If `(a,0)`, agt 0, is the point where the curve `y=sin 2x-sqrt3 sin x` cuts the x-axis first, A is the area bounded by this part of the curve, the origin and the positive x-axis. Then

Options

A. `4A+8 cos a=7`
B. `4+8sin a=7`
C. `4A-8sin a=7`
D. `4A-8cos a=7`

Answer

A

Explanation

(a, 0) lies on the given curve
`therefore" "0=sin 2a-sqrt3 sina`
`rArr" "sin a=0 or cos a = sqrt3//2`
`rArr" "a=(pi)/(6)" (as a gt 0 and the first point of intersection with positive X-axis)"`
Required area `A=int_(0)^(pi//6)(sin2x-sqrt3 sin x)dx`
`=(-(cos2x)/(2)+sqrt3 cosx)_(0)^(pi//6)`
`=(-(1)/(4)+(3)/(2))-(-(1)/(2)+sqrt3)`
`=(7)/(4)-2 cos a`
`rArr" "4A+8 cos a=7`