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Railways Quantitative Aptitude Simplification Sample Paper Simplification Sample Test Paper-3

  • question_answer
    If a is a natural number then \[{{a}^{2}}+\frac{1}{{{a}^{2}}}\] is always greater than or equal to

    A)  5                    

    B)  4         

    C)  3                    

    D)  2

    Correct Answer: D

    Solution :

    [d] ?a? is a natural number. \[\therefore \] \[{{a}^{2}}+\frac{1}{{{a}^{2}}}={{a}^{2}}+\frac{1}{{{a}^{2}}}-2+2\] \[=\,\,{{a}^{2}}+\frac{1}{{{a}^{2}}}-2.a.\frac{1}{a}+2\] \[{{a}^{2}}+\frac{1}{{{a}^{2}}}={{\left( a-\frac{1}{a} \right)}^{2}}+2\] Now, \[{{\left( a-\frac{1}{a} \right)}^{2}}\] is always greater than or equal to zero. \[\therefore \] \[{{a}^{2}}+\frac{1}{{{a}^{2}}}\ge 2\]


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