JEE Advanced Mathematics Question (2026) — Solution

Step-by-step solution

Given f(x) = x _e(x) - x + 1 Differentiating with respect to x: f'(x) = 1 2 x _e(x) + x ( 1 x ) - 1 f'(x) = _e(x) 2 x + 1 x - 1 Differentiating again with respect to x: f''(x) = d dx ( 1 2 x^ -1/2 _e(x) + x^ -1/2 - 1 ) f''(x) = 1 2 ( - 1 2 x^ -3/2 _e(x) + x^ -1/2 1 x ) - 1 2 x^ -3/2 f''(x) = - _e(x) 4x^ 3/2 + 1 2x^ 3/2 - 1 2x^ 3/2 f''(x) = - _e(x) 4x^ 3/2 For x (0, 1), _e(x) 0. Thus, f'(x) is strictly increasing in (0, 1). For x (1, ), _e(x) > 0, which implies f''(x) Therefore, f'(x) attains its maximum value at x = 1. Maximum value of f'(x) = f'(1) = _e(1) 2 + 1 - 1 = 0. Since the maximum value of f'(x) is 0, we have f'(x) 0 for all x (0, ), with equality holding only at x = 1. This means f(x) is a strictly decreasing function on (0, ). Hence, f(x) has neither a point of local maximum nor a point of local minimum in the interval (0, ). Answer: The function f has NEITHER a point of local maximum NOR a point of local minimum in the interval (0, )