Domain Of Functions
The real function defined by the rule f(x) = the square root of (x - 3) added to the square root of (7 - x) has a domain consisting of which closed interval?
Select the correct option:
Solution
[3,7]
The domain of a real square-root function requires each radicand to be non-negative, a basic but frequently tested JEE Advanced constraint. For the first term, x - 3 ≥ 0 demands x ≥ 3. For the second term, 7 - x ≥ 0 demands x ≤ 7. The domain is the intersection of these conditions, namely all x with 3 ≤ x ≤ 7, written as the closed interval [3, 7]. Both endpoints are included because each radicand can equal zero there, keeping the expression real. Option (3, 7) wrongly excludes the endpoints where the roots are simply zero. Option [-7, -3] confuses the sign of the inequalities. Option [3, ∞) ignores the upper bound imposed by the second radical. Hence the domain is [3, 7]. Plausibility check: at x = 3 the value is 0 + 2 = 2 and at x = 7 it is 2 + 0 = 2, both real, while x = 8 makes 7 - x negative, confirming the interval boundaries are exactly correct.
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About This Question
- Subject
- mathematics
- Chapter
- sets, relations and functions
- Topic
- domain of functions
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
[3,7]
The domain of a real square-root function requires each radicand to be non-negative, a basic but frequently tested JEE Advanced constraint. For the first term, x - 3 ≥ 0 demands x ≥ 3. For the second term, 7 - x ≥ 0 demands x ≤ 7. The domain is the intersection of these conditions, namely all x with 3 ≤ x ≤ 7, written as the closed interval [3, 7]. Both endpoints are included because each radicand can equal zero there, keeping the expression real. Option (3, 7) wrongly excludes the endpoints where the roots are simply zero. Option [-7, -3] confuses the sign of the inequalities. Option [3, ∞) ignores the upper bound imposed by the second radical. Hence the domain is [3, 7]. Plausibility check: at x = 3 the value is 0 + 2 = 2 and at x = 7 it is 2 + 0 = 2, both real, while x = 8 makes 7 - x negative, confirming the interval boundaries are exactly correct.
This easy difficulty mathematics question is from the chapter sets, relations and functions, covering the topic of domain of functions. It appeared in the 2025 exam.
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