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

Answer 1

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

We have, ${\int }_a^0{3}^{-x}(3^{-x}-2)dx\ge 0$
Put $3^{-x}=t$
$\Rightarrow 3^{-x}\ln 3dx=-dt\Rightarrow \frac{1}{\ln 3}\underset{1}{\overset{3^{-a}}{\int }}(t-2)dt\ge 0\Rightarrow {[\frac{t^2}{2}-2t]}_1^{3^{-a}}\ge 0\Rightarrow (\frac{3^{-2a}}{2}-2\cdot 3^{-a})-(\frac{1}{2}-2)\ge 0\Rightarrow 3^{-2a}-4\times 3^{-a}+3\ge 0\Rightarrow (3^{-a}-3)(3^{-a}-1)\ge 0\Rightarrow ({3.3}^a-1)(3^a-1)\ge 0\Rightarrow 3^a\le \frac{1}{3}\text{or}{3}^a\ge 1$
Thus, $a\in (-\infty ,-1]\cup [0,\infty )$