Difference between maximum and minimum values of f(x) = x^4e^ -x^… - Vidyadip

MCQ

Difference between maximum and minimum values of $f(x) = x^4e^{-x^2} \ \ \forall x \in R,$ is -

A
$\frac{4}{e^2} - \frac{2}{e}$
B
$\frac{4}{e} - \frac{2}{e^2}$
✓
$\frac{4}{e^2}$
D
$\frac{2}{e}$

✓

Answer

Correct option: C.

$\frac{4}{e^2}$

c
$f^{\prime}(x)=0$ at $x=0, \pm \sqrt{2}$

$f(\mathrm{x})_{\max }=\frac{4}{\mathrm{e}^{2}} ; f(\mathrm{x})_{\min }=0$

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

More "M.C.Q (1 Marks)" from 6. Application of derivatives→6. Application of derivatives — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

The set of all those points, where the function $f(x) = \frac{x}{{1 + |x|}}$ is differentiable, is

If a function $\text{f}:[2,\infty)\rightarrow\ \text{B}$ defined by f(x) = x² - 4x + 5 is a bijection, then B =

$\text{R}$
$[1,\infty)$
$[4,\infty)$
$[5,\infty)$

Let $k$ and $m$ be positive real numbers such that the function $\quad f ( x )=\left\{\begin{array}{cc}3 x ^2+ k \sqrt{ x +1}, & 0< x <1 \\ mx ^2+ k ^2, & x \geq 1\end{array}\right.$ is differentiable for all $x > 0$. Then $\frac{8 f^{\prime}(8)}{f^{\prime}\left(\frac{1}{8}\right)}$ is equal to $.............$.

Let $\quad \overrightarrow{ a }=\alpha \hat{ i }+3 \hat{ j }-\hat{ k }, \overrightarrow{ b }=3 \hat{ i }-\beta \hat{ j }+4 \hat{ k } \quad$ and $\overrightarrow{ c }=\hat{ i }+2 \hat{ j }-2 \hat{ k }$ where $\alpha, \beta \in R$, be three vectors. If the projection of $\vec{a}$ on $\vec{c}$ is $\frac{10}{3}$ and $\overrightarrow{ b } \times \overrightarrow{ c }=-6 \hat{ i }+10 \hat{ j }+7 \hat{ k }$, then the value of $\alpha+\beta$ equal to

The differential equation for all the straight lines which are at a unit distance from the origin is

The distance of the point P(a, b, c) from the x-axis is:

$\sqrt{\text{b}^2+\text{c}^2}$
$\sqrt{\text{a}^2+\text{c}^2}$
$\sqrt{\text{a}^2+\text{b}^2}$
$\text{none of these}$

If $\left| {\vec a} \right| = 2,\left| {\vec b} \right| = 3$ and $\left| {2\,\vec a - \vec b} \right| = 5$, then $\left| {2\,\vec a + \vec b} \right|$ equals

If $\tan^{-1}3+\tan^{-1}\text{x}=\tan^{-1}8,$ then x =

$5$
$\frac{1}{5}$
$\frac{5}{14}$
$\frac{14}{5}$

If $A=\left[\begin{array}{ccc}2 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3\end{array}\right]$, then $A^{-1}$ exists only if

Area bounded by the curve $y = \min \{\sin^2x, \cos^2x \}$ and $x-$ axis between the ordinates $x = 0$ and $x = \frac{{5\pi }}{4}$ is