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Continuity — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsContinuity3 Marks
Question
Determine the value of the constant k so that the function
$\text{f}\text{(x)}=\begin{cases}\frac{\text{x}^2-3\text{x}+2}{\text{x}-1}, &\text{if}\text{ x}\neq1\\\text{k}, &\text{if}\text{ x}=1\end{cases}$ is continuous at x = 1
✓

Answer

We have that the function is continuous at x = 1
$\therefore$ LHL = RHL = f(1) ....(1)
Now,
$\text{LHL}=\lim\limits_{\text{x} \rightarrow 1^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(1-\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\frac{(1-\text{h})^2-3(1-\text{h})+2}{(1-\text{h})-1}$
$\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}^2+\text{h}}{-\text{h}}=\lim\limits_{\text{h} \rightarrow 0}-\text{h}-1=-1$
$\text{f}(1)=\text{k}$
from (1), We get,
k = -1

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