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SSC Sample Paper SSC CHSL (10+2) Sample Test Paper-6

  • question_answer
    If \[x+\frac{1}{x}=2,x\cancel{=}0\]then value of \[{{x}^{2}}+\frac{1}{{{x}^{3}}}\]is equal to

    A)  1                                

    B)  2

    C)  3                    

    D)         4

    Correct Answer: B

    Solution :

     \[x+\frac{1}{x}=2\] \[\Rightarrow \]\[{{x}^{2}}+1=2x\Rightarrow {{x}^{2}}-2x+1=0\] \[\Rightarrow \]\[{{(x-1)}^{2}}=0\Rightarrow x=1\] \[\therefore \]\[{{x}^{2}}+\frac{1}{{{x}^{3}}}=1+1=2\]


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