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Câu hỏi: Cho hàm đa thức bậc bốn $y=f\left( x \right)$. Biết rằng hàm số $g\left( x \right)={{e}^{f\left( x \right)}}$ có bảng biến thiên như sau:
Diện tích hình phẳng giới hạn bởi các đường $y={f}'\left( x \right)$ và $y={g}'\left( x \right)$ thuộc khoảng nào dưới đây?
A. $\left( 26;27 \right)$.
B. $\left( 27;28 \right)$.
C. $\left( 28;29 \right)$.
D. $\left( 29;30 \right)$.
A. $\left( 26;27 \right)$.
B. $\left( 27;28 \right)$.
C. $\left( 28;29 \right)$.
D. $\left( 29;30 \right)$.
Ta có $g\left( x \right)={{e}^{f\left( x \right)}}\Rightarrow {g}'\left( x \right)={f}'\left( x \right){{e}^{f\left( x \right)}}$.
Ta có ${g}'\left( x \right)=0\Leftrightarrow {f}'\left( x \right)=0$, khi đó ${f}'\left( x \right)=0\Leftrightarrow \left[ \begin{matrix}
x={{x}_{1}} \\
x={{x}_{2}} \\
x={{x}_{3}} \\
\end{matrix} \right.$
Phương trình hoành độ giao điểm ${f}'\left( x \right)={g}'\left( x \right)\Leftrightarrow {f}'\left( x \right)={f}'\left( x \right){{e}^{f\left( x \right)}}$
Do $f\left( x \right)>0,\forall x\in \mathbb{R}$ nên ${f}'\left( x \right)={f}'\left( x \right){{e}^{f\left( x \right)}}\Leftrightarrow {f}'\left( x \right)=0\Leftrightarrow \left[ \begin{matrix}
x={{x}_{1}} \\
x={{x}_{2}} \\
x={{x}_{3}} \\
\end{matrix} \right.$.
Ta có $S=\int\limits_{{{x}_{1}}}^{{{x}_{3}}}{\left| {f}'\left( x \right)-{g}'\left( x \right) \right|\text{d}x}=\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left| {f}'\left( x \right)-{g}'\left( x \right) \right|\text{d}x}+\int\limits_{{{x}_{2}}}^{{{x}_{3}}}{\left| {f}'\left( x \right)-{g}'\left( x \right) \right|\text{d}x}$
$=\left| \int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left[ {f}'\left( x \right)\left( 1-{{e}^{f\left( x \right)}} \right) \right]\text{d}x} \right|+\left| \int\limits_{{{x}_{2}}}^{{{x}_{3}}}{\left[ {f}'\left( x \right)\left( 1-{{e}^{f\left( x \right)}} \right) \right]\text{d}x} \right|=\left| \left. f\left( x \right)-{{e}^{f\left( x \right)}} \right|_{{{x}_{1}}}^{{{x}_{2}}} \right|-\left| \left. f\left( x \right)-{{e}^{f\left( x \right)}} \right|_{{{x}_{2}}}^{{{x}_{3}}} \right|$
$=\left| \left. f\left( x \right)-{{e}^{f\left( x \right)}} \right|_{{{x}_{1}}}^{{{x}_{2}}} \right|+\left| \left. f\left( x \right)-{{e}^{f\left( x \right)}} \right|_{{{x}_{2}}}^{{{x}_{3}}} \right|=\left| 3-{{e}^{3}}-2+{{e}^{2}} \right|+\left| \dfrac{1}{2}-\sqrt{e}-3+{{e}^{3}} \right|\approx 27,63$.
Ta có ${g}'\left( x \right)=0\Leftrightarrow {f}'\left( x \right)=0$, khi đó ${f}'\left( x \right)=0\Leftrightarrow \left[ \begin{matrix}
x={{x}_{1}} \\
x={{x}_{2}} \\
x={{x}_{3}} \\
\end{matrix} \right.$
Phương trình hoành độ giao điểm ${f}'\left( x \right)={g}'\left( x \right)\Leftrightarrow {f}'\left( x \right)={f}'\left( x \right){{e}^{f\left( x \right)}}$
Do $f\left( x \right)>0,\forall x\in \mathbb{R}$ nên ${f}'\left( x \right)={f}'\left( x \right){{e}^{f\left( x \right)}}\Leftrightarrow {f}'\left( x \right)=0\Leftrightarrow \left[ \begin{matrix}
x={{x}_{1}} \\
x={{x}_{2}} \\
x={{x}_{3}} \\
\end{matrix} \right.$.
Ta có $S=\int\limits_{{{x}_{1}}}^{{{x}_{3}}}{\left| {f}'\left( x \right)-{g}'\left( x \right) \right|\text{d}x}=\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left| {f}'\left( x \right)-{g}'\left( x \right) \right|\text{d}x}+\int\limits_{{{x}_{2}}}^{{{x}_{3}}}{\left| {f}'\left( x \right)-{g}'\left( x \right) \right|\text{d}x}$
$=\left| \int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left[ {f}'\left( x \right)\left( 1-{{e}^{f\left( x \right)}} \right) \right]\text{d}x} \right|+\left| \int\limits_{{{x}_{2}}}^{{{x}_{3}}}{\left[ {f}'\left( x \right)\left( 1-{{e}^{f\left( x \right)}} \right) \right]\text{d}x} \right|=\left| \left. f\left( x \right)-{{e}^{f\left( x \right)}} \right|_{{{x}_{1}}}^{{{x}_{2}}} \right|-\left| \left. f\left( x \right)-{{e}^{f\left( x \right)}} \right|_{{{x}_{2}}}^{{{x}_{3}}} \right|$
$=\left| \left. f\left( x \right)-{{e}^{f\left( x \right)}} \right|_{{{x}_{1}}}^{{{x}_{2}}} \right|+\left| \left. f\left( x \right)-{{e}^{f\left( x \right)}} \right|_{{{x}_{2}}}^{{{x}_{3}}} \right|=\left| 3-{{e}^{3}}-2+{{e}^{2}} \right|+\left| \dfrac{1}{2}-\sqrt{e}-3+{{e}^{3}} \right|\approx 27,63$.
Đáp án B.
