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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:
$\text{f(x)}=\begin{cases}\frac{2\text{x}^2-3\text{x}-2}{\text{x}-2},&\text{if x}\neq2\\5,&\text{if x}=2\end{cases}$
at x = 2
✓

Answer

We have, $\text{f(x)}=\begin{cases}\frac{2\text{x}^2-3\text{x}-2}{\text{x}-2},&\text{if x}\neq2\\5,&\text{if x}=2\end{cases}$ at x = 2.
At x = 2, $\text{L.H.L}=\lim\limits_{\text{x}\rightarrow2^-}\frac{2\text{x}^2-3\text{x}-2}{\text{x}-2}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2(2-\text{h})^2-3(2-\text{h})-2}{(2-\text{h})-2}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{8+2\text{h}^2-8\text{h}-6+3\text{h}-2}{-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2\text{h}^2-5\text{h}}{-\text{h}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}(2\text{h}-5)}{-\text{h}}=5$
$\text{R.H.L}=\lim\limits_{\text{h}\rightarrow2^+}\frac{2\text{x}^2-3\text{x}-2}{\text{x}-2}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2(2+\text{h})^2-3(2+\text{h})-2}{(2+\text{h})-2}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{8+2\text{h}^2+8\text{h}-6-3\text{h}-2}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2\text{h}^2+5\text{h}}{\text{h}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}(2\text{h}+5)}{\text{h}}=5$
and f(2) = 5
$\therefore$ L.H.L = R.H.L = f(2)
So, f(x) is continuous at x = 2.

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