Let α = ∑ _ k = 1 ^ ∞ ^ 2k ((π)/(6)) . Let g: [0, 1] → ℝ be the...

Given,

$\alpha = \mathop{∑}_{k = 1}^{\infty}\sin^{2k}\left(\dfrac{\pi}{6}\right)$ and g\left(x\right) = {2}^{ax} + {2}^{a\left(1 - x\right)}$. </p><p>Now solving,</p><p>$\alpha = \mathop{∑}_{k = 1}^{\infty}{\left(\dfrac{1}{2}\right)}^{2k} = \mathop{∑}_{k = 1}^{\infty}{\left(\dfrac{1}{4}\right)}^{k} = \dfrac{\dfrac{1}{4}}{1 - \dfrac{1}{4}} = \dfrac{1}{3}$</p><p>Now putting the value of$\alpha$in$g\left(x\right)$we get,</p><p>$g\left(x\right) = 2\dfrac{x}{3} + 2\dfrac{\left(1 - x\right)}{3}$</p><p>Now,${g}^{'}\left(x\right) = \dfrac{\ln 2}{3}\dfrac{\left({2}^{\dfrac{2x}{3}} - {2}^{\dfrac{1}{3}}\right)}{{2}^{\dfrac{x}{3}}}$</p><p>Now finding the critical point by${g}^{'}\left(x\right) = 0$$⇒ x = \dfrac{1}{2}$

And, derivative changes sign from negative to positive at $x = \dfrac{1}{2}$, hence $x = \dfrac{1}{2}$ is point of local minimum as well as absolute minimum of $g\left(x\right)$ for $x \in \left[0, 1\right]$

Hence, minimum value of $g\left(x\right) = g\left(\dfrac{1}{2}\right)$

$= 2\dfrac{1}{6} + 2\dfrac{1}{6} = 2\dfrac{7}{6}$

$⇒$ Option \left(A\right)$is correct</p><p>Now maximum value of$g\left(x\right)$is either equal to$g\left(0\right)$or$g\left(1\right)$.</p><p>$g\left(0\right) = 1 + 2\dfrac{1}{3}$</p><p>$g\left(1\right) = 2\dfrac{1}{3} + 1$

Hence (B) and (C) are also correct.