Let `f(x)="max"{sin x,cos x,1/2}`, then determine the area of region bounded by the curves `y=f(x)`, X-axis, Y-axis and `x=2pi`.

Options

A. `sqrt2-sqrt3+(5pi)/(12)`
B. `sqrt2+(sqrt3)/(2)+(5pi)/(12)`
C. `sqrt2+sqrt3+(5pi)/(12)`
D. none of these

Answer

b

Explanation

`f(x)=max{sinx,cosx,1/2} ={{:(cos",",0lexle(pi)/(4)),(sinx",",(pi)/(4)lexle(5pi)/(6)),((1)/(2)",",(5pi)/(6)lexle(5pi)/(3)):}`
Let A be the required area. Then,
`A=underset(0)overset(5pi//3)(int)f(x)dx`
`impliesA=underset(0)overset(pi//4)(int)cosx dx+underset(pi//4)overset(5pi//6)(int)sinx dx+underset(5pi//6)overset(5pi//3)(int)1/2dx`
`impliesA=[sinx]_(0)^(pi//4)-[cosx]_(pi//4)^(5pi//6)+1/6((5pi)/(3)-(5pi)/(6))`
`impliesA=(1)/(sqrt2)-(-(sqrt3)/(2)-(1)/(sqrt2))+(5pi)/(12)=(5pi)/(12)+(sqrt3)/(2)+sqrt2`