Discuss the number of roots of the equation `e(k-xlogx)=1` for different value of `kdot`

Explanation

We have `e(k-x log_(e)x)=1`
`rarr k-(1)/(e )=x log_(e)x`
Then equation (1) has solution where graph of y =x loge x
and `y=k-1/e` interesect
Consider the function `f(X) =xlog_(e)x`
`f(x) =1 + log_(e)x`
`f(X) =0 rarr x=1//e`
`f(X)=1//x rarr f(1//e)=egt0`
Also `underset(xrarr0)lo in x log_(e) = underset(xrarr0)lim (log_(e)x)/(1/x)=underset(xrarr0)lim (1)/(x) (1)/(1) (-x^(2))=-underset(xrarr0) lim x=0`
From the above information the graph of `f(X) =x log_(e)` x is as follows

From graph equation (1) has two has two distinct roots if line `y =(1)/(e)` cuts the graph between `-(1)/(e)` and 0
i.e `-(1)/(e) ltk-(1)/(e)lt0 or 0lt k lt (1)/(e)`
Equation has no roots if `k-(1)/(e)lt-(1)/(e)rarrk lt 0`
Equation has one root if `k-(1)/(e)=-(1)/(e)`
or `k-(1)/(e) ge 0 i.e jk =0 or kge (1)/(e)`