Let α = ∑ k = 1 ∞ sin 2 k π 6 . Let g : 0 , 1 → ℝ be the function defined…

Given,

$\alpha =\sum _{k=1}^{\infty }{\sin }^{2k}\left(\frac{\pi }{6}\right)$ and $g\left(x\right)=2^{ax}+2^{a\left(1-x\right)}$. 

Now solving,

$\alpha =\sum _{k=1}^{\infty }{\left(\frac{1}{2}\right)}^{2k}=\sum _{k=1}^{\infty }{\left(\frac{1}{4}\right)}^k=\frac{{\displaystyle \frac{1}{4}}}{1-\frac{1}{4}}=\frac{1}{3}$

Now putting the value of $\alpha$ in $g\left(x\right)$ we get,

$g\left(x\right)=2^{\frac{x}{3}}+2^{\frac{\left(1-x\right)}{3}}$

Now, $g^{\text{'}}\left(x\right)=\frac{\ln 2}{3}\frac{\left(2^{{\displaystyle \frac{2x}{3}}}-2^{{\displaystyle \frac{1}{3}}}\right)}{2^{{\displaystyle \frac{x}{3}}}}$

Now finding the critical point by $g^{\text{'}}\left(x\right)=0$$\Rightarrow x=\frac{1}{2}$

And, derivative changes sign from negative to positive at $x=\frac{1}{2}$, hence $x=\frac{1}{2}$ 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 $g\left(x\right)=g\left(\frac{1}{2}\right)$

$=2^{\frac{1}{6}}+2^{\frac{1}{6}}=2^{\frac{7}{6}}$

$\Rightarrow$ Option $\left(A\right)$ is correct

Now maximum value of $g\left(x\right)$ is either equal to $g\left(0\right)$ or $g\left(1\right)$.

$g\left(0\right)=1+2^{\frac{1}{3}}$

$g\left(1\right)=2^{\frac{1}{3}}+1$

Hence (B) and (C) are also correct.