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Since we are told that \(x\) is an integer and there is no integer in the range \(0 \lt x \lt 1\) then \(x \lt -1\), so \(x\) is a negative integer. Sufficient.

(2) \(x \lt x^2\).

\(x^2-x \gt 0\);

\(x(x-1) \gt 0\);

\(x \lt 0\) or \(x \gt 1\), so integer \(x\) could be negative as well as positive. Not sufficient.

Hi, Bunuel I understand everything you said except x<0 in st-2 and x>0 in st-1. Please help me to understand.

Actually, I think integer 0 is not possible for both. I need either +ve or -ve integers.

Now in st1 - no positive integer will satisfy it. So we are sure that integers here are negative. in st2 whatever we put negative or positive we can satisfy the condition. so not sufficient.

My confusion is in 2 if I think it other way. We can express x^2> x in mod x> x. Now this condition is sufficient only if x is negative. I think I am making mistake in thinking in the 2nd way. Where am I making mistake in my 2nd thought?

Hi, Bunuel I understand everything you said except x<0 in st-2 and x>0 in st-1. Please help me to understand.

Actually, I think integer 0 is not possible for both. I need either +ve or -ve integers.

Now in st1 - no positive integer will satisfy it. So we are sure that integers here are negative. in st2 whatever we put negative or positive we can satisfy the condition. so not sufficient.

My confusion is in 2 if I think it other way. We can express x^2> x in mod x> x. Now this condition is sufficient only if x is negative. I think I am making mistake in thinking in the 2nd way. Where am I making mistake in my 2nd thought?

x^2 > x cannot be written as |x| > x.
_________________

I am having problems with this question and I would like to get your opinion.

When I see this type of questions in which exponents with no coefficients are involved and I am dealing with a DS question then I use "the zones" method. That is, I fix my attention on -1, 0 and 1 but not evaluating those exact numbers, rather I randomly pick a number less than -1, a number betwen -1 and 0, a number between 0 and 1 and a number greater than one: The above would look something like this: If x=-2, x^2= 4; x^3= -8 --> summary x^3<x<x^2 If x= -1/2, x^2= 1/4 ; x^3= -1/8 --> summary x<x^3<x^2 If x= 1/2; x^2 = 1/4; x^3= 1/8 --> summary x^3<x^2<x If x=2; x^2=4; x^3=8 --> summary x<x^2<x^3

Once I have this I can evaluate statements. S-1 that x>x^3 is true for x=-2 and x= 1/2. Not sufficient S-2 that x<x^2 is true for x=-2, x=-1/2, x=2. Not sufficient If we combine the information, we have sufficiency

I am having problems with this question and I would like to get your opinion.

When I see this type of questions in which exponents with no coefficients are involved and I am dealing with a DS question then I use "the zones" method. That is, I fix my attention on -1, 0 and 1 but not evaluating those exact numbers, rather I randomly pick a number less than -1, a number betwen -1 and 0, a number between 0 and 1 and a number greater than one: The above would look something like this: If x=-2, x^2= 4; x^3= -8 --> summary x^3<x<x^2 If x= -1/2, x^2= 1/4 ; x^3= -1/8 --> summary x<x^3<x^2 If x= 1/2; x^2 = 1/4; x^3= 1/8 --> summary x^3<x^2<x If x=2; x^2=4; x^3=8 --> summary x<x^2<x^3

Once I have this I can evaluate statements. S-1 that x>x^3 is true for x=-2 and x= 1/2. Not sufficient S-2 that x<x^2 is true for x=-2, x=-1/2, x=2. Not sufficient If we combine the information, we have sufficiency

Notice that the stem specifies that x is an integer.
_________________

Since we are told that \(x\) is an integer and there is no integer in the range \(0 \lt x \lt 1\) then \(x \lt -1\), so \(x\) is a negative integer. Sufficient.

(2) \(x \lt x^2\).

\(x^2-x \gt 0\);

\(x(x-1) \gt 0\);

\(x \lt 0\) or \(x \gt 1\), so integer \(x\) could be negative as well as positive. Not sufficient.

Answer: A

I understood all but the following, How \(x \lt 0\) or \(x \gt 1\) can be derived from \(x(x-1) \gt 0\) ? Wouldn't it be \(x(x-1) \gt 0\) ---> \(x \gt 0\) OR \((x-1) \gt 0\) ? Would be helpful if someone clarifies this......... _________________

Since we are told that \(x\) is an integer and there is no integer in the range \(0 \lt x \lt 1\) then \(x \lt -1\), so \(x\) is a negative integer. Sufficient.

(2) \(x \lt x^2\).

\(x^2-x \gt 0\);

\(x(x-1) \gt 0\);

\(x \lt 0\) or \(x \gt 1\), so integer \(x\) could be negative as well as positive. Not sufficient.

Answer: A

I understood all but the following, How \(x \lt 0\) or \(x \gt 1\) can be derived from \(x(x-1) \gt 0\) ? Wouldn't it be \(x(x-1) \gt 0\) ---> \(x \gt 0\) OR \((x-1) \gt 0\) ? Would be helpful if someone clarifies this.........

If I put value of x=1 then 1>1^3. Then, this equation fails to prove x=1(which is positive). So, I discarded first option. Can you please help me with correct understanding.

If I put value of x=1 then 1>1^3. Then, this equation fails to prove x=1(which is positive). So, I discarded first option. Can you please help me with correct understanding.

Thank you.

Really not sure what are you trying to say there but 1 = 1^3.
_________________

Solution provided by Bunuel is self-explanatory. But I saw few posters were having issue in understanding how to get to \(x<−1\) or \(0<x<1\) from \(x(x−1)(x+1)<0\)

To break any equality of this type - First, we will find the trigger point. Trigger points are the point on which equality will give output = 0 .

So in this equation, we have 3 i.e. \(x=0\) , \(x=1\) and \(x=-1\)

Now we will mark these points on number line. ___-1____0_____1____

Here we are having 4 regions 1. \(x <-1\) 2. \(-1<x<0\) 3. \(0<x<1\) 4. \(x>1\)

Second step , test one value from each region and see if you get the value < 0

\(x(x−1)(x+1)<0\)

1. \(x <-1\) ---> Put x = -2 ; -6 < 0 ; So this range is valid . 2. -\(1<x<0\) ---> Put x = -1/2 ; Some +ve < 0 ;So this range is not valid. 3. \(0<x<1\) ---> Put x = 1/2 ; Some -ve value < 0 ; So this range is valid . 4.\(x>1\) ---> Put x =2 ; SOme +ve value < 0 ; So this range is not valid.

So from here we get 2 ranges \(x <-1\) or \(0<x<1\)

This question asks us if integer x is positive or not. In questions of this type, we need to evaluate each option.Substitute values for each option and check.

Statement 1 - Put both positive and negative values. We can ascertain if x is positive or not Statement 2 - Put both positive and negative values. We can't ascertain if x is positive or not