Cho hàm số y=f(x) thỏa mãn [f'(x)]^2+f(x).f"(x)=x^3-2x, với mọi x thuộc R và f(0)=f'(0)=2 Tính giá trị của T=f^2(2)

Đáp án A

Ta có: $VT=[f\left(x\right).f\text{'}\left(x\right)]\text{'}=f\text{'}\left(x\right).f\text{'}\left(x\right)+f\left(x\right).f\text{'}\text{'}\left(x\right)={[f\text{'}\left(x\right)]}^2+f\left(x\right).f\text{'}\text{'}\left(x\right)$

$\Rightarrow [f\text{'}\left(x\right).f\left(x\right)]\text{'}=x^3-2x(*)$

Nguyên hàm hai vế của (*) ta được: $f\text{'}\left(x\right).f\left(x\right)=\frac{x^4}{4}-x^2+C\left(1\right)$

Lại có: $f\text{'}\left(0\right)=f\left(0\right)=2\Rightarrow C=2.2=4\Rightarrow \left(1\right)\iff f\left(x\right).f\text{'}\left(x\right)=\frac{x^4}{4}-x^2+4$

$\begin{array}{l}\Rightarrow \int f\left(x\right)f\text{'}\left(x\right)dx=\int (\frac{x^4}{4}-x^2+4)dx\iff \int f\left(x\right)df\left(x\right)=\frac{x^5}{20}-\frac{x^3}{3}+4x+A\\ \iff \frac{f^2\left(x\right)}{2}=\frac{x^5}{20}-\frac{x^3}{3}+4x+A\iff f^2\left(x\right)=\frac{x^5}{10}-\frac{2x^3}{3}+8x+2A\end{array}$

Có $f\left(0\right)=2\Rightarrow 4=2A\iff A=2\Rightarrow f^2\left(x\right)=\frac{x^5}{10}-\frac{2x^3}{3}+8x+4$

$\Rightarrow f^2\left(x\right)=\frac{2^5}{10}-\frac{{2.2}^3}{3}+8.2+4=\frac{268}{15}$