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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)

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5 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
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2 years ago
```							Ans 1.   f(x+ay,x-ay) = axy    …......(1)Let x+ay = p & x-ay =qSolving, we getx = (p+q)/2 & y = (p-q)/2aSubstitute the values of p & q in (1)So, f(p,q) = a {(p+q)/2} {(p-q)/2a}                = (p^2 – q^2)/4Replace p by x & q by y to getf(x,y) = (x^2 – y^2)/4 Ans 2.   f(x+y,x-y) = xy Let x+y = p & x-y = qSolving, we getx = (p+q)/2 & y = (p-q)/2Substitute values of x & y in (1)So f(p,q) = {(p+q)/2} {(p-q)/2}               = (p^2 – q^2)/4Replace p by x & q by y to getf(x,y) = (x^2 – y^2)/4Similarly f(y,x) = (y^2 – x^2)/4Arithmetic 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
```
2 years ago
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