JEE Main 2016 · 10 Apr · Shift 1 · Mathematics Q62

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MathematicsSingle correct+4 / -1Question 62 of 90

If x is a solution of the equation, $\sqrt{2x+1}-\sqrt{2x-1}=1$ , $\left(x\ge\dfrac12\right)$ , then $\sqrt{4x^2-1}$ is equal to :

\dfrac34

\dfrac12

2\sqrt2

Answer and worked solution

Official answer: A

$(\sqrt{2x+1}-\sqrt{2x-1})(\sqrt{2x+1}+\sqrt{2x-1})=2$ , so the sum is 2. Then $\sqrt{2x+1}=\frac32$ , $x=\frac58$ , and $\sqrt{4x^2-1}=\sqrt{\frac{100}{64}-1}=\frac34$ .

Topics:Inequalities and modulus

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JEE Main 2016 · 10 Apr · Shift 1, Mathematics question 62