JEE MainMathematicsContinuity and Differentiability
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
- A8
- B5
- C3
- 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,