The true set of values of a for which the ine | vedclass

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(D) Let $I = \int_{a}^{0} (3^{-2x} - 2 \cdot 3^{-x}) \, dx \geq 0$. Substitute $t = 3^{-x}$,then $dt = -3^{-x} \ln 3 \, dx$,so $dx = -\frac{dt}{t \ln

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$(-\infty, -1] \cup [0, \infty)$

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(D) Let $I = \int_{a}^{0} (3^{-2x} - 2 \cdot 3^{-x}) \, dx \geq 0$. Substitute $t = 3^{-x}$,then $dt = -3^{-x} \ln 3 \, dx$,so $dx = -\frac{dt}{t \ln 3}$. When $x = a$,$t = 3^{-a}$. When $x = 0$,$t = 1$. The integral becomes $\int_{3^{-a}}^{1} (t^2 - 2t) \left( -\frac{dt}{t \ln 3} \right) = \frac{1}{\ln 3} \int_{1}^{3^{-a}} (t - 2) \, dt \geq 0$. Since $\ln 3 > 0$,we have $\int_{1}^{3^{-a}} (t - 2) \, dt \geq 0$. Evaluating the integral: $\left[ \frac{t^2}{2} - 2t \right]_{1}^{3^{-a}} \geq 0$. $\left( \frac{3^{-2a}}{2} - 2 \cdot 3^{-a} \right) - \left( \frac{1}{2} - 2 \right) \geq 0$. $\frac{3^{-2a}}{2} - 2 \cdot 3^{-a} + \frac{3}{2} \geq 0$. Multiplying by $2$: $3^{-2a} - 4 \cdot 3^{-a} + 3 \geq 0$. Let $u = 3^{-a}$. Then $u^2 - 4u + 3 \geq 0$,which factors as $(u - 3)(u - 1) \geq 0$. This implies $u \leq 1$ or $u \geq 3$. Case $1$: $3^{-a} \leq 3^0 \implies -a \leq 0 \implies a \geq 0$. Case $2$: $3^{-a} \geq 3^1 \implies -a \geq 1 \implies a \leq -1$. Thus,$a \in (-\infty, -1] \cup [0, \infty)$.

The true set of values of $a$ for which the inequality $\int_{a}^{0} (3^{-2x} - 2 \cdot 3^{-x}) \, dx \geq 0$ is true is:

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