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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:
$\text{f(x)}\begin{cases}\frac{\text{k}\cos\text{x}}{\pi -2\text{x}}\ \text{if}\ \text{x}\neq \frac{\pi}{2}\\3, \ \ \ \ \ \ \ \ \text{if}\ \text{x} =\frac{\pi}{2}\end{cases}$
$\text{at} \text{x} = \frac{\pi}{2}$
✓

Answer

Here $\text{f(x)}\begin{cases}\frac{\text{k}\cos\text{x}}{\pi -2\text{x}},\ \text{if}\ \text{x}\neq \frac{\pi}{2}\\3, \ \ \ \ \ \ \ \ \text{if}\ \text{x} =\frac{\pi}{2}\end{cases}$
$​​^{\ \ \text{Lt}}_{\text{x}\rightarrow\frac{\pi}{2}}\text{f(x)}= ^{\ \ \text{Lt}}_{\text{x}\rightarrow\frac{\pi}{2}}\frac{\text{k}\cos\text{x}}{\pi - 2\text{x}}$ $\left[ \text{Put}\ \text{x} = \frac{\pi}{2} + \text{h},\text{h} > 0 \ \text{so that h} \rightarrow0\ \text{as x} \rightarrow\frac{\pi}{2}\right]$
$ ^{\ \ \text{Lt}}_{\text{h}\rightarrow0}\frac{\text{k}\cos\Big(\frac{\pi}{2}+\text{h}\Big)}{{\pi}- {2}\Big(\frac{\pi}{2}-\text{h}\Big)}=^{\ \ \text{Lt}}_{\text{h}\rightarrow0}\frac{-\text{k}\sin\text{h}}{-2\text{h}}$
$\frac{\text{k}}{2}\ ^{\ \ \text{Lt}}_{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}} = \frac{\text{k}}{2}\times1 = \frac{\text{k}}{2}$
Also $\text{f}(\frac{\pi}{2})= 3$
Since f is continuous at $\text{x}= \frac{\pi}{2}$
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow{\frac{\pi}{2}}}\text {f(x}) = \text{f}\Big(\frac{\pi}{2}\Big)\Rightarrow \frac{\text{k}}{2} = 3 \Rightarrow\text{k} =6$

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