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If f is a function satisfying f(x)+xf(1x)=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!

 Oct 3, 2022
 #1
avatar+118696 
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Hi Proaop,

 

f(6)+6f(16)=1+36f(6)+6f(5)=37(1) f(5)5f(15)=1+25f(5)5f(6)=265f(6)+f(5)=26(2a)30f(6)+6f(5)=156(2b) now(1)(2b) I expect you can finish it.

 Oct 4, 2022
 #2
avatar+1633 
+2

31f(6) = -119

f(6) = -119/31

119 + 31 = 150

 

Thanks melody!!!

proyaop  Oct 4, 2022
 #3
avatar+118696 
+1

You are welcome. :)

Melody  Oct 5, 2022

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