y = f(x) which has differential equation `y(2xy + e^(x)) dx - e^(x) dy = 0` passing through the point (0,n 1). Then which of the following is/are true about the function?

Options

A. `x = 1 + sqrt(2)` is point of local maxima
B. `x = 1 - sqrt(2)` is point of local minima
C. `underset(x rarr oo)("lim") f (x) = - oo`
D. `underset(x rarr -oo)("lim") f(x) = 0`

Answer

A, B, C, D

Explanation

`" "y(2xy + e^(x))dx - e^(x) dy = 0`
`therefore" "y(2xy + e^(x)) = e^(x) (dy)/(dx)`
`therefore" "(dy)/(dx)-y=(2xy^(2))/(e^(x))`
`therefore" "(1)/(y^(2))(dy)/(dx)-(1)/(y)=(2x)/(e^(x))`
Put `(1)/(y) = v rArr (1)/(y^(2))(dy)/(dx)=-(dv)/(dx)`
`therefore" "-(dv)/(dx)-v=(2x)/(e^(x))`
`therefore" "(dv)/(dx)+b=-(2x)/(e^(x))`
`I.F. = e^(int 1 dx) = e^(x)`
Solution `v*e^(x) = int (-2x)/(e^(x))e^(x) dx + c`
`or" "(e^(x))/(y) = -x^(2) + c`
Curve passes through the point (0, 1)
`rarr" "c = 1`
`therefore" ""Curve is y"^(-1) e^(x) = 1 - x^(2)`
`or" "y=(e^(x))/(1-x^(2))`
Clearly `underset(x rarr oo)("lim") f(x) = -oo and underset(x rarr - oo)("lim") f(x) = 0`
Also `y'=(e^(x)(1-x^(2)+2x))/((1-x^(2))^(2))`
`y' = 0 rArr x = 1 +- sqrt(2)`
From sign scheme, we can check that x = 1 + `sqrt(2)` is the point of local maxima and x = 1 - `sqrt(2)` is point of local minima.