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If the function $f: R \rightarrow R$, defined by $f(x)=\left\{\begin{array}{l}5-3 x, \text { if } x \leq \frac{5}{3} \\ x^2-3 x+20, \text { if } x>\frac{5}{3}\end{array}\right.$, then ' $f$ ' is
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Verified Answer
The correct answer is:
differentiable at $x=2$
We have,
$$
\begin{aligned}
& f(x)=\left\{\begin{array}{l}
5-3 x, \text { of } x \leq \frac{5}{3} \\
x^2-3 x+20, \text { of } x>\frac{5}{3}
\end{array}\right. \\
& \lim _{x \rightarrow \frac{5^{-}}{3}} f(x)=5-3\left(\frac{5}{3}\right)=0 \\
& \lim _{x \rightarrow \frac{5^{+}}{3}} f(x)=\lim _{x \rightarrow \frac{5^{+}}{3}} x^2-3 x+20 \\
& =\left(\frac{5}{3}\right)^2-3\left(\frac{5}{3}\right)+20=\frac{160}{9}
\end{aligned}
$$
$$
\lim _{x \rightarrow \frac{5^{-}}{3}} f(x) \neq \lim _{x \rightarrow \frac{5^{+}}{3}} f(x)
$$
$\therefore f(x)$ is discontinuous of $x=\frac{5}{3}$
Hence $f(x)$ is derivative at $x=2$.
$$
\begin{aligned}
& f(x)=\left\{\begin{array}{l}
5-3 x, \text { of } x \leq \frac{5}{3} \\
x^2-3 x+20, \text { of } x>\frac{5}{3}
\end{array}\right. \\
& \lim _{x \rightarrow \frac{5^{-}}{3}} f(x)=5-3\left(\frac{5}{3}\right)=0 \\
& \lim _{x \rightarrow \frac{5^{+}}{3}} f(x)=\lim _{x \rightarrow \frac{5^{+}}{3}} x^2-3 x+20 \\
& =\left(\frac{5}{3}\right)^2-3\left(\frac{5}{3}\right)+20=\frac{160}{9}
\end{aligned}
$$
$$
\lim _{x \rightarrow \frac{5^{-}}{3}} f(x) \neq \lim _{x \rightarrow \frac{5^{+}}{3}} f(x)
$$
$\therefore f(x)$ is discontinuous of $x=\frac{5}{3}$
Hence $f(x)$ is derivative at $x=2$.
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