Cho hàm số y = f(x) có đạo hàm trên R thỏa mãn f(1) = 1 và

* Hướng dẫn giải

Xét $I=\underset{1}{\overset{2}{\int }}xf\text{'}\left(x\right)dx.$ Đặt $\left|f\left(t\right)+m\right|=1\iff \left[\begin{array}{l}f\left(t\right)+m=1\\ f\left(t\right)+m=-1\end{array}\right.\iff \left[\begin{array}{l}f\left(t\right)=1-m\left(1\right)\\ f\left(t\right)=-1-m\left(2\right)\end{array}\right..$ ta có

$I=xf\left(x\right)\left|\begin{array}{l}2\\ 1\end{array}\right.-\underset{1}{\overset{2}{\int }}f\left(x\right)dx=2f\left(2\right)-f\left(1\right)-\underset{1}{\overset{2}{\int }}f\left(x\right)dx$

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

Ta có: $f\left(2x\right)-xf\left(x^2\right)=5x-2x^3-1.f\left(2x\right)-xf\left(x^2\right)=5x-2x^3-1.$ Thay $x=1\Rightarrow f\left(2\right)-f\left(1\right)=2\Rightarrow f\left(2\right)=3.$

$\Rightarrow I=5-\underset{1}{\overset{2}{\int }}f\left(x\right)dx.$

Ta có:

$f\left(2x\right)-xf\left(x^2\right)=5x-2x^3-1$

$\iff 2f\left(2x\right)-2xf\left(x^2\right)=10x-4x^3-2$

Lấy tích phân 2 vế ta có:

$2\underset{0}{\overset{1}{\int }}f\left(2x\right)dx-\underset{0}{\overset{1}{\int }}2xf\left(x^2\right)dx=\underset{0}{\overset{1}{\int }}\left(10x-4x^3-2\right)dx=2$

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

$\iff \underset{0}{\overset{2}{\int }}f\left(t\right)dt-\underset{0}{\overset{1}{\int }}f\left(u\right)du=2$

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

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

Vậy $I=5-2=3.$

Chọn A.