+\frac{2}{x}+1}{x(1-x)}=\frac{1+1}{x\times \frac{2}{x}}=\frac{2}{2}=1\]


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