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(B) Given function,$f(x) = \begin{cases} 1-x, & x  2 \end{cases}$ For $f(x)$ to be continuous at $x=

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$2$

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(B) Given function,$f(x) = \begin{cases} 1-x, & x  2 \end{cases}$ For $f(x)$ to be continuous at $x=1$,we check the left-hand limit,right-hand limit,and the value of the function. $\lim_{x ightarrow 1^-} f(x) = \lim_{x ightarrow 1^-} (1-x) = 1-1 = 0$. $\lim_{x ightarrow 1^+} f(x) = \lim_{x ightarrow 1^+} (1-x)(2-x) = (1-1)(2-1) = 0 	imes 1 = 0$. $f(1) = (1-1)(2-1) = 0$. Since $\lim_{x ightarrow 1^-} f(x) = \lim_{x ightarrow 1^+} f(x) = f(1)$,the function is continuous at $x=1$. Now,for $f(x)$ to be continuous at $x=2$,we check the limits. $\lim_{x ightarrow 2^-} f(x) = \lim_{x ightarrow 2^-} (1-x)(2-x) = (1-2)(2-2) = (-1) 	imes 0 = 0$. $\lim_{x ightarrow 2^+} f(x) = \lim_{x ightarrow 2^+} (3-x) = 3-2 = 1$. Since $\lim_{x ightarrow 2^-} f(x) 
eq \lim_{x ightarrow 2^+} f(x)$,the function is discontinuous at $x=2$.

The value$(s)$ of $x$ for which the function $f(x) = \begin{cases} 1-x, & x < 1 \\ (1-x)(2-x), & 1 \leq x \leq 2 \\ 3-x, & x > 2 \end{cases}$ fails to be continuous is(are):

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