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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
Discuss the continuity of the function f(x) at the point $\text{x}=\frac{1}{2}$ where
$\text{f}\text{(x)}=\begin{cases}\text{x}, & 0\leq\text{x} < \frac{1}{2}\\\frac{1}{2},&\text{x}=\frac{1}{2}\\1-\text{x}, &\frac{1}{2}< \text{x}\leq 1\end{cases}$
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Answer

We want to discuss the continuity of the function at $\text{x}=\frac{1}{2}.$
$\text{LHL}=\lim\limits_{\text{x} \rightarrow \frac{1}{2}^-}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\text{f}\Big(\frac{1}{2}-\text{h}\Big)=\lim\limits_{\text{x} \rightarrow 0}\frac{1}{2}-\text{h}=\frac{1}{2}$
$\text{RHL}=\lim\limits_{\text{x} \rightarrow \Big(\frac{1}{2}^-\Big)}=\lim\limits_{\text{x} \rightarrow 0}\text{f}\Big(\frac{1}{2}+\text{h}\Big)=\lim\limits_{\text{x} \rightarrow 0}1-\Big(\frac{1}{2}+\text{h}\Big)=\frac{1}{2}$
$\text{f}\Big(\frac{1}{2}\Big)=\frac{1}{2}$
Thus, $\text{LHL}=\text{RHL}=\text{f}\Big(\frac{1}{2}\Big)=\frac{1}{2}$
Hence, the function is continuous at $\text{x}=\frac{1}{2}.$

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