JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Given f(x) = (x-1)^4 + 1 for x 1. To find the inverse function f^ -1 (x), let y = (x-1)^4 + 1. (x-1)^4 = y-1 x-1 = (y-1)^ 1/4 x = (y-1)^ 1/4 + 1. Thus, f^ -1 (x) = (x-1)^ 1/4 + 1. Evaluating the first statement, since f'(x) = 4(x-1)^3 0 for x 1, f(x) is strictly increasing. For a strictly increasing function, the solutions to f(x) = f^ -1 (x) are the same as the solutions to f(x) = x. (x-1)^4 + 1 = x (x-1)^4 - (x-1) = 0. Let t = x-1 0. t^4 - t = 0 t(t-1)(t^2+t+1) = 0. Since t 0, the real roots are t = 0 and t = 1. x-1 = 0 x = 1 x-1 = 1 x = 2 Both x=1 and x=2 lie in [1, ). The set contains exactly two elements, making Statement (I) TRUE. Evaluating the second statement, the equation is f(x) = f^ -1 (x+1). f^ -1 (x+1) = (x+1-1)^ 1/4 + 1 = x^ 1/4 + 1. Equating f(x) and f^ -1 (x+1): (x-1)^4 + 1 = x^ 1/4 + 1 (x-1)^4 - x^ 1/4 = 0. Let g(x) = (x-1)^4 - x^ 1/4 . g(2) = (2-1)^4 - 2^ 1/4 = 1 - 2^ 1/4 g(3) = (3-1)^4 - 3^ 1/4 = 16 - 3^ 1/4 > 0. Since g(x) is a continuous function on [1, ) and changes sign between x=2 and x=3, by the Intermediate Value Theorem, there exists at least one real root in the interval (2, 3). Thus, the set is not empty, making Statement (II) FALSE. Therefore, only (I) is TRUE. Answer: only (I) is TRUE