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:
- A \(-3\)
- B \( - 2\sqrt 2 \)
- C \(2\sqrt 2 \)
- D \(3\)
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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