Question

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?

• A. The function $g\left(x\right)$ attains its maximum at more than one point
• B. The maximum value of $g\left(x\right)$ is $1+2^{\frac{1}{3}}$
• C. The minimum value of$g\left(x\right)$ is $2^{\frac{7}{6}}$
• D. The function $g\left(x\right)$ attains its minimum at more than one point

Answer 1

Answer: (A), (B), (C)

The minimum value of$g\left(x\right)$ is $2^{\frac{7}{6}}$

$\alpha =\sum _{k=1}^{\infty }{\left(\frac{1}{2}\right)}^{2k}=\sum _{k=1}^{\infty }{\left(\frac{1}{4}\right)}^k=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{1}{3}$
Hence, $g\left(x\right)=2^{\frac{x}{3}}+2^{\frac{(1-x)}{3}}$

Now, $g^{\prime }\left(x\right)=\frac{\ln 2}{3}\frac{(2^{\frac{2x}{3}}-2^{\frac{1}{3}})}{2^{\frac{x}{3}}}$
$g^{\prime }\left(x\right)=0$ at $x=\frac{1}{2}$
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 $x\in [0,1]$
Hence, minimum value of $g\left(x\right)=g\left(\frac{1}{2}\right)=2^{\frac{1}{6}}+2^{\frac{1}{6}}=2^{\frac{7}{6}}$

Maximum value of $g\left(x\right)$ is either equal to $g\left(0\right)$ or $g\left(1\right)$
$g\left(0\right)=1+2^{\frac{1}{3}}$

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