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(A) Given $f(x)=\int_{1 / x}^x e^{-\left(t+\frac{1}{t}\right)} \frac{d t}{t}$.Using Leibniz's rule for differentiation under the integral sign:$f^{\pr

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$(A, C, D)$

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(A) Given $f(x)=\int_{1 / x}^x e^{-\left(t+\frac{1}{t}\right)} \frac{d t}{t}$.Using Leibniz's rule for differentiation under the integral sign:$f^{\prime}(x) = e^{-\left(x+\frac{1}{x}\right)} \cdot \frac{1}{x} - e^{-\left(\frac{1}{x}+x\right)} \cdot \left(-\frac{1}{x^2}\right) = \frac{e^{-\left(x+\frac{1}{x}\right)}}{x} + \frac{e^{-\left(x+\frac{1}{x}\right)}}{x} = \frac{2 e^{-\left(x+\frac{1}{x}\right)}}{x}$.$(A)$ For $x \in [1, \infty)$,$f^{\prime}(x) > 0$,so $f(x)$ is monotonically increasing. Thus,$(A)$ is correct.$(B)$ For $x \in (0, 1)$,$f^{\prime}(x) > 0$,so $f(x)$ is monotonically increasing on $(0, 1)$. Thus,$(B)$ is incorrect.$(C)$ $f(1/x) = \int_{x}^{1/x} e^{-\left(t+\frac{1}{t}\right)} \frac{dt}{t}$. Let $t = 1/u$,then $dt = -1/u^2 du$. $f(1/x) = \int_{1/x}^{x} e^{-\left(1/u+u\right)} \cdot u \cdot (-1/u^2) du = -\int_{1/x}^{x} e^{-\left(u+\frac{1}{u}\right)} \frac{du}{u} = -f(x)$. Therefore,$f(x) + f(1/x) = 0$. Thus,$(C)$ is correct.$(D)$ Let $g(x) = f(2^x)$. Then $g(-x) = f(2^{-x}) = f(1/2^x) = -f(2^x) = -g(x)$. Thus,$f(2^x)$ is an odd function. Thus,$(D)$ is correct.The correct options are $(A, C, D)$.

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