Ans 1. f(x+ay,x-ay) = axy …......(1)
Let x+ay = p & x-ay =q
Solving, we get
x = (p+q)/2 & y = (p-q)/2a
Substitute the values of p & q in (1)
So, f(p,q) = a {(p+q)/2} {(p-q)/2a}
= (p^2 – q^2)/4
Replace p by x & q by y to get
f(x,y) = (x^2 – y^2)/4
Ans 2. f(x+y,x-y) = xy
Let x+y = p & x-y = q
Solving, we get
x = (p+q)/2 & y = (p-q)/2
Substitute values of x & y in (1)
So f(p,q) = {(p+q)/2} {(p-q)/2}
= (p^2 – q^2)/4
Replace p by x & q by y to get
f(x,y) = (x^2 – y^2)/4
Similarly f(y,x) = (y^2 – x^2)/4
Arithmetic mean of f(x,y) & f(y,x)
= {f(x,y) + f(y,x)}/2
= [{(x^2 – y^2)/4} + {(y^2 – x^2)/4}]/2
= (x^2 – y^2 + y^2 – x^2)/8
= 0/8 = 0
