The number of real solutions of the equation | vedclass

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(D) Let $f(x) = 	an(e^x)$ and $g(x) = e^x + e^{-x}$.We are looking for the number of intersection points of $y = 	an(e^x)$ and $y = e^x + e^{-x}$ fo

Vedclass · Accepted Answer

none of these

Vedclass · Answer

(D) Let $f(x) = 	an(e^x)$ and $g(x) = e^x + e^{-x}$.We are looking for the number of intersection points of $y = 	an(e^x)$ and $y = e^x + e^{-x}$ for $x > 0$.Note that $e^x + e^{-x} = 2 \cosh(x)$,which is always $\ge 2$ for all real $x$.As $x$ increases from $0$ to $\infty$,$e^x$ increases from $1$ to $\infty$.The function $	an(e^x)$ is periodic with respect to $e^x$ and has vertical asymptotes at $e^x = (2n+1)\frac{\pi}{2}$ for $n = 0, 1, 2, \dots$.Since $e^x$ takes all values in $(1, \infty)$ as $x$ ranges from $(0, \infty)$,the function $	an(e^x)$ will oscillate between $-\infty$ and $+\infty$ infinitely many times.Specifically,in every interval where $e^x$ increases by $\pi$,$	an(e^x)$ covers the range $(-\infty, \infty)$.Since $e^x + e^{-x} \ge 2$,the graph of $y = e^x + e^{-x}$ will intersect each branch of the tangent function $	an(e^x)$ at least once as long as the branch reaches values $\ge 2$.Because there are infinitely many such branches as $x 	o \infty$,there are infinitely many solutions.Therefore,the correct answer is 'none of these'.

The number of real solutions of the equation $\tan(e^x) = e^x + e^{-x}$ for $x > 0$ is:

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