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avatar+1005 

Let \(f(x)=\frac{x}{x-1}\). Then there exist unique real numbers \(p, q,\) and \(r\) such that

 

\(f(5x)=\frac{pf(x)+q}{rf(x)+1}.\)

 

Find \(p + q + r\).

 Oct 4, 2022
 #1
avatar
+3

Hi proyaop!

\(f(x)=\dfrac{x}{x-1}\)

 

So: \(f(5x)=\dfrac{5x}{5x-1}\)

Now equate the given with this:
That is, 

\(\dfrac{5x}{5x-1}=\dfrac{pf(x)+q}{rf(x)+1}\)

But, we know \(f(x)\):

\(\dfrac{5x}{5x-1}=\dfrac{p(\dfrac{x}{x-1})+q}{r(\dfrac{x}{x-1})+1}\)

Multiply the right with (x-1)/(x-1):

\(\dfrac{5x}{5x-1}=\dfrac{p*x+q(x-1)}{r*x+(x-1)}\)

Expand and equate.

\(\dfrac{5x}{5x-1}=\dfrac{(p+q)x-q}{(r+1)x-1}\)

Well by comparing we get:

 \(p+q=5 \\ q=0 \\ r+1=5 \\\)

Hence, \(p=5,q=0,r=4\)

So, \(p+q+r=5+0+4=9\)

I hope this helps!

 Oct 5, 2022
 #3
avatar+1005 
+1

Thanks guest, you have a much smarter way of algebraically solving it --- I just plugged in random values of x and hoped for the best...

proyaop  Oct 5, 2022
 #4
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+1

Actually, your idea also is pretty neat! I learnt something from it :D!

Guest Oct 5, 2022
 #2
avatar+1005 
+2

I got it! The answer is 9!

 

How I did it:

First I plugged in x as 0, f(0) = 0 / -1 = 0

(0p + q)/(0r + 1) = f(5 * 0) = 0 => q/1 = 0 => q = 0

Next I plugged in x as -1, f(-1) = 1/2

(p/2 + 0)/(r/2 + 1) = f(-5) = 5/6

p/(r + 2) = 5/6

6p = 5r + 10  (1)

Next I plugged in x as -2, f(-2) = 2/3

(2p/3)/(2r/3 + 1) = f(-10) = 10/11

2p/(2r + 3) = 10/11

22p = 20r + 30

11p = 10r + 15   (2)

Take equation (1) and multiply by 2 => 12p = 10r + 20 (1b)

Take equation 1b - equation 2 => p = 5

Plug it back into equation 1 => 30 = 5r + 10 => r = 4

p + q + r = 5 + 0 + 4 = 9

 Oct 5, 2022

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