Cho hàm số y = f(x) liên tục và có đạo hàm trên R thỏa mãn

Xét tích phân $I=\underset{0}{\overset{1}{\int }}x.f\text{'}\left(x\right)dx.$

Đặt $\left\{\begin{array}{l}u=x\\ dv=f\text{'}\left(x\right)dx\end{array}\right.\Rightarrow \left\{\begin{array}{l}du=dx\\ v=f\left(x\right)\end{array}\right.,$ khi đó ta có $I=xf\left(x\right)\left|\begin{array}{l}1\\ 0\end{array}\right.-\underset{0}{\overset{1}{\int }}f\left(x\right)dx=f\left(1\right)-\underset{0}{\overset{1}{\int }}f\left(x\right)dx.$

Theo bài ra ta có: $5f\left(x\right)-7f\left(1-x\right)=3\left(x^2-2x\right)$

Thay $x=0\Rightarrow 5f\left(0\right)-7f\left(1\right)=0$

Thay $x=1\Rightarrow 5f\left(1\right)-7f\left(0\right)=-3$

$\Rightarrow f\left(0\right)=\frac{7}{8},f\left(1\right)=\frac{5}{8}.$

Xét tích phân $\underset{0}{\overset{1}{\int }}f\left(x\right)dx.$

Từ $5f\left(x\right)-7f\left(1-x\right)=3\left(x^2-2x\right)$ lấy tích phân từ 0 đến 1 hai vế ta có:

$5\underset{0}{\overset{1}{\int }}f\left(x\right)dx-7\underset{0}{\overset{1}{\int }}f\left(1-x\right)dx=\underset{0}{\overset{1}{\int }}3\left(x^2-2x\right)dx.$

$\iff 5\underset{0}{\overset{1}{\int }}f\left(x\right)dx+7\underset{0}{\overset{1}{\int }}f\left(1-x\right)d\left(1-x\right)=-2$

$\iff 5\underset{0}{\overset{1}{\int }}f\left(x\right)dx+7\underset{1}{\overset{0}{\int }}f\left(x\right)dx=-2$

$\iff 5\underset{0}{\overset{1}{\int }}f\left(x\right)dx-7\underset{0}{\overset{1}{\int }}f\left(x\right)dx=-2$

$\iff -2\underset{0}{\overset{1}{\int }}f\left(x\right)dx=-2$

$\iff \underset{0}{\overset{1}{\int }}f\left(x\right)dx=1$

Suy ra $I=f\left(1\right)-\underset{0}{\overset{1}{\int }}f\left(x\right)dx=\frac{5}{8}-1=-\frac{3}{8}\Rightarrow a=3,b=8.$

Vậy $T=3a-b=3.3-8=1.$

Chọn D.