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Câu hỏi: Cho hàm số $y=f\left( x \right)$ liên tục trên $\mathbb{R}\backslash \left\{ 0 \right\}$ và thỏa mãn $2f\left( 3x \right)+3f\left( \dfrac{2}{x} \right)=-\dfrac{15x}{2}$, $\int\limits_{3}^{9}{f\left( x \right)dx}=k$. Tính $I=\int\limits_{\dfrac{1}{2}}^{\dfrac{3}{2}}{f\left( \dfrac{1}{x} \right)dx}$ theo k.
A. $I=-\dfrac{45+k}{9}$.
B. $I=\dfrac{45-k}{9}$.
C. $I=\dfrac{45+k}{9}$.
D. $I=\dfrac{45-2k}{9}$.
A. $I=-\dfrac{45+k}{9}$.
B. $I=\dfrac{45-k}{9}$.
C. $I=\dfrac{45+k}{9}$.
D. $I=\dfrac{45-2k}{9}$.
Đặt $t=2x\Rightarrow dx=\dfrac{1}{2}dt$. Đổi cận $\left| \begin{aligned}
& x=\dfrac{1}{2}\Rightarrow t=1 \\
& x=\dfrac{3}{2}\Rightarrow t=3 \\
\end{aligned} \right.$.
Khi đó $I=\dfrac{1}{2}\int\limits_{1}^{2}{f\left( \dfrac{2}{t} \right)dx}$.
Mà $2f\left( 3x \right)+3f\left( \dfrac{2}{x} \right)=-\dfrac{15x}{2}\Leftrightarrow f\left( \dfrac{2}{x} \right)=-\dfrac{5x}{2}-\dfrac{2}{3}f\left( 3x \right)$
Nên $I=\dfrac{1}{2}\int\limits_{1}^{3}{\left[ -\dfrac{5x}{2}-\dfrac{2}{3}f\left( 3x \right) \right]dx}=-\dfrac{5}{4}\int\limits_{1}^{3}{xdx}-\dfrac{1}{3}\int\limits_{1}^{3}{f\left( 3x \right)dx}=-5-\dfrac{1}{3}\int\limits_{1}^{3}{f\left( 3x \right)dx} \left( * \right)$
Đặt $u=3x\Rightarrow dx=\dfrac{1}{3}dx$. Đổi cận $\left| \begin{aligned}
& x=1\Rightarrow u=3 \\
& x=3\Rightarrow t=9 \\
\end{aligned} \right.$.
Khi đó $I=-5-\dfrac{1}{9}\int\limits_{3}^{9}{f\left( t \right)dt}=-5-\dfrac{k}{9}=-\dfrac{45+k}{9}$.
& x=\dfrac{1}{2}\Rightarrow t=1 \\
& x=\dfrac{3}{2}\Rightarrow t=3 \\
\end{aligned} \right.$.
Khi đó $I=\dfrac{1}{2}\int\limits_{1}^{2}{f\left( \dfrac{2}{t} \right)dx}$.
Mà $2f\left( 3x \right)+3f\left( \dfrac{2}{x} \right)=-\dfrac{15x}{2}\Leftrightarrow f\left( \dfrac{2}{x} \right)=-\dfrac{5x}{2}-\dfrac{2}{3}f\left( 3x \right)$
Nên $I=\dfrac{1}{2}\int\limits_{1}^{3}{\left[ -\dfrac{5x}{2}-\dfrac{2}{3}f\left( 3x \right) \right]dx}=-\dfrac{5}{4}\int\limits_{1}^{3}{xdx}-\dfrac{1}{3}\int\limits_{1}^{3}{f\left( 3x \right)dx}=-5-\dfrac{1}{3}\int\limits_{1}^{3}{f\left( 3x \right)dx} \left( * \right)$
Đặt $u=3x\Rightarrow dx=\dfrac{1}{3}dx$. Đổi cận $\left| \begin{aligned}
& x=1\Rightarrow u=3 \\
& x=3\Rightarrow t=9 \\
\end{aligned} \right.$.
Khi đó $I=-5-\dfrac{1}{9}\int\limits_{3}^{9}{f\left( t \right)dt}=-5-\dfrac{k}{9}=-\dfrac{45+k}{9}$.
Đáp án A.