Easy MCQ +4 / -1 PYQ · JEE Mains 2024

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

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

Solution

<p>$f(x)=\frac{x}{x^2-6 x-16}$</p> <p>Now,</p> <p>$$\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=\frac{-\left(\mathrm{x}^2+16\right)}{\left(\mathrm{x}^2-6 \mathrm{x}-16\right)^2} \\ & \mathrm{f}^{\prime}(\mathrm{x})<0 \end{aligned}$$</p> <p>Thus $f(x)$ is decreasing in</p> <p>$(-\infty,-2) \cup(-2,8) \cup(8, \infty)$</p>

About this question

Subject: Mathematics · Chapter: Application of Derivatives · Topic: Increasing and Decreasing Functions

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