Question

If $x\in R,$ the numbers $2^{1+x}+2^{1-x},b/2,{36}^x+{36}^{-x}$ form an A.P. , then $b$ may lie in the interval

• A. $\left[16,\infty \right)$
• B. $\left[6,\infty \right)$
• C. $\left[\infty ,-6\right)$
• D. $\left[6,12\right)$

Answer 1

The correct option is B $\left[6,\infty \right)$

Given

$2^{1+x}+2^{1-x},\frac{b}{2},{36}^x+{36}^{-x}$ form an AP

The condition to be in AP is

$2\times \left(\frac{b}{2}\right)=2^{1+x}+2^{1-x}+{36}^x+{36}^{-x}$

$b={2.2}^x+2.\frac{1}{2^x}+{36}^x+\frac{1}{{36}^x}$

$b=2(2^x+\frac{1}{2^x})+({36}^x+\frac{1}{{36}^x})$

Let $2^x=y{36}^x=k$

$b=2(y+\frac{1}{y})+(k+\frac{1}{k})$

The min value of $f\left(x\right)+\frac{1}{f\left(x\right)}$ is $2$

The max value is $\infty$

$⟹2\left(2\right)+2\le b\le 2\left(\infty \right)+\left(\infty \right)$

$⟹6\le b\le \infty$

$⟹b\in [6,\infty )$