What is the real domain of f(x) = √(9 − x²)?
Answer and explanation
Correct answer: [−3, 3]
For the square root to be real, its radicand must be non-negative: 9 − x² ≥ 0. This gives x² ≤ 9, so −3 ≤ x ≤ 3. The endpoints are included because at x = 3 and x = −3 the radicand equals zero, and √0 is defined. Hence the real domain is [−3, 3], making option A correct.
Frequently asked questions
What is the correct answer to this question?
[−3, 3]
Why is this the correct answer?
For the square root to be real, its radicand must be non-negative: 9 − x² ≥ 0. This gives x² ≤ 9, so −3 ≤ x ≤ 3. The endpoints are included because at x = 3 and x = −3 the radicand equals zero, and √0 is defined. Hence the real domain is [−3, 3], making option A correct.
Which subject and chapter does this question cover?
This is a Class 11 Mathematics question. Chapter: Relations and Functions. Topic: Real-Valued Functions, Their Domain and Range.
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