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Câu hỏi: Cho biết $f\left( x \right)=\int\limits_{{{e}^{x}}}^{{{e}^{2\text{x}}}}{t{{\ln }^{20}}t\text{d}t}$, hàm số $y=f\left( x \right)$ đạt giá trị cực trị khi
A. $x=\dfrac{21}{2}\ln 2$
B. $x=-\dfrac{21}{2}\ln 2$
C. $x=-\dfrac{21}{2\ln 2}$
D. $x=\dfrac{21}{2\ln 2}$
A. $x=\dfrac{21}{2}\ln 2$
B. $x=-\dfrac{21}{2}\ln 2$
C. $x=-\dfrac{21}{2\ln 2}$
D. $x=\dfrac{21}{2\ln 2}$
Giả sử $F\left( t \right)$ là một nguyên hàm của $t{{\ln }^{20}}t$, ta có: ${F}'\left( t \right)=t{{\ln }^{20}}t$.
Khi đó: $f\left( x \right)=F\left( {{e}^{2\text{x}}} \right)-F\left( {{e}^{x}} \right)\Rightarrow {f}'\left( x \right)=2{{\text{e}}^{2\text{x}}}{F}'\left( {{e}^{2\text{x}}} \right)-{{e}^{x}}{F}'\left( {{e}^{x}} \right)=2{{e}^{2\text{x}}}{{e}^{2\text{x}}}{{\ln }^{20}}\left( {{e}^{2\text{x}}} \right)-{{e}^{x}}{{e}^{x}}{{\ln }^{20}}\left( {{e}^{x}} \right)$
$\Leftrightarrow {f}'\left( x \right)={{2}^{21}}{{e}^{4\text{x}}}{{x}^{20}}-{{e}^{2\text{x}}}{{x}^{20}}={{e}^{2\text{x}}}{{x}^{20}}\left( {{2}^{21}}{{e}^{2\text{x}}}-1 \right)=0\Rightarrow 2\text{x}=\ln \dfrac{1}{{{2}^{21}}}\Leftrightarrow x=-\dfrac{21}{2}\ln 2$.
Khi đó: $f\left( x \right)=F\left( {{e}^{2\text{x}}} \right)-F\left( {{e}^{x}} \right)\Rightarrow {f}'\left( x \right)=2{{\text{e}}^{2\text{x}}}{F}'\left( {{e}^{2\text{x}}} \right)-{{e}^{x}}{F}'\left( {{e}^{x}} \right)=2{{e}^{2\text{x}}}{{e}^{2\text{x}}}{{\ln }^{20}}\left( {{e}^{2\text{x}}} \right)-{{e}^{x}}{{e}^{x}}{{\ln }^{20}}\left( {{e}^{x}} \right)$
$\Leftrightarrow {f}'\left( x \right)={{2}^{21}}{{e}^{4\text{x}}}{{x}^{20}}-{{e}^{2\text{x}}}{{x}^{20}}={{e}^{2\text{x}}}{{x}^{20}}\left( {{2}^{21}}{{e}^{2\text{x}}}-1 \right)=0\Rightarrow 2\text{x}=\ln \dfrac{1}{{{2}^{21}}}\Leftrightarrow x=-\dfrac{21}{2}\ln 2$.
Đáp án B.