The following figure shows the graph of a differentiable function $y=f(x)$ on the interval $[a, b]$ (not containing $0$). Let $g(x)=\frac{f(x)}{x}$. Which of the following is a possible graph of $y=g(x)$?

  • A
    Fig $1$
  • B
    Fig $2$
  • C
    Fig $3$
  • D
    Fig $4$

Explore More

Similar Questions

$A$ cone of maximum volume is inscribed in a sphere of radius $R$. The ratio of the height of the cone to the diameter of the sphere is:

Difficult
View Solution

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 $x \in (1/4, 1)$

Let $f : R \rightarrow R$ be a function defined by $f(x) = ||x+2|-2|x||$. If $m$ is the number of points of local minima and $n$ is the number of points of local maxima of $f$,then $m+n$ is

The values of $x$ at the stationary points of $f(x)=x^3+3x^2-2$ are

Find the semi-vertical angle of a right circular cone of a given slant height,if the volume of the cone is maximum.

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