JEE MAIN - Mathematics (2024 - 29th January Evening Shift - No. 6)
The function $$f(x)=\frac{x}{x^2-6 x-16}, x \in \mathbb{R}-\{-2,8\}$$
decreases in $$(-\infty,-2) \cup(-2,8) \cup(8, \infty)$$
increases in $$(-\infty,-2) \cup(-2,8) \cup(8, \infty)$$
decreases in $$(-2,8)$$ and increases in $$(-\infty,-2) \cup(8, \infty)$$
decreases in $$(-\infty,-2)$$ and increases in $$(8, \infty)$$
Explanation
$$f(x)=\frac{x}{x^2-6 x-16}$$
Now,
$$\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=\frac{-\left(\mathrm{x}^2+16\right)}{\left(\mathrm{x}^2-6 \mathrm{x}-16\right)^2} \\ & \mathrm{f}^{\prime}(\mathrm{x})<0 \end{aligned}$$
Thus $$f(x)$$ is decreasing in
$$(-\infty,-2) \cup(-2,8) \cup(8, \infty)$$
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