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# If x^2 < x, then x must be

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Re: If x^2 < x, then x must be [#permalink]
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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

Given: x² < x
Subtract x from both sides to get: x² - x < 0
Factor: x(x - 1) < 0

We can already see that, when x = 0, we get: 0(0 - 1) < 0, which results in 0 < 0, which doesn't work. ELIMINATE B
Likewise, when x = 1, we get: 1(1 - 1) < 0, which results in 0 < 0, which doesn't work. ELIMINATE D

At this point we can apply a little bit of number sense.

If x is between 0 and 1, then x is POSITIVE, and (x - 1) is NEGATIVE, which means x(x - 1) < 0 becomes POSITIVE(NEGATIVE) < 0, which works!!

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Brent
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Re: If x^2 < x, then x must be [#permalink]
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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

x^2 < x

i.e. x^2 - x < 0
i.e. x (x-1) < 0

- Product of two things x and (x-1) is negative i,e, one of them must be negative and one must be positive
- Since x-1 is smaller than x therefore x-1 must be negative and x must be positive

x > 0 and x-1 < 0
x > 0 and x < 1

i.e. 0 < x < 1

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Re: If x^2 < x, then x must be [#permalink]
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

In order for x^2 to be less than x, then x must be a number between 0 and 1. For example, if x = 1/2, then x^2 = (½)^2 = 1/4. Since 1/4 < 1/2, we see that x^2 < x for this example, and the equation holds true for any positive fraction (i.e., a value between 0 and 1).

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Re: If x^2 < x, then x must be [#permalink]
Ravixxx wrote:
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

$$x^{2}<x$$

$$x^{2}- x<0$$

$$x(x-1)<0$$

$$0<x<1$$

If it was x^2>x
Then what would have been the answer?

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Re: If x^2 < x, then x must be [#permalink]
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Re: If x^2 < x, then x must be [#permalink]
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