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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm, liên tục trên $R$, $f\left( 0 \right)=1$ và thỏa mãn $\dfrac{3{f}'\left( x \right).{{\left[ f\left( x \right) \right]}^{2}}}{{{e}^{{{x}^{2}}+1}}}-\dfrac{2x}{{{e}^{{{\left[ f\left( x \right) \right]}^{3}}}}}=0$. Giá trị của $I=\int\limits_{0}^{\sqrt{26}}{x.f\left( x \right)} \text{d}x$ thuộc khoảng nào sau đây?
A. $\left( 3;5 \right)$.
B. $\left( 5;7 \right)$.
C. $\left( 7;9 \right)$.
D. $\left( 1;3 \right)$.
A. $\left( 3;5 \right)$.
B. $\left( 5;7 \right)$.
C. $\left( 7;9 \right)$.
D. $\left( 1;3 \right)$.
Ta có $\dfrac{3{f}'\left( x \right).{{\left[ f\left( x \right) \right]}^{2}}}{{{e}^{{{x}^{2}}+1}}}-\dfrac{2x}{{{e}^{{{\left[ f\left( x \right) \right]}^{3}}}}}=0\Leftrightarrow 3{f}'\left( x \right).{{\left[ f\left( x \right) \right]}^{2}}.{{e}^{{{\left[ f\left( x \right) \right]}^{3}}}}=2x.{{e}^{{{x}^{2}}+1}}$.
$\Rightarrow \int{3{f}'\left( x \right).{{\left[ f\left( x \right) \right]}^{2}}.{{e}^{{{\left[ f\left( x \right) \right]}^{3}}}}\text{d}x}=\int{2x.{{e}^{{{x}^{2}}+1}}}\text{d}x$
$\Rightarrow \int{{{e}^{{{\left[ f\left( x \right) \right]}^{3}}}}\text{d}\left( {{\left[ f\left( x \right) \right]}^{3}} \right)}=\int{{{e}^{{{x}^{2}}+1}}}\text{d}\left( {{x}^{2}}+1 \right)$
$\Rightarrow {{e}^{{{\left[ f\left( x \right) \right]}^{3}}}}={{e}^{{{x}^{2}}+1}}+C\left( 1 \right)$
Thay $x=0$ vào hai vế của $\left( 1 \right)$ và sử dụng giả thiết $f\left( 0 \right)=1$ ta có:
Hay $f\left( x \right)=\sqrt[3]{{{x}^{2}}+1}$.
Ta có: $I=\int\limits_{0}^{\sqrt{26}}{\dfrac{x}{f\left( x \right)}} \text{d}x=\int\limits_{0}^{\sqrt{26}}{\dfrac{x}{\sqrt[3]{{{x}^{2}}+1}} }\text{d}x=\dfrac{1}{2}\int\limits_{0}^{\sqrt{26}}{{{\left( {{x}^{2}}+1 \right)}^{-\dfrac{1}{3}}}}\text{d}\left( {{x}^{2}}+1 \right)$
$I=\left. \dfrac{3}{4}{{\left( {{x}^{2}}+1 \right)}^{\dfrac{2}{3}}} \right|_{0}^{\sqrt{26}}=\left. \dfrac{3}{4}\sqrt[3]{{{\left( {{x}^{2}}+1 \right)}^{2}}} \right|_{0}^{\sqrt{26}}=\dfrac{3}{4}\left( 9-1 \right)=6$
$\Rightarrow \int{3{f}'\left( x \right).{{\left[ f\left( x \right) \right]}^{2}}.{{e}^{{{\left[ f\left( x \right) \right]}^{3}}}}\text{d}x}=\int{2x.{{e}^{{{x}^{2}}+1}}}\text{d}x$
$\Rightarrow \int{{{e}^{{{\left[ f\left( x \right) \right]}^{3}}}}\text{d}\left( {{\left[ f\left( x \right) \right]}^{3}} \right)}=\int{{{e}^{{{x}^{2}}+1}}}\text{d}\left( {{x}^{2}}+1 \right)$
$\Rightarrow {{e}^{{{\left[ f\left( x \right) \right]}^{3}}}}={{e}^{{{x}^{2}}+1}}+C\left( 1 \right)$
Thay $x=0$ vào hai vế của $\left( 1 \right)$ và sử dụng giả thiết $f\left( 0 \right)=1$ ta có:
${{e}^{{{\left[ f\left( 0 \right) \right]}^{3}}}}=e+C\Leftrightarrow C=0$
Vậy ${{e}^{{{\left[ f\left( x \right) \right]}^{3}}}}={{e}^{{{x}^{2}}+1}}\Leftrightarrow {{\left[ f\left( x \right) \right]}^{3}}={{x}^{2}}+1$.Hay $f\left( x \right)=\sqrt[3]{{{x}^{2}}+1}$.
Ta có: $I=\int\limits_{0}^{\sqrt{26}}{\dfrac{x}{f\left( x \right)}} \text{d}x=\int\limits_{0}^{\sqrt{26}}{\dfrac{x}{\sqrt[3]{{{x}^{2}}+1}} }\text{d}x=\dfrac{1}{2}\int\limits_{0}^{\sqrt{26}}{{{\left( {{x}^{2}}+1 \right)}^{-\dfrac{1}{3}}}}\text{d}\left( {{x}^{2}}+1 \right)$
$I=\left. \dfrac{3}{4}{{\left( {{x}^{2}}+1 \right)}^{\dfrac{2}{3}}} \right|_{0}^{\sqrt{26}}=\left. \dfrac{3}{4}\sqrt[3]{{{\left( {{x}^{2}}+1 \right)}^{2}}} \right|_{0}^{\sqrt{26}}=\dfrac{3}{4}\left( 9-1 \right)=6$
Đáp án B.