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

माना \(f(x)=x^{2}+\frac{1}{x^{2}}\) तथा \(g(x)=x-\frac{1}{x}, x \in R-(-1,0,1)\) है। यदि \(h(x)=\frac{f(x)}{g(x)}\) है, तो \(h(x)\) का स्थानीय न्यूनतम मान है

  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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