Limit Continuity and Differentiability 5 Question 5
5. Let $f:[-1,3] \rightarrow R$ be defined as
$$ f(x)=\begin{array}{cc} |x|+[x], & -1 \leq x<1 \\ x+|x|, & 1 \leq x<2 \\ x+[x], & 2 \leq x \leq 3, \end{array} $$
(2019 Main, 8 April II) where, $[t]$ denotes the greatest integer less than or equal to $t$. Then, $f$ is discontinuous at
(a) four or more points
(b) only two points
(c) only three points
(d) only one point
Show Answer
Answer:
Correct Answer: 5. (b)
Solution:
- Since, $f(x)$ is continuous at $x=0$.
$$ \begin{array}{ll} \Rightarrow & \lim _{x \rightarrow 0} f(x)=f(0) \\ \Rightarrow & f\left(0^{+}\right)=f\left(0^{-}\right)=f(0)=0 \end{array} $$
To show, continuous at $x=k$
$$ \begin{aligned} & \text { RHL }=\lim _{h \rightarrow 0} f(k+h)=\lim _{h \rightarrow 0}[f(k)+f(h)]=f(k)+f\left(0^{+}\right) \\ & =f(k)+f(0) \\ & \text { LHL }=\lim _{h \rightarrow 0} f(k-h)=\lim _{h \rightarrow 0}[f(k)+f(-h)] \\ & =f(k)+f\left(0^{-}\right)=f(k)+f(0) \\ & \therefore \quad \lim _{x \rightarrow k} f(x)=f(k) \end{aligned} $$
$\Rightarrow f(x)$ is continuous for all $x \in R$.
