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Câu hỏi: Biết rằng $x{{\operatorname{e}}^{x}}$ là một nguyên hàm của $f\left( -x \right)$ trên khoảng $\left( -\infty ;+\infty \right)$. Gọi $F\left( x \right)$ là một nguyên hàm của ${f}'\left( x \right){{\operatorname{e}}^{x}}$ thỏa mãn $F\left( 0 \right)=1$, giá trị của $F\left( -1 \right)$ bằng
A. $\dfrac{7}{2}$.
B. $\dfrac{5-\operatorname{e}}{2}$.
C. $\dfrac{7-\operatorname{e}}{2}$.
D. $\dfrac{5}{2}$.
A. $\dfrac{7}{2}$.
B. $\dfrac{5-\operatorname{e}}{2}$.
C. $\dfrac{7-\operatorname{e}}{2}$.
D. $\dfrac{5}{2}$.
Ta có $f\left( -x \right)={{\left( x{{\operatorname{e}}^{x}} \right)}^{\prime }}={{\operatorname{e}}^{x}}+x{{\operatorname{e}}^{x}}$, $\forall x\in \left( -\infty ;+\infty \right)$.
Do đó $f\left( -x \right)={{\operatorname{e}}^{-\left( -x \right)}}-\left( -x \right){{\operatorname{e}}^{-\left( -x \right)}}$, $\forall x\in \left( -\infty ;+\infty \right)$.
Suy ra $f\left( x \right)={{\operatorname{e}}^{-x}}\left( 1-x \right)$, $\forall x\in \left( -\infty ;+\infty \right)$.
Nên ${f}'\left( x \right)={{\left[ {{\operatorname{e}}^{-x}}\left( 1-x \right) \right]}^{\prime }}={{\operatorname{e}}^{-x}}\left( x-2 \right)$ $\Rightarrow {f}'\left( x \right){{\operatorname{e}}^{x}}={{\operatorname{e}}^{-x}}\left( x-2 \right).{{\operatorname{e}}^{x}}=x-2$.
Bởi vậy $F\left( x \right)=\int{\left( x-2 \right)\operatorname{d}x}=\dfrac{1}{2}{{\left( x-2 \right)}^{2}}+C$.
Từ đó $F\left( 0 \right)=\dfrac{1}{2}{{\left( 0-2 \right)}^{2}}+C=C+2$ ; $F\left( 0 \right)=1\Rightarrow C=-1$.
Vậy $F\left( x \right)=\dfrac{1}{2}{{\left( x-2 \right)}^{2}}-1\Rightarrow F\left( -1 \right)=\dfrac{1}{2}{{\left( -1-2 \right)}^{2}}-1=\dfrac{7}{2}$.
Do đó $f\left( -x \right)={{\operatorname{e}}^{-\left( -x \right)}}-\left( -x \right){{\operatorname{e}}^{-\left( -x \right)}}$, $\forall x\in \left( -\infty ;+\infty \right)$.
Suy ra $f\left( x \right)={{\operatorname{e}}^{-x}}\left( 1-x \right)$, $\forall x\in \left( -\infty ;+\infty \right)$.
Nên ${f}'\left( x \right)={{\left[ {{\operatorname{e}}^{-x}}\left( 1-x \right) \right]}^{\prime }}={{\operatorname{e}}^{-x}}\left( x-2 \right)$ $\Rightarrow {f}'\left( x \right){{\operatorname{e}}^{x}}={{\operatorname{e}}^{-x}}\left( x-2 \right).{{\operatorname{e}}^{x}}=x-2$.
Bởi vậy $F\left( x \right)=\int{\left( x-2 \right)\operatorname{d}x}=\dfrac{1}{2}{{\left( x-2 \right)}^{2}}+C$.
Từ đó $F\left( 0 \right)=\dfrac{1}{2}{{\left( 0-2 \right)}^{2}}+C=C+2$ ; $F\left( 0 \right)=1\Rightarrow C=-1$.
Vậy $F\left( x \right)=\dfrac{1}{2}{{\left( x-2 \right)}^{2}}-1\Rightarrow F\left( -1 \right)=\dfrac{1}{2}{{\left( -1-2 \right)}^{2}}-1=\dfrac{7}{2}$.
Đáp án A.