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Câu hỏi: Cho $F\left( x \right)$ là một nguyên hàm của hàm số $f\left( x \right)=\dfrac{1}{2\sqrt{x\left( x+3 \right)}}$ trên khoảng $\left( 0;+\infty \right)$ thỏa mãn $F\left( 1 \right)=\ln 3.$ Giá trị của ${{e}^{F\left( 2021 \right)}}-{{e}^{F\left( 2020 \right)}}$ thuộc khoảng nào?
A. $\left( 0;\dfrac{1}{10} \right)$
B. $\left( \dfrac{1}{10};\dfrac{1}{5} \right)$
C. $\left( \dfrac{1}{5};\dfrac{1}{3} \right)$
D. $\left( \dfrac{1}{3};\dfrac{1}{2} \right)$
A. $\left( 0;\dfrac{1}{10} \right)$
B. $\left( \dfrac{1}{10};\dfrac{1}{5} \right)$
C. $\left( \dfrac{1}{5};\dfrac{1}{3} \right)$
D. $\left( \dfrac{1}{3};\dfrac{1}{2} \right)$
Phương pháp:
Đặt $t=x+\dfrac{3}{2}+\sqrt{x\left( x+3 \right)}.$
Cách giải:
Đặt $t=x+\dfrac{3}{2}+\sqrt{x\left( x+3 \right)}.$
$\Rightarrow dt=\left( 1+\dfrac{2x+3}{2\sqrt{x\left( x+3 \right)}} \right)dx$
$\Rightarrow dt=\dfrac{2\sqrt{x\left( x+3 \right)}+2x+3}{2\sqrt{x\left( x+3 \right)}}dx$
$\Rightarrow dt=\dfrac{x+\dfrac{3}{2}+\sqrt{x\left( x+3 \right)}}{\sqrt{x\left( x+3 \right)}}dx$
$\Rightarrow dt=\dfrac{t}{\sqrt{x\left( x+3 \right)}}dx$
$\Rightarrow \dfrac{dx}{\sqrt{x\left( x+3 \right)}}=\dfrac{dt}{t}$
Khi đó ta có:
$F\left( x \right)=\int\limits_{{}}^{{}}{\dfrac{1}{2\sqrt{x\left( x+3 \right)}}dx}=\dfrac{1}{2}\int\limits_{{}}^{{}}{\dfrac{dt}{t}}=\dfrac{1}{2}\ln \left| t \right|+C$
$\Rightarrow F\left( x \right)=\dfrac{1}{2}\ln \left| x+\dfrac{3}{2}+\sqrt{x\left( x+3 \right)} \right|+C$
Lại có $F\left( 1 \right)=\ln 3$
$\Rightarrow \dfrac{1}{2}\ln \left| 1+\dfrac{3}{2}+\sqrt{4} \right|+C=\ln 3$
$\Leftrightarrow \dfrac{1}{2}\ln \dfrac{9}{2}+C=\ln 3$
$\Leftrightarrow \ln 9-\ln 2+2C=\ln 9$
$\Leftrightarrow 2C=\ln 2\Leftrightarrow C=\dfrac{1}{2}\ln 2$
$\Rightarrow F\left( x \right)=\dfrac{1}{2}\ln \left| x+\dfrac{3}{2}+\sqrt{x\left( x+3 \right)} \right|+\dfrac{1}{2}\ln 2$
$\Rightarrow {{e}^{F\left( x \right)}}={{e}^{\dfrac{1}{2}\ln \left| x+\dfrac{3}{2}+\sqrt{x\left( x+3 \right)} \right|+\dfrac{1}{2}\ln 2}}$
$={{\left( {{e}^{\ln \left| x+\dfrac{3}{2}+\sqrt{x\left( x+3 \right)} \right|}} \right)}^{\dfrac{1}{2}}}.{{\left( {{e}^{\ln 2}} \right)}^{\dfrac{1}{2}}}$
$=\sqrt{2}.\sqrt{\left| x+\dfrac{3}{2}+\sqrt{x\left( x+3 \right)} \right|}$
$=\sqrt{\left| 2x+3+2\sqrt{x\left( x+3 \right)} \right|}$
$\Rightarrow \left\{ \begin{aligned}
& {{e}^{F\left( 2021 \right)}}=\sqrt{\left| 2.2021+3+2\sqrt{2021\left( 2021+3 \right)} \right|} \\
& {{e}^{F\left( 2020 \right)}}=\sqrt{\left| 2.2020+3+2\sqrt{2020\left( 2020+3 \right)} \right|} \\
\end{aligned} \right.$
Vậy ${{e}^{F\left( 2021 \right)}}-{{e}^{F\left( 2020 \right)}}\approx 0,022.$
Đặt $t=x+\dfrac{3}{2}+\sqrt{x\left( x+3 \right)}.$
Cách giải:
Đặt $t=x+\dfrac{3}{2}+\sqrt{x\left( x+3 \right)}.$
$\Rightarrow dt=\left( 1+\dfrac{2x+3}{2\sqrt{x\left( x+3 \right)}} \right)dx$
$\Rightarrow dt=\dfrac{2\sqrt{x\left( x+3 \right)}+2x+3}{2\sqrt{x\left( x+3 \right)}}dx$
$\Rightarrow dt=\dfrac{x+\dfrac{3}{2}+\sqrt{x\left( x+3 \right)}}{\sqrt{x\left( x+3 \right)}}dx$
$\Rightarrow dt=\dfrac{t}{\sqrt{x\left( x+3 \right)}}dx$
$\Rightarrow \dfrac{dx}{\sqrt{x\left( x+3 \right)}}=\dfrac{dt}{t}$
Khi đó ta có:
$F\left( x \right)=\int\limits_{{}}^{{}}{\dfrac{1}{2\sqrt{x\left( x+3 \right)}}dx}=\dfrac{1}{2}\int\limits_{{}}^{{}}{\dfrac{dt}{t}}=\dfrac{1}{2}\ln \left| t \right|+C$
$\Rightarrow F\left( x \right)=\dfrac{1}{2}\ln \left| x+\dfrac{3}{2}+\sqrt{x\left( x+3 \right)} \right|+C$
Lại có $F\left( 1 \right)=\ln 3$
$\Rightarrow \dfrac{1}{2}\ln \left| 1+\dfrac{3}{2}+\sqrt{4} \right|+C=\ln 3$
$\Leftrightarrow \dfrac{1}{2}\ln \dfrac{9}{2}+C=\ln 3$
$\Leftrightarrow \ln 9-\ln 2+2C=\ln 9$
$\Leftrightarrow 2C=\ln 2\Leftrightarrow C=\dfrac{1}{2}\ln 2$
$\Rightarrow F\left( x \right)=\dfrac{1}{2}\ln \left| x+\dfrac{3}{2}+\sqrt{x\left( x+3 \right)} \right|+\dfrac{1}{2}\ln 2$
$\Rightarrow {{e}^{F\left( x \right)}}={{e}^{\dfrac{1}{2}\ln \left| x+\dfrac{3}{2}+\sqrt{x\left( x+3 \right)} \right|+\dfrac{1}{2}\ln 2}}$
$={{\left( {{e}^{\ln \left| x+\dfrac{3}{2}+\sqrt{x\left( x+3 \right)} \right|}} \right)}^{\dfrac{1}{2}}}.{{\left( {{e}^{\ln 2}} \right)}^{\dfrac{1}{2}}}$
$=\sqrt{2}.\sqrt{\left| x+\dfrac{3}{2}+\sqrt{x\left( x+3 \right)} \right|}$
$=\sqrt{\left| 2x+3+2\sqrt{x\left( x+3 \right)} \right|}$
$\Rightarrow \left\{ \begin{aligned}
& {{e}^{F\left( 2021 \right)}}=\sqrt{\left| 2.2021+3+2\sqrt{2021\left( 2021+3 \right)} \right|} \\
& {{e}^{F\left( 2020 \right)}}=\sqrt{\left| 2.2020+3+2\sqrt{2020\left( 2020+3 \right)} \right|} \\
\end{aligned} \right.$
Vậy ${{e}^{F\left( 2021 \right)}}-{{e}^{F\left( 2020 \right)}}\approx 0,022.$
Đáp án A.