JEE MAIN - Mathematics (2025 - 4th April Morning Shift - No. 22)
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 _________.
Answer
3
Explanation
$$\begin{aligned} &\text { for } x \geq 1, x^{21} \geq x^{19} \geq \ldots \geq x \text {. }\\ &f(x)=\left\{\begin{array}{lr} x & x<-1 \\ x^{21} & -1 \leq x \leq 0 \\ x & 0< x<1 \\ x^{21} & x \geq 1 \end{array}\right. \end{aligned}$$
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$$\begin{aligned} &\text { Clearly, } f(x) \text { is continuous everywhere. }\\ &\begin{aligned} & \Rightarrow \alpha=0 \\ & f^{\prime}(x)=\left\{\begin{array}{cl} 1 & ; x<-1 \\ 21 x^{20} & ;-1 \leq x \leq 0 \\ 1 & ; 0< x<1 \\ 21 \cdot x^{20} & ; x \geq 1 \end{array}\right. \\ & \Rightarrow \beta=3 \\ & \Rightarrow \alpha+\beta=3 \end{aligned} \end{aligned}$$
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