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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\text{k}(\text{x}^2-2\text{x}),&\text{if}\text{ x}<0\\\cos\text{x},&\text{if}\text{ x}\geq0\end{cases}\text{at x} = 0$
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Answer

Given,
$\text{f(x)}=\begin{cases}\text{k}(\text{x}^2-2\text{x}),&\text{if}\text{ x}<0\\\cos\text{x},&\text{if}\text{ x}\geq0\end{cases}$
We have,
$(\text{LHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(-\text{h})=\lim_\limits{\text{h}\rightarrow0}\text{k}(\text{h}^2+2\text{h})=0$
$(\text{RHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(\text{h})=\lim_\limits{\text{h}\rightarrow0}\cos\text{h}=1$
$\therefore\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}\neq\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}$
Thus, no value of k exists for which f(x) is continuous at x = 0.

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