Question

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

Answer 1

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)^2-4 k >0 \Rightarrow k ^2-6 k -4 k +9>0 \Rightarrow k ^2-10 k +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.