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0 < x < 1
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Updated on: 25 Sep 2023, 06:21
Updated on: 25 Sep 2023, 06:21
Expert Reply
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\(0 < x < 1
\)
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
\)
Quantity A |
Quantity B |
\(x-x^5\) |
\(x^2-x^3 \) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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Re: 0 < x < 1
[#permalink]
24 Sep 2023, 17:52
24 Sep 2023, 17:52
2
I'm a bit confused. Shouldn't the answer be d?
If x = -2
A: (-2) - (-2)^5 = -2 - (-32) = 30
B: (-2)^2 - (-2)^3 = 4 - (-8) = 12
A > B
If x = 2
A: (2) - (2)^5 = 2 - 32 = -30
B: (2)^2 - (2)^3 = 4 - 8 = -4
A < B
If x = -2
A: (-2) - (-2)^5 = -2 - (-32) = 30
B: (-2)^2 - (-2)^3 = 4 - (-8) = 12
A > B
If x = 2
A: (2) - (2)^5 = 2 - 32 = -30
B: (2)^2 - (2)^3 = 4 - 8 = -4
A < B
Re: 0 < x < 1
[#permalink]
25 Sep 2023, 06:21
25 Sep 2023, 06:21
1
Expert Reply
Sorry. It was in the title of the question that x is a fraction. I did not report the same constraint in the question. My bad
Given that 0 < x < 1
Assume x = 0.1
Quantity A: \(x - x^5\)
Now, when x is a positive fraction between 0 and 1 then as we increase power of x then value of the numbers starts decreasing
Ex: x = 0.1 , \(x^2\) = \(0.1^2\) = 0.01
So, \(x^5\) will be a very small number and will be less than x but will be positive
=> \(x - x^5\) ~ x > 0
Quantity A: \(x^2 - x^3\) = 0
Clearly, Quantity A > Quantity B
So, Answer will be A
Given that 0 < x < 1
Assume x = 0.1
Quantity A: \(x - x^5\)
Now, when x is a positive fraction between 0 and 1 then as we increase power of x then value of the numbers starts decreasing
Ex: x = 0.1 , \(x^2\) = \(0.1^2\) = 0.01
So, \(x^5\) will be a very small number and will be less than x but will be positive
=> \(x - x^5\) ~ x > 0
Quantity A: \(x^2 - x^3\) = 0
Clearly, Quantity A > Quantity B
So, Answer will be A
