Question

The equation of the tangent to the curve is $y=2\sin x+\sin 2x$ at the point $x=\frac{\pi }{3}$ is -

• A. $2y=\sqrt{3}$
• B. $3y=\sqrt{2}$
• C. $2y=3\sqrt{3}$
• D. $2y=3$

Answer 1

Answer: (C)

$2y=3\sqrt{3}$

$y=2\sin x+\sin 2x$
$\frac{dy}{dx}=2\cos x+2\cos 2x$
The slope of tangent at $x=\frac{\pi }{3}$ is, $m={\left(\frac{dy}{dx}\right)}_{x=\frac{\pi }{3}}=2\times \frac{1}{2}+2\times (-\frac{1}{2})=1-1=0$
Now at $x=\frac{\pi }{3};y=2\times \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}$
$y=\frac{\sqrt{3}}{2}\times 3$
So the point is $(\frac{\pi }{3},\frac{3\sqrt{3}}{2})$
Therefore, required tangent is,
$(y-\frac{3\sqrt{3}}{2})=0$
$\Rightarrow y=\frac{3\sqrt{3}}{2}$
$\Rightarrow 2y=3\sqrt{3}$
Hence, option 'C' is correct.