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Câu hỏi: Cho hàm số bậc ba $y=f\left( x \right)$ có đồ thị là đường cong trong hình dưới. Biết hàm số $f\left( x \right)$ đạt cực trị tại hai điểm ${{x}_{1}}, {{x}_{2}}$ thỏa mãn ${{x}_{2}}={{x}_{1}}+2$ ; $f\left( {{x}_{1}} \right)+f\left( {{x}_{2}} \right)=0$ và $\int\limits_{{{x}_{1}}}^{{{x}_{1}}+1}{f\left( x \right)\text{d}x}=\dfrac{5}{4}$. Tính $L=\underset{x\to {{x}_{1}}}{\mathop{\lim }} \dfrac{ f\left( x \right)-2 }{{{\left( x-{{x}_{1}} \right)}^{2}}}$.
A. $-1$.
B. $-2$.
C. $-3$.
D. $-4$.
A. $-1$.
B. $-2$.
C. $-3$.
D. $-4$.
Giả sử $f\left( x \right)=a{{x}^{3}}+b{{x}^{2}}+cx+d$ $\left( a\ne 0 \right)$.
Có ${f}'\left( x \right)=3a{{x}^{2}}+2bx+c=0\Leftrightarrow \left[ \begin{aligned}
& x={{x}_{1}} \\
& x={{x}_{2}}={{x}_{1}}+2 \\
\end{aligned} \right.$.
Suy ra: ${f}'\left( x \right)=3a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)$
$\Rightarrow {f}'\left( x \right)=3a\left( x-{{x}_{1}} \right)\left( x-{{x}_{1}}-2 \right)$
$\Rightarrow {f}'\left( x \right)=3a{{\left( x-{{x}_{1}} \right)}^{2}}-6a\left( x-{{x}_{1}} \right)$.
Lấy nguyên hàm hai vế ta có:
$f\left( x \right)=a{{\left( x-{{x}_{1}} \right)}^{3}}-3a{{\left( x-{{x}_{1}} \right)}^{2}}+C$.
Khi đó $f\left( {{x}_{1}} \right)=C$ và $ f\left( {{x}_{2}} \right)=a{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{3}}-3a{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+C=8a-12a+C=C-4a$.
Mà $f\left( {{x}_{1}} \right)+ f\left( {{x}_{2}} \right)=0$, nên $C+C-4a=0$ $\Leftrightarrow C=2a$.
Suy ra $f\left( x \right)=a{{\left( x-{{x}_{1}} \right)}^{3}}-3a{{\left( x-{{x}_{1}} \right)}^{2}}+2a$.
Mặt khác $\int\limits_{{{x}_{1}}}^{{{x}_{1}}+1}{f\left( x \right)\text{d}x=\dfrac{5}{4}} \Leftrightarrow \int\limits_{{{x}_{1}}}^{{{x}_{1}}+1}{\left[ a{{\left( x-{{x}_{1}} \right)}^{3}}-3a{{\left( x-{{x}_{1}} \right)}^{2}}+2a \right]\text{d}x=\dfrac{5}{4}}$
$\Leftrightarrow \left. \left[ \dfrac{a}{4}{{\left( x-{{x}_{1}} \right)}^{4}}-a{{\left( x-{{x}_{1}} \right)}^{3}}+2ax \right]_{{}}^{{}} \right|_{ {{x}_{1}}}^{ {{x}_{1}}+1}=\dfrac{5}{4}$ $\Leftrightarrow \left[ \dfrac{a}{4}-a+2a\left( {{x}_{1}}+1 \right) \right]-2a{{x}_{1}}=\dfrac{5}{4}$ $\Leftrightarrow a=1$.
Do đó: $f\left( x \right)={{\left( x-{{x}_{1}} \right)}^{3}}-3{{\left( x-{{x}_{1}} \right)}^{2}}+2$.
Vậy $L=\underset{x\to {{x}_{1}}}{\mathop{\lim }} \dfrac{f\left( x \right)-2}{{{\left( x-{{x}_{1}} \right)}^{2}}}=\underset{x\to {{x}_{1}}}{\mathop{\lim }} \dfrac{{{\left( x-{{x}_{1}} \right)}^{3}}-3{{\left( x-{{x}_{1}} \right)}^{2}}}{{{\left( x-{{x}_{1}} \right)}^{2}}}=\underset{x\to {{x}_{1}}}{\mathop{\lim }} \left[ \left( x-{{x}_{1}} \right)-3 \right]=- 3$.
Có ${f}'\left( x \right)=3a{{x}^{2}}+2bx+c=0\Leftrightarrow \left[ \begin{aligned}
& x={{x}_{1}} \\
& x={{x}_{2}}={{x}_{1}}+2 \\
\end{aligned} \right.$.
Suy ra: ${f}'\left( x \right)=3a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)$
$\Rightarrow {f}'\left( x \right)=3a\left( x-{{x}_{1}} \right)\left( x-{{x}_{1}}-2 \right)$
$\Rightarrow {f}'\left( x \right)=3a{{\left( x-{{x}_{1}} \right)}^{2}}-6a\left( x-{{x}_{1}} \right)$.
Lấy nguyên hàm hai vế ta có:
$f\left( x \right)=a{{\left( x-{{x}_{1}} \right)}^{3}}-3a{{\left( x-{{x}_{1}} \right)}^{2}}+C$.
Khi đó $f\left( {{x}_{1}} \right)=C$ và $ f\left( {{x}_{2}} \right)=a{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{3}}-3a{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+C=8a-12a+C=C-4a$.
Mà $f\left( {{x}_{1}} \right)+ f\left( {{x}_{2}} \right)=0$, nên $C+C-4a=0$ $\Leftrightarrow C=2a$.
Suy ra $f\left( x \right)=a{{\left( x-{{x}_{1}} \right)}^{3}}-3a{{\left( x-{{x}_{1}} \right)}^{2}}+2a$.
Mặt khác $\int\limits_{{{x}_{1}}}^{{{x}_{1}}+1}{f\left( x \right)\text{d}x=\dfrac{5}{4}} \Leftrightarrow \int\limits_{{{x}_{1}}}^{{{x}_{1}}+1}{\left[ a{{\left( x-{{x}_{1}} \right)}^{3}}-3a{{\left( x-{{x}_{1}} \right)}^{2}}+2a \right]\text{d}x=\dfrac{5}{4}}$
$\Leftrightarrow \left. \left[ \dfrac{a}{4}{{\left( x-{{x}_{1}} \right)}^{4}}-a{{\left( x-{{x}_{1}} \right)}^{3}}+2ax \right]_{{}}^{{}} \right|_{ {{x}_{1}}}^{ {{x}_{1}}+1}=\dfrac{5}{4}$ $\Leftrightarrow \left[ \dfrac{a}{4}-a+2a\left( {{x}_{1}}+1 \right) \right]-2a{{x}_{1}}=\dfrac{5}{4}$ $\Leftrightarrow a=1$.
Do đó: $f\left( x \right)={{\left( x-{{x}_{1}} \right)}^{3}}-3{{\left( x-{{x}_{1}} \right)}^{2}}+2$.
Vậy $L=\underset{x\to {{x}_{1}}}{\mathop{\lim }} \dfrac{f\left( x \right)-2}{{{\left( x-{{x}_{1}} \right)}^{2}}}=\underset{x\to {{x}_{1}}}{\mathop{\lim }} \dfrac{{{\left( x-{{x}_{1}} \right)}^{3}}-3{{\left( x-{{x}_{1}} \right)}^{2}}}{{{\left( x-{{x}_{1}} \right)}^{2}}}=\underset{x\to {{x}_{1}}}{\mathop{\lim }} \left[ \left( x-{{x}_{1}} \right)-3 \right]=- 3$.
Đáp án C.