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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\frac{\text{x}^2-25}{\text{x}-5},&\text{x}\neq5\\\text{k},&\text{x}=5\end{cases}\text{at x} =5$
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Answer

Given,
$\text{f(x)}=\begin{cases}\frac{\text{x}^2-25}{\text{x}-5},&\text{x}\neq5\\\text{k},&\text{x}=5\end{cases}$
$\Rightarrow\text{f(x)}=\begin{cases}\frac{(\text{x}-5)(\text{x}+5)}{\text{x}-5},&\text{x}\neq5\\\text{k},&\text{x}=5\end{cases}$
$\Rightarrow\text{f(x)}=\begin{cases}{\text{x}-5},&\text{x}\neq5\\\text{k},&\text{x}=5\end{cases}$
If f(x) is continuous at x = 5, then,
$\lim_\limits{\text{x}\rightarrow5}\text{f(x)}=\text{f}(5)$
$\Rightarrow\lim_\limits{\text{x}\rightarrow5}\text{(x}+5)=\text{k}$
$\Rightarrow\text{k}=5+5=10$

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