Given f(x)=(c+1)x2+(1−c2)x+2k⋯(i)
& f(x+y)=f(x)+f(y)−xy∀xy∈R
So f′(x)=2(c+1)x+1−c2
Now y→0limyf(x+y)−f(x)=y→0limyf(y)−xy=y→0limyf(y)−x
⇒f′(x)=f′(0)−x
Integrating, we get
f(x)=−2x2+f′(0)x+λ but f(0)=0⇒λ=0
i.e. f(x)=−2x2+(1−c2)x⋯(ii)
∴ as f′(0)=1−c2
Now comparing equation (i) and (ii), we get
c=−23
∴f(x)=−21x2−45x
Now ∣2x=1∑20f(x)∣=x=1∑20x2+25⋅x=1∑20x=620(21)(41)+25(220(21))
=2870+525
=3395