Let $m$ and $n$ be the number of points at which the function $f(\mathrm{x})=\max \left\{\mathrm{x}, \mathrm{x}^3, \mathrm{x}^5, \ldots ., \mathrm{x}^{21}\right\}, \mathrm{x} \in \mathbb{R}$, is not differentiable and not continuous, respectively. Then $\mathrm{m}+\mathrm{n}$ is equal to ________ .

Question

Aicharya · Accepted Answer

$f(x)=\left\{\begin{array}{cc}
x, & x \lt -1 \\ x^{21}, & -1 \leq x \lt 0 \\ x, & 0 \leq x \lt 1 \\ x^{21}, & x \geq 1
\end{array}\right.$
$f(x)$ is continuous everywhere.
$\begin{aligned}
& \therefore \mathrm{n}=0 \\ & \mathrm{f}^{\prime}(\mathrm{x})=\left\{\begin{array}{cc}
1, & \mathrm{x} \lt -1 \\ 21 \mathrm{x}^{20}, & -1 \leq \mathrm{x} \lt 0 \\ 1, & 0 \lt \mathrm{x} \lt 1 \\ 21 \mathrm{x}^{20}, & \mathrm{x} \geq 1
\end{array}\right.
\end{aligned}$
$\therefore \mathrm{f}(\mathrm{x})$ is non-differentiable at $\mathrm{x}=-1,0,1$
$\begin{aligned}
& \therefore \mathrm{m}=3 \\ & \mathrm{~m}+\mathrm{n}=3
\end{aligned}$

Let $m$ and $n$ be the number of points at which the function $f(\mathrm{x})=\max \left\{\mathrm{x},…

Solution