Question
If $\text{f(x)}=\begin{cases}2\text{x}^2+\text{k},&\text{if }\text{ x}\geq0\\-2\text{x}^2+\text{k},&\text{if }\text{ x}<0\end{cases},$ then what should be the value of k so that f(x) is continuous at x = 0.
✓
Answer
It is given that function is continous at x = 0 then,
$\text{LHL}=\text{RHL}=\text{f}(0)\ ....(\text{i})$
Now, $\text{f}(0)=2\times0+\text{k}=\text{k}$
$\text{LHL}=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}\lim_\limits{\text{h}\rightarrow0}=\lim_\limits{\text{h}\rightarrow0}-2(-\text{h})^2+\text{k}=\text{k}$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})=\lim_\limits{\text{h}\rightarrow0}2(\text{h}^2)+\text{k}=\text{k}$
Thus, the function will be continuous for any $\text{k}\in\text{R}$
$\text{LHL}=\text{RHL}=\text{f}(0)\ ....(\text{i})$
Now, $\text{f}(0)=2\times0+\text{k}=\text{k}$
$\text{LHL}=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}\lim_\limits{\text{h}\rightarrow0}=\lim_\limits{\text{h}\rightarrow0}-2(-\text{h})^2+\text{k}=\text{k}$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})=\lim_\limits{\text{h}\rightarrow0}2(\text{h}^2)+\text{k}=\text{k}$
Thus, the function will be continuous for any $\text{k}\in\text{R}$
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