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Continuity — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsContinuity3 Marks
Question
Identify discontinuity for the following function as either a jump or a removable discontinuity on their respective domain:
$
\begin{aligned}
f(x) & =\frac{x^2+x+1}{x+1}, & & \text { for } x \in[0,3) \\
& =\frac{3 x+4}{x^2-5}, & & \text { for } x \in[3,6]
\end{aligned}
$
✓

Answer

For $\mathrm{x} \in[0,3), \mathrm{f}(\mathrm{x})=\frac{x^2+x+1}{x+1}$, being a rational function is continuous except when its denominator $x+1=0$ i.e., at $x=-1$, which does not belong to $[0,3)$
$\therefore \mathrm{f}$ is continuous on $[0,3)$.
For $x \in[3,6], \mathrm{f}(\mathrm{x})=\frac{3 x+4}{x^2-5}$, being a rational function is continuous except when its denominator $\mathrm{x}^2$ $-5=0$ i.e., at $x= \pm \sqrt{5}$ But $\pm \sqrt{5} \notin[3,6]$
$\therefore \mathrm{f}$ is continuous on $[0,6]$ except possibly at $\mathrm{x}=3$
Continuity at $x=3$
$
\begin{aligned}
& f(x)=\frac{x^2+x+1}{x+1}, \text { for } x \in[0,3) \\
& \therefore \lim _{x \rightarrow 3-} f(x)=\lim _{x \rightarrow 3} \frac{x^2+x+1}{x+1} \\
& =\frac{\lim _{x \rightarrow 3}\left(x^2+x+1\right)}{\lim _{x \rightarrow 3}(x+1)} \\
& =\frac{9+3+1}{3+1} \\
& =\frac{13}{4}
\end{aligned}
$
Also, $f(x)=\frac{3 x+4}{x^2-5}$, for $x \in[3,6]$
$
\begin{aligned}
& \therefore \lim _{x \rightarrow 3^{+}} \mathrm{f}(x)=\mathrm{f}(3)=\frac{9+4}{9-5}=\frac{13}{4} \\
& \therefore \mathrm{f}(3)=\lim _{x \rightarrow 3^{+}} \mathrm{f}(x)=\lim _{x \rightarrow 3^{-}} \mathrm{f}(x)
\end{aligned}
$
$\therefore \mathrm{f}$ is continuous at $\mathrm{x}=3$.
Hence, $f$ is continuous on its domain $[0,6]$.

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