Question

The range of $k$, for which the inequality $k{x}^2-3kx+3<0$, $($where $x\in R)$ holds true is

• A. $(-\infty ,0)\cup (\frac{4}{3},\infty )$
• B. $(\frac{4}{3},\infty )$
• C. $R$
• D. $\varphi$

Answer 1

The correct option is D $\varphi$

For, $k{x}^2-3kx+3<0$
$k\le 0$
Now, if $k=0$, we get
$3<0$ Not possible
Now, $k<0\cdots \left(1\right)$
And
$D<0$
$\Rightarrow 9k^2-12k<0$
$\Rightarrow k(3k-4)<0$
$\Rightarrow k\in (0,\frac{4}{3})\cdots \left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$, we get
$k\in \varphi$