If x^3 < x^2, which of the following must be negative?

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If x^3 < x^2, which of the following must be negative?

A. x
B. −x
C. x^5
D. x − 1
E. x^(−1)

Answer: D
Source: Veritas Prep

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BTGModeratorVI wrote:
Fri Apr 03, 2020 9:37 am
If x^3 < x^2, which of the following must be negative?

A. x
B. −x
C. x^5
D. x − 1
E. x^(−1)

Answer: D
Source: Veritas Prep
For \(x^3 < X^2\) to be true, there are \(3\) cases
Case 1: \(x< -1\), in which case except \(B\) all are negative
Case 2: \(-1<x<0\), in which case except \(B\), all are negative
Case 3: 0<x<1, in which case except A,C,E all are negative.

Thus in all the \(3\) case above for \(x-1\) is negative.

Hence, the correct answer is D

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BTGModeratorVI wrote:
Fri Apr 03, 2020 9:37 am
If x^3 < x^2, which of the following must be negative?

A. x
B. −x
C. x^5
D. x − 1
E. x^(−1)

Answer: D
Source: Veritas Prep
Given: x³ < x²
This tells us that x ≠ 0. So, we can be certain that x² is POSITIVE.
Since x² is POSITIVE, we can safely divide both sides of the inequality by x²
When we do this, we get: (x³)/(x²) < 1
Simplify: x < 1
Subtract 1 from both sides to get: x - 1 < 0
In other words, x - 1 is NEGATIVE

Answer: D

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BTGModeratorVI wrote:
Fri Apr 03, 2020 9:37 am
If x^3 < x^2, which of the following must be negative?

A. x
B. −x
C. x^5
D. x − 1
E. x^(−1)

Answer: D
Source: Veritas Prep
Since x^3 is less than x^2, there are two cases for x: 1) x could be a negative number, or 2) x could be a (positive) number whose value is between 0 and 1.

Thus, x - 1 will always be negative.

Answer: D

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