Railways R.R.C. (Patna) Solved Paper Held on 1st_Shift 23-11-2014

  • question_answer
    If a\[=\frac{\sqrt{3}}{2}\]then the value of\[(\sqrt{1+a+}\sqrt{1-a})is\_\_\]

    A) \[\sqrt{3}\]                                

    B) \[\frac{\sqrt{3}}{2}\]

    C) \[(2+\sqrt{3})\]             

    D) \[(2-\sqrt{3})\]

    Correct Answer: A

    Solution :

    \[(\sqrt{1+a}+\sqrt{1-a})\] \[=\sqrt{{{\left( \sqrt{1+a}+\sqrt{1-a} \right)}^{2}}}\] \[=\sqrt{1+a+1-a+2\sqrt{(1+a)}(1-a)}\] \[=\sqrt{2+2\sqrt{1-{{a}^{2}}}}\] \[=\]\[\sqrt{2+2\sqrt{1-{{\left( \frac{\sqrt{5}}{2} \right)}^{2}}}}\] \[=\sqrt{2+2\sqrt{1-\frac{3}{4}}}\] \[=\sqrt{2+1=}\sqrt{3}\]


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