If x=3+2 sqrt{2}, then the value of (sqrt{x}-1/sqrt{x}) is (a) 1 (b) 2 (c) 2 sqrt{2} (d) 3 sqrt{3}

Given

\[
\begin{aligned}
& x=3+2 \sqrt{2} \\
& x=2+1+2 \sqrt{2} \\
& x=(\sqrt{2})^2+(1)^2+2.1 \cdot \sqrt{2} \\
& x=(\sqrt{2}+1)^2 \\
& \sqrt{x}=(\sqrt{2}+1) \\
& \frac{1}{\sqrt{x}}=\frac{1}{\sqrt{2}+1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1}=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1
\end{aligned}
\]

Now, \(\sqrt{x}-\frac{1}{\sqrt{x}}=\sqrt{2}+1-(\sqrt{2}-1)=\sqrt{2}+1-\sqrt{2}+1\)
\[
\sqrt{x}-\frac{1}{\sqrt{x}}=2
\]

Therefore, Correct option is 2.