Let f:(0,1) rightarrow mathbb{R} be the funct | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The function $f(x) = [4x](x-\frac{1}{4})^2(x-\frac{1}{2})$ can be written as:$f(x) = \begin{cases} 0 & 0 < x < \frac{1}{4} \ (x-\frac{1}{4})^2(x-

Vedclass · Accepted Answer

$A, B$

Vedclass · Answer

(B) The function $f(x) = [4x](x-\frac{1}{4})^2(x-\frac{1}{2})$ can be written as:$f(x) = \begin{cases} 0 & 0 $1$. Continuity: The function is discontinuous at $x = \frac{1}{2}$ and $x = \frac{3}{4}$ because the limits from the left and right do not match at these points. Thus,statement $(A)$ is false.$2$. Differentiability: The function is continuous at $x = \frac{1}{4}$ but not differentiable there. It is discontinuous at $x = \frac{1}{2}$ and $x = \frac{3}{4}$. Thus,there is exactly one point $(x = \frac{1}{4})$ where it is continuous but not differentiable. Statement $(B)$ is true.$3$. Non-differentiability: The function is non-differentiable at $x = \frac{1}{4}, \frac{1}{2}, \frac{3}{4}$. This is exactly three points. Statement $(C)$ is false.$4$. Minimum value: For $x \in [\frac{1}{4}, \frac{1}{2}]$,$f(x) = (x-\frac{1}{4})^2(x-\frac{1}{2})$. Let $t = x-\frac{1}{4}$,then $f(t) = t^2(t-\frac{1}{4}) = t^3 - \frac{1}{4}t^2$. $f'(t) = 3t^2 - \frac{1}{2}t = t(3t - \frac{1}{2})$. Setting $f'(t) = 0$,we get $t = \frac{1}{6}$. The value is $(\frac{1}{6})^2(\frac{1}{6}-\frac{1}{4}) = \frac{1}{36}(-\frac{1}{12}) = -\frac{1}{432}$. Thus,statement $(D)$ is false.

Similar Questions

If $f(x) = \operatorname{sgn} \left( 3\cos x - \frac{a}{3} \right)$ is continuous for all $x$,then the least positive integral value of $a$ is - (where $\operatorname{sgn}(x)$ denotes the signum function of $x$)

If the function $f$ defined on $\left(\frac{\pi}{6}, \frac{\pi}{3}\right)$ by $f(x)=\begin{cases} \frac{\sqrt{2} \cos x-1}{\cot x-1}, & x \neq \frac{\pi}{4} \\ k, & x=\frac{\pi}{4} \end{cases}$ is continuous,then $k$ is equal to

If $f(x) = \frac{2 - \sqrt{x + 4}}{\sin 2x}, (x \neq 0),$ is a continuous function at $x = 0$,then $f(0)$ equals:

Let $f: R \rightarrow R$ be such that $f(2x-1) = f(x)$ for all $x \in R$. If $f$ is continuous at $x = 1$ and $f(1) = 1$, then:

Let $f:[0, \pi] \rightarrow R$ be defined as $f(x)=\begin{cases} \sin x, & \text{if } x \text{ is irrational and } x \in[0, \pi] \\ \tan^2 x, & \text{if } x \text{ is rational and } x \in[0, \pi] \end{cases}$. The number of points in $[0, \pi]$ at which the function $f$ is continuous is

Vedclass Test Series

Exam Paper Generator

Online Exam Module