Question

Let $a,b,c$ be three non-zero real numbers such that the equation $\sqrt{3}acosx+2bsinx=c, \, x\in \left[- \frac{\pi }{2} , \frac{\pi }{2}\right]$ has two distinct real roots $\alpha \, \& \, \beta $ with $\alpha +\beta =\frac{\pi }{3}$ . Then the value of $\frac{4 b}{a}$ is:

Answer 1

$\sqrt{3}cos x+\frac{2 b}{a}sin ⁡ x=\frac{c}{a}$
Now, $\sqrt{3}cos \alpha +\frac{2 b}{a}sin ⁡ \alpha =\frac{c}{a}$ .....(i)
$\sqrt{3}cos \beta +\frac{2 b}{a}sin ⁡ \beta =\frac{c}{a}$ .....(ii)
$\sqrt{3}\left[cos \alpha - cos ⁡ \beta \right]+\frac{2 b}{a}\left(sin ⁡ \alpha - sin ⁡ \beta \right)=0$
$\sqrt{3}\left[- 2 sin \left(\frac{\alpha + \beta }{2}\right) sin ⁡ \left(\frac{\alpha - \beta }{2}\right)\right]+\frac{2 b}{a}\left[2 cos ⁡ \left(\frac{\alpha + \beta }{2}\right) sin ⁡ \left(\frac{\alpha - \beta }{2}\right)\right]$ =0
$-\sqrt{3}+2\sqrt{3}.\frac{b}{a}=0$
$\frac{4 b}{a}=2$