The value $(s)$ of $x$ for which the function $f(x)=\left\{\begin{array}{cc} 1-x, & x 2 \end{array}\right.$…
The value $(s)$ of $x$ for which the function
$f(x)=\left\{\begin{array}{cc}
1-x, & x < 1 \\
(1-x)(2-x), & 1 \leq x \leq 2 \\
3-x, & x>2
\end{array}\right.$
fails to be continuous is(are)
$1$
$2$
$3$
all real numbers
Solution
Given function,
$f(x)=\left\{\begin{array}{cc}
1-x, & x < 1 \\
(1-x)(2-x), & 1 \leq x \leq 2 \\
3-x & x>2
\end{array}\right.$
For $f(x)$ to be continuous at $x=1$.
$\begin{aligned}
& \lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(1) \\
& \Rightarrow \quad \lim _{x \rightarrow 1^{+}}(1-x)=\lim _{x \rightarrow 1^{+}}(1-x)(2-x) \\
& (1-1)=(1-1)(2-1)=0
\end{aligned}$
and $f(1)=(1-1)(2-1)=0$
$\therefore f(x)$ is continuous at $x=1$.
Now, at $x=2, \lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}} f(x)=f(2)$
$\begin{aligned}
& \Rightarrow \lim _{x \rightarrow 2^{-}}(1-x)(2-x)=\lim _{x \rightarrow 2^{+}}(3-x) \\
& \Rightarrow \quad(1-2)(2-2)=(3-2) \\
& \Rightarrow \quad 0 \neq 1 \\
& \therefore \quad f(x) \text { is discontinuous at } x=2
\end{aligned}$
$\therefore f(x)$ is discontinuous at $x=2$.