It is given the expression $y=\frac{x^2-2x+1}{x^2-2x+2}$, where $x$ is a variable. Prove that: (a) if $x_1$ and $x_2$ are two values of $x$, the $y_1$ and $y_2$ are the respective values of $y$ only if $x_1<x_2$, $y_1<y_2$; (b) when $x$ is varying $y$ attains all possible values for which $0\le y<1$.
Problem
Source: Bulgaria 1962 P1
Tags: algebra
Bluecloud123
25.06.2024 12:38
I dont really quite understand question a But here is the solution for b) : Let $x^2-2x+2=k$ we’ll get the discriminant equals $4k-4$ thus for every $k \geqslant 1$ there exist $x \in \mathbb{R}$ which $x^2-2x+2=k$ since $y=1- \frac{1}{x^2-2x+2}$ We can conclude that for all $0\leqslant y <1$ there exist $x \in \mathbb{R}$ which $y=\frac{x^2-2x+1}{x^2-2x+2}$
Mathzeus1024
26.06.2024 15:16
Bluecloud123 wrote: I dont really quite understand question a But here is the solution for b) : Let $x^2-2x+2=k$ we’ll get the discriminant equals $4k-4$ thus for every $k \geqslant 1$ there exist $x \in \mathbb{R}$ which $x^2-2x+2=k$ since $y=1- \frac{1}{x^2-2x+2}$ We can conclude that for all $0\leqslant y <1$ there exist $x \in \mathbb{R}$ which $y=\frac{x^2-2x+1}{x^2-2x+2}$ Property (b) can also be readily proven via differential calculus as well.
damyan
25.12.2024 14:14
a) is false because $y$ is symmetric across the horizontal line $x=-1$ and a minimum of $0$ at that point. Alternatively for the sake of a) being a part of the problem let's assume the question is for $x\geq-1$. Now after differentiating once we get $y'=\frac{2(x+1)}{((x+1)^2+1)^2}$. Now $y'\geq0$ on the given interval, so a) is proven. b) requires to show $\lim_{x\to\infty} y=1$ which is very straightforward after comparing coefficients of the leading term in the numerator and denominator.