Question

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

• A. $(0,\frac{1}{2})$
• B. $(-\infty ,0)\cup (\frac{1}{2},\infty )$
• C. $(\frac{1}{2},\infty )$
• D. $[0.\frac{1}{8})$

Answer 1

The correct option is A $(0,\frac{1}{2})$

Rewriting the equation ;$\Rightarrow x^2+2x+\frac{1}{2k}>0$

$\Rightarrow$descriminant <0 and coefficient of $x^2$$\Rightarrow k>0$

$\Rightarrow 4-2k<0\Rightarrow k<1/2$

$\therefore k\in (0,\frac{1}{2})$