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Câu hỏi: Cho hàm số $y=f\left( x \right)$ luôn dương và thỏa mãn $\dfrac{{f}'\left( x \right)}{\sqrt{f\left( x \right)}}=3{{\text{x}}^{2}}+1$. Biết $f\left( 0 \right)=1$. Tính giá trị $f\left( 1 \right)$.
A. $f\left( 1 \right)=4$
B. $f\left( 1 \right)=16$
C. $f\left( 1 \right)=3$
D. $f\left( 1 \right)=9$
A. $f\left( 1 \right)=4$
B. $f\left( 1 \right)=16$
C. $f\left( 1 \right)=3$
D. $f\left( 1 \right)=9$
Ta có $\int{\left( 3{{\text{x}}^{2}}+1 \right)d\text{x}}=\int{\dfrac{{f}'\left( x \right)}{\sqrt{f\left( x \right)}}d\text{x}}=\int{\dfrac{1}{\sqrt{f\left( x \right)}}d\left( f(x) \right)}\Rightarrow {{x}^{3}}+x+C=2\sqrt{f\left( x \right)}$.
$f\left( 0 \right)=1\Rightarrow C=2\Rightarrow f\left( x \right)={{\left( \dfrac{{{x}^{3}}+x+2}{2} \right)}^{2}}\Rightarrow f\left( 1 \right)=4$.
$f\left( 0 \right)=1\Rightarrow C=2\Rightarrow f\left( x \right)={{\left( \dfrac{{{x}^{3}}+x+2}{2} \right)}^{2}}\Rightarrow f\left( 1 \right)=4$.
Đáp án A.