JEE MAIN - Mathematics (2025 - 3rd April Evening Shift - No. 18)

Let $f: \mathrm{R} \rightarrow \mathrm{R}$ be a function defined by $f(x)=||x+2|-2| x \|$. If $m$ is the number of points of local minima and $n$ is the number of points of local maxima of $f$, then $m+n$ is
3
4
2
5

Explanation

$$\begin{aligned} & f(x)=\left\{\begin{array}{cc} |-x-2+2 x| & x \leq-2 \\ |x+2+2 x| & -2 \leq x \leq 0 \\ |x+2-2 x| & x \geq 0 \end{array}\right. \\ & f(x)=\left\{\begin{array}{lc} |-x-2+2 x| & x \leq-2 \\ |x+2+2 x| & -2 \leq x \leq 0 \\ |x+2-2 x| & x \geq 0 \end{array}\right. \\ & f(x)=\left\{\begin{array}{cc} |x-2| & x \leq-2 \\ |3 x+2| & -2< x \leq 0 \\ |2-x| & x>0 \end{array}\right. \end{aligned}$$

$$\begin{gathered} f(x)=\left\{\begin{array}{cc} 2-x & x \leq-2 \\ -3 x-2 & -2< x \leq-\frac{2}{3} \\ 3 x+2 & -\frac{2}{3}< x \leq 0 \end{array}\right. \\ f(x)=\left\{\begin{array}{cc} 2-x & x \leq-2 \\ -3 x-2 & -2< x \leq-\frac{2}{3} \\ 3 x+2 & -\frac{2}{3}< x \leq 0 \\ 2-x & 0< x<2 \\ x-2 & x \geq 2 \end{array}\right. \end{gathered}$$

JEE Main 2025 (Online) 3rd April Evening Shift Mathematics - Application of Derivatives Question 2 English Explanation

$$\begin{aligned} & \text { No. of maxima }=1 \\ & \text { No. of minima }=2 \\ & m=2 \\ & n=1 \\ & m+n=3 \end{aligned}$$

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