MCQ
The function $f(x)=\frac{x}{x^2-6 x-16}, x \in \mathbb{R}-\{-2,8\}$
  • A
     decreases in $(-2,8)$ and increases in $(-\infty,-2) \cup(8, \infty)$
  •  decreases in $(-\infty,-2) \cup(-2,8) \cup(8, \infty)$
  • C
     decreases in $(-\infty,-2)$ and increases in $(8, \infty)$
  • D
    increases in $(-\infty,-2) \cup(-2,8) \cup(8, \infty)$

Answer

Correct option: B.
 decreases in $(-\infty,-2) \cup(-2,8) \cup(8, \infty)$
b
$f(x)=\frac{x}{x^2-6 x-16}$

Now,

$ \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$

Thus $\mathrm{f}(\mathrm{x})$ is decreasing in

$(-\infty,-2) \cup(-2,8) \cup(8, \infty)$

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