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29 Jul 2024, 03:57
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Difficulty:
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Question Stats:
13% (01:31) correct
88%
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wrong
based on 8
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The domain of the function \(f(x) = \frac{\sqrt{|x| - 2}}{\sqrt{\sqrt{x + 9}- \sqrt{3 - x}}}\) is the set of real numbers \(x\) such that:
A. \(-3 < x ≤ -2\) or \(2 ≤ x < 3\)
B. \(-3 < x ≤ -2\) or \(2 ≤ x ≤ 3\)
C. \(-3 ≤ x ≤ -2\) or \(2 ≤ x ≤ 3\)
D. \(-9 ≤ x < -3\) or \(-3 < x ≤ -2\) or \(2 ≤ x ≤ 3\)
E. \(-9 ≤ x ≤ -2\) or \(2 ≤ x ≤ 3\)
A. \(-3 < x ≤ -2\) or \(2 ≤ x < 3\)
B. \(-3 < x ≤ -2\) or \(2 ≤ x ≤ 3\)
C. \(-3 ≤ x ≤ -2\) or \(2 ≤ x ≤ 3\)
D. \(-9 ≤ x < -3\) or \(-3 < x ≤ -2\) or \(2 ≤ x ≤ 3\)
E. \(-9 ≤ x ≤ -2\) or \(2 ≤ x ≤ 3\)
29 Jul 2024, 03:57
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Official Solution:
The domain of the function \(f(x) = \frac{\sqrt{|x| - 2}}{\sqrt{\sqrt{x + 9}- \sqrt{3 - x}}}\) is the set of real numbers \(x\) such that:
A. \(-3 < x ≤ -2\) or \(2 ≤ x < 3\)
B. \(-3 < x ≤ -2\) or \(2 ≤ x ≤ 3\)
C. \(-3 ≤ x ≤ -2\) or \(2 ≤ x ≤ 3\)
D. \(-9 ≤ x < -3\) or \(-3 < x ≤ -2\) or \(2 ≤ x ≤ 3\)
E. \(-9 ≤ x ≤ -2\) or \(2 ≤ x ≤ 3\)
Given that the expression contains a fraction and square roots, the domain of the function will be defined when the square roots are greater than or equal to 0 (since even roots are not defined for negative numbers) and the denominator of the fraction is not equal to 0 (since division by 0 is not allowed). Thus, the following conditions must be satisfied simultaneously:
1. The expression inside the square root in the numerator, \(|x| - 2\), must be non-negative:
\(|x| - 2 ≥ 0\)
\(|x| ≥ 2\)
This means \(x ≤ -2\) or \(x ≥ 2\).
--------------------------(-2)--------(2)-------
2. The individual expressions inside the square roots in the denominator, \(x + 9\), and \(3 - x\) must be non-negative:
\(x + 9 ≥ 0\) and \(3 - x ≥ 0\)
\(x ≥ -9\) and \(x ≤ 3\)
This means \(-9 ≤ x ≤ 3\).
--(-9)-------------------------------------(3)--
3. The whole expression inside the square roots in the denominator must be positive:
\(\sqrt{x + 9} - \sqrt{3 - x} > 0\)
\(\sqrt{x + 9} > \sqrt{3 - x}\)
\(x + 9 > 3 - x\)
\(2x > -6\)
\(x > -3\)
-------------------(-3)---------------------------
The overlap of these ranges is \(-3 < x ≤ -2\) and \(2 ≤ x ≤ 3\).
--(-9)------------(-3)--(-2)--------(2)--(3)--
Answer: B
The domain of the function \(f(x) = \frac{\sqrt{|x| - 2}}{\sqrt{\sqrt{x + 9}- \sqrt{3 - x}}}\) is the set of real numbers \(x\) such that:
A. \(-3 < x ≤ -2\) or \(2 ≤ x < 3\)
B. \(-3 < x ≤ -2\) or \(2 ≤ x ≤ 3\)
C. \(-3 ≤ x ≤ -2\) or \(2 ≤ x ≤ 3\)
D. \(-9 ≤ x < -3\) or \(-3 < x ≤ -2\) or \(2 ≤ x ≤ 3\)
E. \(-9 ≤ x ≤ -2\) or \(2 ≤ x ≤ 3\)
Given that the expression contains a fraction and square roots, the domain of the function will be defined when the square roots are greater than or equal to 0 (since even roots are not defined for negative numbers) and the denominator of the fraction is not equal to 0 (since division by 0 is not allowed). Thus, the following conditions must be satisfied simultaneously:
1. The expression inside the square root in the numerator, \(|x| - 2\), must be non-negative:
\(|x| - 2 ≥ 0\)
\(|x| ≥ 2\)
This means \(x ≤ -2\) or \(x ≥ 2\).
--------------------------(-2)--------(2)-------
2. The individual expressions inside the square roots in the denominator, \(x + 9\), and \(3 - x\) must be non-negative:
\(x + 9 ≥ 0\) and \(3 - x ≥ 0\)
\(x ≥ -9\) and \(x ≤ 3\)
This means \(-9 ≤ x ≤ 3\).
--(-9)-------------------------------------(3)--
3. The whole expression inside the square roots in the denominator must be positive:
\(\sqrt{x + 9} - \sqrt{3 - x} > 0\)
\(\sqrt{x + 9} > \sqrt{3 - x}\)
\(x + 9 > 3 - x\)
\(2x > -6\)
\(x > -3\)
-------------------(-3)---------------------------
The overlap of these ranges is \(-3 < x ≤ -2\) and \(2 ≤ x ≤ 3\).
--(-9)------------(-3)--(-2)--------(2)--(3)--
Answer: B
22 Sep 2024, 16:43
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If for numerator we take x >=2, will it not become 0.?
