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Official Solution:
Question: is \(a \gt 0\)?
(1) \(x^2-2x+a\) is positive for all \(x\).
\(f(x)=x^2-2x+a\) is a function of an upward parabola (as coefficient of \(x^2\) is positive). We are told that it's positive for all \(x\), so \(f(x)=x^2-2x+a \gt 0\), which means that this function is "above" X-axis OR in other words parabola has no intersections with X -axis OR equation \(x^2-2x+a=0\) has no real roots.
In order for a quadratic equation to have no real roots its discriminant must be negative: \(D=2^2-4a=4-4a \lt 0\), which simplifies to \(1-a \lt 0\) and finally to \(a \gt 1\). Sufficient.
(2) \(ax^2+1\) is positive for all \(x\):
\(ax^2+1 \gt 0\). Now, when \(a \ge 0\) this expression is positive for all \(x\). So, \(a\) can be zero too. Not sufficient.
Answer: A
I have trouble solving this question to get the correct statement.
If x = 3,
then 9-2(3)+a > 0 --> 3+a > 0 (per statement 1). Now , here a can be positive or negative and the expression will still be positive.
Just wondering, how can this statement be sufficient then.
Please help.
Thanks !