Given equation

$\sqrt{2x+1}-\sqrt{2x-1}=1, \qquad x\ge \frac12$

We need to find:

$\sqrt{4x^2-1}$

Use identity by squaring smartly

Let $a=\sqrt{2x+1},\qquad b=\sqrt{2x-1}$ Then $a-b=1$

Also, $a^2-b^2=(2x+1)-(2x-1)=2$

But $a^2-b^2=(a-b)(a+b)$ So, $2=1\cdot(a+b) \implies a+b=2$

Solve for $a$ and $b$

We have $a-b=1,\qquad a+b=2$

Adding, $2a=3 \implies a=\frac32$

Subtracting, $2b=1 \implies b=\frac12$

Find $x$

Since $b=\sqrt{2x-1}=\frac12$ squaring: $2x-1=\frac14$ $2x=\frac54$ $x=\frac58$

Compute $\sqrt{4x^2-1}$

$x=\frac58$

Then $4x^2-1=4\left(\frac{25}{64}\right)-1=\frac{100}{64}-1=\frac{25}{16}-1=\frac{9}{16}$

Therefore, $\sqrt{4x^2-1}=\sqrt{\frac{9}{16}}=\frac34$

Check options

$\frac34$ matches Option A.

Comparison with stored answer

Stored correct answer: A

Derived answer: A

So they agree.