If x^2 < x, which of the following could be a value of x?

Upon simplyfying we can have these two options
x<\(\sqrt{x}\) or X < 1

We know that for numbers between 0 to 1 only , \(\sqrt{x}\) > x
Hence looking into the options we can have it as 6/7

If X^2<x it means x lies between 0 &1. Answer is C i.e. 6/7

Property: x^2<x ---> 0<x<1

Option C

x^2<x, will be true for a positive number between 1 and 0 or a proper fraction. Hence only option C fits here. 6/7

Hi All,

Already mathematical approach has been discussed in the above explanations,

So let me show you the plugging in or POE way,

Here plugging in or using POE-Process of Elimination is also actually very easy,

Also note this a could be question, so plugging in is a good option,

In the answer choices, straight away we can eliminate answer choice A and B, because x^2 is always positive for all the values of x (except 0),

So positive cannot be less than negative, so we can eliminate A and B. Answer options A and B both have negative values.

Looking at the other options, E is definitely out, because 25^2 is greater than 25.

Between C and D,

Option D is an improper fraction. So try to substitute the option in the given expression x^2<x,

25/16 > 5/4. So eliminated.

Option C is a proper fraction (numerator less than denominator), this has to be the answer, because all the other options are out.

Remember, for the proper fraction (for positive), x^2 < x always,

So option C, should be the answer.

So the answer has to be C here.

Hope this helps.

two root 0 and 1

from -infinity to 0 +ive
0 to 1 -ve
1 to infinity +ive
Only C is between 0 to 1

In order for x^2 to be less than x, x must be a positive proper fraction. Thus, 6/7 could be a value of x.

Answer: C

As x^2<x

Hence x => (0,1)

The only option that lies in the range is 6/7 which is 0.85something....


Hence C.

Plug in the answer choices:

Only (C) satisfies the given inequality.

\(x^2 < x\)

\(x^2 - x < 0\)

\(x*(x-1) < 0\)

x cannot be negative since \(x*(x-1)\) is negative and negative*negative will be positive.

Thus, \(x > 0\) and \(x-1 < 0\) => \(x < 1\)

\(0<x<1\)

Only option C. fits this criteria.

Answer C.

Two things to remember when pushing through with logic:

1. negative times a negative negates each other
2. fraction times a fraction makes a smaller fraction if it's under 1
   

That means:
A: higher number because it's negative
B: higher number because it's negative
C: lower number because it's a fraction under 1
D: higher number since it's over 1
E: higher number since it's over 1

Hope that helps.