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)#?

Question

1180 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-02T15:25:51

#6#

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#

Answer 2 · 2017-06-02T16:30:07

#6#

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#