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Continuity — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsContinuity3 Marks
Question
If $\text{f(x)}=\begin{cases}\frac{\text{x}^2}{2},&\text{if }0\leq\text{ x}\leq1\\2\text{x}^2-3\text{x}+\frac{3}{2},&\text{if }1<\text{x}\leq2\end{cases}$ Show that f is continuous at x = 1.
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Answer

We want to discuss the continuity of the function at x = 1
We need to prove that
$​​\text{LHL}=​​\text{RHL}=\text{f}(1)$
$\text{f}(1)=\frac{1^2}{2}=\frac{1}{2}$
$\text{LHL}=\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(1-\text{h})=\lim_\limits{\text{h}\rightarrow0}\frac{(1-\text{h})^2}{2}=\frac{1}{2}$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow1^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}(1+\text{h})=\lim_\limits{\text{h}\rightarrow0}2(1+\text{h}^2)-3(1+\text{h})+\frac{3}{2}$
$=2-3+\frac{3}{2}=\frac{1}{2}$
Thus, $\text{LHL}=\text{RHL}=\text{f}(1)=\frac{1}{2}$
Hence, function is continuous at x = 1

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