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SSC Sample Paper SSC CHSL (10+2) Sample Test Paper-23

  • question_answer
    \[\frac{1}{1+{{4}^{a-b}}}+\frac{1}{1+{{2}^{2b-2a}}}\] is equal to -

    A)  0                                

    B)  1

    C)  2                                

    D)  \[~-1\]

    Correct Answer: B

    Solution :

     \[\frac{1}{1+{{4}^{a-b}}}+\frac{1}{1+{{2}^{2b-2a}}}\] \[=\,\,\,\,\frac{1}{1+{{4}^{(a-b)}}}+\frac{1}{1+{{4}^{(b-a)}}}\] \[=\,\,\,\frac{1}{1+\frac{{{4}^{a}}}{{{4}^{b}}}}+\frac{1}{1+\frac{{{4}^{b}}}{{{4}^{a}}}}\] \[=\,\,\,\frac{{{4}^{b}}}{{{4}^{a}}+{{4}^{b}}}+\frac{{{4}^{a}}}{{{4}^{a}}+{{4}^{b}}}=\underline{\mathbf{1}}\]


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