Functional Equation

This is how I went about it.

The crux step is this: If for any t_1, t_2 \in \mathbb{R}^+ we can show that f(t_1) = f(t_2), we can conclude that f(x) is a constant function.

So we argue that given t_1, t_2 \in \mathbb{R}^+ we can find x,y \in \mathbb{R}^+ such that x+y = t_1; x^2+y^2 = t_2 and therefore from the problem statement f(t_1) =f(t_2)

But we have the inequality 2(x^2+y^2) \ge (x+y)^2 and so 2t_1 \ge t_2^2 must hold for us to directly use the method outlined above.

In case this does happen, then, to guarantee the existence of such x,y, a geometric argument will suffice. x^2+y^2 = t_2 is a circle and x+y=t_1 is a line and the intersection of these two will take place in the first quadrant, and so we have our (x,y)

Three cases therefore need to be sorted out:

1) 2<t_1; 0 < t_2 \le 2

Its obvious that 2t_1 \ge t_2^2 holds

2) 0 < t_1,t_2 <2

I will argue that either 2t_1 \ge t_2^2 or 2t_2 \ge t_1^2.

If to the contrary, we have both 2t_1 < t_2^2 and 2t_2 < t_1^2, then multiplying, we obtain t_1t_2>4 which is not possible as 0 < t_1,t_2 <2

3) 2<t_1, t_2

Now there is no guarantee that either of the inequalities will hold. But, we use here the result obtained by some of you that f(2t) = f(2t^2) or written in another way f(t) = f\left(\frac{t^2}{2} \right)

When t>2, the sequence t_{n} = \frac{t_{n-1}^2}{2} is a strictly increasing sequence

This tells us that we can substitute t₁ with a large enough t to satisfy 2t \ge t_2^2

Thus all possible cases have been disposed off.

Hence, to summarise, given any t_1, t_2 \in \mathbb{R}^+, we can find equivalently t_1', t_2' \in \mathbb{R}^+ such that

1) f(t_1) = f(t_1'); f(t_2) = f(t_2')

2) \exists \ x,y \in \mathbb{R}^+ with x+y = t_1'; x^2+y^2 = t_2'

so that we may conclude that f(t_1) = f(t_2)

Since this is true for any arbitrary t₁, t₂, we must have f as the constant function

u got f(2x)= f(2x²)

u must have,

x + y = x² + y²

x -x²=y²-y

x(1 - x) = y(y - 1)

so , x = y f(2x)=f(2x²)

but wt about x = y - 1 , 1- x = y & 1-x =y-1 . they are also possible..

how they can be neglected...

tell me dude.........

sir nothing is entering my mind...plz explain

if x+ y = x²+ y²

put x + y =t

( x + y)² =t ²

x²+ y² +2xy = t²

t + 2 xy = t²

t² - t -2xy = 0

so 1+ 8 xy ≥ 0

xy ≤ 18

now wt to do ... can anyone continue.....

equating x+y = x² + y²

=> x+y= (x+y)² - 2xy

putting x+y=t ,

t² - (2x+1)t + 2x² = 0

putting disc ≥ 0,

4x² - 4x - 1 ≤ 0, which gives 1-√22 ≤ x ≤ 1 + √22

But since x,y Σ R⁺, the relation x+y = x² + y² can only be satisfied the ordered pair (1,1)

hence either f(x) = constant

or f(x) be any function which satisfies given relation for (1,1)

cool down ashish bhai, bp high ho jana hai.

jaini maine solve to kiya tujhse to ans. bhi nahi nikla.........

profit bhaiya waise answer hai kya?coz i hav got f(2x)=f(2x²) ,after dis i don no wat to do?

I will respond only to a more detailed answer. One liners, guesses etc. will be contemptuously ignored

prophet answer kya hai??????/