If f is a function satisfying f(x)+xf(1−x)=1+x2 for all real x, then f(6) may be written in the form −pq for relatively prime positive integers p and q. Find p+q.
Any help is much appreciated --- thanks all!
Hi Proaop,
f(6)+6f(1−6)=1+36f(6)+6f(−5)=37(1) f(−5)−5f(1−−5)=1+25f(−5)−5f(6)=26−5f(6)+f(−5)=26(2a)−30f(6)+6f(−5)=156(2b) now(1)−(2b) I expect you can finish it.
31f(6) = -119
f(6) = -119/31
119 + 31 = 150
Thanks melody!!!
You are welcome. :)