Q104 / 160 · TS EAMCET 2019 (04 May Shift 2) · Mathematics

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

  • A
    continuous at $x=\frac{5}{3}$
  • B
    differentiable at $x=\frac{5}{3}$
  • Correct answer
    C
    differentiable at $x=2$
  • D
    discontinuous at $x=-2$

Solution

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 $x=2$.

Hence, the correct answer is option: 'C'.

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