If -1

Question

If -1 < x < 1 , which of the following must be true? A. x > x^3 B. x + x^2 > 0 C. x^2 > x^3 D. (x+\frac{1}{2})(x-\frac{1}{2}) < 0 E. (x^2+1)(x-1) < 0 Screenshot 2024-02-19 112719.png

Bunuel · Accepted Answer

The "must be true" condition implies that the correct option must hold true for all possible values of x. Substituting x = 0 quickly reveals that only options D and E hold true. Substituting 3/4 shows that only E remains.

Answer E.

Alternatively, we could notice that since  x^2 + 1  is always positive, we can reduce it in  (x^2+1)(x-1) < 0  to get  x-1 < 0 , which yields  x < 1 . Since this is given to be true, then E must be true.

Answer: E.

egmat · Answer

This question is of the format -   If condition, then what MUST be true  . In Quant, such questions can be asked in the context of Number Properties or Algebra concepts. In this question, the context is Algebra.

To answer this question comfortably, one needs to completely understand the approach to solve this question. Translating the question statement with the given condition in your words helps to solidify the approach and guards against getting confused in the process of analyzing the choice.

And then one needs to carefully simplify the choices presented – one at a time. So, one clear faltering point is if the student rushes through this processing. In fact, in this question, one needs to process each choice patiently.

Other faltering points in this question include not considering all cases - such as in choice C, failing to consider that x cannot be 0.

Watch the solution and identify which step you faltered at.

https://www.youtube.com/watch?v=Ng0mN97f8Rc

MeghaYadav01 · Answer

How is Option C incorrect, in what cases will C be untrue, Is it in case of x = 0?

Bunuel · Answer

Yes. x^2 > x^3 implies x < 0 or 0 < x < 1. As you can see, x = 0 does not satisfy this inequality, so if x = 0, x^2 > x^3 is not true.

KarishmaB · Answer

First check out this write up on the tricky must be true questions: https://anaprep.com/algebra-game-must-b ... questions/ Given: -1 < x < 1 So x can take values -0.9 ... -0.48 ... 0 ... 0.3 ...0.5... 0.8 .. Which of the following must be true for all these values given? x > x^3 Not true for 0. Eliminate x + x^2 > 0 Not true for 0. Eliminate x^2 > x^3 Not true for 0. Eliminate (x+\frac{1}{2})(x-\frac{1}{2}) < 0 Not true for 1/2. Eliminate. (x^2+1)(x-1) < 0 (x^2+1) is always positive. So as long as x < 1 (true for all our given values), this will be negative. Correct. Answer (E) Method 2: -1 < x < 1 should be a subset of the values satisfying the given option. (If A, then B implies A must be a subset of B). SO given option should be a superset of -1 < x < 1. x > x^3 x(x^2 - 1) < 0 So x < -1 or 0 < x < 1. Not a super set x + x^2 > 0 x > 0 or x < -1. Not a super set x^2 > x^3 x^2(x -1) < 0 x < 1 but not 0. Not a super set (x+\frac{1}{2})(x-\frac{1}{2}) < 0 -1/2 < x < 1/2. Not a super set. (x^2+1)(x-1) < 0 x < 1. It is a super set.

eaat · Answer

For D, doesn't the wavy line method imply that x is between -0.5 and 0.5 and hence lies within the range stated in the question?

Bunuel · Answer

Yes, option D implies that x is between -0.5 and 0.5. But that does not make D correct. The question asks which statement must be true if -1 < x < 1. Since the given condition also allows values such as x = 0.75 or x = -0.8, and those do not satisfy D, D is not something that must always be true.

If -1 < x < 1, which of the following must be true

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