Question

If $kx ^2+2 kx +\frac{1}{2}>0, \forall x \in R$, then complete set of values of $k$ is

• A. $\left(0, \frac{1}{2}\right)$
• B. $(-\infty, 0) \cup\left(\frac{1}{2}, \infty\right)$
• C. $\left(\frac{1}{2}, \infty\right)$
• D. $\left[0, \frac{1}{2}\right)$

Answer 1

Answer: (D)

$\text { Case-I: } k \neq 0 $
$k >0 \,\,\,\,\, D <0 $
$4 k ^2-4( k ) \cdot \frac{1}{2}<0 $
$\Rightarrow 4 k ^2-2 k <0 $
$\Rightarrow 2 k (2 k -1)<0 $
$\therefore k \in\left(0, \frac{1}{2}\right)$
$\text { Case-II: } k =0 $
$\frac{1}{2}>0 \text {, which is true } $
$\therefore \text { solution set for } k $