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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
Question
Find all points of discontinuity of $\mathrm{f},$ where $\mathrm{f}$ is defined by: $f(x)=\left\{\begin{array}{l}2 x+3, x \leq 2 \\ 2 x-3, x>2\end{array}\right.$
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Answer

Given: $f\left( x \right) = \left\{ \begin{gathered} 2x + 3,\,\,\,if\,\,x \leq 2 \hfill \\ 2x - 3,\,\,if\,\,\,x > 2 \hfill \\ \end{gathered} \right.$
Here $f(x)$ is defined for $x \leq 2$
i.e., on $\left( { - \infty ,2} \right]$ and also for $x > 2$ i.e., on $\left( {2,\infty } \right)$
$\therefore$ Domain of $f$ is $\left( { - \infty ,2} \right] \cup \left( {2,\infty } \right) = \left( { - \infty ,\infty } \right) = R$
$\therefore$ For all $x < 2,f\left( x \right) = 2x + 3$ is a polynomial and hence continuous and
for all $x > 2,f\left( x \right) = 2x - 3$ is a continuous and hence $f(x)$ is continuous on $R – {2}$.
Now Left Hand limit at $x=2= \mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {2x + 3} \right) $
$= 2 \times 2 + 3 = 4 + 3 = 7$
Right Hand limit at $x=2= \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {2x - 3} \right) $
$= 2 \times 2 - 3 = 4 - 3 = 1$
Since $ \mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right)$
Therefore, $\mathop {\lim }\limits_{x \to {2}} f\left( x \right)$ does not exist and hence $f(x)$ is discontinuous at $x =2 \ ($only$)$

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