The function f(x) = begin{cases} 2x + 1, & x | vedclass

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

Question

(D) For a function defined as $f(x) = g(x)$ when $x \in \mathbb{Q}$ and $f(x) = h(x)$ when $x 
otin \mathbb{Q}$,the function is continuous at $x = a$

Vedclass · Accepted Answer

differentiable and continuous only at one point and discontinuous elsewhere

Vedclass · Answer

(D) For a function defined as $f(x) = g(x)$ when $x \in \mathbb{Q}$ and $f(x) = h(x)$ when $x 
otin \mathbb{Q}$,the function is continuous at $x = a$ if and only if $g(a) = h(a)$.Here,$g(x) = 2x + 1$ and $h(x) = x^2 - 2x + 5$.Setting $g(x) = h(x)$ gives $2x + 1 = x^2 - 2x + 5$,which simplifies to $x^2 - 4x + 4 = 0$,or $(x - 2)^2 = 0$.Thus,$x = 2$ is the only point where the function is continuous.For differentiability at $x = a$,the function must be continuous at $x = a$,and the derivatives must satisfy $g'(a) = h'(a)$.Here,$g'(x) = 2$ and $h'(x) = 2x - 2$.At $x = 2$,$g'(2) = 2$ and $h'(2) = 2(2) - 2 = 2$.Since $g'(2) = h'(2)$,the function is differentiable at $x = 2$.Therefore,the function is continuous and differentiable only at $x = 2$ and discontinuous elsewhere.

The function $f(x) = \begin{cases} 2x + 1, & x \in \mathbb{Q} \\ x^2 - 2x + 5, & x \notin \mathbb{Q} \end{cases}$ is

Similar Questions

If $f(x) = |x|,$ then $f'(0) = $

The function $y = e^{-|x|}$ is

The function that is not differentiable at $x=1$ is

The function $f(x) = |\cos x|$ is

Which of the following functions is differentiable at $x = 0$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module