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Continuity — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsContinuity5 Marks
Question
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\frac{\text{x}^2+\text{x}^2-16\text{x}+20}{(\text{x}-2)^2},&\text{ x}\neq2\\\text{k},&\text{x}=2\end{cases}$
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Answer

Given,
$\text{f(x)}=\begin{cases}\frac{\text{x}^2+\text{x}^2-16\text{x}+20}{(\text{x}-2)^2},&\text{ x}\neq2\\\text{k},&\text{x}=2\end{cases}$
$\Rightarrow\text{f(x)}=\begin{cases}\frac{\text{x}^2+\text{x}^2-16\text{x}+20}{\text{x}^2-4\text{x}+4},&\text{ x}\neq2\\\text{k},&\text{x}=2\end{cases}$
$\Rightarrow\text{f(x)}=\begin{cases}\text{x}+5,&\text{ x}\neq2\\\text{k},&\text{x}=2\end{cases}$
If f(x) is continuous at x = 2, then
$\lim_\limits{\text{x}\rightarrow2}\text{f(x)}=\text{f}(2)$
$\Rightarrow\lim_\limits{\text{x}\rightarrow2}\text{(x}+5)=\text{k}$
$\Rightarrow\text{k}=2+5=7$

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