The maximum value of function `f(x)=sinx(1+cosx),xiniR` is

Options

A. `(3^(3//2))/(4)`
B. `(3^(5//3))/(4)`
C. `(3)/(2)`
D. `(3^(7//5))/(4)`

Answer

A

Explanation

Given, `f(x)=sinx(1+cosx)`
`impliesf(x)="sin x"(1)/(2)sin2x`
On differentiating w.r.t. x, we get
`f'(x)=cosx+cos2x`
Put f'(x)=0
cosx+cos2x=0.
`implies2cos((3x)/(2))cos((x)/(2))=0`
`implies"cos"(x)/(2)=0,"cos"(3x)/(2)=0`
`impliesx=pi,x=(pi)/(3)`
Now, `f"(x)=-sinx-2sin2x`
At `x=(pi)/(3),f"(x)=-"sin"(pi)/(3)-2"sin"(2pi)/(3)`
`=-(sqrt(3))/(2)-sqrt(3)lt0`, maxima
`therefore` Maximum value
`f((pi)/(3))=sin((pi)/(3))(1+"cos"(pi)/(3))`
`=(sqrt(3))/(2)(1+(1)/(2))=(3sqrt(3))/(4)=(3^(3//2))/(4)`