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Find the domain of definition of each function ([*] is greatest integer, {*} is fractional part): (i) f(x) = sqrt(x² - |x|) + 1/sqrt(9 - x²); (ii) f(x) = sqrt((x² - 3x - 10)*(ln(x-3))²); (iii) f(x) = sqrt((5x - 6 - x²)*{[ln x]}) + sqrt(7x - 5 - 2x²) + (ln(7/2 - x))⁻¹.

1. (i) (-3,-1] U {0} U [1,3); (ii) {x > 5} U {x = 4}; (iii) [1, 5/2)
2. (i) [-3,3]; (ii) x > 3; (iii) (1, 7/2)
3. (i) (-3,3); (ii) x >= 5; (iii) [1,3]
4. (i) [-1,1]; (ii) x > 5; (iii) (5/2, 7/2)

Correct answer: (i) (-3,-1] U {0} U [1,3); (ii) {x > 5} U {x = 4}; (iii) [1, 5/2)

Solution

(i) Need x² - |x| >= 0 -> |x| >= 1 or x = 0; and 9 - x² > 0 -> -3 < x < 3. Intersection: (-3,-1] U {0} U [1,3). (ii) Need (x² - 3x - 10)*(ln(x-3))² >= 0 with x - 3 > 0. x² - 3x - 10 = (x-5)(x+2) >= 0 for x >= 5 (within x>3); the square term is >= 0 always and = 0 at x = 4, which is admissible. So domain {x > 5} U {x = 4}. (iii) 7x - 5 - 2x² >= 0 -> 1/2 <= x <= 5? Solve 2x² - 7x + 5 <= 0 -> (2x-5)(x-1) <= 0 -> 1 <= x <= 5/2. Combined with x > 0 for ln, 7/2 - x > 0 and != 1 (so x != 5/2), and the {[ln x]} factor non-negativity, the domain reduces to [1, 5/2).

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