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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

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

Consider, $\text{f(x)}=\begin{cases}\frac{\text{x}-4}{|\text{x}-4|}+\text{a},&\text{if x}<4\\\text{a+}\text{b},&\text{if x}=4\\\frac{\text{x}-4}{|\text{x}-4|}+\text{b},&\text{if x}>4\end{cases}$
At x = 4 $\text{L.H.L}=\lim\limits_{\text{x}\rightarrow4^-}\frac{\text{x}-4}{|\text{x}-4|}+\text{a}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{4-\text{h}-4}{|4-\text{h}-4|}+\text{a}=\lim\limits_{\text{h}\rightarrow0}\frac{-\text{h}}{\text{h}}+\text{a}$
$=-1+\text{a}$
$\text{R.H.L}=\lim\limits_{\text{x}\rightarrow4^+}\frac{\text{x}-4}{|\text{x}-4|}+\text{b}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{4+\text{h}-4}{|4+\text{h}-4|}+\text{b}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}}{\text{h}}+\text{b}=1+\text{b}$
$\text{f}(4)=\text{a}+\text{b}$
Since, f(x) is continuous at x = 0
$\therefore$ L.H.L = R.H.L = f(0)
⇒ -1 + a = 1 + b = a + b
⇒ -1 + a = a + b and 1 + b = a + b
$\therefore$ b = -1 and a = 1

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