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2017 AMC 12B Problem 2

Problem 2 of 25EasierAlgebra

Real numbers x,x, y,y, and zz satisfy the inequalities 0<x<1,1<y<0,0 \lt x \lt 1, \quad -1 \lt y \lt 0, and 1<z<2.1 \lt z \lt 2. Which of the following numbers is necessarily positive?

Answer choices

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Solution

Adding y>1y \gt -1 and z>1z \gt 1 gives y+z>0,y + z \gt 0, so y+zy + z is always positive. Each of the other four choices can be made negative: with x=18,x = \tfrac18, y=14,y = -\tfrac14, z=32,z = \tfrac32, every one of y+x2,y + x^2, y+xz,y + xz, y+y2,y + y^2, and y+2y2y + 2y^2 is negative. Thus, the correct answer is E.

More practice

Concepts: inequality · counterexample

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.