Cho hàm số f(x) nhận giá trị dương và có đạo hàm cấp một không âm trên

* Hướng dẫn giải

$\frac{3}{x^2}f\left(x\right)f\text{'}\left(x\right){\left[xf\left(x\right)\right]}^{\text{'}}+\frac{1}{x^3}\ln \left(1+\frac{xf\text{'}\left(x\right)}{f\left(x\right)}\right)+{\left[f\text{'}\left(x\right)\right]}^3=0$

$\iff \frac{3}{x^2}f\left(x\right)f\text{'}\left(x\right)\left[f\left(x\right)+xf\text{'}\left(x\right)\right]+\frac{1}{x^3}\ln \left(1+\frac{xf\text{'}\left(x\right)}{f\left(x\right)}\right)+{\left[f\text{'}\left(x\right)\right]}^3=0$

Do:

$f\left(x\right)>0,f\text{'}\left(x\right)\ge 0\forall x>0$

+) $f\left(x\right)+xf\text{'}\left(x\right)\ge f\left(x\right)\Rightarrow f\left(x\right)+x.f\text{'}\left(x\right)>0$

Nên ta có: $\frac{3}{x^2}.f\left(x\right)f\text{'}\left(x\right)\left[f\left(x\right)+xf\text{'}\left(x\right)\right]\ge 0$

+) $\ln \left(1+\frac{xf\text{'}\left(x\right)}{f\left(x\right)}\right)\ge \ln 1\Rightarrow \ln \left(1+\frac{xf\text{'}\left(x\right)}{f\left(x\right)}\right)\ge 0$

+) ${\left[f\text{'}\left(x\right)\right]}^3\ge 0$

Suy ra: $\frac{3}{x^2}f\left(x\right)f\text{'}\left(x\right)\left[f\left(x\right)+xf\text{'}\left(x\right)\right]+\frac{1}{x^3}\ln \left(1+\frac{xf\text{'}\left(x\right)}{f\left(x\right)}\right)+{\left[f\text{'}\left(x\right)\right]}^3\ge 0\forall x>0$

Dấu bằng xảy ra $\iff f\text{'}\left(x\right)=0\forall x>0\Rightarrow f\text{'}\left(2021\right)=0$

Do đó: $P=2019+2020f\text{'}\left(2021\right)=2019$

Chọn B.