Đặt: \(t = - x \Rightarrow dt = - dx. \)
Đổi cận: \(\left\{ \begin{array}{l} x = \alpha \Rightarrow t = - \alpha \\ x = 0 \Rightarrow t = 0 \end{array} \right.\)
Ta có:
\(\begin{array}{l}\int\limits_\alpha ^0 {\left( {{3^{ - 2x}} - {{2.3}^{ - x}}} \right)dx} = - \int\limits_{ - \alpha }^0 {({3^{2t}} - {{2.3}^t})dt} \\ = \int\limits_0^{ - \alpha } {({3^{2t}} - {{2.3}^t})dt} = \int\limits_0^{ - \alpha } {{3^{2t}}dt} - 2.\int\limits_0^{ - \alpha } {{3^t}dt} \\ = \frac{1}{2}.\left. {\frac{{{3^{2t}}}}{{\ln 3}}} \right|_0^{ - \alpha } - 2.\left. {\frac{{{3^t}}}{{\ln 3}}} \right|_0^{ - \alpha } = \frac{1}{{2\ln 3}}\left( {{3^{ - 2\alpha }} - {{4.3}^{ - \alpha }} + 3} \right)\end{array}\)
Theo đề bài ta có:
\(\int\limits_\alpha ^0 {\left( {{3^{ - 2x}} - {{2.3}^{ - x}}} \right)dx} \ge 0 \Leftrightarrow \frac{1}{{2\ln 3}}\left( {{3^{ - 2\alpha }} - {{4.3}^{ - \alpha }} + 3} \right) \ge 0 \Leftrightarrow {3^{ - 2\alpha }} - {4.3^{ - \alpha }} + 3 \ge 0\)
Đặt: \(t = {3^{ - \alpha }},t > 0\) Bất phương trình trở thành:
\({t^2} - 4t + 3 \ge 0 \Leftrightarrow \left[ \begin{array}{l} t \ge 3\\ t \le 1 \end{array} \right.\)
\( \Rightarrow \left[ {\begin{array}{*{20}{c}}{0 < t \le 1}\\{t \ge 3}\end{array}} \right. \Rightarrow \left[ {\begin{array}{*{20}{c}}{0 < {{\left( {\frac{1}{3}} \right)}^\alpha } \le 1}\\{{{\left( {\frac{1}{3}} \right)}^\alpha } \ge 3}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\alpha \ge 0}\\{\alpha \le - 1}\end{array}} \right.\)
