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Câu hỏi: Cho hàm số $f\left( x \right)=\dfrac{{{e}^{2x}}}{{{e}^{2x}}+e}$. Đặt $S=f\left( \dfrac{1}{2021} \right)+f\left( \dfrac{2}{2021} \right)+f\left( \dfrac{3}{2021} \right)+...+f\left( \dfrac{2021}{2021} \right)$. Khi đó giá trị của $P=\log S$ thuộc khoảng nào dưới đây?
A. $\left( 1;2 \right)$.
B. $\left( 2;3 \right)$.
C. $\left( 3;4 \right)$.
D. $\left( 4;5 \right)$.
Xét hai số dương $a$ và $b$ sao cho $a+b=1$, ta có
$f\left( a \right)+f\left( b \right)=\dfrac{{{e}^{2a}}}{{{e}^{2a}}+e}+\dfrac{{{e}^{2b}}}{{{e}^{2b}}+e}=\dfrac{{{e}^{2a}}({{e}^{2b}}+e)+{{e}^{2b}}({{e}^{2a}}+e)}{\left( {{e}^{2a}}+e \right)\left( {{e}^{2b}}+e \right)}$
$\dfrac{{{e}^{2\left( a+b \right)}}+{{e}^{2\left( a+b \right)}}+e\left( {{e}^{2a}}+{{e}^{2b}} \right)}{{{e}^{2\left( a+b \right)}}+e\left( {{e}^{2a}}+{{e}^{2b}} \right)+{{e}^{2}}}=\dfrac{{{e}^{2\left( a+b \right)}}+{{e}^{2}}+e\left( {{e}^{2a}}+{{e}^{2b}} \right)}{{{e}^{2\left( a+b \right)}}+e\left( {{e}^{2a}}+{{e}^{2b}} \right)+{{e}^{2}}}=1$ (vì $a+b=1$ )
Do đó $S=\left[ f\left( \dfrac{1}{2021} \right)+f\left( \dfrac{2020}{2021} \right) \right]+\left[ f\left( \dfrac{2}{2021} \right)+f\left( \dfrac{2019}{2021} \right) \right]+...+f\left( 1 \right)$
$=1010+f\left( 1 \right)=1010+\dfrac{e}{1+e}$
Vây $P=\log S=\log \left( \dfrac{1010+1011e}{1+e} \right)\approx 3,005$.
A. $\left( 1;2 \right)$.
B. $\left( 2;3 \right)$.
C. $\left( 3;4 \right)$.
D. $\left( 4;5 \right)$.
Xét hai số dương $a$ và $b$ sao cho $a+b=1$, ta có
$f\left( a \right)+f\left( b \right)=\dfrac{{{e}^{2a}}}{{{e}^{2a}}+e}+\dfrac{{{e}^{2b}}}{{{e}^{2b}}+e}=\dfrac{{{e}^{2a}}({{e}^{2b}}+e)+{{e}^{2b}}({{e}^{2a}}+e)}{\left( {{e}^{2a}}+e \right)\left( {{e}^{2b}}+e \right)}$
$\dfrac{{{e}^{2\left( a+b \right)}}+{{e}^{2\left( a+b \right)}}+e\left( {{e}^{2a}}+{{e}^{2b}} \right)}{{{e}^{2\left( a+b \right)}}+e\left( {{e}^{2a}}+{{e}^{2b}} \right)+{{e}^{2}}}=\dfrac{{{e}^{2\left( a+b \right)}}+{{e}^{2}}+e\left( {{e}^{2a}}+{{e}^{2b}} \right)}{{{e}^{2\left( a+b \right)}}+e\left( {{e}^{2a}}+{{e}^{2b}} \right)+{{e}^{2}}}=1$ (vì $a+b=1$ )
Do đó $S=\left[ f\left( \dfrac{1}{2021} \right)+f\left( \dfrac{2020}{2021} \right) \right]+\left[ f\left( \dfrac{2}{2021} \right)+f\left( \dfrac{2019}{2021} \right) \right]+...+f\left( 1 \right)$
$=1010+f\left( 1 \right)=1010+\dfrac{e}{1+e}$
Vây $P=\log S=\log \left( \dfrac{1010+1011e}{1+e} \right)\approx 3,005$.
Đáp án C.