Let (2^{1-a} + 2^{1+a}), f(a), (3^a + 3^{-a}) | vedclass

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(B) Given that $(2^{1-a} + 2^{1+a})$, $f(a)$, and $(3^a + 3^{-a})$ are in $A$.$P$.,we have $2f(a) = (2^{1-a} + 2^{1+a}) + (3^a + 3^{-a})$.Since $2^{1-

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$\frac{1}{4}\log_e(\frac{4}{3})$

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(B) Given that $(2^{1-a} + 2^{1+a})$, $f(a)$, and $(3^a + 3^{-a})$ are in $A$.$P$.,we have $2f(a) = (2^{1-a} + 2^{1+a}) + (3^a + 3^{-a})$.Since $2^{1-a} + 2^{1+a} = 2(2^{-a} + 2^a)$, and by the $AM$-$GM$ inequality, $2^a + 2^{-a} \geq 2$, the minimum value of $2^a + 2^{-a}$ is $2$.Similarly, the minimum value of $3^a + 3^{-a}$ is $2$.Thus, $2f(a) \geq 2(2) + 2 = 6$, which implies $f(a) \geq 3$. Therefore, $\alpha = 3$.The integral becomes $I = \int_{\log_e 2}^{\log_e 3} \frac{dx}{e^{2x} - e^{-2x}} = \int_{\log_e 2}^{\log_e 3} \frac{e^{2x} dx}{e^{4x} - 1}$.Let $u = e^{2x}$, then $du = 2e^{2x} dx$, so $e^{2x} dx = \frac{du}{2}$.When $x = \log_e 2$, $u = e^{2\log_e 2} = 4$. When $x = \log_e 3$, $u = e^{2\log_e 3} = 9$.$I = \frac{1}{2} \int_{4}^{9} \frac{du}{u^2 - 1} = \frac{1}{2} \cdot \frac{1}{2} \log_e |\frac{u-1}{u+1}| \Big|_4^9 = \frac{1}{4} (\log_e \frac{8}{10} - \log_e \frac{3}{5}) = \frac{1}{4} \log_e (\frac{4}{5} \cdot \frac{5}{3}) = \frac{1}{4} \log_e (\frac{4}{3})$.

Let $(2^{1-a} + 2^{1+a})$, $f(a)$, $(3^a + 3^{-a})$ be in $A$.$P$. and $\alpha$ be the minimum value of $f(a)$. Then the value of the integral $\int_{\log_e(\alpha-1)}^{\log_e(\alpha)} \frac{dx}{(e^{2x} - e^{-2x})}$ is:

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