Cho hàm số y = f(x) là hàm chẵn, liên tục trên R và

* Hướng dẫn giải

Chọn B

Ta có: 

$M=\underset{-2}{\overset{2}{\int }}\frac{f\left(x\right)}{{2020}^x+1}d\text{x}=\underset{-2}{\overset{0}{\int }}\frac{f\left(x\right)}{{2020}^x+1}d\text{x}+\underset{0}{\overset{2}{\int }}\frac{f\left(x\right)}{{2020}^x+1}d\text{x}$

Xét $N=\underset{-2}{\overset{0}{\int }}\frac{f\left(x\right)}{{2020}^x+1}d\text{x}$, đặt $t=-x\Rightarrow x=-t$, suy ra $d\text{x}=-dt$.

Đổi cận: $x=-2\Rightarrow t=2;x=0\Rightarrow t=0$. Khi đó 

$$\begin{gathered} N=\underset{2}{\overset{0}{\int }}\frac{f\left(-t\right)}{{2020}^{-t}+1}\left(-dt\right)=\underset{0}{\overset{2}{\int }}\frac{f\left(t\right)}{{2020}^{-t}+1}dt \\ =\underset{0}{\overset{2}{\int }}\frac{{2020}^t.f\left(t\right)}{{2020}^t+1}dt=\underset{0}{\overset{2}{\int }}\frac{{2020}^x.f\left(x\right)}{{2020}^x+1}d\text{x} \end{gathered}$$

Do đó: 

$$\begin{gathered} \underset{0}{\overset{2}{\int }}f\left(x\right)d\text{x}=\underset{0}{\overset{2}{\int }}\frac{{2020}^x.f\left(x\right)}{{2020}^x+1}d\text{x}+\underset{0}{\overset{2}{\int }}\frac{f\left(x\right)}{{2020}^x+1}d\text{x} \\ =\underset{-2}{\overset{0}{\int }}\frac{f\left(x\right)}{{2020}^x+1}d\text{x}+\underset{0}{\overset{2}{\int }}\frac{f\left(x\right)}{{2020}^x+1}d\text{x}=M=29 \end{gathered}$$