Fun with Functions!

Algebra Level 4

Let f : R → R f: \mathbb {R \to R} such that:

{ f ( 1 ) = 1 f ( x + 5 ) ≥ f ( x ) + 5 f ( x + 1 ) ≤ f ( x ) + 1 \begin{cases} f(1)=1 \\ f(x+5) \ge f(x)+5 \\ f(x+1) \le f(x)+1 \end {cases}

Let g ( x ) = f ( x ) + 1 − x g(x)=f(x)+1-x . Then find g ( 2002 ) g(2002) .

Source: This is a problem from the China Mathematics Competition (2002).


The answer is 1.

Rishabh Tiwari by Rishabh Tiwari , 20, India

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1 solution

Rishabh Tiwari
Jun 17, 2016

First we will find f ( 2002 ) f (2002) .

From the conditions we have ,

f ( x ) + 5 ≤ f ( x + 5 ) ≤ f ( x + 4 ) + 1 ≤ f ( x + 3 ) + 2 ≤ f ( x + 2 ) + 3 ≤ f ( x + 1 ) + 4 ≤ f ( x ) + 5 f (x)+5 \ \le f (x+5) \ \le f (x+4)+1 \\ \le f (x+3)+2 \ \le f (x+2)+3 \\ \le f (x+1)+4 \ \le f (x)+5

Thus the equality holds for all.

So we have f ( x + 1 ) = f ( x ) + 1 f (x+1)=f (x)+1

Hence from f ( 1 ) = 1 f (1)=1 ,

We get, f ( 2 ) = 2 ; f ( 3 ) = 3 ; … ; f ( 2002 ) = 2002 f (2)=2 \ ; \ f (3)=3 \ ; \ \ldots \ ; \ f (2002)=2002 .

Therefore , g ( 2002 ) = f ( 2002 ) + 1 − 2002 = 2002 + 1 − 2002 = 1 g (2002) \ = \ f (2002)+1-2002 \\ =2002+1-2002 \\ = \color{#20A900}{\boxed {1}} .

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