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JEE Mains · Maths · STD 12 - 6. Application of derivatives

Let \(f\left( x \right) = {x^2} + \frac{1}{{{x^2}}}\) and \(g\left( x \right) = x - \frac{1}{x},\;x \in R - \left\{ { - 1,1,0} \right\}\). If \(h\left( x \right) = \frac{{f\left( x \right)}}{{g\left( x \right)}}\) then the local minimum value of  \(h\left( x \right)\) is:

  1. A \(-3\)
  2. B \( - 2\sqrt 2 \)
  3. C \(2\sqrt 2 \)
  4. D \(3\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(2\sqrt 2 \)

Step-by-step Solution

Detailed explanation

Here, \(h\left( x \right) = \frac{{{x^2} + \frac{1}{{{x^2}}}}}{{x - \frac{1}{x}}} = \left( {x - \frac{1}{x}} \right) + \frac{2}{{x - \frac{1}{x}}}\) when \(x - \frac{1}{x} < 0\) \(x - \frac{1}{x} + \frac{2}{{x - \frac{1}{x}}} \le - 2\sqrt 2 \) Hence, \( - 2\sqrt 2 \) will be…
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