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SSC Sample Paper SSC CHSL (10+2) Sample Test Paper-24

  • question_answer
    If \[x+\frac{1}{x}=2,\] then find the value of \[{{x}^{2019}}+\frac{1}{{{x}^{2021}}}.\]

    A) \[4040\]                        

    B)  \[-2\]

    C)  2                                

    D)  3

    Correct Answer: C

    Solution :

     \[x+\frac{1}{x}=2\] If we put the value of \[x=1,\]then the equation satisfies. So, \[{{x}^{2019}}+\frac{1}{{{x}^{2021}}}={{1}^{2018}}+\frac{1}{{{1}^{2021}}}\] \[=\,\,\,\,\,1+1=\underline{\mathbf{2}}\]


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