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

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(C) The given equation is $2{e^{\left| x ight|}}{	an ^{ - 1}}\left| x ight| = 1$.This can be rewritten as $2{	an ^{ - 1}}\left| x ight| = \fra

Vedclass · Accepted Answer

$2$

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(C) The given equation is $2{e^{\left| x ight|}}{	an ^{ - 1}}\left| x ight| = 1$.This can be rewritten as $2{	an ^{ - 1}}\left| x ight| = \frac{1}{{{e^{\left| x ight|}}}}$,which is equivalent to $2{	an ^{ - 1}}\left| x ight| = {e^{ - \left| x ight|}}$.Let $f(x) = 2{	an ^{ - 1}}\left| x ight|$ and $g(x) = {e^{ - \left| x ight|}}$.Both functions $f(x)$ and $g(x)$ are even functions,meaning they are symmetric about the $y$-axis.For $x \ge 0$,$f(x) = 2{	an ^{ - 1}}x$ is a strictly increasing function starting from $f(0) = 0$ and approaching $\pi$ as $x 	o \infty$.For $x \ge 0$,$g(x) = {e^{ - x}}$ is a strictly decreasing function starting from $g(0) = 1$ and approaching $0$ as $x 	o \infty$.Since $f(x)$ is increasing and $g(x)$ is decreasing for $x > 0$,and $f(0)  \lim_{x 	o \infty} g(x)$ (as $\pi > 0$),there must be exactly one intersection point for $x > 0$.Due to symmetry,there will also be exactly one intersection point for $x Thus,there are $2$ solutions in total.

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

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