If $\operatorname{Lt}_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=e^x(x+1)$ and $f(0)=0$,then $\frac{d}{d x}\left(f(x) e^{-x}\right)+\frac{d}{d x}\left(\frac{f(x)}{x}\right)=$

  • A
    $e^x+1$
  • B
    $x^2 e^x+x$
  • C
    $x e^x+1$
  • D
    $x^2 e^x$

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Similar Questions

Let $f: R \rightarrow R$ be a function defined by $f(x) = \begin{cases} \max_{t \leq x} \{t^3 - 3t\} & x \leq 2 \\ x^2 + 2x - 6 & 2 < x < 3 \\ [x-3] + 9 & 3 \leq x \leq 5 \\ 2x + 1 & x > 5 \end{cases}$ where $[t]$ is the greatest integer less than or equal to $t$. Let $m$ be the number of points where $f$ is not differentiable and $I = \int_{-2}^{2} f(x) dx$. Then the ordered pair $(m, I)$ is equal to:

Let $f_1: R \rightarrow R, f_2:\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \rightarrow R, f_3:\left(-1, e^{\frac{\pi}{2}}-2\right) \rightarrow R$ and $f_4: R \rightarrow R$ be functions defined by:
$(i)$ $f_1(x)=\sin \left(\sqrt{1-e^{-x^2}}\right)$
$(ii)$ $f_2(x)=\begin{cases} \frac{|\sin x|}{\tan^{-1} x} & \text{if } x \neq 0 \\ 1 & \text{if } x=0 \end{cases}$,where the inverse trigonometric function $\tan^{-1} x$ assumes values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
$(iii)$ $f_3(x)=\left[\sin \left(\log_e(x+2)\right)\right]$,where,for $t \in R, [t]$ denotes the greatest integer less than or equal to $t$.
$(iv)$ $f_4(x)=\begin{cases} x^2 \sin \left(\frac{1}{x}\right) & \text{if } x \neq 0 \\ 0 & \text{if } x=0 \end{cases}$
$LIST-I$ $LIST-II$
$P$. The function $f_1$ is $1$. $NOT$ continuous at $x=0$
$Q$. The function $f_2$ is $2$. Continuous at $x=0$ and $NOT$ differentiable at $x=0$
$R$. The function $f_3$ is $3$. Differentiable at $x=0$ and its derivative is $NOT$ continuous at $x=0$
$S$. The function $f_4$ is $4$. Differentiable at $x=0$ and its derivative is continuous at $x=0$

The correct option is:

If $f(x) = \begin{cases} ax+b, & \text{if } x \leq 1 \\ ax^2+c, & \text{if } 1 < x \leq 2 \\ \frac{dx^2+1}{x}, & \text{if } x > 2 \end{cases}$ is differentiable on $\mathbb{R}$,then $ad-bc = $

The number of real roots of the equation $e^{6x} - e^{4x} - 2e^{3x} - 12e^{2x} + e^{x} + 1 = 0$ is:

The number of points,where the curve $f(x) = e^{8x} - e^{6x} - 3e^{4x} - e^{2x} + 1$,$x \in R$ cuts the $x$-axis,is equal to

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