BITSAT 2015 Paper

Let f(x) = $e^{x–1}$ + x – 2 check for x = 1 Then, f (1) = $e^0$ + 1 – 2 = 0 So, x = 1 is a real root of the equation f(x) = 0 Let x = α be the other root such that α > 1 or α < 1. Consider the interval [1,α] or [α,1]. Clearly f(1) = f(α) = 0 By Rolle’s theorem f ' (x) = 0 has a root in (1, α) or in (α, 1). But f ' (x) = $e^{x – 1}$ + 1 > 0, for all x. Thus, f ' (x) ≠ 0 , for any x ∊ (1,α) or x ∊ (α,1) , which is a contradiction. Hence, f(x) = 0 has no real root other than 1.