The vertices of a triangle are (0,0), `(x ,cosx),` and `(sin^3x ,0),w h e r e0

Options

A. `3sqrt(3)/(32)`
B. `(sqrt(3))/(32)`
C. `(4)/(32)`
D. `6 sqrt(3)/(32)`

Answer

1

Explanation

f(x)=`(sin^(3)xcosx)/(2)`
`therefore f(x)=(3 sin^(2)xcos^(23)x-sin^(4)x)/(2)`
`f(x) =0 rarr 3 sin^(2)x cos^(2)x - sin^(4)x=0`
or `3 cos^(2)x-sin^(2)x=0`
or `4cos^(2)x-1=0`
or `cosx=1/2`
or `x=(pi)/(3)` which is the point of maxima
`therefore f_(max)=(sqrt(3)//2)^(3)(1//2)/(2)=(3sqrt(3))/(32)`