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Câu hỏi: Cho hàm số f(x) liên tục trên $\mathbb{R}\backslash \left\{ 0 \right\}$ và thỏa mãn $\int\limits_{2}^{4}{f\left( x \right)}=58$ và $f\left( 2x \right)+2.f\left( \dfrac{1}{x} \right)=12{{x}^{2}}+3+\dfrac{6}{{{x}^{2}}}.$ Tính tích phân $I=\int\limits_{2}^{4}{f\left( \dfrac{2}{x} \right)dx}.$
A. 4.
B. 7.
C. 6.
D. 5.
A. 4.
B. 7.
C. 6.
D. 5.
Đặt $\dfrac{2}{x}=\dfrac{1}{t}\Rightarrow x=2t\Rightarrow I-\int\limits_{1}^{2}{f\left( \dfrac{1}{t} \right)d\left( 2t \right)}=2\int\limits_{1}^{2}{f\left( \dfrac{1}{x} \right)dx}$.
Từ $f\left( 2x \right)+2.f\left( \dfrac{1}{x} \right)=12{{x}^{2}}+3+\dfrac{6}{{{x}^{2}}}$
$\Rightarrow I=\int\limits_{1}^{2}{\left( 12{{x}^{2}}+3+\dfrac{6}{{{x}^{2}}}-f\left( 2x \right) \right)dx}=\left. \left( 4{{x}^{3}}+3x-\dfrac{6}{x} \right) \right|_{1}^{2}-\int\limits_{1}^{2}{f\left( 2x \right)dx}$
$=34-\int\limits_{2}^{4}{f\left( u \right)d\left( \dfrac{u}{2} \right)}=34-\dfrac{1}{2}\int\limits_{2}^{4}{f\left( x \right)dx}=34-\dfrac{1}{2}.58=5$.
Từ $f\left( 2x \right)+2.f\left( \dfrac{1}{x} \right)=12{{x}^{2}}+3+\dfrac{6}{{{x}^{2}}}$
$\Rightarrow I=\int\limits_{1}^{2}{\left( 12{{x}^{2}}+3+\dfrac{6}{{{x}^{2}}}-f\left( 2x \right) \right)dx}=\left. \left( 4{{x}^{3}}+3x-\dfrac{6}{x} \right) \right|_{1}^{2}-\int\limits_{1}^{2}{f\left( 2x \right)dx}$
$=34-\int\limits_{2}^{4}{f\left( u \right)d\left( \dfrac{u}{2} \right)}=34-\dfrac{1}{2}\int\limits_{2}^{4}{f\left( x \right)dx}=34-\dfrac{1}{2}.58=5$.
Đáp án D.