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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Find the value of a and b so that the function f given by $\text{f(x)}=\begin{cases}1,&\text{if }\text{ x}\leq3\\\text{ax}+\text{b},&\text{if }3<\text{x}<5\\7,&\text{if }\text{ x}\geq5\end{cases}$ is continuous x = 3 and x = 5.
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Answer

Given,
$\text{f(x)}=\begin{cases}1,&\text{if }\text{ x}\leq3\\\text{ax}+\text{b},&\text{if }3<\text{x}<5\\7,&\text{if }\text{ x}\geq5\end{cases}$
We have,
$(\text{LHL at x}= 3)=\lim_\limits{\text{x}\rightarrow3^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(3-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(3-\text{h})=\lim_\limits{\text{h}\rightarrow0}(1)=1$
$(\text{RHL at x}= 3)=\lim_\limits{\text{x}\rightarrow3+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(3+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{a}(3+\text{h})+\text{b}=3\text{a}+\text{b}$
$(\text{LHL at x}= 5)=\lim_\limits{\text{x}\rightarrow5^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(5-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}(\text{a}(5-\text{h})+\text{b})=5\text{a}+\text{b}$
$(\text{RHL at x}= 5)=\lim_\limits{\text{x}\rightarrow5+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(5+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}7=7$
If f(x) is continuous at x = 3 and 5, then
$\therefore\ \lim_\limits{\text{x}\rightarrow3^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow3^+}\text{f(x)}$ and $\lim_\limits{\text{x}\rightarrow5^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow5^+}\text{f(x)}$
$\Rightarrow1=3\text{a}+\text{b}\ .... (\text{i})$ and $5\text{a}+\text{b}=7\ .... (\text{ii})$
On solving eqs. (i) and (ii) we get
$\text{a}=3$ and $\text{b}=-8$

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