Cho hàm số y = f(x) có đạo hàm liên tục trên [0; 1] thỏa mãn

* Hướng dẫn giải

Chọn C.

Cách 1:

Ta có

${\left[f\text{'}\left(x\right)\right]}^2=4\left[2x^2+1-f\left(x\right)\right]\iff {\left[f\text{'}\left(x\right)\right]}^2+4f\left(x\right)=4\left(2x^2+1\right)$

$\Rightarrow \underset{0}{\overset{1}{\int }}{\left[f\text{'}\left(x\right)\right]}^{2}dx+\underset{0}{\overset{1}{\int }}4f\left(x\right)dx=\underset{0}{\overset{1}{\int }}4\left(2x^2+1\right)dx\iff \underset{0}{\overset{1}{\int }}{\left[f\text{'}\left(x\right)\right]}^{2}dx+4xf\left(x\right)\left|\begin{array}{l}1\\ 0\end{array}\right.-4\underset{0}{\overset{1}{\int }}xf\text{'}\left(x\right)dx=\frac{20}{3}$

$\iff \underset{0}{\overset{1}{\int }}{\left[f\text{'}\left(x\right)\right]}^{2}dx-4\underset{0}{\overset{1}{\int }}xf\text{'}\left(x\right)dx+4f\left(1\right)=\frac{20}{3}$

$\iff \underset{0}{\overset{1}{\int }}{\left[f\text{'}\left(x\right)\right]}^{2}dx-4\underset{0}{\overset{1}{\int }}xf\text{'}\left(x\right)dx+4\underset{0}{\overset{1}{\int }}{x}^{2}dx=\frac{20}{3}-8+4\underset{0}{\overset{1}{\int }}{x}^{2}dx$

$\iff \underset{0}{\overset{1}{\int }}{\left[f\text{'}\left(x\right)-2x\right]}^{2}dx=0\iff f\text{'}\left(x\right)-2x=0\Rightarrow f\left(x\right)=x^2+C.$

$f\left(1\right)=2\iff C=1\iff f\left(x\right)=x^2+1.$

Vậy $I=\underset{0}{\overset{1}{\int }}xf\left(x\right)dx=\frac{3}{4}.$

Cách 2:

Đặt $f\left(x\right)=a{x}^2+bx+c,$ ta có:

${\left[f\text{'}\left(x\right)\right]}^2=4\left[2x^2+1-f\left(x\right)\right]\iff {\left(2ax+b\right)}^2=4\left(2x^2+1-a{x}^2-bx-c\right)\iff \left\{\begin{array}{l}4a^2=4\left(2-a\right)\\ 4ab=-4b\\ b^2=4\left(1-c\right)\end{array}\right..$

Kết hợp với điều kiện $f\left(1\right)=2\iff a+b+c=2$ ta có nghiệm $\left\{\begin{array}{l}a=1\\ b=0\\ c=1\end{array}\right..$ Vậy $f\left(x\right)=x^2+1.$