Answer:
Option A,B,C
Explanation:
Given, ∫31x2F′(x)dx=−12
⇒ [x2F(x)]31−∫312x.F(x)dx=−12
⇒9F(3)−F(1)−∫31f(x)dx=−12
[∴
\Rightarrow -36-0-2\int_{1}^{3} f(x)dx=-12
\therefore \int_{1}^{3} f(x) dx=-12
and \therefore \int_{1}^{3} x^{3} F"(x) dx=40
\Rightarrow [x^{3} F'(x)]_1^3-\int_{1}^{3} 3x^{2}F'(x)dx=40
\Rightarrow [x^{2}(xF'(x)]_1^3-3\times(-12)=40
\Rightarrow \left\{x^{2}.[f'(x)-F(x)]\right\}_1^3=4
\Rightarrow 9[f'(3)-F(3)]-[f'(1)-F(1)]=4
\Rightarrow 9[f'(3)+4]-[f'(1)-0]=4
\Rightarrow 9f'(3)-f'(1)=-32