Question

The true set of values of $a$ for which the inequality $\underset{a}{\overset{0}{\int }}(3^{-2x}-{2\cdot 3}^{-x})dx\ge 0$ is true is

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

Answer 1

The correct option is D $(-\infty ,-1]\cup [0,\infty )$

$\underset{a}{\overset{0}{\int }}(3^{-2x}-{2\cdot 3}^{-x})dx\ge 0$
$\Rightarrow {[\frac{3^{-2x}}{-2\ln 3}+\frac{2\cdot 3^{-x}}{\ln 3}]}_a^0\ge 0$
$\Rightarrow 3^{-2a}-4\cdot 3^{-a}+3\ge 0$
$\Rightarrow (3^{-a}-3)(3^{-a}-1)\ge 0$
Let $t=3^{-a}>0$
Then, $(t-3)(t-1)\ge 0$

$\therefore t\in (0,1]\cup [3,\infty )$
$\Rightarrow 3^{-a}\in (0,1]\cup [3,\infty )$
$\therefore a\in (-\infty ,-1]\cup [0,\infty )$