Minimum integral value of k for which the equation `e^(x)=kx^(2)` has exactly three real distinct solution,

Explanation

Given equation is `e^(x)=kx^(2)`
`k=(e^(x))/(x^(2)) =f(x)` say
`F(x)=((x-2)e^(x))/(.^(3))`
f(x)=0
x=2 which is point minima
f(2)=`(e^(2))/(4)`
`underset(xrarr00)lim(e^(x))/(x^(2))=00`
`underset(xrarr-00)lim(e^(x))/(x^(2))=0` and `underset(xrarr0)(lim(e^(x))/(x^(2))=00`
Further `f(x)gt 0, forall` x in R
From this information graph of f(x) is as shown below

For three real distinct solutions of equation(1) line y = k must intersect the graph above `(e(2))/(4)`
So `kgt (e^(2))/(4)`