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# Tough function question --- help please!

+1
172
3
+1315

If $$f$$ is a function satisfying $$f(x)+xf(1-x)=1+x^2$$ for all real $$x$$, then $$f(6)$$ may be written in the form $$-{p\over q}$$ for relatively prime positive integers $$p$$ and $$q$$. Find $$p + q$$.

Any help is much appreciated --- thanks all!

Oct 3, 2022

#1
+118448
+3

Hi Proaop,

$$f(6)+6f(1-6)=1+36\\ f(6)+6f(-5)=37\qquad (1)\\~\\ f(-5)-5f(1--5)=1+25\\ f(-5)-5f(6)=26\\ -5f(6)+f(-5)=26\qquad (2a)\\ -30f(6)+6f(-5)=156\qquad (2b)\\~\\ now\;\;\;(1)-(2b)\\~\\ \text{I expect you can finish it.}$$

Oct 4, 2022
#2
+1315
+2

31f(6) = -119

f(6) = -119/31

119 + 31 = 150

Thanks melody!!!

proyaop  Oct 4, 2022
#3
+118448
+1

You are welcome. :)

Melody  Oct 5, 2022