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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Find which of the function:
$\text{f(x)}=\begin{cases}\frac{\text{x}^2}{2},&\text{if }0\leq\text{x}\leq1\\2\text{x}^2-3\text{x}+\frac{3}{2},&\text{if }1<\text{x}\leq2\end{cases}$
at x = 1
✓

Answer

We have, $\text{f(x)}=\begin{cases}\frac{\text{x}^2}{2},&\text{if }0\leq\text{x}\leq1\\2\text{x}^2-3\text{x}+\frac{3}{2},&\text{if }1<\text{x}\leq2\end{cases}$ at x = 1.
At x = 1, $\text{L.H.L}=\lim\limits_{\text{h}\rightarrow1^-}\frac{\text{x}^2}{2}=\lim\limits_{\text{h}\rightarrow0}\frac{(1-\text{h})^2}{2}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1+\text{h}^2-2\text{h}}{2}=\frac{1}{2}$
$\text{R.H.L}=\lim\limits_{\text{h}\rightarrow1^+}\Big(2\text{x}^2-3\text{x}+\frac{3}{2}\Big)$
$=\lim\limits_{\text{h}\rightarrow0}\Big[2(1+\text{h})^2-3(1+\text{h})+\frac{3}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\Big(2+2\text{h}^2+4\text{h}-3-3\text{h}+\frac{3}{2}\Big)$ $=-1+\frac{3}{2}=\frac{1}{2}$
And $\text{f}(1)=\frac{1^2}{2}=\frac{1}{2}$
$\therefore$ L.H.L = R.H.L = f(1)
Hence, f(x) is continuous at x = 1.

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