A variant of 2003 CMO Problem 3

Algebra Level 3

Given the function f ( x ) = 1 1 + x + 1 1 + a + a x a x + 8 f(x)=\dfrac{1}{\sqrt{1+x}}+\dfrac{1}{\sqrt{1+a}}+\sqrt{\dfrac{ax}{ax+8}} , x ∈ ( 0 , + ∞ ) x \in (0,+\infty) .

Is it always true that 1 < f ( x ) < 2 1<f(x)<2 for all a ∈ ( 0 , + ∞ ) a \in (0,+\infty) ?

Yes No

Alice Smith by Alice Smith , 20, China

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1 solution

Tom Engelsman
Apr 8, 2021

Quick -n- dirty check with limits:

l i m x → 0 + , a → 0 + f ( x ) = 1 1 + 0 + 1 1 + 0 + 1 − 8 0 2 + 8 → 1 + 1 + 0 = 2 lim_{x \rightarrow 0^{+}, a \rightarrow 0^{+}} f(x) = \frac{1}{\sqrt{1+0}} + \frac{1}{\sqrt{1+0}} + \sqrt{1 - \frac{8}{0^2+8}} \rightarrow 1 + 1 + 0 = 2 ;

l i m x → ∞ , a → ∞ f ( x ) = 1 ∞ + 1 ∞ + 1 − 8 ∞ → 0 + 0 + 1 = 1 lim_{x \rightarrow \infty, a \rightarrow \infty} f(x) = \frac{1}{\infty} + \frac{1}{\infty} + \sqrt{1 - \frac{8}{\infty}} \rightarrow 0 + 0 + 1 = 1 .

Thus, it is TRUE that f ( x ) ∈ ( 1 , 2 ) f(x) \in (1,2) for all x , a ∈ ( 0 , ∞ ) . x,a \in (0, \infty).

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