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

Mathematics · Limits, Continuity and Differentiability

JEE Main 2025 — 4 April, Morning Shift — Question 40

Let $m$ and $n$ be the number of points at which the function $f(x)=\max \left\{x, x^{3}, x^{5}, \ldots x^{21}\right\}, x \in \mathbb{R}$ , is

not differentiable and not continuous, respectively. Then $m+n$ is equal to $\_\_\_\_$

Numerical answer — enter this value.

\text{for } x \geq 1, x^{21} \geq x^{19} \geq \dots \geq x.

f(x) = \begin{cases} x & x < -1 \\ x^{21} & -1 \leq x \leq 0 \\ x & 0 < x < 1 \\ x^{21} & x \geq 1 \end{cases}

Clearly, f(x) is continuous everywhere.

\implies \alpha = 0

f'(x) = \begin{cases} 1 & ; x < -1 \\ 21x^{20} & ; -1 < x < 0 \\ 1 & ; 0 < x < 1 \\ 21x^{20} & ; x > 1 \end{cases}

\implies \beta = 3

\implies \alpha + \beta = 3

Answer key and solution verified before publishing.

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.