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
Explanation:
Given:
#b^2f(a) = a^2f(b)#
Let
Then, putting
#4f(a) = a^2*f(2) = 4ka^2#
Dividing both ends by
#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
Explanation:
Making
NOTE: