A function `f:R rarrR` is defined as
`f(x)=lim_(nrarroo) (ax^(2)+bx+c+e^(nx))/(1+c.e^(nx))` where f is continuous on R, then

Options

A. point (a, b, c) lies on line in space
B. point (a, b) represents the 2-dimensional Cartesian plane
C. Locus of point (a, c) and (c, b) intersect at one point
D. point (a, b, c) lies on the plane in space

Answer

A, B, C

Explanation

`f(x)=underset(nrarroo)(lim)(ax^(2)+bx+c+e^(nx))/(1+c.e^(nx))`
`={{:(underset(nrarroo)(lim)(ax^(2)+bx+c+(e^(x))^(n))/(1+c.(e^(x))^(n)),,xlt0),(underset(nrarroo)(lim)(ax^(2)+bx+c+(e^(x))^(n))/(1+c.(e^(x))^(n)),,x=0),(underset(nrarroo)(lim)((ax^(2)+bx+c)/((e^(x))^(n)))/((1)/((e^(x))^(n))+c),,xgt0):}`
`={{:(ax^(2)+bx+c,,xlt0),(1,,x=0),((1)/(c),,xgt0):}`
since f(x) is continuous function `AA x in R`
`therefore" "underset(xrarr0^(+))(lim)f(x)=underset(xrarr0^(-))(lim)f(x)=f(0)`
`rArr" "underset(xrarr0^(+))(lim)((1)/(c))=underset(xrarr0^(-))(lim)(ax^(2)+bx+c)=1`
`rArr" "(1)/(c)=1=c rArr c=1`
`therefore " "c=1, a, b in R`