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CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
Determine the value of the constant k so that the function
$\text{f}\text{(x)}=\begin{cases}\frac{\sin2\text{x}}{5\text{x}}, &\text{if}\text{ x}\neq0\\\text{k}, &\text{if}\text{ x}=0\end{cases}$ is continuous at x = 0.
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Answer

We have given that the funtion is continuous at x = 0
So, LHL = RHL = f(0) ....(i)
Now,
$\text{LHL}=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0-\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\frac{\sin2(-\text{h})}{5(-\text{h})}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{-\sin2\text{h}}{-5\text{h}}=\lim\limits_{\text{h} \rightarrow 0}\frac{\sin2\text{h}}{2\text{h}}\times\frac{2\text{h}}{5\text{h}}=\frac{2}{5}$
$\text{f}(0)=\text{k}$
Using(i), $\text{k}=\frac{2}{5}$

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