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4-2.Quadratic Equations and Inequations
hard
જો $x$ એ સમીકરણ $\sqrt {2x + 1} - \sqrt {2x - 1} = 1, \left( {x \ge \frac{1}{2}} \right)$ નો ઉકેલ હોય તો $\sqrt {4{x^2} - 1} $ ની કિમત મેળવો.
A
$\frac{3}{4}$
B
$\frac{1}{2}$
C
$2\sqrt 2 $
D
$2$
(JEE MAIN-2016)
Solution
$\sqrt{2 x+1}-\sqrt{2 x-1}=1……….(1)$
$\Rightarrow \quad 2 x+1+2 x-1-2 \sqrt{4 x^{2}-1}=1$
$\Rightarrow 4 x-1=2 \sqrt{4 x^{2}-1}$
$\Rightarrow 16 x^{2}-8 x+1=16 x^{2}-4$
$\Rightarrow 8 x=5$
$\Rightarrow \quad x=\frac{5}{8}$ which satisfies equation $(1)$
So, $\sqrt{4 x^{2}-1}=\frac{3}{4}$
Standard 11
Mathematics
