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Let f(x) = [ x^2 - 3x ] + | x - [x] | , where x R and [t] denotes the greatest integer less than or equal to t . The number of points in the open interval (0, 4) where f(x) is discontinuous is

Options

  1. A8
  2. B5
  3. C3
  4. D6

Correct answer

D. 6

Step-by-step solution

Given f(x) = [ x^2 - 3x ] + | x - [x] | . Since x - [x] = x 0 , we can write f(x) = [ x^2 - 3x ] + x . Let g(x) = x^2 - 3x . Discontinuities of f(x) can only occur where g(x) is an integer or where x is an integer. In the interval (0, 4) , the minimum value of g(x) occurs at x = 1.5 , where g(1.5) = -2.25 . At the endpoints, g(0) = 0 and g(4) = 4 . Thus, g(x) takes values in [-2.25, 4) . The integer values g(x) can take are -2, -1, 0, 1, 2, 3 . Solving x^2 - 3x = k for these integers in (0, 4) : For k = -2 , x = 1,

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