Domain Of Functions

Question

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?

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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.

Domain Of Functions

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