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Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn ${{\left[ f'\left( x \right) \right]}^{2}}+f\left( x \right).f''\left( x \right)=40{{x}^{3}}+16x,\forall x\in \mathbb{R}$ và $f\left( 0 \right)=f'\left( 0 \right)=1$. Giá trị của ${{f}^{2}}\left( 3 \right)$ bẳng:
A. 559.
B. 562.
C. 1117.
D. 1123.
A. 559.
B. 562.
C. 1117.
D. 1123.
Ta có: ${{\left( f'\left( x \right) \right)}^{2}}+f\left( x \right).f''\left( x \right)=40{{x}^{3}}+16x$.
$\Rightarrow \left[ f'\left( x \right).f\left( x \right) \right]'=40{{x}^{3}}+16x\Rightarrow f'\left( x \right).f\left( x \right)=\int{\left( 40{{x}^{3}}+16x \right)dx}=10{{x}^{4}}+8{{x}^{2}}+{{C}_{1}}$.
Bài ra $f\left( 0 \right)=f'\left( 0 \right)=1\Rightarrow {{C}_{1}}=1\Rightarrow f'\left( x \right).f\left( x \right)=10{{x}^{4}}+8{{x}^{2}}+1$.
$\begin{aligned}
& \Rightarrow \int{f'\left( x \right).f\left( x \right)dx}=\int{\left( 10{{x}^{4}}+8{{x}^{2}}+1 \right)dx}=\int{f\left( x \right)d\left[ f\left( x \right) \right]}=2{{x}^{5}}+\dfrac{8{{x}^{3}}}{3}+x+{{C}_{2}} \\
& \Rightarrow \dfrac{{{\left[ f\left( x \right) \right]}^{2}}}{2}=2{{x}^{5}}+\dfrac{8{{x}^{3}}}{3}+x+{{C}_{2}}\Rightarrow {{\left[ f\left( x \right) \right]}^{2}}=4{{x}^{5}}+\dfrac{16{{x}^{3}}}{3}+2x+2{{C}_{2}} \\
\end{aligned}$
Bài ra $f\left( 0 \right)=1\Rightarrow {{C}_{2}}=\dfrac{1}{2}\Rightarrow {{\left[ f\left( x \right) \right]}^{2}}=4{{x}^{5}}+\dfrac{16{{x}^{3}}}{3}+2x+1\Rightarrow {{f}^{2}}\left( 3 \right)=1123$.
$\Rightarrow \left[ f'\left( x \right).f\left( x \right) \right]'=40{{x}^{3}}+16x\Rightarrow f'\left( x \right).f\left( x \right)=\int{\left( 40{{x}^{3}}+16x \right)dx}=10{{x}^{4}}+8{{x}^{2}}+{{C}_{1}}$.
Bài ra $f\left( 0 \right)=f'\left( 0 \right)=1\Rightarrow {{C}_{1}}=1\Rightarrow f'\left( x \right).f\left( x \right)=10{{x}^{4}}+8{{x}^{2}}+1$.
$\begin{aligned}
& \Rightarrow \int{f'\left( x \right).f\left( x \right)dx}=\int{\left( 10{{x}^{4}}+8{{x}^{2}}+1 \right)dx}=\int{f\left( x \right)d\left[ f\left( x \right) \right]}=2{{x}^{5}}+\dfrac{8{{x}^{3}}}{3}+x+{{C}_{2}} \\
& \Rightarrow \dfrac{{{\left[ f\left( x \right) \right]}^{2}}}{2}=2{{x}^{5}}+\dfrac{8{{x}^{3}}}{3}+x+{{C}_{2}}\Rightarrow {{\left[ f\left( x \right) \right]}^{2}}=4{{x}^{5}}+\dfrac{16{{x}^{3}}}{3}+2x+2{{C}_{2}} \\
\end{aligned}$
Bài ra $f\left( 0 \right)=1\Rightarrow {{C}_{2}}=\dfrac{1}{2}\Rightarrow {{\left[ f\left( x \right) \right]}^{2}}=4{{x}^{5}}+\dfrac{16{{x}^{3}}}{3}+2x+1\Rightarrow {{f}^{2}}\left( 3 \right)=1123$.
Đáp án D.