An algebra problem by A Former Brilliant Member

Algebra Level 4

Let f : ( 0 , ∞ ) → R f:(0 ,\infty) \rightarrow R be a function such that

(a) f f is strictly increasing,

(b) f ( x ) > − 1 x f (x) > -\dfrac 1x for all x > 0 x>0 and

(c) f ( x ) f ( f ( x ) + 1 x ) = 1 f(x)f(f(x) + \frac 1x)= 1 for all x > 0. x>0.

f ( 1 ) = a − b c f(1)= \dfrac {a - \sqrt b}{c} where a , b , c a, b, c are coprime positive integers and b b is square free.

Find a + b + c . a + b + c.


The answer is 8.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

Let t = f ( 1 ) t = f(1)

Put x = 1 x=1 in (c) we get, t f ( t + 1 ) = 1 ⇒ t ≠ 0 t f(t + 1) = 1 \Rightarrow t \neq 0

f ( 1 + t ) = 1 t f(1 +t) = \dfrac 1t

Put x = t + 1 x = t + 1 in (c) we get, f ( t + 1 ) f ( f ( t + 1 ) + 1 t + 1 ) = 1. f(t +1)f(f(t+1) + \frac {1}{t+1}) = 1.

Then, f ( 1 t + 1 t + 1 ) = t = f ( 1 ) . f(\frac 1t + \frac {1}{t+1} ) = t = f(1).

Since, f ( x ) f(x) is strictly increasing, 1 t + 1 t + 1 = 1 \dfrac 1t + \dfrac {1}{t+1} = 1

Solving, we get t = 1 + 5 2 , 1 − 5 2 t = \dfrac {1 + \sqrt 5}{2} , \dfrac {1 - \sqrt 5}{2}

If t = 1 + 5 2 > 0 t = \dfrac {1 + \sqrt 5}{2} >0 then 1 < t , f ( t ) < f ( 1 + t ) = 1 t < 1 1 < t , f(t) < f(1 + t) = \dfrac 1t <1 which is a contradiction.

Hence, f ( 1 ) = t = 1 − 5 2 = a − b c f(1) = t = \dfrac {1 - \sqrt 5}{2} = \dfrac {a - \sqrt b}{c}

Therefore, a + b + c = 1 + 5 + 2 = 8 a + b + c = 1 + 5 + 2 = \boxed{8}

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