Lời giải

\( f'(x) f(x) = x^4 + x^2 \)

\( \Rightarrow \int f'(x) f(x) \, dx = \frac{x^5}{5} + \frac{x^3}{3} + C \)

\( \Rightarrow \frac{(f(x))^2}{2} = \frac{x^5}{5} + \frac{x^3}{3} + C \)

\( f(0) = 2  \Rightarrow C = 2. \)

 \( \Rightarrow (f(x))^2 = \frac{2x^5}{5} + \frac{2x^3}{3} + 4 \)

 \( (f(2))^2 = \frac{332}{15} \quad \Rightarrow \boxed{A} \)

page1

Lời giải

\( f'(x) = -e^x f^2(x)\Leftrightarrow  \frac{f'(x)}{f^2(x)} = -e^x \)  

\(\Rightarrow \int \frac{f'(x)}{f^2(x)} \, dx = \int -e^x \, dx  \Rightarrow -\frac{1}{f(x)} = -e^x + C \)  

\( \Leftrightarrow f(x) = \frac{1}{e^x + C}. \)  

 \( f(0) = \frac{1}{2} \Leftrightarrow \frac{1}{1 + C} = \frac{1}{2} \Rightarrow C = 1\)  

\( f(x) = \frac{1}{e^x + 1} \Rightarrow f(\ln 2) = \frac{1}{3} \Rightarrow \boxed{C} \)

page2

Lời giải

•  \( \frac{f'(x)}{f(x)} = 2 - 2x  \Rightarrow \ln f(x) = 2x - x^2 + C \)  

•  \( f(0) = 1 \Rightarrow C = 0 \)  

 \( \ln |f(x)| = 2x - x^2 \quad \Rightarrow f(x) = e^{2x - x^2}\)  

\( f'(x) = (2 - 2x) e^{2x - x^2} = 0 \Leftrightarrow  x = 1 \)

\(\begin{array}{c|c c c}
x &  & 1 & \\
\hline
f'(x) & + & 0 & - \\
\hline
f(x) & 0 \nearrow & e & \searrow 0 \\
\end{array}\)

Phương trình \( f(x) = m \) có 2 nghiệm \( \Leftrightarrow 0 < m < e \Rightarrow \boxed{B}\)

page 3

Lời giải

•  \(\frac{ f''(x)}{\left[f'(x)\right]^2} = 1  \implies -\frac{1}{f'(x)} = x + c \implies f'(x) = \frac{-1}{x + c} \)

•   \( f'(0) = -1 \implies c = 1: f'(x) = \frac{-1}{x + 1} \)

 \(\implies f(x) = -\ln|x+1|  + c \)

 \( \implies f(1) - f(0) =  -\ln(2) = -0.6931 \in (-1, 0) \) =>  \( \boxed{B} \)

page 4

Lời giải

\( \int_0^1 f(x) \, dx = \int_0^1 6x^2 f(x^3) \, dx - \int_0^1 \frac{6}{\sqrt{3x+1}} \, dx \).  

\( \int_0^1 f(x) \, dx = 2 \int_0^1 f(t) \, dt - 4\sqrt{3x+1}\Big|_0^1 \) 

\( \left(t = x^3 \quad vì  \quad (\sqrt{3x+1})' =  \frac{3}{2\sqrt{3x+1}} \right) \)

\( \Rightarrow \int_0^1 f(x) \, dx =4\sqrt{3x+1}\Big|_0^1= 4 \Rightarrow \boxed{B} \)

page 5