Match List I with List II regarding the domain of functions:List I (Function)List II (Domain)A. f(x) = 1/√(x-2) + 1/√(3-x) I. [2, 3] B. f(x) = √(x-2/3 - Mathematics MCQ Mathematics Mock Test - 11 2026 | ExamTest.live

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Match List I with List II regarding the domain of functions:List I (Function)List II (Domain)A. f (x) = \frac{1}{x - 2} + \frac{1}{3 - x} I. [2, 3] B. f (x) = \frac{x - 2}{3 - x} II. (2, 3] C. f (x) = \frac{3 - x}{x - 2} III. [2, 3) D. f (x) = (x - 2) (3 - x) IV. (2, 3) Choose the correct answer:

Solution & Step-by-step Explanation

A: For 1/ x - 2 to be defined, x - 2 > 0 ⟹ x > 2 . For 1/ 3 - x to be defined, 3 - x > 0 ⟹ x < 3 . Intersection: x \in (2, 3) . (A - IV)B: For (x - 2) / (3 - x) to be defined, (x - 2) / (3 - x) \geq 0 and 3 - x \neq = 0 .Using wavy curve: (x - 2) (x - 3) \leq 0 with x \neq = 3 . So x \in [2, 3) . (B - III)C: For (3 - x) / (x - 2) to be defined, (3 - x) / (x - 2) \geq 0 and x - 2 \neq = 0 .Using wavy curve: (x - 3) (x - 2) \leq 0 with x \neq = 2 . So x \in (2, 3] . (C - II)D: For (x - 2) (3 - x) to be defined, (x - 2) (3 - x) \geq 0 . (x - 2) (x - 3) \leq 0 ⟹ x \in [2, 3] . (D - I)Correct sequence: A-IV, B-III, C-II, D-I.

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Match List I with List II regarding the domain of functions:List I (Function)List II (Domain)A. f (x) = \frac{1}{x - 2} + \frac{1}{3 - x} I. [2, 3] B. f (x) = \frac{x - 2}{3 - x} II. (2, 3] C. f (x) = \frac{3 - x}{x - 2} III. [2, 3) D. f (x) = (x - 2) (3 - x) IV. (2, 3) Choose the correct answer:

A - I, B - III, C - II, D - IV

A - IV, B - III, C - II, D - I

A - I, B - II, C - III, D - IV

A - IV, B - II, C - III, D - I

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