The number of solutions of the equation 2e^{| | vedclass

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(C) The given equation is $2e^{|x|} 	an^{-1}|x| = 1$.This can be rewritten as $2 	an^{-1}|x| = e^{-|x|}$.Let $f(x) = 2 	an^{-1}|x|$ and $g(x) = e^{

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$2$

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(C) The given equation is $2e^{|x|} 	an^{-1}|x| = 1$.This can be rewritten as $2 	an^{-1}|x| = e^{-|x|}$.Let $f(x) = 2 	an^{-1}|x|$ and $g(x) = e^{-|x|}$.Since both functions are even,we can analyze the behavior for $x \ge 0$.For $x \ge 0$,$f(x) = 2 	an^{-1}x$ and $g(x) = e^{-x}$.At $x = 0$,$f(0) = 2 	an^{-1}(0) = 0$ and $g(0) = e^0 = 1$. Thus,$f(0) As $x 	o \infty$,$f(x) 	o 2(\frac{\pi}{2}) = \pi \approx 3.14$ and $g(x) 	o 0$.Since $f(x)$ is a strictly increasing function for $x \ge 0$ and $g(x)$ is a strictly decreasing function for $x \ge 0$,there must be exactly one intersection point for $x > 0$.Due to symmetry about the $y$-axis,there will also be exactly one intersection point for $x Therefore,there are $2$ solutions in total.

List $I$	List $II$
$A. 3x^4 - 2x^3 - 6x^2 + 6x + 1$	$(I)$ has minimum value at $x = 4$
$B. x + \frac{1}{x}, \forall x < 0$	$(II)$ has maximum value at $x = -1$
$C. x^4(7 - x)^3$	$(III)$ has maximum value at $x = 4$
$D. x^4 + (8 - x)^4$	$(IV)$ is decreasing in $[2, \infty)$
	$(V)$ is increasing in $[2, \infty)$

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

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