3sir i have done this before here.
wat i done is x=y give f(0)2=(f(x)-x)2 so f(x)=x±f(0)
again putting x=0 yields us f(0)=±f(0) which is possible when f(0)=0
Again from Askiitians.com (this problem was an oasis in the mind-numbingly boring queries there)
Find all continuous functions f: R→R satisfying
f[(x-y)2] = f2(x) - 2x f(y) + y2
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12 Answers
62substitute x=y=0 to get f(0)
f(0) = f2(0)
f(0)=0 or 1
Hint: Can you convert this to a perfect square?
3
341almost there, you have made a small mistake with finding the value of f(0)
341Its just this: Putting x=y=0, we get f(0) = f2(0). So that f(0) = 0 or f(0) = 1. Mabye you made a mistake in placing the exponent in its right position.
By the way, you guys have cracked it in a much simpler manner than yours truly :D
What i did was:
Swapping x and y we get LHS same as before but RHS is f2(y) - 2yf(x) +x2
So that f2(x) - 2xf(y) +y2= f2(y) - 2yf(x) +x2
Now put y = 0 to obtain f2(x) = f2(0) + 2x f(0) + x2
The rest is similar, and so we we have the solutions f(x) =x or f(x) = x+1
3though i gave the soln,i have this dbt when started the functional eqns.Sir while putting y=x, isnt dat the functional eqns is proved only when y=x,but actually these two are independent variable sir.so how can they always be equal.
62sankara.. your question is a good one in a way....
See the thing is that such equations hold for all values of x and y
So they hold for each and every value of y
in particular for the case when y=x
341sorry, i just saw my chatbox and remembered this thread.
for msp: to add to what Nishant sir said, in qns where it clearly says f:R→R, we have no restrictions in choosing x and y as we please. There are some qns where the functional equation features an element from the range of the function. In such cases either sometimes we willl be given the function is onto and we can breathe easy. For JEE it wont go beyond this level of difficulty
for shreyan: f(x) = -x does not satisfy the given equation. f(x) = -x-1 is also ruled out, because in this case we need f(0) = 1.
3the point where i got is "in particular for the case when y=x" i have proved,but wat abt the other cases if y=x+1, i havent proved for dat case.
62Sankara... I dont know how to explain this.. but let me try...
The point is not this case or that case..
Functional equations are valid in any case. In each and every case. So if you can use some trick/ (generally a substitution) to get f(x) then you have found f(x) for all cases.....
If after finding such a f(x) you tend to realize that it does not work for some other case, you can safely say that f(x) does not exists!