Let f(x) = cases ( x 2 ), & x where [t] is the greatest integer less than or equal to t , and t is the fractional part of t . Let m be the number of points where f is not continuous, and n be the number of points where f is not differentiable. Then the value of m + n is

Step-by-step solution

For x For 0 x f(x) = (x, x^2) . Since 0 x For 1 x f(x) = (x - 1, x^2 - 1) . Since x 1 , x^2 x x^2 - 1 x - 1 , so f(x) = x^2 - 1 . For x 2 , f(x) = x^2 - x + 1 . Thus, f(x) = cases ( x 2 ), & x Checking continuity and differentiability at junction points: At x = 0 : f(0^-) = (0) = 0 f(0) = f(0^+) = 0 . Continuous. LHD = 2 (0) = 2 RHD = 1 Since LHD RHD, f(x) is not differentiable at x = 0 . At x = 1 : f(1^-) = 1 f(1) = f(1^+) = 1^2 - 1 = 0 Since f(1^-) f(1^+) , f(x) is not continuous at x = 1 . Hence, it is also not