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

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Answer:

Correct Answer: 5. (b)

Solution:

  1. 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 $x \in R$.



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