The equation ex−1 + x - 2 = 0 has
The correct option is A one real root
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
ϵ (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 no real root other than 1.