Let h(x) = ⁻¹ ( [x] a ) + _ 1/2 (x^2 - 6x + b) , where a and b are positive integers and [t] denotes the greatest integer function. If the domain of h(x) is exactly [2, 3) , then the value of a + b is

Step-by-step solution

For h(x) to be defined, both terms must be defined. For the square root and logarithm to be defined: _ 1/2 (x^2 - 6x + b) 0 Since the base is 1/2 0 0 For the domain to have an open boundary at x = 3 , the expression (x - 3)^2 + b - 9 must be exactly 0 at x = 3 . This gives: b - 9 = 0 b = 9 Substituting b = 9 into the inequality: 0 This means x 3 and -1 x - 3 1 , which gives 2 x 4 . Thus, the domain of the second term is [2, 3) (3, 4] . For the inverse cosine term to be defined: -1 [x] a 1 -a [x] a Using the propert