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Câu hỏi: Gọi F(x) là nguyên hàm trên $\mathbb{R}$ của hàm số $f\left( x \right)={{x}^{2}}{{e}^{ax}}\left( a\ne 0 \right)$, sao cho $F\left( \dfrac{1}{a} \right)=F\left( 0 \right)+1.$ Chọn mệnh đề đúng trong các mệnh đề sau:
A. $0<a\le 1.$
B. $a<-2.$
C. $a\ge 3.$
D. $1<a<2.$
A. $0<a\le 1.$
B. $a<-2.$
C. $a\ge 3.$
D. $1<a<2.$
$F\left( x \right)=\int{{{x}^{2}}{{e}^{ax}}dx}.$
Đặt $\left\{ \begin{aligned}
& u={{x}^{2}} \\
& dv={{e}^{ax}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=2xdx \\
& v=\dfrac{1}{a}{{e}^{ax}} \\
\end{aligned} \right..$
$\Rightarrow F\left( x \right)=\dfrac{1}{a}{{x}^{2}}{{e}^{ax}}-\dfrac{2}{a}\int{x{{e}^{ax}}dx}=\dfrac{1}{a}{{x}^{2}}{{e}^{ax}}-\dfrac{2}{a}.A \left( 1 \right)$
Xét $A=\int{x{{e}^{ax}}dx}.$ Đặt $\left\{ \begin{aligned}
& u=x \\
& dv={{e}^{ax}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=dx \\
& v=\dfrac{1}{a}{{e}^{ax}} \\
\end{aligned} \right..$
$\Rightarrow A=\dfrac{1}{a}x{{e}^{ax}}-\dfrac{1}{a}\int{{{e}^{ax}}dx} \left( 2 \right)$
Từ (1) và (2) suy ra $F\left( x \right)=\dfrac{1}{a}{{x}^{2}}{{e}^{ax}}-\dfrac{2}{{{a}^{2}}}x{{e}^{ax}}+\dfrac{2}{{{a}^{2}}}\int{{{e}^{ax}}dx}=\dfrac{1}{a}{{x}^{2}}{{e}^{ax}}-\dfrac{2}{{{a}^{2}}}x{{e}^{ax}}+\dfrac{2}{{{a}^{3}}}{{e}^{ax}}+C.$
Mà $F\left( \dfrac{1}{a} \right)=F\left( 0 \right)+1\Rightarrow \dfrac{1}{{{a}^{3}}}e-\dfrac{2}{{{a}^{3}}}e+\dfrac{2}{{{a}^{3}}}e+C=\dfrac{2}{{{a}^{3}}}+1+C$
$\Rightarrow {{a}^{3}}=e-2\Rightarrow a=\sqrt[3]{e-2}\Rightarrow 0<a\le 1.$
Đặt $\left\{ \begin{aligned}
& u={{x}^{2}} \\
& dv={{e}^{ax}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=2xdx \\
& v=\dfrac{1}{a}{{e}^{ax}} \\
\end{aligned} \right..$
$\Rightarrow F\left( x \right)=\dfrac{1}{a}{{x}^{2}}{{e}^{ax}}-\dfrac{2}{a}\int{x{{e}^{ax}}dx}=\dfrac{1}{a}{{x}^{2}}{{e}^{ax}}-\dfrac{2}{a}.A \left( 1 \right)$
Xét $A=\int{x{{e}^{ax}}dx}.$ Đặt $\left\{ \begin{aligned}
& u=x \\
& dv={{e}^{ax}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=dx \\
& v=\dfrac{1}{a}{{e}^{ax}} \\
\end{aligned} \right..$
$\Rightarrow A=\dfrac{1}{a}x{{e}^{ax}}-\dfrac{1}{a}\int{{{e}^{ax}}dx} \left( 2 \right)$
Từ (1) và (2) suy ra $F\left( x \right)=\dfrac{1}{a}{{x}^{2}}{{e}^{ax}}-\dfrac{2}{{{a}^{2}}}x{{e}^{ax}}+\dfrac{2}{{{a}^{2}}}\int{{{e}^{ax}}dx}=\dfrac{1}{a}{{x}^{2}}{{e}^{ax}}-\dfrac{2}{{{a}^{2}}}x{{e}^{ax}}+\dfrac{2}{{{a}^{3}}}{{e}^{ax}}+C.$
Mà $F\left( \dfrac{1}{a} \right)=F\left( 0 \right)+1\Rightarrow \dfrac{1}{{{a}^{3}}}e-\dfrac{2}{{{a}^{3}}}e+\dfrac{2}{{{a}^{3}}}e+C=\dfrac{2}{{{a}^{3}}}+1+C$
$\Rightarrow {{a}^{3}}=e-2\Rightarrow a=\sqrt[3]{e-2}\Rightarrow 0<a\le 1.$
Đáp án A.