A function f(x) satisfies f(x) = f(frac{c}{x} | vedclass

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(D) Let $I = \int_{1}^{c} \frac{f(x)}{x} dx$. We can split the integral as $I = \int_{1}^{\sqrt{c}} \frac{f(x)}{x} dx + \int_{\sqrt{c}}^{c} \frac{f(x)

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$6$

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(D) Let $I = \int_{1}^{c} \frac{f(x)}{x} dx$. We can split the integral as $I = \int_{1}^{\sqrt{c}} \frac{f(x)}{x} dx + \int_{\sqrt{c}}^{c} \frac{f(x)}{x} dx$. For the second integral,let $x = \frac{c}{u}$. Then $dx = -\frac{c}{u^2} du$. When $x = \sqrt{c}$,$u = \sqrt{c}$,and when $x = c$,$u = 1$. So,$\int_{\sqrt{c}}^{c} \frac{f(x)}{x} dx = \int_{\sqrt{c}}^{1} \frac{f(c/u)}{c/u} (-\frac{c}{u^2}) du = \int_{1}^{\sqrt{c}} \frac{f(c/u)}{c/u} \frac{c}{u^2} du = \int_{1}^{\sqrt{c}} \frac{f(u)}{u} du$. Since $f(x) = f(c/x)$,we have $\int_{\sqrt{c}}^{c} \frac{f(x)}{x} dx = \int_{1}^{\sqrt{c}} \frac{f(u)}{u} du = 3$. Therefore,$I = 3 + 3 = 6$.

$A$ function $f(x)$ satisfies $f(x) = f(\frac{c}{x})$ for some real number $c$ $(c > 1)$ and $\forall\, x > 0$. If $\int_{1}^{\sqrt{c}} \frac{f(x)}{x} dx = 3$,then the value of $\int_{1}^{c} \frac{f(x)}{x} dx$ is

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