Let $\alpha =\sum _{k=1}^{\infty }{\sin }^{2k}\left(\frac{\pi }{6}\right)$. Let $g:[0,1]\to ℝ$ be the function defined by $g\left(x\right)=2^{\alpha x}+2^{\alpha (1-x)}$. Then, which of the following statements is/are TRUE ?

To determine the correct statements about the function $g\left(x\right)$, we will first find the exact numerical value of the constant $\alpha$.

Step 1: Evaluate the value of $\alpha$
The constant $\alpha$ is given by the infinite geometric series:

$\alpha =\sum _{k=1}^{\infty }{\sin }^{2k}\left(\frac{\pi }{6}\right)$

We know that:

$\sin \left(\frac{\pi }{6}\right)=\frac{1}{2}$

Therefore, the term inside the summation becomes:

${\sin }^2\left(\frac{\pi }{6}\right)={\left(\frac{1}{2}\right)}^2=\frac{1}{4}$

Expanding the infinite series for $\alpha$:

$\alpha ={\left(\frac{1}{4}\right)}^1+{\left(\frac{1}{4}\right)}^2+{\left(\frac{1}{4}\right)}^3+\dots$

This is an infinite geometric progression with first term $a=\frac{1}{4}$ and common ratio $r=\frac{1}{4}$.
Using the formula for the sum of an infinite geometric series $S=\frac{a}{1-r}$:

$\alpha =\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$

Step 2: Simplify the function $g\left(x\right)$
Substituting $\alpha =\frac{1}{3}$ into $g\left(x\right)$:

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

for $x\in [0,1]$.

Step 3: Find the minimum value of $g\left(x\right)$
By applying the Arithmetic Mean - Geometric Mean (AM-GM) inequality to the positive terms $2^{\frac{1}{3}x}$ and $2^{\frac{1}{3}(1-x)}$:

$\frac{2^{\frac{1}{3}x}+2^{\frac{1}{3}(1-x)}}{2}\ge \sqrt{2^{\frac{1}{3}x}·2^{\frac{1}{3}(1-x)}}$

Simplifying the term under the square root:

$2^{\frac{1}{3}x}·2^{\frac{1}{3}(1-x)}=2^{\frac{1}{3}x+\frac{1}{3}-\frac{1}{3}x}=2^{\frac{1}{3}}$

So, the inequality becomes:

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

$\frac{g\left(x\right)}{2}\ge 2^{\frac{1}{6}}$

Multiplying both sides by 2:

$g\left(x\right)\ge 2·2^{\frac{1}{6}}=2^{1+\frac{1}{6}}=2^{\frac{7}{6}}$

The equality holds when the two terms are equal:
$2^{\frac{1}{3}x}=2^{\frac{1}{3}(1-x)}\Rightarrow x=1-x\Rightarrow x=\frac{1}{2}$
Since $x=\frac{1}{2}\in [0,1]$, the minimum value of $g\left(x\right)$ is indeed $2^{\frac{7}{6}}$ and it occurs at a unique point ($x=\frac{1}{2}$).
Hence, the first statement is TRUE, and the statement that it attains its minimum at more than one point is FALSE.

Step 4: Find the maximum value of $g\left(x\right)$
Since $g\left(x\right)$ is a strictly convex function on the closed interval $[0,1]$, its maximum value occurs at the boundary endpoints $x=0$ and $x=1$.

Evaluating $g\left(x\right)$ at the endpoints:
At $x=0$:

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

At $x=1$:

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

Thus, the maximum value of $g\left(x\right)$ is $1+2^{\frac{1}{3}}$, and it is attained at two distinct points, namely $x=0$ and $x=1$.
Hence, both the second and third statements are TRUE.

Conclusion:
The TRUE statements are:
1. The minimum value of $g\left(x\right)$ is $2^{\frac{7}{6}}$
2. The maximum value of $g\left(x\right)$ is $1+2^{\frac{1}{3}}$
3. The function $g\left(x\right)$ attains its maximum at more than one point