Google
Câu hỏi: Cho hàm số f(x) thỏa mãn ${{\left[ {f}'\left( x \right) \right]}^{3}}+{{x}^{2}}.{f}'\left( x \right)=2{{x}^{3}}+4{{x}^{2}}+3x+1,\forall x\in \mathbb{R}$ và $f\left( 0 \right)=2.$ Tích phân $\int\limits_{0}^{6}{f\left( x \right)dx}$ bằng
A. 26.
B. 66.
C. 42.
D. 102.
A. 26.
B. 66.
C. 42.
D. 102.
Ta có ${{\left[ {f}'\left( x \right) \right]}^{3}}+{{x}^{2}}.{f}'\left( x \right)=2{{x}^{3}}+4{{x}^{2}}+3x+1={{\left( x+1 \right)}^{3}}+{{x}^{2}}\left( x+1 \right)$
$\Rightarrow {{\left[ {f}'\left( x \right) \right]}^{3}}-{{\left( x+1 \right)}^{3}}+{{x}^{2}}\left[ {f}'\left( x \right)-x-1 \right]=0$
$\Rightarrow \left[ {f}'\left( x \right)-x-1 \right].\left[ {{\left( {f}'\left( x \right) \right)}^{2}}+\left( x+1 \right).{f}'\left( x \right).{{\left( x+1 \right)}^{2}}+{{x}^{2}} \right]=0$ (1)
Lại có ${{\left( {f}'\left( x \right) \right)}^{2}}+\left( x+1 \right).{f}'\left( x \right).{{\left( x+1 \right)}^{2}}+{{x}^{2}}={{\left[ {f}'\left( x \right)+\dfrac{x+1}{2} \right]}^{2}}+\dfrac{3}{4}{{\left( x+1 \right)}^{2}}+{{x}^{2}}\ge 0,\forall x\in \mathbb{R}.$
Dấu "=" xảy ra $\Leftrightarrow {{\left[ {f}'\left( x \right)+\dfrac{x+1}{2} \right]}^{2}}=\dfrac{3}{4}{{\left( x+1 \right)}^{2}}={{x}^{2}}=0.$
Đây là điều kiện vô lý nên dấu "=" không xảy ra $\Rightarrow {{\left[ {f}'\left( x \right)+\dfrac{x+1}{2} \right]}^{2}}+\dfrac{3}{4}{{\left( x+1 \right)}^{2}}+{{x}^{2}}>0,\forall x\in \mathbb{R}$
Do đó (1) $\Leftrightarrow {f}'\left( x \right)=x+1\Rightarrow f\left( x \right)=\int{\left( x+1 \right)dx}=\dfrac{{{x}^{2}}}{2}+x+C.$
Mà $f\left( 0 \right)=2\Rightarrow C=2\Rightarrow f\left( x \right)=\dfrac{{{x}^{2}}}{2}+x+2\Rightarrow \int\limits_{0}^{6}{f\left( x \right)dx}=\left. \left( \dfrac{{{x}^{3}}}{6}+\dfrac{{{x}^{2}}}{2}+2x \right) \right|_{0}^{6}=66.$
$\Rightarrow {{\left[ {f}'\left( x \right) \right]}^{3}}-{{\left( x+1 \right)}^{3}}+{{x}^{2}}\left[ {f}'\left( x \right)-x-1 \right]=0$
$\Rightarrow \left[ {f}'\left( x \right)-x-1 \right].\left[ {{\left( {f}'\left( x \right) \right)}^{2}}+\left( x+1 \right).{f}'\left( x \right).{{\left( x+1 \right)}^{2}}+{{x}^{2}} \right]=0$ (1)
Lại có ${{\left( {f}'\left( x \right) \right)}^{2}}+\left( x+1 \right).{f}'\left( x \right).{{\left( x+1 \right)}^{2}}+{{x}^{2}}={{\left[ {f}'\left( x \right)+\dfrac{x+1}{2} \right]}^{2}}+\dfrac{3}{4}{{\left( x+1 \right)}^{2}}+{{x}^{2}}\ge 0,\forall x\in \mathbb{R}.$
Dấu "=" xảy ra $\Leftrightarrow {{\left[ {f}'\left( x \right)+\dfrac{x+1}{2} \right]}^{2}}=\dfrac{3}{4}{{\left( x+1 \right)}^{2}}={{x}^{2}}=0.$
Đây là điều kiện vô lý nên dấu "=" không xảy ra $\Rightarrow {{\left[ {f}'\left( x \right)+\dfrac{x+1}{2} \right]}^{2}}+\dfrac{3}{4}{{\left( x+1 \right)}^{2}}+{{x}^{2}}>0,\forall x\in \mathbb{R}$
Do đó (1) $\Leftrightarrow {f}'\left( x \right)=x+1\Rightarrow f\left( x \right)=\int{\left( x+1 \right)dx}=\dfrac{{{x}^{2}}}{2}+x+C.$
Mà $f\left( 0 \right)=2\Rightarrow C=2\Rightarrow f\left( x \right)=\dfrac{{{x}^{2}}}{2}+x+2\Rightarrow \int\limits_{0}^{6}{f\left( x \right)dx}=\left. \left( \dfrac{{{x}^{3}}}{6}+\dfrac{{{x}^{2}}}{2}+2x \right) \right|_{0}^{6}=66.$
Đáp án B.