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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
Find the values of k so that the function f is continuous at the indicated point:
$\text{f(x)}= \begin{cases}\text{k}\text{x}+1,\ \text{if}\ \text{x}\leq{\pi}\\ \cos\text{x}, \ \ \ \ \text{if}\ \text{x} >{\pi}\end{cases}$
$\text{at}\ \text{x} = {\pi}$
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Answer

Here $\text{f(x)}= \begin{cases}\text{k}\text{x}+1,\ \text{if}\ \text{x}\leq{\pi}\\ \cos\text{x}, \ \ \ \ \text{if}\ \text{x} >{\pi}\end{cases}$

$​​^{\ \ \text{Lt}}_{\text{x}\rightarrow{\pi}^{-}}\text{f(x)}= ^{\ \ \text{Lt}}_{\text{x}\rightarrow{\pi}^{-}}(\text{k}\text{x}+1)$ 

$\left[\text{Put}\ \text{x} = \pi -\text{h}, \text{h}>0\ \text{so that}\ \text{h}\rightarrow0\ \text{as}\ \text{x}\rightarrow\pi^{-}\right]$

$= ^{\ \ \text{Lt}}_{\text{x}\rightarrow0}\left\{\text{k}(\pi - \text{h})+1\right\}={\text{k}}(\pi - 0) = 1 = \text{k}\pi + 1$

$^{\ \ \text{Lt}}_{\text{x}\rightarrow\pi^{+}}\text{f(x)} =^{\ \ \text{Lt}}_{\text{x}\rightarrow\pi^{+}}\cos\pi$

$\left[\text{Put}\ \text{x} = \pi +\text{h}, \text{h}>0\ \text{so that}\ \text{h}\rightarrow0\ \text{as}\ \text{x}\rightarrow\pi^{+}\right]$

$^{\ \ \text{Lt}}_{\text{h}\rightarrow0}\cos(\pi + \text{h}) = ^{\ \ \text{Lt}}_{\text{h}\rightarrow0}(\cos\pi\cos\text{h} - \sin\pi\sin\text{h})$

$\cos \pi. 1 - \sin \pi. 0 = \cos\pi = -1$

Since f(x) is continuous at $\text{x} = \pi$

$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow{\pi}^{-}}\text {f(x}) =^{\ \ \text{Lt}}_{\text{x}\rightarrow{\pi}^{+}}\text{f(x)}$

$\therefore \text{k}\pi = -1 \Rightarrow \text{k} = -\frac{1}{\pi}$

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