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Continuity — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsContinuity5 Marks
Question
If the functions f(x), defined below is continuous at x = 0, find the value of k.
$\text{f(x)}=\begin{cases}\frac{1-\cos2\text{x}}{2\text{x}^2},&\text{x}<0\\\text{k},&\text{x}=0\\\frac{\text{x}}{|\text{x}|},&\text{x}>0\end{cases}$
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Answer

Given, $\text{f(x)}=\begin{cases}\frac{1-\cos2\text{x}}{2\text{x}^2},&\text{x}<0\\\text{k},&\text{x}=0\\\frac{\text{x}}{|\text{x}|},&\text{x}>0\end{cases}$
$\text{f(x)}=\begin{cases}\frac{1-\cos2\text{x}}{2\text{x}^2},&\text{x}<0\\\text{k},&\text{x}=0\\1,&\text{x}>0\end{cases}$
We have,
$(\text{LHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow 0}\Big(\frac{1-\cos2(-\text{h})}{2(-\text{h})^2}\Big)$
$=\lim_\limits{\text{h}\rightarrow 0}\Big(\frac{1-\cos2\text{h}}{2\text{h}^2}\Big)$
$=\frac{1}{2}\lim_\limits{\text{h}\rightarrow 0}\Big(\frac{2\sin^2\text{h}}{\text{h}^2}\Big)$
$=\frac{2}{2}\lim_\limits{\text{h}\rightarrow 0}\Big(\frac{\sin^2\text{h}}{\text{h}^2}\Big)$
$=\frac{2}{2}\lim_\limits{\text{h}\rightarrow 0}\Big(\frac{\sin^2\text{h}}{\text{h}}\Big)^2$
$=1\times1$
$=1$
$(\text{RHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow 0}\text{f(h)}=\lim_\limits{\text{h}\rightarrow 0}(1)=1$
Also, $\text{f}(0)=\text{k}$
If f(x) is continuous at x = 0, then
$\lim_\limits{\text{x}\rightarrow 0^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow 0^+}=\text{f}(0)$
$\Rightarrow1=1=\text{k}$
Hence, the required value of k is 1

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