Question

The range of the function $f\left(x\right)=(8^x+4^x+8^{-x}+4^{-x}+5)$ is

• A. $\left(\frac{7}{4},\infty \right)$
• B. $\left[\frac{7}{4},\infty \right)$
• C. $\left(9,\infty \right)$
• D. $\left[9,\infty \right)$

Answer 1

The correct option is D $\left[9,\infty \right)$

Given that,

$f\left(x\right)=(8^x+4^x+8^{-x}+4^{-x}+5)$

$\because A.M.\ge G.M.$

$\therefore \frac{4^x+4^{-x}}{2}\ge \sqrt{4^x\times 4^{-x}}$ $\therefore \frac{8^x+8^{-x}}{2}\ge \sqrt{8^x\times 8^{-x}}$

$\therefore \frac{4^x+4^{-x}}{2}\ge \sqrt{4^x\times \frac{1}{4^x}}$ $\therefore \frac{8^x+8^{-x}}{2}\ge \sqrt{8^x\times \frac{1}{8^x}}$

$\therefore \frac{4^x+4^{-x}}{2}\ge \sqrt{1}$ $\therefore \frac{8^x+8^{-x}}{2}\ge \sqrt{1}$

$\therefore 4^x+4^{-x}\ge 2$ $\therefore 8^x+8^{-x}\ge 2$

$\therefore f\left(x\right)=(8^x+4^x+8^{-x}+4^{-x}+5)\ge 2+2+5$

$f\left(x\right)\ge 4+5$

$\therefore f\left(x\right)\ge 9$

$Range:[9,\infty )$