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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
If $\text{f}\text{(x)}=\begin{cases}\frac{\text{x}-4}{\text{|x}-4|}+\text{a}, &\text{if x} <4\\\text{a}+\text{b},&\text{if x}=4\\\frac{\text{x}-4}{\text{|x}-4|}+\text{b}, & \text{if x}>4\end{cases}$ is continuous at x = 4. Find a, b.
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Answer

Given,
f(x) is continuous at x = 4 & f(4) = a + b
For f(x) to be continuous at $x = 4, f(4)^- = f(4)^+= f(4)$
$\text{L.H.L}=\text{f(4)}^-=\lim\limits_{\text{x} \rightarrow 4}\frac{\text{x}-4}{|\text{x}-4|}+\text{a}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{(4-\text{h})-4}{|(4-\text{h})-4|}+\text{a}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{(4-\text{h}-4)}{|(4-\text{h}-4)|}+\text{a}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{(-\text{h})}{|(-\text{h})|}+\text{a}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{(-\text{h})}{\text{h}}+\text{a}$
$\Rightarrow\text{a}-1$
$\text{L.H.L}=\text{f(4)}^+=\lim\limits_{\text{x} \rightarrow 0}\frac{\text{x}-4}{|\text{x}-4|}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{(4+\text{h})-4}{|(4+\text{h})-4|}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{4+\text{h}-4}{|4+\text{h}-4|}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}}{|\text{h}|}$
$\Rightarrow\lim\limits_{\text{h} \rightarrow 0}\frac{1}{|1|}$
$\Rightarrow1$
Since, f(x) is is continuous at x = 4 & f(4) = a + b
$\text{f(4)}^-=\text{f(4)}^+=\text{f(4)}$
$\therefore\ \text{a}-1=\text{a}+\text{b}=1$
$\Rightarrow\text{a}-1=1$
$\Rightarrow\text{a}=2$
$\Rightarrow\text{a}+\text{b}=1$
$\Rightarrow\text{b}=1-2$
$\Rightarrow\text{b}=-1$

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