Question

If roots of the equation $2x^2-4x+2sin\theta -1=0$ are of opposite sign, then $\theta$ belongs to

• A. $(\frac{\pi }{6},\frac{5\pi }{6})$
• B. $(0,\frac{\pi }{6})\cup (\frac{5\pi }{6},\pi )$
• C. $(\frac{13\pi }{6},\frac{17\pi }{6})$
• D. none of these

Answer 1

The correct option is B $(0,\frac{\pi }{6})\cup (\frac{5\pi }{6},\pi )$

Solution : $2x^2-4x+2sin\theta -1=0$
coefficient of $x^2$ is positive
$f\left(0\right)<0$
$2sin\theta -1<0$
$sin\theta <\frac{1}{2}$
$\theta \in (0,\frac{\pi }{6})\cup (\frac{5\pi }{6},\pi )$