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Câu hỏi: Cho đường thẳng $d:y=g\left( x \right)$ cắt đồ thị hàm bậc ba $y=f\left( x \right)$ tại ba điểm phân biệt có hoành độ ${{x}_{1}},{{x}_{2}},{{x}_{3}}$, $\left( {{x}_{1}}<{{x}_{2}}<{{x}_{3}} \right)$. Gọi ${{S}_{1}}$ là diện tích hình phẳng giới hạn bởi các đường $y=f\left( x \right)$ ; $y=g\left( x \right)$ ; $x={{x}_{1}};x={{x}_{2}}$. Gọi ${{S}_{2}}$ là diện tích hình phẳng giới hạn bởi các đường $y=f\left( x \right)$ ; $y=g\left( x \right)$ ; $x={{x}_{2}};x={{x}_{3}}$. Khi ${{S}_{1}}$ = 2 ${{S}_{2}}$ thì $\dfrac{{{x}_{1}}-{{x}_{2}}}{{{x}_{2}}-{{x}_{3}}}$ thuộc khoảng nào dưới đây?
A. $\left( 1;\dfrac{4}{3} \right).$
B. $\left( \dfrac{4}{3};\dfrac{3}{2} \right)$.
C. $\left( \dfrac{3}{2};\dfrac{8}{5} \right)$
D. $\left( \dfrac{8}{5};2 \right).$
A. $\left( 1;\dfrac{4}{3} \right).$
B. $\left( \dfrac{4}{3};\dfrac{3}{2} \right)$.
C. $\left( \dfrac{3}{2};\dfrac{8}{5} \right)$
D. $\left( \dfrac{8}{5};2 \right).$
Vì đường đường thẳng $d:y=g\left( x \right)$ cắt đồ thị hàm bậc ba $y=f\left( x \right)$ tại ba điểm phân biệt có hoành độ ${{x}_{1}},{{x}_{2}},{{x}_{3}}$, $\left( {{x}_{1}}<{{x}_{2}}<{{x}_{3}} \right)$ nên giả sử với $a>0$.
Khi đó ta có ${{S}_{1}}=\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left( f\left( x \right)-g(x) \right)dx}=\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left( a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right) \right)dx}$
Xét $J=\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)dx}=\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left( x-{{x}_{1}} \right)\left( \left( x-{{x}_{1}} \right)+{{x}_{1}}-{{x}_{2}} \right)dx}$
$\begin{aligned}
& =\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left( {{\left( x-{{x}_{1}} \right)}^{2}}+\left( x-{{x}_{1}} \right)\left( {{x}_{1}}-{{x}_{2}} \right) \right)dx} \\
& =\left. \dfrac{{{\left( x-{{x}_{1}} \right)}^{3}}}{3} \right|_{{{x}_{1}}}^{{{x}_{2}}}+\left( {{x}_{1}}-{{x}_{2}} \right) \left. \dfrac{{{\left( x-{{x}_{1}} \right)}^{2}}}{2} \right|_{{{x}_{1}}}^{{{x}_{2}}}=-\dfrac{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{3}}}{6} \\
\end{aligned}$
Vậy ${{S}_{1}}=\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left( f\left( x \right)-g(x) \right)dx}=\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left( a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right) \right)dx}$
$=\dfrac{a}{2}\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left( \left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( 2x-\left( {{x}_{1}}+{{x}_{2}} \right)+\left( {{x}_{1}}+{{x}_{2}} \right)-2{{x}_{3}} \right) \right)dx}$ $=\dfrac{a}{2}\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left( \left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( 2x-\left( {{x}_{1}}+{{x}_{2}} \right) \right) \right)dx}+\dfrac{a}{2}\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left( \left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( \left( {{x}_{1}}+{{x}_{2}} \right)-2{{x}_{3}} \right) \right)dx}$
$=\left. \dfrac{a{{\left( \left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right) \right)}^{2}}}{2} \right|_{{{x}_{1}}}^{{{x}_{2}}}-\dfrac{a{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{3}}\left( \left( {{x}_{1}}+{{x}_{2}} \right)-2{{x}_{3}} \right)}{12}$ $=-\dfrac{a{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{3}}\left( \left( {{x}_{1}}+{{x}_{2}} \right)-2{{x}_{3}} \right)}{12}$
Tương tự ta có ${{S}_{2}}=\dfrac{a{{\left( {{x}_{3}}-{{x}_{2}} \right)}^{3}}\left( \left( {{x}_{3}}+{{x}_{2}} \right)-2{{x}_{1}} \right)}{12}$.
Theo bài ra
${{S}_{1}}=2{{S}_{2}}\Leftrightarrow \dfrac{{{S}_{1}}}{{{S}_{2}}}=2$ $\Leftrightarrow \dfrac{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{3}}\left( \left( {{x}_{1}}+{{x}_{2}} \right)-2{{x}_{3}} \right)}{{{\left( {{x}_{2}}-{{x}_{3}} \right)}^{3}}\left( -\left( {{x}_{3}}+{{x}_{2}} \right)+2{{x}_{1}} \right)}=2\Leftrightarrow \dfrac{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{3}}\left( \left( {{x}_{1}}-{{x}_{2}} \right)+2\left( {{x}_{2}}-{{x}_{3}} \right) \right)}{{{\left( {{x}_{2}}-{{x}_{3}} \right)}^{3}}\left( \left( {{x}_{2}}-{{x}_{3}} \right)+2\left( {{x}_{1}}-{{x}_{2}} \right) \right)}=2$.
$\Leftrightarrow \dfrac{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{3}}\left( \dfrac{\left( {{x}_{1}}-{{x}_{2}} \right)}{\left( {{x}_{2}}-{{x}_{3}} \right)}+2 \right)}{{{\left( {{x}_{2}}-{{x}_{3}} \right)}^{3}}\left( 1+2\dfrac{\left( {{x}_{1}}-{{x}_{2}} \right)}{\left( {{x}_{2}}-{{x}_{3}} \right)} \right)}=2$.(*)
Đặt $t=\dfrac{\left( {{x}_{1}}-{{x}_{2}} \right)}{\left( {{x}_{2}}-{{x}_{3}} \right)},t>0$, khi đó (*) trở thành
$\begin{aligned}
& {{t}^{3}}\dfrac{t+2}{2t+1}=2\Leftrightarrow {{t}^{4}}+2{{t}^{3}}-4t-2=0 \\
& \Leftrightarrow \left( {{t}^{2}}+\left( \sqrt{3}-1 \right)t-\left( \sqrt{3}-1 \right) \right)\left( {{t}^{2}}-\left( \sqrt{3}+1 \right)t+\left( \sqrt{3}+1 \right) \right)=0 \\
\end{aligned}$
$\Leftrightarrow {{t}^{2}}-\left( \sqrt{3}-1 \right)t-\left( \sqrt{3}-1 \right)=0\Leftrightarrow \left[ \begin{aligned}
& t=\dfrac{\sqrt{3}-1-\sqrt{2\sqrt{3}}}{2}\left( l \right) \\
& t=\dfrac{\sqrt{3}-1+\sqrt{2\sqrt{3}}}{2}\approx 1.2966 \\
\end{aligned} \right.$.
Khi đó ta có ${{S}_{1}}=\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left( f\left( x \right)-g(x) \right)dx}=\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left( a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right) \right)dx}$
Xét $J=\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)dx}=\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left( x-{{x}_{1}} \right)\left( \left( x-{{x}_{1}} \right)+{{x}_{1}}-{{x}_{2}} \right)dx}$
$\begin{aligned}
& =\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left( {{\left( x-{{x}_{1}} \right)}^{2}}+\left( x-{{x}_{1}} \right)\left( {{x}_{1}}-{{x}_{2}} \right) \right)dx} \\
& =\left. \dfrac{{{\left( x-{{x}_{1}} \right)}^{3}}}{3} \right|_{{{x}_{1}}}^{{{x}_{2}}}+\left( {{x}_{1}}-{{x}_{2}} \right) \left. \dfrac{{{\left( x-{{x}_{1}} \right)}^{2}}}{2} \right|_{{{x}_{1}}}^{{{x}_{2}}}=-\dfrac{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{3}}}{6} \\
\end{aligned}$
Vậy ${{S}_{1}}=\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left( f\left( x \right)-g(x) \right)dx}=\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left( a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right) \right)dx}$
$=\dfrac{a}{2}\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left( \left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( 2x-\left( {{x}_{1}}+{{x}_{2}} \right)+\left( {{x}_{1}}+{{x}_{2}} \right)-2{{x}_{3}} \right) \right)dx}$ $=\dfrac{a}{2}\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left( \left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( 2x-\left( {{x}_{1}}+{{x}_{2}} \right) \right) \right)dx}+\dfrac{a}{2}\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left( \left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( \left( {{x}_{1}}+{{x}_{2}} \right)-2{{x}_{3}} \right) \right)dx}$
$=\left. \dfrac{a{{\left( \left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right) \right)}^{2}}}{2} \right|_{{{x}_{1}}}^{{{x}_{2}}}-\dfrac{a{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{3}}\left( \left( {{x}_{1}}+{{x}_{2}} \right)-2{{x}_{3}} \right)}{12}$ $=-\dfrac{a{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{3}}\left( \left( {{x}_{1}}+{{x}_{2}} \right)-2{{x}_{3}} \right)}{12}$
Tương tự ta có ${{S}_{2}}=\dfrac{a{{\left( {{x}_{3}}-{{x}_{2}} \right)}^{3}}\left( \left( {{x}_{3}}+{{x}_{2}} \right)-2{{x}_{1}} \right)}{12}$.
Theo bài ra
${{S}_{1}}=2{{S}_{2}}\Leftrightarrow \dfrac{{{S}_{1}}}{{{S}_{2}}}=2$ $\Leftrightarrow \dfrac{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{3}}\left( \left( {{x}_{1}}+{{x}_{2}} \right)-2{{x}_{3}} \right)}{{{\left( {{x}_{2}}-{{x}_{3}} \right)}^{3}}\left( -\left( {{x}_{3}}+{{x}_{2}} \right)+2{{x}_{1}} \right)}=2\Leftrightarrow \dfrac{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{3}}\left( \left( {{x}_{1}}-{{x}_{2}} \right)+2\left( {{x}_{2}}-{{x}_{3}} \right) \right)}{{{\left( {{x}_{2}}-{{x}_{3}} \right)}^{3}}\left( \left( {{x}_{2}}-{{x}_{3}} \right)+2\left( {{x}_{1}}-{{x}_{2}} \right) \right)}=2$.
$\Leftrightarrow \dfrac{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{3}}\left( \dfrac{\left( {{x}_{1}}-{{x}_{2}} \right)}{\left( {{x}_{2}}-{{x}_{3}} \right)}+2 \right)}{{{\left( {{x}_{2}}-{{x}_{3}} \right)}^{3}}\left( 1+2\dfrac{\left( {{x}_{1}}-{{x}_{2}} \right)}{\left( {{x}_{2}}-{{x}_{3}} \right)} \right)}=2$.(*)
Đặt $t=\dfrac{\left( {{x}_{1}}-{{x}_{2}} \right)}{\left( {{x}_{2}}-{{x}_{3}} \right)},t>0$, khi đó (*) trở thành
$\begin{aligned}
& {{t}^{3}}\dfrac{t+2}{2t+1}=2\Leftrightarrow {{t}^{4}}+2{{t}^{3}}-4t-2=0 \\
& \Leftrightarrow \left( {{t}^{2}}+\left( \sqrt{3}-1 \right)t-\left( \sqrt{3}-1 \right) \right)\left( {{t}^{2}}-\left( \sqrt{3}+1 \right)t+\left( \sqrt{3}+1 \right) \right)=0 \\
\end{aligned}$
$\Leftrightarrow {{t}^{2}}-\left( \sqrt{3}-1 \right)t-\left( \sqrt{3}-1 \right)=0\Leftrightarrow \left[ \begin{aligned}
& t=\dfrac{\sqrt{3}-1-\sqrt{2\sqrt{3}}}{2}\left( l \right) \\
& t=\dfrac{\sqrt{3}-1+\sqrt{2\sqrt{3}}}{2}\approx 1.2966 \\
\end{aligned} \right.$.
Đáp án A.