The area under the curve `y+sqrt(3) sin x=sin 2x` between the ordinates `x=0` and `x=(pi)/6` is

Options

A. `1/4(sqrt(7)-4)`
B. `1/4(7-4sqrt(3))`
C. `1/4(7-sqrt(3))`
D. None of these

Answer

B

Explanation

The curve `y+sqrt(3)sinx=sin2x` intersects X-axis `(y=0)` at the points where `2sin x cosx-sqrt(3)sin x=0`
or `sinx(2cosx-sqrt(3))=0`
or `sinx=0` or `cos x=(sqrt(3))/2`
`x=0,pi,..,(pi)/6,(11pi)/6`………..
`:.` Curve does not intersect `X`-axis
between 0 and `(pi)/6`
`:.` Required area of the region
`=int_(0)^(pi//6^(-))(sin 2x-sqrt(3)sinx)dx`
`=((7-4sqrt(3)))/4`