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

  • question_answer
    If\[R=\frac{{{a}^{3}}+1}{a-1}\] and \[S=\frac{{{a}^{2}}-a+1}{a+1},\] then \[R+S\]is

    A)  \[\frac{a-1}{a+1}\]

    B)                     \[\frac{1}{{{a}^{2}}+a+1}\]

    C)  \[1\]                            

    D)  \[\frac{{{(a+1)}^{2}}}{(a-1)}\]

    Correct Answer: D

    Solution :

    \[R\div S=\frac{{{a}^{3}}+1}{a-1}+\frac{{{a}^{2}}-a+1}{a+1}=\frac{{{a}^{3}}+1}{a-1}\times \frac{a+1}{({{a}^{2}}-a+1)}\] \[=\frac{(a+1)({{a}^{2}}-a+1)(a+1)}{(a-1)({{a}^{2}}-a+1)}=\frac{{{(a+1)}^{2}}}{(a-1)}\]


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