For every pair of numbers a and b, the function f satisfies b^2 f(a) = a^2 f(b). If f(2) does not equal 0, find the value of [f(5) - f(1)] / f(2)?

2 Answers
Jun 2, 2017

6

Explanation:

Given:

b^2f(a) = a^2f(b)

Let k=f(2)/4 != 0

Then, putting b=2 we find that for any a

4f(a) = a^2*f(2) = 4ka^2

Dividing both ends by 4, we find:

f(a) = ka^2

So:

(f(5)-f(1))/f(2) = (k*5^2-k*1^2)/(k*2^2) = (25-1)/4 = 6

Jun 2, 2017

6

Explanation:

Making b = lambda a

b^2f(a)=a^2f(b) rArr f(lambda a) = lambda^2f(a) now making a=1 we have

f(lambda)=lambda^2f(1) and then

(f(5)-f(1))/f(2)=(5^2f(1)-f(1))/(2^2 f(1))=6

NOTE: f(1) ne 0 because otherwise f(2)=0