Is x > 0 ? (1) x < x^2 (2) x < x^3

Can I deduce as following :
1) \(x < x^2\)
holds true for x > 1 and x < 0
Thus, not sufficient

2) \(x < x^3\)
holds true for x > 0 and 0 < x < 1
Not sufficient

No unique value is determined when combining 1 & 2.
Thus, E.

Please correct me if I am incorrect in the above logic.

You are correct.

Statement 1: Positive fractions will never make this statement true. The square of a positive fraction is always smaller than the original fraction itself.

Statement 2: Only positive numbers greater than 1 make this statement true. Therefore, by combining them, there is only one possibility to make both true: Numebrs greater than 1, which allows us to answer the original question.

Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.

So you can not divide both parts of inequality \(x<x^2\) by \(x\) as you don't know the sign of this unknown: if \(x>0\) you should write \(1<x\) BUT if \(x<0\) you should write \(1>x\).

Hope it helps.

No your ranges for (2) are not correct, it should be: \(-1<x<0\) or \(x>1\). Negative fractions from the range (-1,0) also make statement (2) true. Below is complete solution:

Is \(x>0\)?

(1) \(x<x^2\) --> \(x*(x-1)>0\) --> \(x<0\) or \(x>1\) --> ---------0----1----. Not sufficient as if \(x\) is in the green zone answer is YES but if \(x\) is in the red zone answer is NO.

(2) \(x<x^3\) --> \(x(x-1)(x+1)>0\) --> \(-1<x<0\) or \(x>1\) --> ----(-1)----(0)----(1)----. Not sufficient as if \(x\) is in the green zone answer is YES but if \(x\) is in the red zone answer is NO.

(1)+(2) Intersection of the ranges from (1) and (2) is:
----(-1)----(0)----(1)----
Again not sufficient as if \(x\) is in the green zone answer is YES but if \(x\) is in the red zone answer is NO.

Answer: E.

Hope it helps.

You're right. I hadn't thought of negative fractions. -1/2 ^3 = -1/8, which is greater than -1/2. Thanks.

No your ranges for (2) are not correct, it should be: \(-1<x<0\) or \(x>1\). Negative fractions from the range (-1,0) also make statement (2) true. Below is complete solution:

Is \(x>0\)?

(1) \(x<x^2\) --> \(x*(1-x)>0\) --> \(x<0\) or \(x>1\) --> ---------0----1----. Not sufficient as if \(x\) is in the green zone answer is YES but if \(x\) is in the red zone answer is NO.

(2) \(x<x^3\) --> \(x(x-1)(x+1)>0\) --> \(-1<x<0\) or \(x>1\) --> ----(-1)----(0)----(1)----. Not sufficient as if \(x\) is in the green zone answer is YES but if \(x\) is in the red zone answer is NO.

(1)+(2) Intersection of the ranges from (1) and (2) is:
----(-1)----(0)----(1)----
Again not sufficient as if \(x\) is in the green zone answer is YES but if \(x\) is in the red zone answer is NO.

Answer: E.

Hope it helps.[/quote]

Hi Bunuel,

Could you please explain why you're considering only the negative factors while considering \(x<x^2\) --> \(x*(1-x)>0\) --> \(x<0\) or \(x>1\).. From \(x*(1-x)>0\) can't we consider both x and (x-1) to be positive also ?

Similarly for \(x(x-1)(x+1)>0\) --> \(-1<x<0\) or \(x>1\), I seem to be missing your logic.

Thanks

Actually, both cases were considered.

\(x<x^2\) --> \(x*(x-1)>0\)
x>0 and x-1>0 (x>1) --> x>1;
x<0 and x-1<0 (x<1) --> x<0.

Thus, \(x<x^2\) holds true for \(x<0\) and \(x>1\).

Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html
xy-plane-71492.html

Hope it helps.

Hi Bunnuel

With ref to your above post i would like to contradictory expalnation from DS section of OG 13 question no 52
TO make things easier i ll post the question

If X and Y are positive , Is X<10<Y?
1. X<Y and XY = 100
2. X^2 <100< Y^2

In the explanation part
2nd stem is directly reduced to X<10<Y

Isnt your explanation contradictory to above expalantion of OG
Pls help in getting my concept right.

First of all, they are not reducing, they are taking square root. Also, notice that the stem says that x and y are positive numbers.

P.S. This question is discussed here: if-x-and-y-are-positive-is-x-10-y-139873.html

Thanks BUnnel
really missed the statement.Sorry

But is it ok to take square root,considering signs are unknown in above scenario.

If we were not told that x and y are positive numbers, then from x^2<100<y^2, we would have |x|<10<|y| (since \(\sqrt{x^2}=|x|\)).

Generally:

A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
\(2<4\) --> we can square both sides and write: \(2^2<4^2\);
\(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);

But if either of side is negative then raising to even power doesn't always work.
For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we can not square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
\(-2<-1\) --> we can raise both sides to third power and write: \(-2^3=-8<-1=-1^3\) or \(-5<1\) --> \(-5^2=-125<1=1^3\);
\(x<y\) --> we can raise both sides to third power and write: \(x^3<y^3\).

Hope it helps.

1. x < x^2
Test x = 2: 2 < 4 OK!
Test x = -1: -1 < 1 OK!
Test x = -1/4: -1/4 < -1/16 OK!
INSUFFICIENT!

2. x < x^3
Test x = 2: 2 < 8 OK!
Test x = -1: -1 = -1 NOT OK!
Test x = -1/4: -1/4 < -1/64 OK!
INSUFFICIENT!

Together: When x is negative fraction or x is greater than 2, both inequalities work!

Answer: E

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is x > 0 ?

(1) x < x^2
(2) x < x^3

If the range of the question covers that of the condition, the condition is sufficient
There is one variable (x) in the original condition and 2 equations are given by the conditions, so there is high chance (D) will be the answer
For condition 1, x<x^2, 0<x^2-x, 0<x(x-1)--> x<0, 1<x, and this is insufficient as this range is not included in the question,
For condition 2, x<x^3, 0<x^3-x, 0<x(x-1)(x+1) --> -1<x<0, 1<x is also not sufficient as this is also not included in the question
Looking at the conditions together, 1<x<0, 1<x is also insufficient, as the range of the question does not include that of the conditions
Hence, the answer becomes (E).

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.

S1 -> x < \(x^2\)
x = -2, \(x^2\) = 4
x = 2, \(x^2\) = 4
Insufficient.

S2 -> x < \(x^3\)
x = 2, \(x^3\) = 8
x = - 1/2, \(x^3\) = - 1/8
Insufficient.

(1+2)
x can be a negative fraction leading to a negative cube less greater than x, or a positive square greater than x.
Insufficient.

E.

Is x > 0 ?

(1) x < x^2
(2) x < x^3

From A we can say that
x^2>x
if X>0 means
than X is integer or fraction
if X >0 and integer the it satisfy that
if X>0 and fraction then it is not satisfy as x=0.2 then x^2=0.04
so From A we cant say.

From B
If X^3>x
means same happens here as well
so we cant say that X^3>X.

from A and B
we can say that of X lies between 0 and 1 then it is not true
So Answer is E

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