Correct answer: 27
\(f(x)=\int\limits_{-1}^x\left(\left(e^t-1\right)^{11}(2 t-1)^5(t-2)^7(t-3)^{12}(2 t-10)^{61}\right) d t\)
Using Lebnitz
\(f^{\prime}(x)=\left(e^x-1\right)^{11}(2 x-1)^5(x-2)^7(x-3)^{12}(2 x-10)^{61}\)
Point of maxima at \(\mathrm{x}=0,2\)
Point of minima at \(\mathrm{x}=\frac{1}{2}, 5\)
\(\mathrm{p}=0^2+2^2=4 \)
\(\mathrm{q}=\frac{1}{2}+5=\frac{11}{2} \)
\(\mathrm{p}^2+2 \mathrm{q}=16+11=27\)