1finding it is very easy
but you need to prove it step by step
hence you have missed another function satisfying it
again there are just two steps
(as this a subjective one!!)
find all functions f:R->R
such that for all x,y belonging to - R
f(f(x)+y)=f(x2-y) + 4yf(x)
this has a two line answer!!
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1me too got jus f(x)=x^2. ... jus replaced y=0.
...another thing an even fuction.. but if its x^2 .. it is already an even function...
1put .. y=0,
then, f(f(x)) = f(x2) +0
so, f(x)=x2
now, put x=0,
so we have,
f(f(0)+y) = f(-y) + 4yf(0)
now, f(0)=0 [since f(x)=x^2]
so, f(y) = f(-y).
1skygirl ,
you cant do that !!
it is because then we get
f(f(x))=f(x2)
it is NOT MENTIONED that the function is one-one etc.
nothing is mentioned as you can see
even that it is differentiable or not!!
you CANT equate the two expressions in the brackets
hence you too have missed another solution!!
1ok now you have two solutions but how will you prove that
no other solution exists?
beacuase it is mentioned find ALL functions
1OK I AM NOW POSTING THE SOLUTION
as it is valid for all x,y ε R and f:R->R
We can carry out the following substitutions
step 1=>
put y=a2 and x=a
for any real number a
we get
f(f(a)+a2)-4a2f(a)=f(0)
step 2=
put y=( - f(a)) and x=a
we get =>
f(0)=f(f(a)+a2)-4*f(a)*f(a)
equating the above two expressions we get
4a2f(a)=4(f(a))2
thus f(a)=0 or f(a)=a2
WASN'T IT A TWO STEP ANSWER??...
33U should be little more dependent on instincts.....
Xam me itna time nahi hota.....ki pahli baar question dekha aur y=a2 x=a..
raat me 4 baje sapna aega kya...:)
1actually in these types of questions you need to figure out the substitution by little thinking
in other words you may need to play around with the question!!
this was a question from bosnia-herzegovinia IMO selection test
