Question

Solution set of the inequality $2^{2x+1}-5{\left(6\right)}^x+3^{2x+1}\le 0$ is

• A. $[-1,0]$
• B. $(0,1)$
• C. $(1,\infty )$
• D. $(-\infty ,-1)$

Answer 1

The correct option is A $[-1,0]$

$2^{2x+1}-5{\left(6\right)}^x+3^{2x+1}\le 0$

$\Rightarrow 2^{2x+1}-{5.2}^x{.3}^x+3^{2x+1}\le 0$

Divide by $3^{2x+1}$

$\Rightarrow \frac{2}{3}.{\left(\frac{2}{3}\right)}^{2x}-\frac{5}{3}.{\left(\frac{2}{3}\right)}^x+1\le 0$

$\Rightarrow 2y^2-5y+3\le 0$ where $y={\left(\frac{2}{3}\right)}^x$

$\Rightarrow (y-1)(2y-3)\le 0$

$\Rightarrow y\in [\frac{3}{2},1]$

$\Rightarrow {\left(\frac{2}{3}\right)}^x\in [\frac{3}{2},1]$

$\Rightarrow x\in [-1,0]$