The function $f(x) = x e^{-x}$ for all $x \in R$ attains a maximum value at $x = k$, then $k = $

  • A
    $1$
  • B
    $2$
  • C
    $\frac{1}{e}$
  • D
    $3$

Explore More

Similar Questions

Let $f:[0,1] \rightarrow \mathbb{R}$ be a function. Suppose $f$ is twice differentiable,$f(0)=f(1)=0$ and satisfies $f^{\prime \prime}(x)-2 f^{\prime}(x)+f(x) \geq e^x$ for $x \in[0,1]$.
$1.$ Which of the following is true for $0 < x < 1$?
$(A)$ $0 < f(x) < \infty$
$(B)$ $-\frac{1}{2} < f(x) < \frac{1}{2}$
$(C)$ $-\frac{1}{4} < f(x) < 1$
$(D)$ $-\infty < f(x) < 0$
$2.$ If the function $g(x) = e^{-x} f(x)$ assumes its minimum in the interval $[0,1]$ at $x=\frac{1}{4}$,which of the following is true?
$(A)$ $f^{\prime}(x) < f(x)$ for $x \in (0, 1/4)$
$(B)$ $f^{\prime}(x) > f(x)$ for $x \in (0, 1/4)$
$(C)$ $f^{\prime}(x) < f(x)$ for $x \in (1/4, 1)$
$(D)$ $f^{\prime}(x) > f(x)$ for $x \in (1/4, 1)$

The maximum area of the rectangle that can be inscribed in a circle of radius $r$ is

The function $y = a(1 - \cos x)$ is maximum when $x = $

Find the local maximum and local minimum values for the function $f(x) = \sin x - \cos x$ where $0 < x < 2\pi$.

Difficult
View Solution

The minimum value of $e^{(2x^2 - 2x - 1)\sin^2 x}$ is .......

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo