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JEE Main & Advanced JEE Main Paper (Held On 10 April 2016)

  • question_answer
    If x is a solution of the equation, \[\sqrt{2x+1}-\sqrt{2x-1}=1\,,\,\,\left( x\le \frac{1}{2} \right)\] , then \[\sqrt{4{{x}^{2}}-1}\] is equal to   JEE Main Online Paper (Held On 10 April 2016)

    A) \[2\]                                               

    B) \[\frac{3}{4}\]

    C) \[2\sqrt{2}\]

    D) \[\frac{1}{2}\]

    Correct Answer: B

    Solution :

    \[\sqrt{2x+1}=1+\sqrt{2x-1}\]                 Squaring on both sides \[2x+1=1+2x-1+2\sqrt{2x-1}\]                    \[1=2\sqrt{2x-1}\]                    \[1=4\sqrt{2x-1}\]                    \[x=5/8\] Now \[\sqrt{4{{x}^{2}}-1}\] at \[x=5/8\]  =  \[\sqrt{4\times \frac{25}{64}-1}=3/4\]


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