Question

Find the number of positive integral values of $k$ for which $k{x}^2+(k-3)x+1<0$ for atleast one positive $x$.

Answer 1

Answer: 0

Let $f(x)=k^2+(k-3)x+1,k\in N$.
Here, $f(0)=1$ and product of roots $=\frac{1}{k}>0$
Now, for $f(x)<0$ for atleast one positive $x$, sum of roots $>0$
$\Rightarrow \frac{-(k-3)}{k}>0\Rightarrow \frac{k-3}{k}<0\Rightarrow k\in (0,3)$ ....(1)

Also, $D>0$
$\Rightarrow (k-3{)}^2-4k>0\Rightarrow k^2-6k-4k+9>0\Rightarrow k^2-10k+9>0$
$\Rightarrow (k-9)(k-1)>0\Rightarrow k\in (-\infty ,-1)\cup (9,\infty )$....(2)
$\therefore (1)\cap (2)$
$\Rightarrow k\in (0,1)$
Hence, number of positive integral values of $k$ is zero.