Let f(x) = begin{cases} operatorname{sgn}(x^2 | vedclass

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

Question

(A) The function is defined as $f(x) = \operatorname{sgn}(x^2 - 3x + 2)$ for $x \in \mathbb{Q}$ and $f(x) = 0$ for $x 
otin \mathbb{Q}$.We know that

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) The function is defined as $f(x) = \operatorname{sgn}(x^2 - 3x + 2)$ for $x \in \mathbb{Q}$ and $f(x) = 0$ for $x 
otin \mathbb{Q}$.We know that $\operatorname{sgn}(x^2 - 3x + 2) = \operatorname{sgn}((x-1)(x-2))$.The signum function takes values $\{-1, 0, 1\}$.For $f(x)$ to be continuous at a point $x = a$,the limit $\lim_{x 	o a} f(x)$ must exist and equal $f(a)$.Since $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ are dense in $\mathbb{R}$,for the limit to exist,the function must take the same value for both rational and irrational sequences approaching $a$.Thus,we require $\operatorname{sgn}(a^2 - 3a + 2) = 0$.This occurs when $a^2 - 3a + 2 = 0$,which gives $a = 1$ or $a = 2$.At $x = 1$,$f(1) = \operatorname{sgn}(0) = 0$. For any sequence $x_n 	o 1$,$f(x_n)$ will approach $0$ because the values of $\operatorname{sgn}(x^2 - 3x + 2)$ are $0$ at $x=1$ and $x=2$,and the function is $0$ for irrational numbers.Specifically,near $x=1$,the function values are $\{-1, 0, 1\}$ for rationals and $0$ for irrationals. The limit exists only if the function value is $0$ for all sequences,which happens at the roots.Thus,$f(x)$ is continuous at $x=1$ and $x=2$.There are $2$ such points.

Let $f(x) = \begin{cases} \operatorname{sgn}(x^2 - 3x + 2) & x \in \mathbb{Q} \\ 0 & x \notin \mathbb{Q} \end{cases}$,then the number of points where $f(x)$ is continuous is (where $\operatorname{sgn}(x)$ denotes the signum function of $x$).

Similar Questions

The function $f(x) = x + |x|$ is continuous for:

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

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,

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:

Find all points of discontinuity of the function $f$ defined by $f(x) = |x| - |x + 1|$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module