JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The critical points of a function are the points in its domain where the derivative is zero or does not exist. First, consider x = 0. The function is continuous at x = 0 since _ x 0 | x x | = 1 = f(0). For x (- , ), x x > 0, so f(x) = x x . The derivative at x = 0 is given by f'(0) = _ x 0 x x - 1 x = _ x 0 x - x x^2 = 0. Since f'(0) = 0, x = 0 is a critical point. Next, consider the points where f(x) = 0, which occurs at x = and x = - in the interval (-2 , 2 ). At x = , the inner function g(x) = x x has g( ) = 0 and g'( ) = - 1 0. Since the derivative of the inner function is non-zero, the absolute value function f(x) = |g(x)| is not differentiable at x = . By symmetry, f(x) is also not differentiable at x = - . Thus, x = and x = - are critical points. For x 0, , - , the function is differentiable and f'(x) = x x - x x^2 . Setting f'(x) = 0 gives x x - x = 0 x = x. We analyze the roots of x = x in (-2 , 2 ): In (0, ), x > x for x (0, /2) and x In ( , 2 ), x increases from 0 to on ( , 3 /2). Since = 0 Since x = x is an odd equation, there is exactly one symmetric root in (-2 , - ). This gives 2 additional critical points where f'(x) = 0. The total number of critical points in (-2 , 2 ) is 1 (at x=0) + 2 (at x= ) + 2 (roots of x = x) = 5. Answer: 5