Cho hàm số f(x) thỏa mãn (f'(x))^2

* Hướng dẫn giải

Đáp án C

$\begin{array}{l}{\left(f\text{'}\left(x\right)\right)}^2+f\left(x\right).f\text{'}\text{'}\left(x\right)=x^3-2x\\ \iff \left[f\left(x\right).f\text{'}\left(x\right)\right]\text{'}=x^3-2x\\ \iff f\left(x\right).f\text{'}\left(x\right)=\frac{x^4}{4}-x^2+C\end{array}$

Ta có $f\left(0\right)=f\text{'}\left(0\right)=1$ nên C=1

$\Rightarrow f\left(x\right).f\text{'}\left(x\right)=\frac{x^4}{4}-x^2+1\iff \left[\frac{1}{2}{f}^2\left(x\right)\right]\text{'}=\frac{x^4}{4}-x^2+1\iff \frac{1}{2}{f}^2\left(x\right)=\frac{x^5}{20}-\frac{x^2}{3}+x+C_1$

Ta có f(0)=1 nên $C_1=\frac{1}{2}$

$\Rightarrow \frac{1}{2}{f}^2\left(x\right)=\frac{x^5}{20}-\frac{x^2}{3}+x+\frac{1}{2}\iff f^2\left(x\right)=\frac{x^5}{10}-\frac{2x^3}{3}+2x+1\Rightarrow f^2\left(2\right)=\frac{43}{15}$

Vậy $f^2\left(2\right)=\frac{43}{15}.$