Functional Inequality

Algebra Level 4

Let f : R R f:R→R be a function such that f ( x + 19 ) f ( x ) + 19 f(x+19)≤f(x)+19 and f ( x + 94 ) f ( x ) + 94 f(x+94)≥f(x)+94 for all real x x .Find the value of f ( 625 ) f ( 0 ) f ( 25 ) f ( 0 ) \sqrt{\frac{f(625)-f(0)}{f(25)-f(0)}}


The answer is 5.

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2 solutions

Souryajit Roy
Jul 15, 2014

Write the inequality in the form f ( x ) f ( x + 19 ) 19 f(x)≥f(x+19)-19 .

Replacing x x by x 19 x-19 , we get f ( x 19 ) f ( x ) 19 f(x-19)≥f(x)-19 .

Similarly, f ( x 94 ) f ( x ) 94 f(x-94)≤f(x)-94 .

Write the inequality in the form f ( x ) f ( x + 19 ) 19 f(x)≥f(x+19)-19 .

Replacing x x by x 19 x-19 , we get f ( x 19 ) f ( x ) 19 f(x-19)≥f(x)-19 .

Similarly, f ( x 94 ) f ( x ) 94 f(x-94)≤f(x)-94 .

Using induction, one can get

a. f ( x + 19 n ) f ( x ) + 19 n f(x+19n)≤f(x)+19n and f ( x 19 n ) f ( x ) 19 n f(x-19n)≥f(x)-19n .

b. f ( x + 94 n ) f ( x ) + 94 n f(x+94n)≥f(x)+94n and f ( x 94 n ) f ( x ) 94 n f(x-94n)≤f(x)-94n where n n is an integer.

Now, 1 = 5 1=5 x 19 94 19-94

So, f ( x + 1 ) = f ( x 94 + 5 f(x+1)=f(x-94+5 x 19 ) 19) f ( x 94 ) + 5 ≤f(x-94)+5 x 19 19 f ( x ) 94 + 5 ≤f(x)-94+5 x 19 19 = = f ( x ) + 1 f(x)+1 .

Similarly,using 1 = 18 1=18 x 94 89 94-89 x 19 19 ,show that f ( x + 1 ) f ( x ) + 1 f(x+1)≥f(x)+1

Hence, f ( x + 1 ) = f ( x ) + 1 f(x+1)=f(x)+1 .

So, f ( n ) = f ( 0 ) + n f(n)=f(0)+n for any natural number n n .

the multiplication symbol may have gone a little down...however 5 5 x 19 19 means 5 multiplied for 19

Souryajit Roy - 6 years, 11 months ago
Prabhat Sharma
Jul 15, 2014

Conflict in equations leads to conclude that eqation is f(x)=x ,now its easy to give ans

Basically, you are wrong since at first one will get f(x+1)=f(x)+1.Hence f(x)=x+f(0).

Souryajit Roy - 6 years, 11 months ago

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