If function f(x) = begin{cases} x, & text{if | vedclass

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

Question

(B) For $f(x)$ to be continuous at a point $x = a$,the limit $\lim_{x 	o a} f(x)$ must exist and be equal to $f(a)$.Consider the point $x = a$. For a

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) For $f(x)$ to be continuous at a point $x = a$,the limit $\lim_{x 	o a} f(x)$ must exist and be equal to $f(a)$.Consider the point $x = a$. For any neighborhood of $a$,there exist both rational and irrational numbers.If $x$ is rational,$f(x) = x$,and if $x$ is irrational,$f(x) = 1 - x$.For the limit to exist at $x = a$,we must have $a = 1 - a$,which implies $2a = 1$,or $a = 1/2$.Now,check continuity at $a = 1/2$:$f(1/2) = 1/2$ (since $1/2$ is rational).However,in any neighborhood of $1/2$,there are irrational numbers $x$ such that $f(x) = 1 - x$,which approaches $1 - 1/2 = 1/2$.Wait,let us re-evaluate: $\lim_{x 	o 1/2} f(x)$ exists if the values of $f(x)$ for rational and irrational numbers approach the same value.For rational $x$,$f(x) = x 	o 1/2$.For irrational $x$,$f(x) = 1 - x 	o 1 - 1/2 = 1/2$.Since both limits are equal to $1/2$,the limit exists at $x = 1/2$.Thus,$f(x)$ is continuous only at $x = 1/2$.Therefore,the number of points of continuity is $1$.

If function $f(x) = \begin{cases} x, & \text{if } x \text{ is rational} \\ 1 - x, & \text{if } x \text{ is irrational} \end{cases}$,then $f(x)$ is continuous at how many points?

Similar Questions

The function $f(x) = \frac{\log(1 + ax) - \log(1 - bx)}{x}$ is not defined at $x = 0$. The value which should be assigned to $f$ at $x = 0$ so that it is continuous at $x = 0$ is:

If $f(x) = \begin{cases} \frac{a}{2}(x - |x|), & \text{for } x < 0 \\ 0, & \text{for } x = 0 \\ bx^2 \sin \left( \frac{1}{x} \right), & \text{for } x > 0 \end{cases}$ is continuous at $x = 0$,then

Let $f: R \rightarrow R$ be defined as $f(x) = \begin{cases} 2 \sin \left(-\frac{\pi x}{2}\right), & \text{if } x < -1 \\ |ax^2 + x + b|, & \text{if } -1 \leq x \leq 1 \\ \sin(\pi x), & \text{if } x > 1 \end{cases}$. If $f(x)$ is continuous on $R$,then $a + b$ equals ..... .

If $f(x) = \begin{cases} e^{1/x}, & x \ne 0 \\ 0, & x = 0 \end{cases}$,then:

For $a \neq 0$ and $b \neq 0$,if the real valued function $f(x) = \frac{\sqrt[5]{a(625+x)} - 5}{\sqrt[4]{625+bx} - 5}$ is continuous at $x = 0$,then $f(0) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module