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

Hi, can u plz explain this problem using wavy line method ?
Using that method I got confused between options C & E.

There is no need to use the wavy curve method, just take any value between -1 and 1 and put in the options given.

Given,

x lies between -1 and 1, can be 0, as 0 is in between these.
So to eliminate some of the options, just put x=0.
This leaves us with options D&E.
In D, \( x+1/2\) will be always positive for any value above 0. But if x is above \(1/2\), the second part of option D, which is \(x-1/2\) will also be positive, making it a positive which will be wrong.

In option E, due to \(x^2\), first part is always positive, and due to 1, second part, which is (x-1) is always negative, which means this equation will be always negative.

We can also check that inequality at x=0 is only correct in option E to decide between C and E options

No. That wouldn't be a must be case, that can be a may be case, for a must be you need to check all possibilities.

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

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.

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. \((x^2+1)(x-1) < 0\)

Here "0" can help fast identify which one holds and which one doesn't.
A, B and C are out immediately.
D has problem when x = -1/2

Answer E.

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.