If `y=f(x)` is the solution of equation `ydx+dy=-e^(x)y^(2)`dy, f(0)=1 and area bounded by the curve `y=f(x), y=e^(x)` and x=1 is A, then

Options

A. curve y=f(x) is passing through `(-2,e)`.
B. Curve `y=f(x)` is passing through `(1,1//e)`
C. curve `y=f(x)` is passing through `(1,1//3)`
D. `A=e+2/sqrt(e )-3`

Answer

A, D

Explanation

`ydx+dy=-e^(x)y^(2)dy`
`rArr (e^(-x)ydx+e^(-x)dy)/(y^(2))=-dy`
`rArr d(e^(-x)/y)=dy`
`rArr e^(-x)=y^(2)+cy`
`therefore f(0)=1, therefore c=0`
`rArr e^(-x)=y^(2)`
`rArr y=e^(-x//2)`
`rArr A = int_(0)^(1)(e^(x)-e^(-x//2))dx=e+2/sqrt(e)-3`