Q102 / 160 · AP EAMCET 2024 (22 May Shift 1) · Mathematics
The function $f(x)=|x-24|$ is
- Adifferentiable at $[0,25]$
- Bnot continuous at $x=24$
- Cneither continuous nor differentiable on [0,25].
- Correct answerDcontinuous on $[0,25]$, but not differentiable on $[0,25]$.
Solution
$\begin{gathered}f(x)=|x-24|=\left\{\begin{array}{cc}-x+24, & x \lt 24 \\ x-24, & x \geq 24\end{array}\right. \\ \lim _{x \rightarrow 24^{-}} f(x)=\lim _{x \rightarrow 24}(-x+24)=0\end{gathered}$
$\lim _{x \rightarrow 24^{+}} f(x)=\lim _{x \rightarrow 24}(x-24)=0 \text { and } f(24)=0$ $\therefore f$ is continuous on [ 0,25$]$. Also, $\lim _{x \rightarrow 24^{-}} \frac{f(x)-f(24)}{x-24}=\lim _{x \rightarrow 24} \frac{-x+24+0}{x-24}=-1$ $\lim _{x \rightarrow 24^{+}} \frac{f(x)-f(24)}{x-24}=\lim _{x \rightarrow 24} \frac{x-24}{x-24}=1$ $\therefore f$ is not differentiable on $[0,25]$.
$\lim _{x \rightarrow 24^{+}} f(x)=\lim _{x \rightarrow 24}(x-24)=0 \text { and } f(24)=0$ $\therefore f$ is continuous on [ 0,25$]$. Also, $\lim _{x \rightarrow 24^{-}} \frac{f(x)-f(24)}{x-24}=\lim _{x \rightarrow 24} \frac{-x+24+0}{x-24}=-1$ $\lim _{x \rightarrow 24^{+}} \frac{f(x)-f(24)}{x-24}=\lim _{x \rightarrow 24} \frac{x-24}{x-24}=1$ $\therefore f$ is not differentiable on $[0,25]$.
Hence, the correct answer is option: 'D'.