Let f:R rightarrow R be defined as f(x) = beg | vedclass

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

Question

(D) function $f(x)$ defined as $f(x) = \begin{cases} g(x), & x \in \mathbb{Q} \ h(x), & x 
otin \mathbb{Q} \end{cases}$ is continuous at $x = a$ if

Vedclass · Accepted Answer

$f$ is continuous at $x = k\pi$ where $k$ is an integer

Vedclass · Answer

(D) function $f(x)$ defined as $f(x) = \begin{cases} g(x), & x \in \mathbb{Q} \ h(x), & x 
otin \mathbb{Q} \end{cases}$ is continuous at $x = a$ if and only if $g(a) = h(a)$. Here, $g(x) = \sin |x|$ and $h(x) = 0$. For continuity, we require $\sin |x| = 0$. This occurs when $|x| = n\pi$ for some integer $n$, which implies $x = n\pi$ for $n \in \mathbb{Z}$. At any point $x = k\pi$ (where $k \in \mathbb{Z}$), $f(x) = \sin |k\pi| = 0$. For any other point $x 
eq k\pi$, $\sin |x| 
eq 0$, so the function is discontinuous because the values for rational and irrational numbers do not coincide. Thus, $f$ is continuous only at $x = k\pi$ and discontinuous everywhere else. Therefore, option $D$ is correct.

Let $f:R \rightarrow R$ be defined as $f(x) = \begin{cases} 0, & x \text{ is irrational} \\ \sin |x|, & x \text{ is rational} \end{cases}$. Then, which of the following is true?

Similar Questions

Let $f(x) = \frac{1-\tan x}{4x-\pi}$ for $x \neq \frac{\pi}{4}$ and $x \in [0, \frac{1}{2}]$. If $f(x)$ is continuous in $[0, \frac{\pi}{2}]$,then $f(\frac{\pi}{4})$ is

If $f(x) = \frac{1+\cos \pi x}{\pi(1-x)^2}$ for $x \neq 1$ is continuous at $x=1$,then $f(1)$ is equal to

Consider the function $f:(0,2) \rightarrow R$ defined by $f(x)=\frac{x}{2}+\frac{2}{x}$ and the function $g(x)$ defined by $g(x)=\begin{cases} \min \{f(t) : 0 < t \leq x\}, & 0 < x \leq 1 \\ \frac{3}{2}+x, & 1 < x < 2 \end{cases}$. Then,

Let $f(x) = \begin{cases} x^p \sin \left( \frac{1}{x} \right) + x|x^3|, & x \neq 0 \\ 0, & x = 0 \end{cases}$. Then the complete set of values of $p$ for which $f''(x)$ is continuous at $x = 0$ is:

Let $f(x) = \begin{cases} x^2 \sin \left(\frac{1}{x}\right) & , x \neq 0 \\ 0 & , x=0 \end{cases}$. Then at $x=0$:

Vedclass Test Series

Exam Paper Generator

Online Exam Module