SATHEE: Application of Derivatives 4 Question 11

Application of Derivatives 4 Question 11

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

(a) 3

(b) -3

(d) $2 \sqrt{2}$

(2018 Main)

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Answer:

Correct Answer: 12. (c)

Solution:

We have,

$\begin{aligned} f (x) = x^{2} + \frac{1}{x^{2}} and g (x) = x - \frac{1}{x} \Rightarrow h (x) = \frac{f (x)}{g (x)} \\ ∴ h (x) = \frac{x^{2} + \frac{1}{x^{2}}}{x - \frac{1}{x}} = \frac{x - \frac{1}{x} + 2}{x - \frac{1}{x}} \\ \Rightarrow h (x) = x - \frac{1}{x} + \frac{2}{x - \frac{1}{x}} \\ x - \frac{1}{x} > 0, x - \frac{1}{x} + \frac{2}{x - \frac{1}{x}} \in [2 \sqrt{2}, \infty) \\ x - \frac{1}{x} < 0, x - \frac{1}{x} + \frac{2}{x - \frac{1}{x}} \in (- \infty, 2 \sqrt{2}] \end{aligned}$

$∴$ Local minimum value is $2 \sqrt{2}$ .

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Application of Derivatives 4 Question 11

Answer:

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