Consider the two curves `C_(1):y=1+cos x and C_(2): y=1 + cos (x-alpha)" for "alpha in (0,(pi)/(2))," where "x in [0,pi].` Also the area of the figure bounded by the curves `C_(1),C_(2), and x=0` is same as that of the figure bounded by `C_(2),y=1, and x=pi`.
The value of `alpha` is

Options

A. `(pi)/(4)`
B. `(pi)/(3)`
C. `(pi)/(6)`
D. `(pi)/(8)`

Answer

C

Explanation

`1+ cos x =1 + cos (x-alpha)`
`"or "x=alpha-x or x=(alpha)/(2)`
`"Now "overset(alpha//2)underset(0)int((1+cos x)-(1+cos (x-alpha)))dx`
`=-underset((pi)/(2)+alpha)overset(pi)int(1-(1+cos (x-alpha)))dx`
`"or "[ sin x - sin (x-a)]_(0)^(alpha//2)=[sin (x-alpha)]_(pi)^((pi)/(2)+alpha)`
`"or "[sin""(alpha)/(2)-sin (-(alpha)/(2))]-[0-sin (-alpha)]`
`=sin ((pi)/(2))-sin (pi-alpha)`
`"or "2 sin""(alpha)/(2)-sin alpha=1- sin alpha`
`"Hence, "2sin""(alpha)/(2)=1 or alpha=(pi)/(3).`