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Câu hỏi: Cho hàm số $f(x)$ liên tục trên $\mathbb{R}$ thỏa mãn các điều kiện: $f(0)=\sqrt{2}, f(x)>0, \forall x \in \mathbb{R}$ và $f^2(x) \cdot\left[f^{\prime}(x)\right]^2=\left(\dfrac{x^2+x}{2}\right)^2\left(2+f^2(x)\right)^3, \forall x \in \mathbb{R}$. Khi đó, giá trị $f(1)$ bằng
A. 42 .
B. $\sqrt{142}$.
C. 142.
D. $\sqrt{42}$.
A. 42 .
B. $\sqrt{142}$.
C. 142.
D. $\sqrt{42}$.
Ta có: $f^2(x) \cdot\left[f^{\prime}(x)\right]^2=\left(\dfrac{x^2+x}{2}\right)^2\left(2+f^2(x)\right)^3, \forall x \in \mathbb{R} \Leftrightarrow \dfrac{2 f(x) . f^{\prime}(x)}{\sqrt{\left(2+f^2(x)\right)^3}}=x^2+x, \forall x \in \mathbb{R}$.
Suy ra: $\int \dfrac{2 f(x) \cdot f^{\prime}(x)}{\sqrt{\left(2+f^2(x)\right)^3}} d x=\int\left(x^2+x\right) d x \Leftrightarrow \int \dfrac{d\left(2+f^2(x)\right)}{\sqrt{\left(2+f^2(x)\right)^3}}=\int\left(x^2+x\right) d x$
$
\Leftrightarrow-\dfrac{2}{\sqrt{2+f^2(x)}}=\dfrac{x^3}{3}+\dfrac{x^2}{2}+C .
$
Theo giả thiết $f(0)=\sqrt{2}$ suy ra $-\dfrac{2}{\sqrt{2+(\sqrt{2})^2}}=C \Leftrightarrow C=-1$.
Với $C=-1$ thì $-\dfrac{2}{\sqrt{2+f^2(x)}}=\dfrac{x^3}{3}+\dfrac{x^2}{2}-1 \Rightarrow-\dfrac{2}{\sqrt{2+f^2(1)}}=\dfrac{1^3}{3}+\dfrac{1^2}{2}-1 f(1)=\sqrt{142}$.
Vậy $f(1)=\sqrt{142}$.
Suy ra: $\int \dfrac{2 f(x) \cdot f^{\prime}(x)}{\sqrt{\left(2+f^2(x)\right)^3}} d x=\int\left(x^2+x\right) d x \Leftrightarrow \int \dfrac{d\left(2+f^2(x)\right)}{\sqrt{\left(2+f^2(x)\right)^3}}=\int\left(x^2+x\right) d x$
$
\Leftrightarrow-\dfrac{2}{\sqrt{2+f^2(x)}}=\dfrac{x^3}{3}+\dfrac{x^2}{2}+C .
$
Theo giả thiết $f(0)=\sqrt{2}$ suy ra $-\dfrac{2}{\sqrt{2+(\sqrt{2})^2}}=C \Leftrightarrow C=-1$.
Với $C=-1$ thì $-\dfrac{2}{\sqrt{2+f^2(x)}}=\dfrac{x^3}{3}+\dfrac{x^2}{2}-1 \Rightarrow-\dfrac{2}{\sqrt{2+f^2(1)}}=\dfrac{1^3}{3}+\dfrac{1^2}{2}-1 f(1)=\sqrt{142}$.
Vậy $f(1)=\sqrt{142}$.
Đáp án B.