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If x^2 < x, then x must be
[#permalink]
22 May 2020, 05:55
22 May 2020, 05:55
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If x^2 < x, then x must be
A. less than 0
B. equal to 0
C. between 0 and 1
D. equal to 1
E. greater than 1
PS21231
Where does p lie if p² < p?
A. Between -1 and 0
B. Between -1 and 1
C. Between 0 and 1
D. It is always less than 0
E. It is always greater than 1
A. less than 0
B. equal to 0
C. between 0 and 1
D. equal to 1
E. greater than 1
PS21231
Where does p lie if p² < p?
A. Between -1 and 0
B. Between -1 and 1
C. Between 0 and 1
D. It is always less than 0
E. It is always greater than 1
Re: If x^2 < x, then x must be
[#permalink]
22 May 2020, 06:07
22 May 2020, 06:07
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Bookmarks
Bunuel wrote:
If x^2 < x, then x must be
A. less than 0
B. equal to 0
C. between 0 and 1
D. equal to 1
E. greater than 1
PS21231
A. less than 0
B. equal to 0
C. between 0 and 1
D. equal to 1
E. greater than 1
PS21231
\(x^{2}<x\)
\(x^{2}- x<0\)
\(x(x-1)<0\)
\(0<x<1\)
Answer C
General Discussion
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Re: If x^2 < x, then x must be
[#permalink]
22 May 2020, 06:11
22 May 2020, 06:11
Let's put values of x one by one
A) less than zero = -1/2
=> (-1/2)^2 & -1/2
=> 1/4>-1/2 (incorrect)
B) equal to zero
=> 0^2 & 0
=> 0=0 (incorrect)
C) between 0 & 1 = 1/2
=> (1/2)^2 & (1/2)
=> 1/4<1/2 ((correct))
D) equal to 1
=> 1^2 & 1
=> 1=1 (incorrect)
E) more than one = 2
=> (2)^2 & 2
=> 4>2 (incorrect)
Posted from my mobile device
A) less than zero = -1/2
=> (-1/2)^2 & -1/2
=> 1/4>-1/2 (incorrect)
B) equal to zero
=> 0^2 & 0
=> 0=0 (incorrect)
C) between 0 & 1 = 1/2
=> (1/2)^2 & (1/2)
=> 1/4<1/2 ((correct))
D) equal to 1
=> 1^2 & 1
=> 1=1 (incorrect)
E) more than one = 2
=> (2)^2 & 2
=> 4>2 (incorrect)
Posted from my mobile device
