Let for some R, f : R R be a function satisfying f(x+y)=f(x)+2y2+y+ xy for all x, y R. If f(0)=-1 and f(1)=2, then the value of n=15 ( +f(n) ) is :
- 110
- 140
- 150
- 170
Answer
(2) 140
Solution
Put x=0: f(y)=f(0)+2y2+y=2y2+y-1
(Check: f(1)=2+1-1=2, as given.)
f(x+y)=2(x+y)2+(x+y)-1= (2x2+x-1 )+2y2+y+4xy
=f(x)+2y2+y+4xy =4
+f(n)=4+2n2+n-1=2n2+n+3
n=15 (2n2+n+3 )=2(55)+15+15=140
Answer: (2)