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Let f(x) = ⁻¹ ( 2x^2 - 3x + 1 x^2 + x - 2 ) _e(x^2 - 4x + 5) . The sum of all integers in the domain of f(x) is

Options

  1. A5
  2. B4
  3. C3
  4. D6

Correct answer

C. 3

Step-by-step solution

For the numerator to be defined, the argument of the inverse cosine must lie in [-1, 1] . We have 2x^2 - 3x + 1 x^2 + x - 2 = (2x - 1)(x - 1) (x + 2)(x - 1) . For x 1 , this simplifies to 2x - 1 x + 2 . Solving -1 2x - 1 x + 2 1 : Case 1: 2x - 1 x + 2 1 x - 3 x + 2 0 x (-2, 3] . Case 2: 2x - 1 x + 2 -1 3x + 1 x + 2 0 x (- , -2) [- 1 3 , ) . The intersection of these conditions gives x [- 1 3 , 3 ] . Since x 1 , the domain of the numerator is [- 1 3 , 3 ] - 1 . For the denominator to be defined, the argument of the

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