The equation ex−1+x−2=0 as

Clearly, x = 1 satisfies the given equation.  Assume that f(x)= ex−1 + x−2  =  0 has a real root α  other than x = 1. We may suppose that α>1 (the case α<1 is exactly similar).  Applying Rolle's theorem on [1,α] (if α<1 apply the  theorem on [α,1], we get β∈(1,α) such that f′(β)=0. But f′(β)=eβ−1+1, so that eβ−1=−1, which is not possible. Hence there is noreal root other than 1.