let $x<0$ such that $kx^2-2x>3k-1$ then find the range of $k$

Question

let $x<0$ such that $kx^2-2x>3k-1$ then find the range of $k$

My Try :

$$kx^2-2x>3k-1 \\kx^2-2x-3k+1 >0$$

So $k>0$ and $$\Delta =(-2)^2-4(k)(-3k+1)<0 \\ 3k^2+1 < k$$

that no solution exist . Thus $k>0$ . it is right ?

Answer 1

You need that for $x< 0$, $$kx^2-2x-3k+1 > 0$$

This means that either of the following must be satisfied

1. $k>0 \land \frac{2}{2k} > 0 \land f(0) >0$ ($k>0$ means its parabola opens upward and$\frac{2}{2k}$ is abscissa of vertex of parabola.) From here you get $k \in (0, \tfrac{1}{3})$
2. $k > 0 \land \Delta \lt 0$. So we get $$4+4k(3k-1) <0 \\3k^2-k+1 < 0$$ This is not true for any $k$ as $\Delta <0$ and coefficient of $k^2$ that is $3$ is positive. So this condition is useless.

When you take union of these conditions, you get $k\in (0, \tfrac{1}{3}) $