Mathematics · Application of Derivatives

JEE Main 2024 — 29 January, Shift 2 — Question 16

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

Option A:
decreases in $(-2,8)$ and increases in $(-\infty,-2) \cup(8, \infty)$
Option B:
decreases in $(-\infty,-2) \cup(-2,8) \cup(8, \infty)$
Correct
Option C:
decreases in $(-\infty,-2)$ and increases in $(8, \infty)$
Option D:
increases in $(-\infty,-2) \cup(-2,8) \cup(8, \infty)$

Answer: B

Step-by-step solution

$f(x)=\frac{x}{x^{2}-6 x-16}$

Now, $f^{\prime}(x)=\frac{-\left(x^{2}+16\right)}{\left(x^{2}-6 x-16\right)^{2}}$

$\mathrm{f}^{\prime}(\mathrm{x})<0$

Thus $f(x)$ is decreasing in $(-\infty,-2) \cup(-2,8) \cup(8, \infty)$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 2
Subject: Mathematics
Chapter: Application of Derivatives
Topic: Monotonicity

The function f(x)=2 x+3(x)^2/3, x in mathbbR , has

The function f(x)=frac x x 2 -6 x-16 , x in mathbb R -\ -2,8\ | JEE Main 2024 PYQ with Solution · DhiX AI