JEE MainMathematicsContinuity and Differentiability
Let f(x) = [2x^3 - 15x^2 + 36x] , where x (0, 4) and [t] denotes the greatest integer less than or equal to t . The number of points where f(x) is not differentiable is
Correct answer
32
Step-by-step solution
Let g(x) = 2x^3 - 15x^2 + 36x . Differentiating with respect to x , we get: g'(x) = 6x^2 - 30x + 36 = 6(x-2)(x-3) The critical points are x = 2 and x = 3 . Evaluating g(x) at the critical points and endpoints: g(0) = 0 g(2) = 2(8) - 15(4) + 36(2) = 28 g(3) = 2(27) - 15(9) + 36(3) = 27 g(4) = 2(64) - 15(16) + 36(4) = 32 We analyze the intervals and critical points separately: 1. In the interval (0, 2) , g(x) strictly increases from 0 to 28 . It takes integer values from 1 to 27 , contributing 27 points of non-differ