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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm ${f}'\left( x \right)=\dfrac{2}{{{x}^{3}}}-1, \forall x\ne 0$ và $f\left( 1 \right)=0$. Biết $F\left( x \right)$ là nguyên hàm của $f\left( x \right)$ trên $\left( 0; +\infty \right)$ thỏa mãn $F\left( 1 \right)=\dfrac{3}{2}$, khi đó $F\left( 2 \right)$
A. $\dfrac{5}{2}$.
B. $-\dfrac{3}{4}$.
C. $\dfrac{7}{2}$.
D. $\dfrac{3}{2}$.
A. $\dfrac{5}{2}$.
B. $-\dfrac{3}{4}$.
C. $\dfrac{7}{2}$.
D. $\dfrac{3}{2}$.
Ta có $f\left( x \right)=\int{{f}'\left( x \right)\text{d}x}=\int{\left( \dfrac{2}{{{x}^{3}}}-1 \right)\text{d}x}=-\dfrac{1}{{{x}^{2}}}-x+C$
Mà $f\left( 1 \right)=0\Rightarrow C=2\Rightarrow f\left( x \right)=-\dfrac{1}{{{x}^{2}}}-x+2$.
Khi đó $F\left( x \right)=\int{\left( -\dfrac{1}{{{x}^{2}}}-x+2 \right)\text{d}x}=\dfrac{1}{x}-\dfrac{{{x}^{2}}}{2}+2x+{C}'$.
Mặt khác $F\left( 1 \right)=\dfrac{3}{2}\Rightarrow {C}'=-1\Rightarrow F\left( x \right)=\dfrac{1}{x}-\dfrac{{{x}^{2}}}{2}+2x-1$.
Vậy $F\left( 2 \right)=\dfrac{3}{2}$.
Mà $f\left( 1 \right)=0\Rightarrow C=2\Rightarrow f\left( x \right)=-\dfrac{1}{{{x}^{2}}}-x+2$.
Khi đó $F\left( x \right)=\int{\left( -\dfrac{1}{{{x}^{2}}}-x+2 \right)\text{d}x}=\dfrac{1}{x}-\dfrac{{{x}^{2}}}{2}+2x+{C}'$.
Mặt khác $F\left( 1 \right)=\dfrac{3}{2}\Rightarrow {C}'=-1\Rightarrow F\left( x \right)=\dfrac{1}{x}-\dfrac{{{x}^{2}}}{2}+2x-1$.
Vậy $F\left( 2 \right)=\dfrac{3}{2}$.
Đáp án D.