If $f: \mathbf{R} \rightarrow \mathbf{R}$ is defined by $$ f(x)=\left\{\begin{array}{cl} x-1, & \text { for…
If $f: \mathbf{R} \rightarrow \mathbf{R}$ is defined by
$$
f(x)=\left\{\begin{array}{cl}
x-1, & \text { for } x \leq 1 \\
2-x^2, & \text { for } 1 < x \leq 3 \\
x-10, & \text { for } 3 < x < 5 \\
2 x, & \text { for } x \geq 5
\end{array}\right.
$$
then the set of points of discontinuity of $f$ is
$\mathbf{R}-\{1,5\}$
$\{1,3,5\}$
$\{1,5\}$
$\mathbf{R}-\{1,3,5\}$
Solution
Given function
$
f(x)= \begin{cases}x-1, & \text { for } x \leq 1 \\ 2-x^2, & \text { for } 1 < x \leq 3 \\ x-10, & \text { for } 3 < x < 5 \\ 2 x, & \text { for } x \geq 5\end{cases}
$
So, $f(x)$ will be continuous in the intervals $(-\infty, 1),(1,3),(3,5)$ and $(5, \infty)$
Now, let us check the continuity at $x=1,3$ and 5
Here,
(i) At $x=1$,
$
\lim _{x \rightarrow 1^{-}} f(x)=1-1=0 \text { and } \lim _{x \rightarrow 1^{+}} f(x)=2-1=1
$
therefore $\mathrm{F}$ is not continuous at $x=1$
(ii)
$
\begin{aligned}
& \text { At } x=3, \\
& \lim _{x \rightarrow 3^{-}} f(x)=2-9=-7 \text { and } \\
& f(3)=-7 \text { and } \lim _{x \rightarrow 3^{+}} f(x)=3-10=-7
\end{aligned}
$
therefore $f$ is continuous at $x=3$
(iii) At $x=5, \lim _{x \rightarrow 5^{-}} f(x)=5-10=-5$ and,
$
\lim _{x \rightarrow 5^{+}} f(x)=2 \times 5=10
$
therefore $f$ is not continuous at $x=5$ Hence, points of discontinuity of $f$ are $\{1,5\}$