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Câu hỏi: Cho hàm số $y=f(x)$ có đạo hàm là ${{f}^{\prime }}(x)=\ln \left( x+a \right),\forall x>-a, \ a$ là số thực dương và $f(0)=a\ln a$. Biết $\int\limits_{0}^{a}{f\left( x \right)\text{d}x}=0,$ khi đó mệnh đề nào sau đây đúng?
A. $a\in \left( 2;e \right).$
B. $a\in \left( 0;1 \right).$
C. $a\in \left( 1;\sqrt{2} \right).$
D. $a\in \left( \dfrac{e}{2};2 \right).$
A. $a\in \left( 2;e \right).$
B. $a\in \left( 0;1 \right).$
C. $a\in \left( 1;\sqrt{2} \right).$
D. $a\in \left( \dfrac{e}{2};2 \right).$
Ta có: ${{f}^{\prime }}(x)=\ln \left( x+a \right)\Rightarrow f\left( x \right)=\left( x+a \right)\ln \left( x+a \right)-x+C$.
Vì $f\left( 0 \right)=a\ln a\Rightarrow \left( 0+a \right)\ln \left( 0+a \right)-0+C=a\ln a\Rightarrow C=0$ $\Rightarrow f\left( x \right)=\left( x+a \right)\ln \left( x+a \right)-x$
$\int\limits_{0}^{a}{f\left( x \right)\text{d}x}=\int\limits_{0}^{a}{\left( \left( x+a \right)\ln \left( x+a \right)-x \right)\text{d}x}=\int\limits_{0}^{a}{\left( x+a \right)\ln \left( x+a \right)\text{d}x}-\int\limits_{0}^{a}{x\text{d}x}={{I}_{1}}-{{I}_{2}}$. $\begin{aligned}
& {{I}_{1}}=\int\limits_{0}^{a}{\left( x+a \right)\ln \left( x+a \right)\text{d}x}=\left. \left( \dfrac{{{x}^{2}}}{2}+a x \right).\ln \left( x+a \right) \right|_{0}^{a}-\int\limits_{0}^{a}{\left( \dfrac{{{x}^{2}}}{2}+a x \right).\dfrac{1}{x+a}dx} \\
& =\dfrac{3{{a}^{2}}}{2}\ln \left( 2a \right)-\int\limits_{0}^{a}{\left( \dfrac{1}{2}\left( x+a \right)-\dfrac{1}{2}{{a}^{2}}.\dfrac{1}{x+a} \right)dx}=\dfrac{3{{a}^{2}}}{2}\ln \left( 2a \right)-\left. \left[ \dfrac{1}{2}\left( \dfrac{{{x}^{2}}}{2}+a x \right)-\dfrac{1}{2}{{a}^{2}}.\ln \left( x+a \right) \right] \right|_{0}^{a} \\
& =2{{a}^{2}}\ln \left( 2a \right)-\dfrac{1}{2}{{a}^{2}}\ln a-\dfrac{3{{a}^{2}}}{4} \\
\end{aligned}$
${{I}_{2}}=\int\limits_{0}^{a}{x\text{d}x}=\dfrac{{{a}^{2}}}{2}$ $\Rightarrow \int\limits_{0}^{a}{f\left( x \right)\text{d}x}=2{{a}^{2}}\ln 2a-\dfrac{1}{2}{{a}^{2}}\ln a-\dfrac{3{{a}^{2}}}{4}-\dfrac{{{a}^{2}}}{2}=2{{a}^{2}}\ln 2a-\dfrac{1}{2}{{a}^{2}}\ln a-\dfrac{5{{a}^{2}}}{4}$
Theo giả thiết $\int\limits_{0}^{a}{f\left( x \right)\text{d}x}=0$
$\begin{aligned}
& \Rightarrow 2{{a}^{2}}\ln 2a-\dfrac{1}{2}{{a}^{2}}\ln a-\dfrac{5{{a}^{2}}}{4}=0\Rightarrow 2\ln 2a-\dfrac{1}{2}\ln a=\dfrac{5}{4} \left( a>0 \right) \\
& \Rightarrow 8\ln 2a-2\ln a=5\Rightarrow \ln 256{{a}^{6}}=5\Rightarrow a=\sqrt[6]{\dfrac{{{e}^{5}}}{256}}\approx 0,58\in \left( 0,1 \right) \\
\end{aligned}$
Vì $f\left( 0 \right)=a\ln a\Rightarrow \left( 0+a \right)\ln \left( 0+a \right)-0+C=a\ln a\Rightarrow C=0$ $\Rightarrow f\left( x \right)=\left( x+a \right)\ln \left( x+a \right)-x$
$\int\limits_{0}^{a}{f\left( x \right)\text{d}x}=\int\limits_{0}^{a}{\left( \left( x+a \right)\ln \left( x+a \right)-x \right)\text{d}x}=\int\limits_{0}^{a}{\left( x+a \right)\ln \left( x+a \right)\text{d}x}-\int\limits_{0}^{a}{x\text{d}x}={{I}_{1}}-{{I}_{2}}$. $\begin{aligned}
& {{I}_{1}}=\int\limits_{0}^{a}{\left( x+a \right)\ln \left( x+a \right)\text{d}x}=\left. \left( \dfrac{{{x}^{2}}}{2}+a x \right).\ln \left( x+a \right) \right|_{0}^{a}-\int\limits_{0}^{a}{\left( \dfrac{{{x}^{2}}}{2}+a x \right).\dfrac{1}{x+a}dx} \\
& =\dfrac{3{{a}^{2}}}{2}\ln \left( 2a \right)-\int\limits_{0}^{a}{\left( \dfrac{1}{2}\left( x+a \right)-\dfrac{1}{2}{{a}^{2}}.\dfrac{1}{x+a} \right)dx}=\dfrac{3{{a}^{2}}}{2}\ln \left( 2a \right)-\left. \left[ \dfrac{1}{2}\left( \dfrac{{{x}^{2}}}{2}+a x \right)-\dfrac{1}{2}{{a}^{2}}.\ln \left( x+a \right) \right] \right|_{0}^{a} \\
& =2{{a}^{2}}\ln \left( 2a \right)-\dfrac{1}{2}{{a}^{2}}\ln a-\dfrac{3{{a}^{2}}}{4} \\
\end{aligned}$
${{I}_{2}}=\int\limits_{0}^{a}{x\text{d}x}=\dfrac{{{a}^{2}}}{2}$ $\Rightarrow \int\limits_{0}^{a}{f\left( x \right)\text{d}x}=2{{a}^{2}}\ln 2a-\dfrac{1}{2}{{a}^{2}}\ln a-\dfrac{3{{a}^{2}}}{4}-\dfrac{{{a}^{2}}}{2}=2{{a}^{2}}\ln 2a-\dfrac{1}{2}{{a}^{2}}\ln a-\dfrac{5{{a}^{2}}}{4}$
Theo giả thiết $\int\limits_{0}^{a}{f\left( x \right)\text{d}x}=0$
$\begin{aligned}
& \Rightarrow 2{{a}^{2}}\ln 2a-\dfrac{1}{2}{{a}^{2}}\ln a-\dfrac{5{{a}^{2}}}{4}=0\Rightarrow 2\ln 2a-\dfrac{1}{2}\ln a=\dfrac{5}{4} \left( a>0 \right) \\
& \Rightarrow 8\ln 2a-2\ln a=5\Rightarrow \ln 256{{a}^{6}}=5\Rightarrow a=\sqrt[6]{\dfrac{{{e}^{5}}}{256}}\approx 0,58\in \left( 0,1 \right) \\
\end{aligned}$
Đáp án B.