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# 1 -  if f(x+ay , x-ay) = axy , then find f(x,y) 2 – if f(x+y , x-y) = xy , find the arithmetic mean of f(x,y) and f(y,x)

Shubham Patil
15 Points
3 years ago
f(x+ay,x-ay)=axyf(x+ay,x-ay)=(x)(ay) =(2x/2)(2ay/2) =((x+x)/2)((ay+ay)/2) =((x+x+ay-ay)/2)((x+ay-(x-ay))/2) =(x+ay+x-ay)(x+ay-(x-ay))/4 =(x+y)(x-y)/4 =(x^2-y^2)/4
Samyak Jain
333 Points
3 years ago
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