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

Thay $x=\frac{\pi }{2}$ vào đẳng thức $f\left(x\right)+f\left(\frac{\pi }{2}-x\right)=\sin x.\cos x\Rightarrow f\left(\frac{\pi }{2}\right)+f\left(0\right)=0\iff f\left(\frac{\pi }{2}\right)=0.$

Xét $I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}x.f\text{'}\left(x\right)dx$

Đặt $\left\{\begin{array}{l}u=x\\ dv=f\text{'}\left(x\right)dx\end{array}\right.\Rightarrow \left\{\begin{array}{l}du=dx\\ v=f\left(x\right)\end{array}\right.$

$\Rightarrow I=x.f\left(x\right)\left|{}_0^{\frac{\pi }{2}}\right.-\underset{0}{\overset{\frac{\pi }{2}}{\int }}f\left(x\right)dx=-\underset{0}{\overset{\frac{\pi }{2}}{\int }}f\left(x\right)dx\left(1\right)$

Lại có: $\underset{0}{\overset{\frac{\pi }{2}}{\int }}f\left(x\right)dx+\underset{0}{\overset{\frac{\pi }{2}}{\int }}f\left(\frac{\pi }{2}-x\right)dx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\sin x.\cos xdx$

$\iff 2\underset{0}{\overset{\frac{\pi }{2}}{\int }}f\left(x\right)dx=\frac{1}{2}\underset{0}{\overset{\frac{\pi }{2}}{\int }}\sin 2xdx\iff 2\underset{0}{\overset{\frac{\pi }{2}}{\int }}f\left(x\right)dx=\frac{1}{2}\iff \underset{0}{\overset{\frac{\pi }{2}}{\int }}f\left(x\right)dx=\frac{1}{4}.$

Vậy $I=-\frac{1}{4}.$

Chọn C.