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Câu hỏi: Cho hàm số f(x) có đạo hàm liên tục trên đoạn [0;1], thỏa mãn f(0) = 0, f(1) = 1 và $\int\limits_{0}^{1}{\dfrac{{{\left[ f'\left( x \right) \right]}^{2}}}{{{\text{e}}^{\text{x}}}}}dx=\dfrac{1}{e-1}$. Tích phân $\int\limits_{0}^{1}{f\left( x \right)}dx$ bằng
A. $\dfrac{e-2}{e-1}$.
B. 1.
C. $\dfrac{1}{\left( e-1 \right)\left( e-2 \right)}$.
D. $\dfrac{e-1}{e-2}$.
A. $\dfrac{e-2}{e-1}$.
B. 1.
C. $\dfrac{1}{\left( e-1 \right)\left( e-2 \right)}$.
D. $\dfrac{e-1}{e-2}$.
Lời giải: Chọn k sao cho ${{\int\limits_{0}^{1}{\left[ \dfrac{f'\left( x \right)}{\sqrt{{{e}^{x}}}}+k\sqrt{{{e}^{x}}} \right]}}^{2}}dx=\int\limits_{0}^{1}{\dfrac{{{\left[ f'\left( x \right) \right]}^{2}}}{{{e}^{x}}}}dx+2k\int\limits_{0}^{1}{f'\left( x \right)dx+{{k}^{2}}\int\limits_{0}^{1}{{{e}^{x}}dx}=0}$
$\Leftrightarrow \dfrac{1}{e-1}+2k+{{k}^{2}}\left( e-1 \right)\Leftrightarrow k=-\dfrac{1}{e-1}=\dfrac{1}{1-e}$
Khi đó $\dfrac{f'\left( x \right)}{\sqrt{{{e}^{x}}}}=-k\sqrt{{{e}^{x}}}\Rightarrow f'\left( x \right)=-k\sqrt{{{e}^{x}}}=\dfrac{1}{e-1}{{e}^{x}}\Rightarrow f\left( x \right)=\dfrac{{{e}^{x}}}{e-1}+C$
Do $f\left( 0 \right)=0\Rightarrow f\left( x \right)=\dfrac{{{e}^{x}}-1}{e-1}$
Do đó $\int\limits_{0}^{1}{f\left( x \right)dx}=\dfrac{1}{e-1}\int\limits_{0}^{1}{\left( {{e}^{x}}-1 \right)dx}=\left. \dfrac{1}{e-1}\left( {{e}^{x}}-1 \right) \right|_{0}^{1}=1-\dfrac{1}{e-1}=\dfrac{e-2}{e-1}$.
$\Leftrightarrow \dfrac{1}{e-1}+2k+{{k}^{2}}\left( e-1 \right)\Leftrightarrow k=-\dfrac{1}{e-1}=\dfrac{1}{1-e}$
Khi đó $\dfrac{f'\left( x \right)}{\sqrt{{{e}^{x}}}}=-k\sqrt{{{e}^{x}}}\Rightarrow f'\left( x \right)=-k\sqrt{{{e}^{x}}}=\dfrac{1}{e-1}{{e}^{x}}\Rightarrow f\left( x \right)=\dfrac{{{e}^{x}}}{e-1}+C$
Do $f\left( 0 \right)=0\Rightarrow f\left( x \right)=\dfrac{{{e}^{x}}-1}{e-1}$
Do đó $\int\limits_{0}^{1}{f\left( x \right)dx}=\dfrac{1}{e-1}\int\limits_{0}^{1}{\left( {{e}^{x}}-1 \right)dx}=\left. \dfrac{1}{e-1}\left( {{e}^{x}}-1 \right) \right|_{0}^{1}=1-\dfrac{1}{e-1}=\dfrac{e-2}{e-1}$.
Đáp án A.