The function f(x)=x x^2-6 x-16 , x R- -2,8 - Vidyadip

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

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

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

The odd natural number $a,$ such that the area of the region bounded by $y =1, y =3, x =0, x = y ^{a}$ is $\frac{364}{3}$, equal to.

If for all real triplets $(a, b, c), f(x)=a+b x+c x^{2}$ then $\int \limits_{0}^{1} f(\mathrm{x}) \mathrm{d} \mathrm{x}$ is equal to

If $\int {{x^5}{e^{ - 4{x^3}}}\,dx = \frac{1}{{48}}{e^{ - 4{x^3}}}f\left( x \right) + C} $, where $C$ is a constant of integration, then $f(x)$ is equal to

If ${{{x^3} - 6{x^2} + 10x - 2} \over {{x^2} - 5x + 6}} = f(x) + {A \over {(x - 2)}} + {B \over {(x - 3)}}$, then $f(x) = $

If the domain of the function $f(x)=\log _e$ $\left(\frac{2 x+3}{4 x^2+x-3}\right)+\cos ^{-1}\left(\frac{2 x-1}{x+2}\right)$ is $(\alpha, \beta]$, then the value of $5 \beta-4 \alpha$ is equal to

$\lim _{t \rightarrow 0}\left(1^{\frac{1}{\sin ^2 t}}+2^{\frac{1}{\sin ^2 t}}+\ldots .+n^{\frac{1}{\sin ^2 t}}\right)^{\sin ^2 t}$ is equal to $.......$

Let $f(\mathrm{x})=\mathrm{x} \cos ^{-1}(-\sin |\mathrm{x}|), \quad \mathrm{x} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right],$ then which of the following is true?

The area of the region, enclosed by the circle $\mathrm{x}^{2}+\mathrm{y}^{2}=2$ which is not common to the region bounded by the parabola $y^{2}=x$ and the straight line $\mathrm{y}=\mathrm{x},$ is

The number of points at which the function $f(x) = |x - 0.5| + |x - 1| + \tan x$ does not have a derivative in the interval $(0, 2),$ is

If the area (in sq. units) bounded by the parabola $y^2 =4\lambda x$ and the line $y = \lambda x$, $\lambda > 0$, is $\frac{1}{9}$, then $\lambda $ is equal to