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

Options

A. 1
B. 2
C. 3
D. 4

Answer

B

Explanation

Given equation is `e^(x)=kx^(2)`
`rArr" "k=(e^(x))/(x^(2))`
Now let `f(x)=(e^(x))/(x^(2)).`
`therefore" "f'(x)=((x-2)e^(x))/(x^(3))`
`f'(x)=0 rArr x=2`, which is point of minima.
Also `f(2)=(e^(2))/(4)`
`underset(xrarroo)(lim)(e^(x))/(x^(2))=oo" (using L'Hopital Rule twice)"`
`underset(xrarr-oo)(lim)(e^(x))/(x^(2))=0`
`underset(xrarr0)(lim)(e^(x))/(x^(2))=oo`
Futher `f(x)gt0, AA x in R.`
From this information, graph of f(x) is as shown below.

Thus, three real distinct solutions `kgt(e^(2))/(4),k in I.`
So, `k_("min")=2`.