MathDB
Vieta Theorem

Source: 2004 National High School Mathematics League, Exam One, Problem 15

March 18, 2020
function

Problem Statement

α,β\alpha,\beta are two different solutions to the equation 4x2−4tx+1=0(t∈R)4x^2-4tx+1=0(t\in\mathbb{R}), the domain of definition of the function f(x)=2x−tx2+1f(x)=\frac{2x-t}{x^2+1} is <aclass=′latex−hyperlink′href=′α<β′>α,β</a><a class='latex-hyperlink' href='\alpha<\beta'>\alpha,\beta</a>. (a) Find g(t)=max⁡f(x)−min⁡f(x)g(t)=\max f(x)-\min f(x). (b) Prove: for ui∈(0,π2)(i=1,2,3)u_i\in\left(0,\frac{\pi}{2}\right)(i=1,2,3), if sin⁡u1+sin⁡u2+sin⁡u3=1\sin u_1+\sin u_2+\sin u_3=1, then 1g(tan⁡u1)+1g(tan⁡u2)+1g(tan⁡u3)<346\frac{1}{g(\tan u_1)}+\frac{1}{g(\tan u_2)}+\frac{1}{g(\tan u_3)}<\frac{3}{4}\sqrt6.
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