Question

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

• A. $[12,\infty )$
• B. $[6,\infty )$
• C. $(-\infty ,6]$
• D. $[6,12]$

Answer 1

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

$2^{1+x}\ast$? $1^{-x},\frac{b}{2},{36}^4+{36}^{-x}$

$2\frac{b}{2}={2.2}^x+\frac{2}{2^x}+{36}^4+\frac{1}{{36}^x}$

$b={2.2}^x+\frac{2}{2^x}+3^{2x}{.2}^{2x}+\frac{1}{3^{2x}{.2}^{2x}}$

$b\in [6,\infty ]$