Let [x] denote the greatest integer less than or equal to x . Let f(x) = (x^2 - a) [x] for x (-2, 2) , where a 0 . If it is given that f(x) is discontinuous at exactly one point in this interval, then the number of points in (-2, 2) where f(x) is non-differentiable is :

Step-by-step solution

The function is f(x) = (x^2 - a) [x] . The potential points of discontinuity in the interval (-2, 2) are the integers x = -1, 0, 1 . For f(x) to be continuous at an integer k , the jump caused by [x] must be nullified by the other factor being zero. That is, we must have k^2 - a = 0 . If a = 0 , then k^2 - 0 = 0 k = 0 . Thus, f(x) would be continuous at x = 0 but discontinuous at x = -1 and x = 1 . This gives 2 points of discontinuity, which contradicts the given condition. If a = 1 , then k^2 - 1 = 0 k = 1 . Thus,