DSSSB TGT CS Mock 04 – Question 052

Question:

If \(x = 3 + 2\sqrt{2}\), find the value of \(\sqrt{x} – \frac{1}{\sqrt{x}}\).
यदि \(x = 3 + 2\sqrt{2}\) है, तो \(\sqrt{x} – \frac{1}{\sqrt{x}}\) का मान क्या होगा?

Options:

  • 2
  • \(\sqrt{2}\)
  • 4
  • \(2\sqrt{2}\)
  • 2
  • \(\sqrt{2}\)
  • 4
  • \(2\sqrt{2}\)

Correct Answer: a)

Explanation:

\(x = 2 + 1 + 2\sqrt{2} = (\sqrt{2} + 1)^2 \implies \sqrt{x} = \sqrt{2} + 1\).
\(\frac{1}{\sqrt{x}} = \frac{1}{\sqrt{2} + 1} = \sqrt{2} – 1\).
\(\sqrt{x} – \frac{1}{\sqrt{x}} = (\sqrt{2} + 1) – (\sqrt{2} – 1) = 2\).
\(\sqrt{x} = \sqrt{2} + 1\), \(1/\sqrt{x} = \sqrt{2} – 1\)। घटाने पर मान 2 प्राप्त होता है।

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