Limit Continuity And Differentiability Ques 47
- Determine the values of $x$ for which the following function fails to be continuous or differentiable $f(x)=\left\{\begin{array}{cc}1-x, & x<1 \\ (1-x)(2-x), & 1 \leq x \leq 2\\ 3-x, & x>2\end{array}\right.$
Justify your answer.
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Answer:
Correct Answer: 47.$(1,2)$
Solution: It is clear that the given function
$ f(x)= \begin{cases}(1-x), & x<1 \\ (1-x)(2-x), & 1 \leq x \leq 2 \\ (3-x), & x>2\end{cases} $
continuous and differentiable at all points except possibly at $x=1$ and 2 .
Continuity at $x=1$,
$ \begin{aligned} \mathrm{LHL} & =\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(1-x) \\ & =\lim _{h \rightarrow 0}[1-(1-h)]=\lim _{h \rightarrow 0} h=0 \end{aligned} $
and
$ \begin{aligned} \mathrm{RHL} & =\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(1-x)(2-x) \\ & =\lim _{h \rightarrow 0}[1-(1+h)][2-(1+h)] \\ & =\lim _{h \rightarrow 0}-h \cdot(1-h)=0 \end{aligned} $
$\therefore \quad \mathrm{LHL}=\mathrm{RHL}=f(1)=0$
Therefore, $f$ is continuous at $x=1$
Differentiability at $x=1$,
$ \begin{aligned} L f^{\prime}(1) & =\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h} \\ & =\lim _{h \rightarrow 0} \frac{1-(1-h)-0}{-h}=\lim _{h \rightarrow 0}\left(\frac{h}{-h}\right)=-1 \end{aligned} $
and
$ \begin{aligned} R f^{\prime}(1) & =\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} \\ & =\lim _{h \rightarrow 0} \frac{[1-(1+h)][(2-(1+h)]-0}{h} \\ & =\lim _{h \rightarrow 0} \frac{-h(1-h)}{h}=\lim _{h \rightarrow 0}(h-1)=-1 \end{aligned} $
Since, $L\left[f^{\prime}(1)\right]=R f^{\prime}(1)$, therefore $f$ is differentiable at $x=1$.
Continuity at $x=2$,
$ \begin{aligned} \mathrm{LHL} & =\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}}(1-x)(2-x) \\ & =\lim _{h \rightarrow 0}[1-(2-h)][(2-(2-h)] \end{aligned} $
$ \quad\quad =\lim _{h \rightarrow 0}(-1+h) h=0 $
and
$ \begin{aligned} \text { RHL } & =\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}(3-x) \\ & =\lim _{h \rightarrow 0}[3-(2+h)]=\lim _{h \rightarrow 0}(1-h)=1 \end{aligned} $
Since, $\mathrm{LHL} \neq \mathrm{RHL}$, therefore $f$ is not continuous at $x=2$ as such $f$ cannot be differentiable at $x=2$.
Hence, $f$ is continuous and differentiable at all points except at $x=2$.
BETA
