Đáp án

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

\(\left( \left( \frac{f(x)}{x} \right)' = \frac{x f'(x) - f(x)}{x^2}\right) \)

\(\implies \left( \frac{f(x)}{x} \right)' = e^x \implies \frac{f(x)}{x} = e^x + C \).

\( f(1) = e  \implies C = 0 \implies f(x) = x e^x \).

\( I =  \int_1^2 x e^x \, dx = e^2 \implies \boxed{C} \).

page26

Đáp án

\( (f'(x)f(x))' = (f'(x))^2 + f''(x)f(x) = 15x^4 + 12x \).

\( \implies  \int (f'(x)f(x))' dx = 3x^5 + 6x^2 + C \).

\( \implies f'(x)f(x) = 3x^5 + 6x^2 + C \).

\( f(0) = f'(0) = 1 \Rightarrow C = 1 \Rightarrow f'(x)f(x) = 3x^5 + 6x^2 + 1 \)

\( \Rightarrow \frac{(f(x))^2}{2} = \frac{x^6}{2} + 2x^3 + x + C \)

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

\( (f(x))^2 = x^6 + 4x^3 + 2x + 1 \Rightarrow (f(1))^2 = 8 \Rightarrow f(1) = 2 \Rightarrow \boxed{D} \)

page27

Đáp án

\( 3f'(x)  e^{f^3(x) - x^2 - 1} = \frac{2x}{f^2(x)} \)

\( \left(Gợi ý: f^3(x))' = 3f'(x) (f(x))^2 , \quad   (x^2 + 1)' = 2x \right)\)  

\( \iff 3f'(x) \cdot (f(x))^2 \cdot e^{f^3(x)} = 2x \cdot e^{x^2+1} \)

\( \iff e^{f^3(x)} = e^{x^2+1} + C \)

\( f(0) = 1 \Rightarrow C = 0 \)

Suy ra: \( (f(x))^3 = x^2 + 1  \implies f(x) = \sqrt[3]{x^2 + 1} \)

\( I = \int_{0}^{\sqrt{7}} x \sqrt[3]{x^2 + 1} \, dx = \frac{45}{8} \implies\boxed{C} \)

page28

Đáp án

\(\frac{f'(x) f(x)}{\sqrt{1 + (f(x))^2}} = 2x\)  
vì \((f(x))^2 + 1)' = 2f'(x)f(x)\)  

\(\Rightarrow \int \frac{f'(x) f(x)}{\sqrt{1 + (f(x))^2}} dx = x^2 + C \quad \left(\frac{1}{2} \int \frac{u'}{\sqrt{u}} dx = \frac{1}{2} \cdot \frac{u^{\frac{1}{2}}}{\frac{1}{2}} = \sqrt{u}\right)\)  

\(\Rightarrow \sqrt{1 + (f(x))^2} = x^2 + C.\)  

\(f(0) = 0 \Rightarrow C = 1\)  

\(1 + (f(x))^2 = (x^2 + 1)^2 = x^4 + 2x^2 + 1\)  
\(\Rightarrow (f(x))^2 = x^4 + 2x^2 \Rightarrow f(x) = \sqrt{x^4 + 2x^2}\)  

\(\text{Max } f(x) \text{ trên } [1, 3] = f(3) = \sqrt{81 + 18} = \sqrt{99} = 3\sqrt{11} \Rightarrow \boxed{D}\)  

\( f(x) > 0, \forall x \in [1, 3] \rightarrow \text{Max } f(x)_{[1, 3]} = f(3) = 3\sqrt{11}\)

\(f'(x) = \frac{4x^3 + 4x}{2\sqrt{x^4 + 2x^2}} = \frac{2x(x^2 + 1)}{\sqrt{x^4 + 2x^2}} = 0 \Leftrightarrow x = 0\)

page29

​​​​​​​Đáp án

\(\frac{[f(x)]^2 [f'(x)]^2}{e^{2x}} = 1 + (f(x))^2\)  

\(\Rightarrow \frac{f'(x) f(x)}{e^x} = \sqrt{1 + (f(x))^2} \Rightarrow \frac{f'(x) f(x)}{\sqrt{1 + (f(x))^2}} = e^x\)  

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

\(\Rightarrow \sqrt{1 + (f(x))^2} = e^x + C.\)  

\(f(0) = 1 \Rightarrow C = \sqrt{2} - 1\)  

\(\Rightarrow \sqrt{1 + (f(x))^2} = e + (\sqrt{2} - 1)\)  

\(\Rightarrow f(1)^2 = (e+ \sqrt{2} - 1)^2 - 1\)  

\(\Rightarrow f(1) = \sqrt{(e + \sqrt{2} - 1)^2 - 1} \approx 2,96 \Rightarrow \boxed{A}\)

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