Let [x] denote the greatest integer less than | vedclass

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

Question

(D) For the interval $-1 \le x \le 1$,consider the function $g(x) = x \sin \pi x$.Since $\sin \pi x$ has the same sign as $x$ for $x \in (-1, 1)$,the

Vedclass · Accepted Answer

All the above

Vedclass · Answer

(D) For the interval $-1 \le x \le 1$,consider the function $g(x) = x \sin \pi x$.Since $\sin \pi x$ has the same sign as $x$ for $x \in (-1, 1)$,the product $x \sin \pi x$ is always non-negative.Specifically,for $x \in (-1, 1)$,$0 \le x \sin \pi x Therefore,the greatest integer function $[x \sin \pi x] = 0$ for all $x \in [-1, 1]$.Since $f(x) = 0$ is a constant function on the interval $[-1, 1]$,it is continuous everywhere in this interval,including at $x = 0$ and in the sub-interval $(-1, 0)$.Furthermore,a constant function is differentiable everywhere,so $f(x)$ is differentiable in $(-1, 1)$.Thus,all the given statements are correct.

Let $[x]$ denote the greatest integer less than or equal to $x$. If $f(x) = [x \sin \pi x]$,then $f(x)$ is

Similar Questions

If $f(x) = \begin{cases} ax^2 - b, & 0 \le x < 1 \\ 2, & x = 1 \\ x + 1, & 1 < x \le 2 \end{cases}$ is continuous at $x = 1$,then the most suitable values of $a$ and $b$ are:

If $f(x) = \begin{cases} -x^2, & \text{when } x \le 0 \\ 5x - 4, & \text{when } 0 < x \le 1 \\ 4x^2 - 3x, & \text{when } 1 < x < 2 \\ 3x + 4, & \text{when } x \ge 2 \end{cases}$,then:

If the real valued function $f(x) = \begin{cases} \frac{\cos 3x - \cos x}{x \sin x} & \text{if } x < 0 \\ p & \text{if } x = 0 \\ \frac{\log(1 + q \sin x)}{x} & \text{if } x > 0 \end{cases}$ is continuous at $x = 0$, then $p + q =$

If the function $f$ defined on $\left(-\frac{1}{3}, \frac{1}{3}\right)$ by $f(x) = \begin{cases} \frac{1}{x} \log_{e}\left(\frac{1+3x}{1-2x}\right) & x \neq 0 \\ k & x = 0 \end{cases}$ is continuous,then $k$ is equal to

If the function $f(x)=\begin{cases} \frac{1-\cos x}{x^{2}}, & \text{for } x \neq 0 \\ k, & \text{for } x=0 \end{cases}$ is continuous at $x=0$,then the value of $k$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module