Properties and Extremum Values of Exponential Function of Infinite Series Sum

To determine the correct statements, we first evaluate the value of $\alpha$ and then analyze the properties of the function $g(x)$ on the interval $[0, 1]$.

Step 1: Calculate $\alpha$

We are given: $\alpha = \sum_{k=1}^{\infty} \sin^{2k} \left(\frac{\pi}{6}\right)$

Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$, we have: $\sin^2\left(\frac{\pi}{6}\right) = \frac{1}{4}$

Thus, $\alpha$ is the sum of an infinite geometric series with first term $a = \frac{1}{4}$ and common ratio $r = \frac{1}{4}$: $\alpha = \sum_{k=1}^{\infty} \left(\frac{1}{4}\right)^k = \frac{\frac{1}{4}}{1 - \frac{1}{4}} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}$

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Step 2: Define and Analyze $g(x)$

Substituting $\alpha = \frac{1}{3}$ into the expression for $g(x)$, we get: $g(x) = 2^{\frac{x}{3}} + 2^{\frac{1-x}{3}} \quad \text{for } x \in [0, 1]$

To find the critical points, we compute the first derivative of $g(x)$: $g'(x) = \frac{\mathrm{d}}{\mathrm{d}x} \left( 2^{\frac{x}{3}} + 2^{\frac{1-x}{3}} \right) = \frac{\ln 2}{3} \left( 2^{\frac{x}{3}} - 2^{\frac{1-x}{3}} \right)$

Setting $g'(x) = 0$: $2^{\frac{x}{3}} = 2^{\frac{1-x}{3}} \implies \frac{x}{3} = \frac{1-x}{3} \implies x = \frac{1}{2}$

Next, we compute the second derivative of $g(x)$: $g''(x) = \left(\frac{\ln 2}{3}\right)^2 \left( 2^{\frac{x}{3}} + 2^{\frac{1-x}{3}} \right)$

Since $2^y > 0$ for all real $y$, $g''(x) > 0$ for all $x \in [0, 1]$. This implies that $g(x)$ is strictly convex on $[0, 1]$.

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Step 3: Evaluate Minimum and Maximum Values

1. Minimum Value: Since $g'(x) = 0$ at $x = \frac{1}{2}$ and $g''(x) > 0$, $g(x)$ attains a unique absolute minimum at $x = \frac{1}{2}$. $g\left(\frac{1}{2}\right) = 2^{\frac{1}{6}} + 2^{\frac{1}{6}} = 2 \cdot 2^{\frac{1}{6}} = 2^{1 + \frac{1}{6}} = 2^{\frac{7}{6}}$

   • Statement (A) is TRUE.
   • Statement (D) is FALSE because the minimum is attained at only one point, $x = \frac{1}{2}$.

2. Maximum Value: For a strictly convex function defined on a closed interval $[0, 1]$, the maximum value must occur at the boundary points $x = 0$ or $x = 1$. Evaluating $g(x)$ at the endpoints: $g(0) = 2^0 + 2^{\frac{1}{3}} = 1 + 2^{\frac{1}{3}}$ $g(1) = 2^{\frac{1}{3}} + 2^0 = 1 + 2^{\frac{1}{3}}$

   Thus, the maximum value of $g(x)$ is $1 + 2^{\frac{1}{3}}$, which is attained at two distinct points, $x = 0$ and $x = 1$.

   • Statement (B) is TRUE.
   • Statement (C) is TRUE.

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Conclusion

The correct options are A, B, and C.