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The function $$f(x)=\frac{x}{x^2-6 x-16}, x \in \mathbb{R}-\{-2,8\}$$

JEE · Math · previous-year question

  1. A.decreases in $$(-\infty,-2) \cup(-2,8) \cup(8, \infty)$$correct
  2. B.increases in $$(-\infty,-2) \cup(-2,8) \cup(8, \infty)$$
  3. C.decreases in $$(-2,8)$$ and increases in $$(-\infty,-2) \cup(8, \infty)$$
  4. D.decreases in $$(-\infty,-2)$$ and increases in $$(8, \infty)$$

Answer

A. decreases in $$(-\infty,-2) \cup(-2,8) \cup(8, \infty)$$

Explanation

$$f(x)=\frac{x}{x^2-6 x-16}$$ Now, $$\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=\frac{-\left(\mathrm{x}^2+16\right)}{\left(\mathrm{x}^2-6 \mathrm{x}-16\right)^2} \\ & \mathrm{f}^{\prime}(\mathrm{x}) Thus $$f(x)$$ is decreasing in $$(-\infty,-2) \cup(-2,8) \cup(8, \infty)$$

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