Cho hai hàm số y =f(x(;y=(gx) có đồ thị hàm số y =f(x(;y=(gx) như hình vẽ sau:

* Hướng dẫn giải

Đáp án A

Ta có: $h\left(x\right)=f\left(x\right)-g\left(x\right)\Rightarrow h\text{'}\left(x\right)=f\text{'}\left(x\right)-g\text{'}\left(x\right)$.

Ta có bảng biến thiên $y=h\left(x\right)$:

Từ bảng biến thiên ta có$\underset{\left[-5;5\right]}{\min }h\left(x\right)\in \left\{h\left(-5\right);h\left(2\right)\right\},\underset{\left[-5;5\right]}{\max }h\left(x\right)\in \left\{h\left(-2\right);h\left(5\right)\right\}$.

Từ đồ thị hàm số ta có:

$\begin{array}{l}{S}_1=S_3\Rightarrow \underset{-5}{\overset{-2}{\int }}\left[f\text{'}\left(x\right)-g\text{'}\left(x\right)\right]dx=\underset{2}{\overset{5}{\int }}\left[f\text{'}\left(x\right)-g\text{'}\left(x\right)\right]dx\Rightarrow \left(f\left(x\right)-g\left(x\right)\right)\left|\begin{array}{l}{}^{-2}\\ {}_{-5}\end{array}\right.=\left(f\left(x\right)-g\left(x\right)\right)\left|\begin{array}{l}{}^5\\ {}_2\end{array}\right.\\ \Rightarrow h\left(-2\right)-h\left(-5\right)=h\left(5\right)-h\left(2\right).\end{array}$

Diện tích $S_2=\underset{-2}{\overset{2}{\int }}\left[g\text{'}\left(x\right)-f\text{'}\left(x\right)\right]dx=-h\left(x\right)\left|{}_{-2}^2\right.=h\left(-2\right)-h\left(2\right).$

Ta có $S_2<S_1\Rightarrow h\left(-2\right)-h\left(2\right)<h\left(-2\right)-h\left(-5\right)\Rightarrow h\left(-5\right)<h\left(2\right)$. Suy ra $\underset{\left[-5;5\right]}{\min }h\left(x\right)=h\left(-5\right).$

Lại có $S_2<S_3\Rightarrow h\left(-2\right)-h\left(2\right)<h\left(5\right)-h\left(2\right)\Rightarrow h\left(-2\right)<h\left(5\right)$. Suy ra $\underset{\left[-5;5\right]}{\max }h\left(x\right)=h\left(5\right).$