Question

If $(a,0);a>0$ is the first point where the curve $y=\sin 2x-\sqrt{3}\sin x$ cuts the $x-$axis and $A$ is the area bounded by this part of the curve, the origin and the positive $x-$axis, then

• A. $4A+8\cos a=7$
• B. $4A+8\sin a=7$
• C. $4A-8\sin a=7$
• D. $4A-8\cos a=7$

Answer 1

The correct option is A $4A+8\cos a=7$

$(a,0)$ lies on the given curve.
$\therefore \sin 2a-\sqrt{3}\sin a=0$
$\Rightarrow \sin a(2\cos a-\sqrt{3})=0$
$\Rightarrow \sin a=0$ or $\cos a=\frac{\sqrt{3}}{2}$
$\Rightarrow a=\frac{\pi }{6}$ as $a>0$ and $a$ is the first point of intersection with positive $x-$axis.
$A=\underset{0}{\overset{\pi /6}{\int }}(\sin 2x-\sqrt{3}\sin x)dx$
$={(-\frac{\cos 2x}{2}+\sqrt{3}\cos x)}_0^{\pi /6}=(-\frac{1}{4}+\frac{3}{2})-(-\frac{1}{2}+\sqrt{3})=\frac{7}{4}-\sqrt{3}=\frac{7}{4}-2\cos a(\because 2\cos a=\sqrt{3})$
$\therefore 4A+8\cos a=7$