JEE Main20262 April 2026Morning ShiftMathematicsApplication of DerivativesActual
The number of critical points of the function f(x) = cases | x x |, & x 0 1, & x = 0 cases in the interval (-2 , 2 ) is equal to :
Options
- A1
- B3
- C5
- D7
Correct answer
C. 5
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(