Play with functions

Algebra Level 5

Let $f: \mathbb R \to \mathbb R$ be a function satisfying $f(x-f(y)) = f(f(y))+xf(y)+f(x)-1$ where $x$ and $y$ are real numbers.

Find the value of $f(1) - f(2)$ .

The answer is 1.500.

4 solutions

Surya Prakash
Dec 13, 2015

Let $k$ be a real number such that $f(k) = 0$ .

Now, let $y=k$ .

Then, $f(x-f(k)) = f(f(k)) + xf(k) + f(x) - 1 \implies f(x) = f(0) + f(x) - 1 \implies f(0) = 1$ .

Now, let $y = 0$ . Then, $f(x-f(0)) = f(f(0)) + xf(0) + f(x) - 1 \implies f(x-1) = f(1) +x + f(x) -1$

In the above equation, let $x=1$ , then,

$f(0) = f(1) +1 + f(1) - 1 \implies f(1) = \dfrac{1}{2}$

So, $f(x-1) - f(x) = f(1) + x - 1 = x - 1 + \dfrac{1}{2} = x - \dfrac{1}{2}$

Take $x=2$ in the above equation. So, we get

$f(1) - f(2) = 2 - \dfrac{1}{2} = \boxed{\dfrac{3}{2}}$

How do you know there exists a real number $k$ such that $f(k) = 0$ ? If you are going to use this result, you must prove it.

Jon Haussmann - 5 years, 6 months ago

Because the function is clearly onto. U can vary x and y over all reals

Shrihari B - 5 years, 5 months ago

Why is the function "clearly onto"? You must prove this.

In fact, it can be shown that $f(x) = 1 - \frac{x^2}{2}.$ This function is not onto, so your assertion that the function is onto is completely unsupportable.

Jon Haussmann - 5 years, 5 months ago

@Jon Haussmann – sir @Jon Haussmann please can prove this for some k f(k)=0 ? plz @Pi Han Goh help me out !!!

Rudraksh Sisodia - 4 years, 9 months ago

Same method!

Adarsh Kumar - 5 years, 6 months ago

prove that the function is surjective

rajdeep brahma - 2 years, 11 months ago

Amrit Anand
Dec 14, 2015

 First Time I Am Doing This Type Of Substitution. Is It Right Way To Solve?

This is all right. Note that the substitution $x= 2 f(y)$ implicitly implies that $x/2$ belongs to the range of the function $f$ . Thus, the conclusion $f(x)=1-x^2/2$ is valid only for those values of $x$ , for which $x/2$ is in range of $1-x^2/2$ . It can be easily seen that both $1/2$ and $1$ are in range of $1-x^2/2$ and that's all we require here.

Abhishek Sinha - 5 years, 6 months ago

I do not get why is it mandatory to have a y for every x such that f(y)=x/2

rajdeep brahma - 2 years, 11 months ago

go through comment below by Abhishek Sinha .......

Amrit Anand - 2 years, 11 months ago

Jon Haussmann
Dec 14, 2015

This is from the 1999 IMO .

Yeah it is given in Rajeev Manocha

Aniswar S K - 4 years, 1 month ago

Andreas Wendler
Dec 17, 2015

Set f(y)=x => f(0)=2 * f(x) + x^2 - 1. Now first for x=1 obtain f(0)=2 * f(1) and for x=2 get f(0)=2 * f(2) + 3. Calculate the difference of these equations to determine 3/2.

Prove f is onto first

rajdeep brahma - 2 years, 11 months ago

Play with functions

The answer is 1.500.

4 solutions

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