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02 Dec 2021, 09:45
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
68% (02:16) correct
32%
(02:19)
wrong
based on 222
sessions
History
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Time
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Not Attempted Yet
What value of x satisfy \(x^{\frac{2}{3}} + x^{\frac{1}{3}} - 2 ≤ 0\) ?
(A) \(–8 ≤ x ≤ 1\)
(B) \(–1 ≤ x ≤ 8\)
(C) \(1 < x < 8\)
(D) \(1 ≤ x ≤ 8\)
(E) \(–8 ≤ x ≤ 8\)
(A) \(–8 ≤ x ≤ 1\)
(B) \(–1 ≤ x ≤ 8\)
(C) \(1 < x < 8\)
(D) \(1 ≤ x ≤ 8\)
(E) \(–8 ≤ x ≤ 8\)
02 Dec 2021, 10:30
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Assume x^1/3 =y
Y^2+y-1 <=0
Y-1 * y+2 <=0
Y<=1
Y>= -2
Put y with x^1/3
We need to consider the other sign off y+2
Hence answer is e
I originally made an error when I chose A
Posted from my mobile device
Y^2+y-1 <=0
Y-1 * y+2 <=0
Y<=1
Y>= -2
Put y with x^1/3
We need to consider the other sign off y+2
Hence answer is e
I originally made an error when I chose A
Posted from my mobile device
02 Dec 2021, 16:21
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I think the answer is A
x^1/3 = y
x^2/3= (x^1/3)(x^1/3) = y^2
New formula:
y^2+y-2 ≤ 0
(y+2)(y-1) ≤ 0
Then y ≤ -2, 1
The only answer that satisfies that is A. The majority of the other answer choices have the option of y ≤ 8. If you substitute 7 into the new formula (y+2)(y-1) ≤ 0 --- (7+2)(7-1) is not ≤ 0.
x^1/3 = y
x^2/3= (x^1/3)(x^1/3) = y^2
New formula:
y^2+y-2 ≤ 0
(y+2)(y-1) ≤ 0
Then y ≤ -2, 1
The only answer that satisfies that is A. The majority of the other answer choices have the option of y ≤ 8. If you substitute 7 into the new formula (y+2)(y-1) ≤ 0 --- (7+2)(7-1) is not ≤ 0.
