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11 Mar 2024, 00:01
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D
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Difficulty:
85%
(hard)
Question Stats:
43% (01:42) correct
57%
(01:48)
wrong
based on 302
sessions
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If \(\sqrt[10]{y^{10}} = -y\), and \(\sqrt[5]{y^{5}} \geq -y\), then which of the following gives all possible values of y?
A. \(-1 < y < 1\)
B. \(y \geq 0\)
C. \(y \leq 0\)
D. \(y = 0\)
E. \(|y| \geq 0 \)
A. \(-1 < y < 1\)
B. \(y \geq 0\)
C. \(y \leq 0\)
D. \(y = 0\)
E. \(|y| \geq 0 \)
11 Mar 2024, 00:17
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Bunuel
\(\sqrt[10]{y^{10}} = -y\)
The value of \(\sqrt[10]{y^{10}}\) is non negative. Hence, we can infer that \(y \leq 0\)
\(\sqrt[5]{y^{5}} \geq -y\)
Assume y = +ve
- \(\sqrt[5]{y^{5}}\) is positive while -y is negative. Hence, the inequality holds.
- \(\sqrt[5]{y^{5}}\) is negative while -y is positive. Hence, the inequality doesn't hold.
- The inequality holds.
Option D
11 Mar 2024, 00:55
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Bunuel
Firstly, recall that \(\sqrt{x^2}=|x|\). This is because the square root function always returns non-negative values, and |x| ensures that the result is non-negative for all x.For instance, consider x = -5:
\(\sqrt{(-5)^2} = \sqrt{25} = 5\), which is |-5|.
By the similar logic, \(\sqrt[10]{y^{10}} = |y|\). Hence, \(\sqrt[10]{y^{10}} = -y\), yields \(|y| = -y\). This on the other hand implies that \(y \leq 0\).
Moving on to \(\sqrt[5]{y^{5}}\), since y is odd, it simply equals y. Thus, \(\sqrt[5]{y^{5}} \geq -y\) leads to \(y \geq - y\), which consequently yields \(2y \geq 0\), and finally, \(y \geq 0\).
Therefore, we have both \(y \leq 0\) and \(y \geq 0\), which can only hold true if \(y = 0\).
Answer: D.
