Let alpha = sum_{k=1}^{infty} sin^{2k}left(fr | vedclass

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(D) Given $\alpha = \sum_{k=1}^{\infty} \sin^{2k}\left(\frac{\pi}{6}\right) = \sum_{k=1}^{\infty} \left(\frac{1}{2}\right)^{2k} = \sum_{k=1}^{\infty}

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$A, B, C$

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(D) Given $\alpha = \sum_{k=1}^{\infty} \sin^{2k}\left(\frac{\pi}{6}\right) = \sum_{k=1}^{\infty} \left(\frac{1}{2}\right)^{2k} = \sum_{k=1}^{\infty} \left(\frac{1}{4}\right)^k$.This is a geometric series with first term $a = 1/4$ and common ratio $r = 1/4$.$\alpha = \frac{1/4}{1 - 1/4} = \frac{1/4}{3/4} = \frac{1}{3}$.Thus,$g(x) = 2^{x/3} + 2^{(1-x)/3} = 2^{x/3} + \frac{2^{1/3}}{2^{x/3}}$.Let $u = 2^{x/3}$. Since $x \in [0, 1]$,$u \in [2^0, 2^{1/3}] = [1, 2^{1/3}]$.Then $g(u) = u + \frac{2^{1/3}}{u}$.$g'(u) = 1 - \frac{2^{1/3}}{u^2}$. Setting $g'(u) = 0$ gives $u^2 = 2^{1/3}$,so $u = 2^{1/6}$.Since $2^{1/6} \approx 1.12$ and $2^{1/3} \approx 1.26$,the critical point $u = 2^{1/6}$ lies in the interval $[1, 2^{1/3}]$.At $u = 2^{1/6}$,$g(2^{1/6}) = 2^{1/6} + \frac{2^{1/3}}{2^{1/6}} = 2^{1/6} + 2^{1/6} = 2 \cdot 2^{1/6} = 2^{7/6}$. This is the minimum value.At the endpoints $u = 1$ and $u = 2^{1/3}$,$g(1) = 1 + 2^{1/3}$ and $g(2^{1/3}) = 2^{1/3} + \frac{2^{1/3}}{2^{1/3}} = 2^{1/3} + 1$.Thus,the maximum value is $1 + 2^{1/3}$,which is attained at $x = 0$ and $x = 1$.Therefore,statements $(A)$,$(B)$,and $(C)$ are true.

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