Cho hàm số

A

- Hướng dẫn giải

Đáp án A

Ta có $-\frac{1}{30}={\int }_1^2\left(x-1\right)f\left(x\right)dx=\frac{1}{2}{\int }_1^{2}f\left(x\right)d\left({\left(x-1\right)}^2\right)$

$=\frac{1}{2}{\left(x-1\right)}^{2}f\left(x\right)\overline{){}_1^2}-\frac{1}{2}{\int }_1^2{\left(x-1\right)}^{2}f\text{'}\left(x\right)dx$

$\iff {\int }_1^2{\left(x-1\right)}^{2}f\text{'}\left(x\right)dx=\frac{1}{15}$

Ta lại có ${\int }_1^2{\left(x-1\right)}^{4}dx=\frac{1}{5}{\left(x-1\right)}^5\overline{){}_1^2=\frac{1}{5}}$

Từ giả thiết và các kết quả ta có

$9{\int }_1^2{\left[f\text{'}\left(x\right)\right]}^{2}dx-6{\int }_1^2{\left(x-1\right)}^{2}f\text{'}\left(x\right)dx+{\int }_1^2{\left(x-1\right)}^{4}dx=0$

Mặt khác:

$9{\int }_1^2{\left[f\text{'}\left(x\right)\right]}^{2}dx-6{\int }_1^2{\left(x-1\right)}^{2}f\text{'}\left(x\right)dx+{\int }_1^2{\left(x-1\right)}^{4}dx={\int }_1^2{\left[3f\text{'}\left(x\right)-{\left(x-1\right)}^2\right]}^2$

Do vậy xét trên đoạn [1;2] , ta có

$3f\text{'}\left(x\right)-{(x-1)}^2=0\iff f\text{'}\left(x\right)=\frac{1}{3}{\left(x-1\right)}^2\Rightarrow f\left(x\right)=\frac{1}{9}{\left(x-1\right)}^3+c$ 

Lại do f(2) = 0 nên $C+\frac{1}{9}=0\iff C=-\frac{1}{9}\Rightarrow f\left(x\right)=\frac{1}{9}{\left(x-1\right)}^3-\frac{1}{9}$ 

Suy ra $I=\frac{1}{9}{\int }_1^2\left[{\left(x-1\right)}^3-1\right]dx=\frac{1}{36}{\left(x-1\right)}^4\overline{){}_1^2-\frac{1}{9}{\left(x-1\right)}_1^2=-\frac{1}{12}}$