For x 0,let h(x) = begin{cases} frac{1}{q} | vedclass

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

Question

(A) The function $h(x)$ is known as the Thomae's function (or the popcorn function). $1$. For any rational number $x = \frac{p}{q}$ (where $p, q$ are

Vedclass · Accepted Answer

$h(x)$ is discontinuous for all $x$ in $(0, \infty)$.

Vedclass · Answer

(A) The function $h(x)$ is known as the Thomae's function (or the popcorn function). $1$. For any rational number $x = \frac{p}{q}$ (where $p, q$ are relatively prime),$h(x) = \frac{1}{q}$. As $x$ approaches a rational number,the values of $h(x)$ for nearby irrational numbers approach $0$. Since $\frac{1}{q} 
eq 0$,the function is discontinuous at every rational point. Thus,statement $C$ is true. $2$. For any irrational number $x$,$h(x) = 0$. For any $\epsilon > 0$,there are only finitely many rational numbers $\frac{p}{q}$ in any interval such that $\frac{1}{q} \geq \epsilon$. Thus,the limit of $h(x)$ as $x$ approaches any irrational number is $0$. Since $h(x) = 0$ at irrational points,the function is continuous at every irrational point. Thus,statement $B$ is true. $3$. Since the function is continuous at irrational points and discontinuous at rational points,it is not discontinuous for all $x$ in $(0, \infty)$. Therefore,statement $A$ is false. $4$. Since the function is discontinuous at every rational point,it is not differentiable (derivable) at any rational point. Thus,statement $D$ is true. Conclusion: Statement $A$ does not hold good.

For $x > 0$,let $h(x) = \begin{cases} \frac{1}{q} & \text{if } x = \frac{p}{q} \text{ (where } p, q \in \mathbb{N} \text{ are relatively prime)} \\ 0 & \text{if } x \text{ is irrational} \end{cases}$. Which of the following statements does not hold true?

Similar Questions

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

If $f(x) = \begin{cases} x \sin x, & 0 < x \le \frac{\pi}{2} \\ \frac{\pi}{2} \sin(\pi + x), & \frac{\pi}{2} < x < \pi \end{cases}$,then

If $f(x)$ is continuous at $x = 3$ where $f(x) = \begin{cases} ax + 1, & \text{for } x \leq 3 \\ bx + 3, & \text{for } x > 3 \end{cases}$,then

If $f(x) = \begin{cases} x^2 - 1, & \text{if } x \ge 2 \\ x + 1, & \text{if } x < 2 \end{cases}$, then $\lim_{x \to 1} f(x) + \lim_{x \to 2} f(x) =$

If $f(x)$,defined below,is continuous at $x = 4$,then find the values of $a$ and $b$ given that $f(x)$ is continuous on the interval $[0, 8]$.
$f(x) = \begin{cases} x^2 + ax + b, & 0 \leq x < 2 \\ 3x + 2, & 2 \leq x \leq 4 \\ 2ax + 5b, & 4 < x \leq 8 \end{cases}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module