M13-23

Official Solution:


Is \(a\) positive?

(1) \(x^2-2x+a\) is positive for all \(x\):

It's crucial to notice that we are given that the expression is positive not just for some value of \(x\) but for every value of \(x\).

Next, since the coefficient of \(x^2\) is positive, \(f(x)=x^2-2x+a\) represents an upward parabola. Since \(f(x)\) is positive for all \(x\), it means that the parabola lies entirely above the x-axis and has no intersections with it. Therefore, the equation \(x^2-2x+a=0\) has no real roots.

For a quadratic equation not to have real roots, its discriminant must be negative. Thus, we have \(discriminant = 2^2 - 4a = 4 - 4a \lt 0\), which simplifies to \(1-a \lt 0\) and finally to \(a \gt 1\). Sufficient.

(2) \(ax^2+1\) is positive for all \(x\):

Notice that, if \(a > 0\), then \(ax^2+1\) is positive for every \(x\). But the trick here is that \(ax^2+1\) is also positive for every \(x\) when \(a = 0\), giving a NO answer to the question! Not sufficient.


Answer: A