functions!!

try drawing the graphs
it'll be helpful!!

Is the answer option 3 ?

I'm getting the only solution as ln2 .

If you let y=e^x, then the equation is f(y) = y^3 + 2y^2+1-\frac{2}{y} = 0

We note that f'(y)>0. Hence it has at most one solution.

As y \rightarrow 0, f(y) \rightarrow -\infty and as y \rightarrow \infty, f(y) \rightarrow \infty

So there exists a positive root to the given equation ( we need +ve as y=e^x>0)

Hence the given equation has exactly one root