Б
![\int\limits^x_0 {x(3t^2-8t+3) \, dt =t^3-4t^2+3t|^x_{0}=x^3-4x^2+3x\ \textgreater \ 0](https://tex.z-dn.net/?f=+%5Cint%5Climits%5Ex_0+%7Bx%283t%5E2-8t%2B3%29+%5C%2C+dt+%3Dt%5E3-4t%5E2%2B3t%7C%5Ex_%7B0%7D%3Dx%5E3-4x%5E2%2B3x%5C+%5Ctextgreater+%5C+0)
x(x²-4x+3)>0
x(x-1)(x-3)>0
x=0 x=1 x=3
_ + _ +
------------------(0)-----------(1)----------------(3)---------------------
x∈(0;1) U (3;∞)
![\int\limits^x_0 {t^3} \, dt =t^4/4|^x_{0}=x^4/4\ \textless \ 1/4](https://tex.z-dn.net/?f=+%5Cint%5Climits%5Ex_0+%7Bt%5E3%7D+%5C%2C+dt+%3Dt%5E4%2F4%7C%5Ex_%7B0%7D%3Dx%5E4%2F4%5C+%5Ctextless+%5C+1%2F4)
x^4-1<0
(x-1)(x+1)(t²+1)<0
x=1 x=-1
+ _ +
-----------------(-1)-----------------(1)-----------------
x∈(-1;1)
Пусть корни равны
Из условия , тогда
![x_{1}=-4x_{2}\\ ](https://tex.z-dn.net/?f=+x_%7B1%7D%3D-4x_%7B2%7D%5C%5C%0A)
, по теореме Виета получим
![x_{1}+x_{2} = -4x_{2}+x_{2} = -3x_{2} = -p \\ x_{1}x_{2} = 4x_{2}^2=16 \\ \\ x_{2}= \pm2 \\ p= \pm6 ](https://tex.z-dn.net/?f=++++x_%7B1%7D%2Bx_%7B2%7D+%3D+-4x_%7B2%7D%2Bx_%7B2%7D+%3D+-3x_%7B2%7D+%3D++-p+%5C%5C%0A+x_%7B1%7Dx_%7B2%7D+%3D+4x_%7B2%7D%5E2%3D16+++++++++++++++%5C%5C%0A+%5C%5C%0A+x_%7B2%7D%3D+%5Cpm2+%5C%5C+%0A+p%3D+%5Cpm6+%0A)
Формула графика у=кх, подставим заданную точку: 12=-4*к, к=-3, значит у=-3х -ответ