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Continuity (p-1) — Maths (commerce) STD 11 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 11 Commerce / ArtsMaths (commerce)Continuity (p-1)1 Mark
Question
Examine whether the function is continuous at the points indicated against them. $f(x)=x^3-2 x+1$, for $x \leq 2$ $=3 x-2$, for $x>2$, at $x=2$
✓

Answer

$
\begin{aligned}
\lim _{x \rightarrow 2^{-}} \mathrm{f}(x) & =\lim _{x \rightarrow 2^{-}}\left(x^3-2 x+1\right) \\
& =(2)^3-2(2)+1=5 \\
\lim _{x \rightarrow 2^{+}} \mathrm{f}(x) & =\lim _{x \rightarrow 2^{+}}(3 x-2) \\
& =3(2)-2=4 \\
\lim _{x \rightarrow 2^{-}} \mathrm{f}(x) & \neq \lim _{x \rightarrow 2^{+}} \mathrm{f}(x)
\end{aligned}
$
$\therefore$ Function $\mathrm{f}$ is discontinuous at $\mathrm{x}=2$

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