Find the number of positive integral values of $k$ for which $kx^2+(k-3)x+1<0$ for at least one positive $x$.
Step-by-Step Solution
Key Concept: For $k>0$: need discriminant $>0$ and at least one positive root. Disc $=(k-3)^2-4k>0\Rightarrow k>9$. Sum of roots $=-(k-3)/k<0$ for $k>3$, product $=1/k>0$ — both roots negative. So no positive $x$. For $k=0$: $-3x+1<0$ for $x>1/3$ ✓, but $k=0$ not positive integer.
0 positive integral values.
Correct Answer: 0