The ratio of the areas of two regions of the curve `C_(1)-=4x^(2)+pi^(2)y^(2)=4pi^(2)` divided by the curve `C_(2)-=y=-(sgn(x-(pi)/(2)))cosx` (where sgn (x) = signum (x)) is

Options

A. `(pi^(2)-2)/(pi^(2)-2sqrt2)`
B. `(pi^(2)+2)/(pi^(2)-2sqrt2)`
C. `(pi^(2)+6)/(pi^(2)+3sqrt2)`
D. `(pi^(2)-1)/(pi^(2)-sqrt2)`

Answer

A

Explanation

`C_(1): 4x^(2)+pi^(2)y^(2)=4pi^(2)`
`therefore" "C_(1):(pi^(2))/(pi^(2))+(y^(2))/(4)=1`
Curve `C_(2):y'=sgn(x-(pi)/(2))cosx`
`={{:(-cosx",",xgt(pi)/(2)),(0",",x=(pi)/(2)),(cosx",",xlt(pi)/(2)):}`

Area `A_(1) and A_(2)` are equal.
`rArr" Required ratio"=(pi^(2)-2)/(pi^(2)+2)`