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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm liên tục trên $\left[ 0;1 \right]$ thỏa mãn ${{\left[ f'\left( x \right) \right]}^{2}}=4\left[ 2{{x}^{2}}+1-f\left( x \right) \right]$ với mọi $x$ thuộc đoạn $\left[ 0;1 \right]$ và $f\left( 1 \right)=2.$ Giá trị $I=\int\limits_{0}^{1}{xf\left( x \right)dx}$ bằng
A. $\dfrac{4}{3}$
B. $\dfrac{11}{4}$
C. $\dfrac{3}{4}$
D. $\dfrac{5}{3}$
A. $\dfrac{4}{3}$
B. $\dfrac{11}{4}$
C. $\dfrac{3}{4}$
D. $\dfrac{5}{3}$
Cách 1:
Ta có
${{\left[ f'\left( x \right) \right]}^{2}}=4\left[ 2{{x}^{2}}+1-f\left( x \right) \right]\Leftrightarrow {{\left[ f'\left( x \right) \right]}^{2}}+4f\left( x \right)=4\left( 2{{x}^{2}}+1 \right)$
$\Rightarrow \int\limits_{0}^{1}{{{\left[ f'\left( x \right) \right]}^{2}}dx}+\int\limits_{0}^{1}{4f\left( x \right)dx}=\int\limits_{0}^{1}{4\left( 2{{x}^{2}}+1 \right)dx}\Leftrightarrow \int\limits_{0}^{1}{{{\left[ f'\left( x \right) \right]}^{2}}dx}+4xf\left( x \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.-4\int\limits_{0}^{1}{xf'\left( x \right)dx}=\dfrac{20}{3}$
$\Leftrightarrow \int\limits_{0}^{1}{{{\left[ f'\left( x \right) \right]}^{2}}dx}-4\int\limits_{0}^{1}{xf'\left( x \right)dx}+4f\left( 1 \right)=\dfrac{20}{3}$
$\Leftrightarrow \int\limits_{0}^{1}{{{\left[ f'\left( x \right) \right]}^{2}}dx}-4\int\limits_{0}^{1}{xf'\left( x \right)dx}+4\int\limits_{0}^{1}{{{x}^{2}}dx}=\dfrac{20}{3}-8+4\int\limits_{0}^{1}{{{x}^{2}}dx}$
$\Leftrightarrow \int\limits_{0}^{1}{{{\left[ f'\left( x \right)-2x \right]}^{2}}dx}=0\Leftrightarrow f'\left( x \right)-2x=0\Rightarrow f\left( x \right)={{x}^{2}}+C.$
$f\left( 1 \right)=2\Leftrightarrow C=1\Leftrightarrow f\left( x \right)={{x}^{2}}+1.$
Vậy $I=\int\limits_{0}^{1}{xf\left( x \right)dx}=\dfrac{3}{4}.$
Cách 2:
Đặt $f\left( x \right)=a{{x}^{2}}+bx+c,$ ta có:
${{\left[ f'\left( x \right) \right]}^{2}}=4\left[ 2{{x}^{2}}+1-f\left( x \right) \right]\Leftrightarrow {{\left( 2ax+b \right)}^{2}}=4\left( 2{{x}^{2}}+1-a{{x}^{2}}-bx-c \right)\Leftrightarrow \left\{ \begin{aligned}
& 4{{a}^{2}}=4\left( 2-a \right) \\
& 4ab=-4b \\
& {{b}^{2}}=4\left( 1-c \right) \\
\end{aligned} \right..$
Kết hợp với điều kiện $f\left( 1 \right)=2\Leftrightarrow a+b+c=2$ ta có nghiệm $\left\{ \begin{aligned}
& a=1 \\
& b=0 \\
& c=1 \\
\end{aligned} \right.. $ Vậy $ f\left( x \right)={{x}^{2}}+1.$
Ta có
${{\left[ f'\left( x \right) \right]}^{2}}=4\left[ 2{{x}^{2}}+1-f\left( x \right) \right]\Leftrightarrow {{\left[ f'\left( x \right) \right]}^{2}}+4f\left( x \right)=4\left( 2{{x}^{2}}+1 \right)$
$\Rightarrow \int\limits_{0}^{1}{{{\left[ f'\left( x \right) \right]}^{2}}dx}+\int\limits_{0}^{1}{4f\left( x \right)dx}=\int\limits_{0}^{1}{4\left( 2{{x}^{2}}+1 \right)dx}\Leftrightarrow \int\limits_{0}^{1}{{{\left[ f'\left( x \right) \right]}^{2}}dx}+4xf\left( x \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.-4\int\limits_{0}^{1}{xf'\left( x \right)dx}=\dfrac{20}{3}$
$\Leftrightarrow \int\limits_{0}^{1}{{{\left[ f'\left( x \right) \right]}^{2}}dx}-4\int\limits_{0}^{1}{xf'\left( x \right)dx}+4f\left( 1 \right)=\dfrac{20}{3}$
$\Leftrightarrow \int\limits_{0}^{1}{{{\left[ f'\left( x \right) \right]}^{2}}dx}-4\int\limits_{0}^{1}{xf'\left( x \right)dx}+4\int\limits_{0}^{1}{{{x}^{2}}dx}=\dfrac{20}{3}-8+4\int\limits_{0}^{1}{{{x}^{2}}dx}$
$\Leftrightarrow \int\limits_{0}^{1}{{{\left[ f'\left( x \right)-2x \right]}^{2}}dx}=0\Leftrightarrow f'\left( x \right)-2x=0\Rightarrow f\left( x \right)={{x}^{2}}+C.$
$f\left( 1 \right)=2\Leftrightarrow C=1\Leftrightarrow f\left( x \right)={{x}^{2}}+1.$
Vậy $I=\int\limits_{0}^{1}{xf\left( x \right)dx}=\dfrac{3}{4}.$
Cách 2:
Đặt $f\left( x \right)=a{{x}^{2}}+bx+c,$ ta có:
${{\left[ f'\left( x \right) \right]}^{2}}=4\left[ 2{{x}^{2}}+1-f\left( x \right) \right]\Leftrightarrow {{\left( 2ax+b \right)}^{2}}=4\left( 2{{x}^{2}}+1-a{{x}^{2}}-bx-c \right)\Leftrightarrow \left\{ \begin{aligned}
& 4{{a}^{2}}=4\left( 2-a \right) \\
& 4ab=-4b \\
& {{b}^{2}}=4\left( 1-c \right) \\
\end{aligned} \right..$
Kết hợp với điều kiện $f\left( 1 \right)=2\Leftrightarrow a+b+c=2$ ta có nghiệm $\left\{ \begin{aligned}
& a=1 \\
& b=0 \\
& c=1 \\
\end{aligned} \right.. $ Vậy $ f\left( x \right)={{x}^{2}}+1.$
Đáp án C.