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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục và nhận giá trị dương trên $\left[ 0;1 \right]$. Biết $f\left( x \right).f\left( 1-x \right)=1$ với $\forall x\in \left[ 0;1 \right].$ Tích phân $\int\limits_{0}^{1}{\dfrac{dx}{1+f\left( x \right)}}$ bằng
A. $\dfrac{3}{2}.$
B. $\dfrac{1}{2}.$
C. 1.
D. 2.
A. $\dfrac{3}{2}.$
B. $\dfrac{1}{2}.$
C. 1.
D. 2.
Xét $I=\int\limits_{0}^{1}{\dfrac{dx}{1+f\left( x \right)}.}$
Đặt $x=1-t\Rightarrow I=\int\limits_{1}^{0}{\dfrac{d\left( 1-t \right)}{1+f\left( 1-t \right)}=\int\limits_{0}^{1}{\dfrac{dt}{1+f\left( 1-t \right)}=\int\limits_{0}^{1}{\dfrac{dx}{1+f\left( 1-x \right)}.}}}$
Bài ra $f\left( x \right).f\left( 1-x \right)=1\Rightarrow f\left( 1-x \right)=\dfrac{1}{f\left( x \right)}\Rightarrow I=\int\limits_{0}^{1}{\dfrac{dx}{1+\dfrac{1}{f\left( x \right)}}=\int\limits_{0}^{1}{\dfrac{f\left( x \right)}{1+f\left( x \right)}dx}}$
$\Rightarrow I+I=\int\limits_{0}^{1}{\dfrac{dx}{1+f\left( x \right)}+\int\limits_{0}^{1}{\dfrac{f\left( x \right)}{1+f\left( x \right)}dx}=\int\limits_{0}^{1}{\dfrac{1+f\left( x \right)}{1+f\left( x \right)}dx=1\Rightarrow I=\dfrac{1}{2}.}}$
Đặt $x=1-t\Rightarrow I=\int\limits_{1}^{0}{\dfrac{d\left( 1-t \right)}{1+f\left( 1-t \right)}=\int\limits_{0}^{1}{\dfrac{dt}{1+f\left( 1-t \right)}=\int\limits_{0}^{1}{\dfrac{dx}{1+f\left( 1-x \right)}.}}}$
Bài ra $f\left( x \right).f\left( 1-x \right)=1\Rightarrow f\left( 1-x \right)=\dfrac{1}{f\left( x \right)}\Rightarrow I=\int\limits_{0}^{1}{\dfrac{dx}{1+\dfrac{1}{f\left( x \right)}}=\int\limits_{0}^{1}{\dfrac{f\left( x \right)}{1+f\left( x \right)}dx}}$
$\Rightarrow I+I=\int\limits_{0}^{1}{\dfrac{dx}{1+f\left( x \right)}+\int\limits_{0}^{1}{\dfrac{f\left( x \right)}{1+f\left( x \right)}dx}=\int\limits_{0}^{1}{\dfrac{1+f\left( x \right)}{1+f\left( x \right)}dx=1\Rightarrow I=\dfrac{1}{2}.}}$
Đáp án B.