From An Erroneous Problem

Algebra Level 3

Function f : R → R f: \mathbb {R \to R} satisfies

f ( x ) + f ( 1 − 1 x ) = 1 + x \large f(x) + f \left(1-\frac 1x\right) = 1+x

for x ≠ 0 x \ne 0 and x ≠ 1 x \ne 1 . Then 2017 − 2 f ( 2017 ) = 1 k ( k − 1 ) 2017 - 2f(2017) = \dfrac 1{k(k-1)} , where k k is a positive integer. Find k k .

A corrected problem from Priyanshu Mishra .


The answer is 2017.

Chew-Seong Cheong by Chew-Seong Cheong , 63, Malaysia

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1 solution

Chew-Seong Cheong
Aug 13, 2017

{ f ( x ) + f ( x − 1 x ) = 1 + x . . . ( 1 ) f ( x − 1 x ) + f ( 1 1 − x ) = 1 + x − 1 x . . . ( 2 ) f ( 1 1 − x ) + f ( x ) = 1 + 1 1 − x . . . ( 3 ) \begin{cases} f(x) + f \left(\dfrac {x-1}x \right) = 1+x & ...(1) \\ f \left(\dfrac {x-1}x \right) + f\left(\dfrac 1{1-x} \right) = 1+\dfrac {x-1}x & ...(2) \\ f \left(\dfrac 1{1-x} \right) + f(x) = 1+\dfrac 1{1-x} & ...(3) \end{cases}

From ( 1 ) − ( 2 ) + ( 3 ) : (1)-(2)+(3):

2 f ( x ) = 1 + x − 1 − x − 1 x + 1 + 1 1 − x = x − 1 + 1 x + 1 − 1 x − 1 = x + 1 x − 1 x − 1 = x − 1 x ( x − 1 ) \begin{aligned} 2f(x) & = 1+x-1- \frac {x-1}x + 1 + \frac 1{1-x} \\ & = x - 1 + \frac 1x + 1 - \frac 1{x-1} \\ & = x + \frac 1x - \frac 1{x-1} \\ & = x - \frac 1{x(x-1)} \end{aligned}

⟹ 2017 − 2 f ( 2017 ) = 1 2017 ( 2017 − 1 ) \begin{aligned} \implies 2017 - 2f(2017) & = \frac 1{2017(2017-1)}\end{aligned}

⟹ k = 2017 \implies k = \boxed{2017} .

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