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CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}(\text{x}-1)\tan\frac{\pi\text{x}}{2},&\text{if}\text{ x}\neq1\\\text{k},&\text{if}\text{ x}=1\end{cases}\text{at x} = 1$
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Answer

Let $\text{x}-1=\text{y}$
$\Rightarrow\text{x}=\text{y}+1$
Thus,
$\lim_\limits{\text{x}\rightarrow 1}(\text{x}-1)\tan\frac{\pi\text{x}}{2}=\lim_\limits{\text{y}\rightarrow0}\text{y}\tan\frac{\pi(\text{y}+1)}{2}$
$=\lim_\limits{\text{y}\rightarrow0}\text{y}\tan\Big(\frac{\pi\text{y}}{2}+\frac{\pi}{2}\Big)$
$=-\lim_\limits{\text{y}\rightarrow0}\text{y}\cot\frac{\pi\text{y}}{2}$
$=-\lim_\limits{\text{y}\rightarrow0}\text{y}\frac{\cot\frac{\pi\text{y}}{2}}{\sin\frac{\pi\text{y}}{2}}$
$=-\lim_\limits{\text{y}\rightarrow0}\text{y}\frac{\cot\frac{\pi\text{y}}{2}}{\frac{\Big(\sin\frac{\pi\text{y}}{2}\Big)\frac{\pi}{2}}{\frac{\pi}{2}}}$
$=-\lim_\limits{\text{y}\rightarrow0}\frac{\cot\frac{\pi\text{y}}{2}}{\frac{\Big(\sin\frac{\pi\text{y}}{2}\Big)\frac{\pi}{2}}{\frac{\pi\text{y}}{2}}}$
$=-\lim_\limits{\text{y}\rightarrow0}\frac{2}{\pi}\frac{\cot\frac{\pi\text{y}}{2}}{\frac{\Big(\sin\frac{\pi\text{y}}{2}\Big)\frac{\pi}{2}}{\frac{\pi\text{y}}{2}}}$
$=-\frac{2}{\pi}\lim_\limits{\text{y}\rightarrow0}\cos\frac{\pi\text{y}}{2}$
$=-\frac{2}{\pi}$
Since the function is continuous, LHL = RHL.
Thus, $\text{k}=-\frac{2}{\pi}$

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