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Câu hỏi: Cho hàm số $f\left( x \right)$ nhận giá trị dương và có đạo hàm cấp một không âm trên $\left( 0;+\infty \right)$ đồng thời thỏa mãn: $\dfrac{3}{{{x}^{2}}}f\left( x \right)f'\left( x \right){{\left[ xf\left( x \right) \right]}^{\prime }}+\dfrac{1}{{{x}^{3}}}\ln \left( 1+\dfrac{xf'\left( x \right)}{f\left( x \right)} \right)+{{\left[ f'\left( x \right) \right]}^{3}}=0,\forall x>0.$ Giá trị của $P=2019+2020.f'\left( 2021 \right)$ là
A. $P=2020$
B. $P=2019$
C. $P=2021$
D. $P=0$
A. $P=2020$
B. $P=2019$
C. $P=2021$
D. $P=0$
$\dfrac{3}{{{x}^{2}}}f\left( x \right)f'\left( x \right){{\left[ xf\left( x \right) \right]}^{\prime }}+\dfrac{1}{{{x}^{3}}}\ln \left( 1+\dfrac{xf'\left( x \right)}{f\left( x \right)} \right)+{{\left[ f'\left( x \right) \right]}^{3}}=0$
$\Leftrightarrow \dfrac{3}{{{x}^{2}}}f\left( x \right)f'\left( x \right)\left[ f\left( x \right)+xf'\left( x \right) \right]+\dfrac{1}{{{x}^{3}}}\ln \left( 1+\dfrac{xf'\left( x \right)}{f\left( x \right)} \right)+{{\left[ f'\left( x \right) \right]}^{3}}=0$
Do:
$f\left( x \right)>0,f'\left( x \right)\ge 0\forall x>0$
+) $f\left( x \right)+xf'\left( x \right)\ge f\left( x \right)\Rightarrow f\left( x \right)+x.f'\left( x \right)>0$
Nên ta có: $\dfrac{3}{{{x}^{2}}}.f\left( x \right)f'\left( x \right)\left[ f\left( x \right)+xf'\left( x \right) \right]\ge 0$
+) $\ln \left( 1+\dfrac{xf'\left( x \right)}{f\left( x \right)} \right)\ge \ln 1\Rightarrow \ln \left( 1+\dfrac{xf'\left( x \right)}{f\left( x \right)} \right)\ge 0$
+) ${{\left[ f'\left( x \right) \right]}^{3}}\ge 0$
Suy ra: $\dfrac{3}{{{x}^{2}}}f\left( x \right)f'\left( x \right)\left[ f\left( x \right)+xf'\left( x \right) \right]+\dfrac{1}{{{x}^{3}}}\ln \left( 1+\dfrac{xf'\left( x \right)}{f\left( x \right)} \right)+{{\left[ f'\left( x \right) \right]}^{3}}\ge 0\forall x>0$
Dấu bằng xảy ra $\Leftrightarrow f'\left( x \right)=0\forall x>0\Rightarrow f'\left( 2021 \right)=0$
Do đó: $P=2019+2020f'\left( 2021 \right)=2019$
$\Leftrightarrow \dfrac{3}{{{x}^{2}}}f\left( x \right)f'\left( x \right)\left[ f\left( x \right)+xf'\left( x \right) \right]+\dfrac{1}{{{x}^{3}}}\ln \left( 1+\dfrac{xf'\left( x \right)}{f\left( x \right)} \right)+{{\left[ f'\left( x \right) \right]}^{3}}=0$
Do:
$f\left( x \right)>0,f'\left( x \right)\ge 0\forall x>0$
+) $f\left( x \right)+xf'\left( x \right)\ge f\left( x \right)\Rightarrow f\left( x \right)+x.f'\left( x \right)>0$
Nên ta có: $\dfrac{3}{{{x}^{2}}}.f\left( x \right)f'\left( x \right)\left[ f\left( x \right)+xf'\left( x \right) \right]\ge 0$
+) $\ln \left( 1+\dfrac{xf'\left( x \right)}{f\left( x \right)} \right)\ge \ln 1\Rightarrow \ln \left( 1+\dfrac{xf'\left( x \right)}{f\left( x \right)} \right)\ge 0$
+) ${{\left[ f'\left( x \right) \right]}^{3}}\ge 0$
Suy ra: $\dfrac{3}{{{x}^{2}}}f\left( x \right)f'\left( x \right)\left[ f\left( x \right)+xf'\left( x \right) \right]+\dfrac{1}{{{x}^{3}}}\ln \left( 1+\dfrac{xf'\left( x \right)}{f\left( x \right)} \right)+{{\left[ f'\left( x \right) \right]}^{3}}\ge 0\forall x>0$
Dấu bằng xảy ra $\Leftrightarrow f'\left( x \right)=0\forall x>0\Rightarrow f'\left( 2021 \right)=0$
Do đó: $P=2019+2020f'\left( 2021 \right)=2019$
Đáp án B.