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 solutions to $f(x) = g(x)$ for $x > 0$.Since $x > 0$,let $u = e^

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infinitely many

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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 solutions to $f(x) = g(x)$ for $x > 0$.Since $x > 0$,let $u = e^x$. As $x$ ranges from $(0, \infty)$,$u$ ranges from $(1, \infty)$.The equation becomes $	an(u) = u + \frac{1}{u}$.We know that $u + \frac{1}{u} \ge 2$ for $u > 0$.The function $h(u) = 	an(u)$ has vertical asymptotes at $u = \frac{(2n+1)\pi}{2}$ for $n = 0, 1, 2, \dots$.In each interval $(\frac{(2n-1)\pi}{2}, \frac{(2n+1)\pi}{2})$,the function $	an(u)$ increases from $-\infty$ to $+\infty$.Since $g(u) = u + \frac{1}{u}$ is a continuous function that is always $\ge 2$,and $	an(u)$ covers all real values in each branch,there will be at least one intersection point in each interval $(\frac{(2n-1)\pi}{2}, \frac{(2n+1)\pi}{2})$ where $	an(u) > 2$.As $n 	o \infty$,there are infinitely many such intervals,and thus infinitely many solutions.

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

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