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${f^ / }(x) = a{.3^x}\ln 3.$ Giả thiết ${f^ / }\left( 0 \right) = 2$
$ \Leftrightarrow \,\,\,a\ln 3 = 2\,\,\, \Leftrightarrow \,\,\,a = \frac{2}{{\ln 3}}$
$\int\limits_{1}^{2} {\left( {a{{.3}^x} + b} \right)dx = 12\,\,\,
\\\Leftrightarrow \,\,\left[ {{{a{{.3}^x}}}{} + bx} \right]}|
_1^2 = 12\,\,\, \Leftrightarrow 6a+b=12\Leftrightarrow b = 12 - 6a=12-\frac{{12}}{{{{\ln }^2}3}}$
Vậy: $a = \frac{2}{{\ln 3}}\,\,,\,\,b = 12 - \frac{{12}}{{{{\ln }^2}3}}$
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