Let f(x) = [ a x x^2+1 ] , where a is a positive integer and [t] denotes the greatest integer t . If the number of points of discontinuity of f(x) in the open interval (0, 2) is 43 , then the maximum value of a is ______.

Step-by-step solution

Let u(x) = a x x^2+1 . Differentiating u(x) with respect to x , we get u'(x) = a(1-x^2) (x^2+1)^2 . Thus, u(x) is strictly increasing on (0, 1) and strictly decreasing on (1, 2) . The values at the boundaries and the extremum are: u(0) = 0 u(1) = a 2 u(2) = 2a 5 Case 1: If a is even, a 2 is an integer. In (0, 1) , u(x) increases from 0 to a 2 . It takes integer values 1, 2, , a 2 -1 , giving a 2 -1 discontinuities. At x=1 , u(1) = a 2 . Since u(x) In (1, 2) , u(x) decreases from a 2 to 2a 5 . It takes integer value