If for all real triplets (a,b,c),f(x)=a+bx+cx2; then ∫01f(x)dx is equal to:

∫01a+bx+cx2dx=ax+bx22+cx3301=a+b2+c3 We have f(1)=a+b+c,f(0)=a,f12=a+b2+c4 Now 16f(1)+f(0)+4f12=16a+b+c+a+4a+b2+c4                                                        =16(6a+3b+2c)=a+b2+c3Therefore, it is matching with the option 4