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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\text{k}+1,&\text{if}\text{ x}\leq\pi\\\cos\text{x},&\text{if}\text{ x}>\pi\end{cases}\text{at x} = \pi$
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Answer

Given,
$\text{f(x)}=\begin{cases}\text{k}+1,&\text{if}\text{ x}\leq\pi\\\cos\text{x},&\text{if}\text{ x}>\pi\end{cases}$
We have,
$(\text{LHL at x}= \pi)=\lim_\limits{\text{x}\rightarrow\pi^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(\pi-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{k}(\pi-\text{h})+1=\text{k}\pi+1$
$(\text{RHL at x}= \pi)=\lim_\limits{\text{x}\rightarrow\pi^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(\pi+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\cos(\pi+\text{h})=\cos\pi=-1$
If f(x) is continuous at $\text{x}=\pi,$ then
$\lim_\limits{\text{x}\rightarrow\pi^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow\pi^+}\text{f(x)}$
$\Rightarrow\text{k}\pi+1=-1$
$\Rightarrow\text{k}=\frac{-2}{\pi}$

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