The number of roots of the equation x^2 cdot | vedclass

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(B) To find the number of roots of the equation $x^2 \cdot e^{2 - |x|} = 1$,we analyze the function $f(x) = x^2 \cdot e^{2 - |x|}$.Since $f(x)$ is an

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$4$

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(B) To find the number of roots of the equation $x^2 \cdot e^{2 - |x|} = 1$,we analyze the function $f(x) = x^2 \cdot e^{2 - |x|}$.Since $f(x)$ is an even function $(f(x) = f(-x))$,we first analyze $f(x)$ for $x \ge 0$,where $f(x) = x^2 \cdot e^{2 - x}$.Taking the derivative: $f'(x) = 2x \cdot e^{2 - x} - x^2 \cdot e^{2 - x} = x(2 - x) \cdot e^{2 - x}$.Setting $f'(x) = 0$,we find critical points at $x = 0$ and $x = 2$.At $x = 0$,$f(0) = 0$. At $x = 2$,$f(2) = 2^2 \cdot e^{2 - 2} = 4 \cdot 1 = 4$.As $x 	o \infty$,$f(x) 	o 0$. Thus,for $x \ge 0$,the function increases from $0$ to $4$ and then decreases towards $0$.Since the function is even,the graph is symmetric about the $y$-axis. The maximum value is $4$ at $x = 2$ and $x = -2$,and the minimum value is $0$ at $x = 0$.We are looking for the number of solutions to $f(x) = 1$. Since $0  0$ and two points for $x Therefore,there are $4$ roots in total.

The number of roots of the equation $x^2 \cdot e^{2 - |x|} = 1$ is

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