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Algebra Level 5

Let ω ( x ) = a x + b c x + d \omega(x)=\dfrac{ax+b}{cx+d} , where a , b , c , d a,b,c,d are real numbers.

Given that: ω ( ω ( ω ( 1 ) ) ) = 1 \omega(\omega(\omega(1)))=1 and ω ( ω ( ω ( 2 ) ) ) = 2015 \omega(\omega(\omega(2)))=2015 .

Find the sum of possible values of ω ( 1 ) \omega(1) .


The answer is 1.000.

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Khang Nguyen Thanh by Khang Nguyen Thanh , 27, Vietnam

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1 solution

Note that w(x) is strictly monotonous. Suppose that w(x) is strictly crescent (without losing of generality). If w(1)>1 , hence we have w(w(1))>w(1) (because it is strictly crescent) and that implies that w(w(w(1)))>w(1)>1, which is an absurd. Following the same path, if w(1)<1 we also have an absurd. Therefore the only suitable solution is that one where w(1)=1.

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I think your solution is not correct. Because ω ( x ) \omega(x) is not continous function.

Khang Nguyen Thanh - 5 years, 10 months ago

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