AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper

It is given that,
f(x)=x2e2x
Differentiate the above equation.
f′(x)=2e2xx2+2xe2x
0=2e2x(x2+x)
x2+x=0
x=0,1
The maxima and minima is obtained at -2,-1,0,2 as the function bound.
Thus,
f(−2)=(−2)2e−4=4e−4
f(−1)=(1)2e−2
f(0)=0
f(2)=(2)2e4=4e4
Thus,
p=4e4 and q=0
pe−4+qe4=4e4(e−)+0=4e0=4