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

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(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

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$2$

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(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 $x$).

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