+0  
 
+1
90
3
avatar+1007 

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
avatar+118132 
+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
avatar+1007 
+2

31f(6) = -119

f(6) = -119/31

119 + 31 = 150

 

Thanks melody!!!

proyaop  Oct 4, 2022
 #3
avatar+118132 
+1

You are welcome. :)

Melody  Oct 5, 2022

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