Cho hàm số f(x) liên tục trên đoạn [0;4] thỏa mãn f''(x)f(x) + [f(x)]^2/ căn bậc 2 của ( 2x + 1)^3 = [f'(x)]^2 và f(x) lớn hơn 0 với mọi x thuộc 0;4. Biết rằng f'(0) = f(0) = 1 giá...

Đáp án A

Ta có: $f\text{'}\text{'}\left(x\right)f\left(x\right)+\frac{{\left[f\left(x\right)\right]}^2}{\sqrt{{\left(2\text{x}+1\right)}^3}}={\left[f\text{'}\left(x\right)\right]}^2\iff f\text{'}\text{'}\left(x\right)f\left(x\right)-{\left[f\text{'}\left(x\right)\right]}^2=-\frac{{\left[f\left(x\right)\right]}^2}{\sqrt{{\left(2\text{x}+1\right)}^3}}$

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

$\iff \frac{f\text{'}\left(x\right)}{f\left(x\right)}=-\int \frac{1}{\sqrt{{\left(2\text{x}+1\right)}^3}}dx\iff \frac{f\text{'}\left(x\right)}{f\left(x\right)}=-\int {\left(2\text{x}+1\right)}^{\frac{-3}{2}}dx\iff \frac{f\text{'}\left(x\right)}{f\left(x\right)}=\frac{1}{\sqrt{2\text{x}+1}}+C_1$

Thay $x=0$ ta được $C_1=0$

$\Rightarrow \frac{f\text{'}\left(x\right)}{f\left(x\right)}=\frac{1}{\sqrt{2\text{x}+1}}\Rightarrow \underset{}{\overset{}{\int }}\frac{f\text{'}\left(x\right)}{f\left(x\right)}dx=\int \frac{dx}{\sqrt{2\text{x}+1}}\iff \ln \left[f\left(x\right)\right]=\sqrt{2\text{x}+1}+C_2$

Thay $x=0$ ta được $C_2=-1.$

$\Rightarrow \ln \left[f\left(x\right)\right]=\sqrt{2\text{x}+1}-1$

Thay $x=4$ ta được $\ln \left[f\left(4\right)\right]=2\Rightarrow f\left(4\right)=e^2.$