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Continuity and Differentiability — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSContinuity and Differentiability5 Marks
Question
Examine the continuity of function at $x=2$ and $x =3 , f(x)=|x-2|+|x-3| $
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Answer

Given function $f(x)=|x-2|+|x-3|$, It is written as :
$\begin{array}{l} f(x)=\left\{\begin{array}{cc} -(x-2)-(x-3), & 0 \leq x<2 \\ x-2-(x-3), & 2 \leq x<3 \\ x-2+x-3, & x \geq 3 \end{array}\right. \end{array} $
$ \Rightarrow f(x)=\left\{\begin{array}{cl} 5-2 x, & \text { if } 0 \leq x>2 \\ 1, & \text { if } 2 \leq x<3 \\ 2 x-5, & \text { if } x \geq 3 \end{array}\right. $
$(i)$ Examining the continuity at $x=2$.
$\begin{array}{l} \text { R.H.L. }=\lim _{h \rightarrow 0} f(2+h)=\lim _{h \rightarrow 0} 1=1\end{array} $
$ \begin{aligned} \text { L.H.L. } & =\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0}[5-2(2-h)] \end{aligned} $
$ =\lim _{h \rightarrow 0}(5-4+2 h)=1$
at $x=2$
$ f(x)=5-2 x$
$\therefore f(2)=5-2 \times 2$
$=5-4$
$=1$
hence at $x=2$
$f(2)=\lim _{h \rightarrow 0} f(2+h)=\lim _{h \rightarrow 0} f(2-h)=1$
hence function is continuous at $x=2$.
$(ii)$ Examining the continuity at $x=3$
$\text{R.H.L.}$
$\lim _{h \rightarrow 0} f(3+h) =\lim _{h \rightarrow 0}[2(3+h)-5]$
$ =\lim _{h \rightarrow 0}(6+2 h-5)=1$
$\text{L.H.L.}$
$\lim _{h \rightarrow 0} f(3-h)=\lim _{h \rightarrow 0} 1=1$
at $x=3$
$f(x)=2 x-5,$
$\therefore f(3)=2 \times 3-5=6-5=1$
$\therefore f(3)=\lim _{h \rightarrow 0} f(3+h)=\lim _{h \rightarrow 0} f(3-h)=1$
Hence function is continuous at $x=3$.

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