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Is x > x ^3 ? [#permalink]
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28 Jan 2011, 05:59
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Is x > x ^3 ? (1) x < 0 (2) x^2  x^3 > 2
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Re: Friday Algebra DS [#permalink]
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28 Jan 2011, 06:54



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Re: Friday Algebra DS [#permalink]
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17 Mar 2011, 22:49
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I goofed on this because I tricked myself but this is easy...think about it as always before putting pen to paper.. When is x>x^3? ONLY when x = negative integer OR a positive fraction. 1) x= ve, ok but is it a fraction? Insuff 2) x^2  x^3 >2 x^2 (x1)<2 Use gurpreets method to draw the number line. You will see that Statement is positive for all: x>1 = positive, x>0 = positive Therefore between 0 and 1 the statement is negative..hence the original statement holds true. Because X is a positive fraction, B is sufficient.



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Re: Friday Algebra DS [#permalink]
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18 Mar 2011, 00:02
(1) If x is not a fraction, say 2 , then x > x^3 But if x = 1/2, then 1/2 < 1/8 (2) x is not a proper fraction, and is a ve number (3)^2  (3)^3 = 9  (27) = 36 So 2 is sufficient. Answer B.
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Re: Friday Algebra DS [#permalink]
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05 Mar 2013, 06:49
Bunuel wrote: rxs0005 wrote: Is x > x ^3
(2) x^2x^3>2 > \(x^2(1x)>2\) > only true for \(x<1\) (note that if \(x>1\) then \(x^2(1x)\) is negative so this range is not good and if \(1\leq{x}\leq{1}\) then \(x^2(1x)\leq{2}\) so this range is also not good). Sufficient.
Answer: B. Hi Bunuel, For x^2x^3>2 how did you directly arrive at the intervals <1 between 1 and 1 and >1. When we deal with (1x) >2 we get x <1. But how did you chose the other point 1?



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Re: Friday Algebra DS [#permalink]
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06 Mar 2013, 02:55



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Re: Friday Algebra DS [#permalink]
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07 Mar 2013, 00:08
Bunuel wrote: Check here:
Hi Bunuel, Thanks for your reply and reference to some wonderful materials around quadratic inequalities. May I ask you one more question  going back to the basics now. What would be the roots of the eqn we have in hand in this post: x^2  x^3 >2. I solved the roots to be 0 and 1 by following the below steps: x^2(1x)>2 X^2 implies 0 is a root. 1x>2 implies x<1, so 1 is a root. From your explanations the roots seem to be 1 and 1. Where am I going wrong? (I also saw another post of yours where I was not able to solve the correct roost when x^3 was involved. Please help.)
Last edited by Jaisri on 07 Mar 2013, 06:28, edited 1 time in total.



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Re: Is x > x ^3 ? [#permalink]
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07 Mar 2013, 02:03
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rxs0005 wrote: Is x > x ^3 ?
(1) x < 0 (2) x^2  x^3 > 2 Nothing to add after Bunuel's explanation. But just writing down how to get the inequality x<1 from F.S 2 for those who couldn't get it. F.S 1 clearly not sufficient. Take x= 1,you get a NO for stem, but for 0.5, you get a YES. F.S 2, states \(x^2x^32>\)0 or\(( x^21)(x^3+1)>0\) or\((x+1)[(x1)  (x^2+1x)]>0\) \(or (x+1)[x^2+2x2]>0\) or \((x+1)[(x^22x+2)]>0\) or \((x+1)[(x1)^2+1]>0\) as\((x1)^2+1\) will always be positive, thus (x+1) has to be negative. or x+1<0 or x<1 Sufficient. B.
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Last edited by mau5 on 08 Mar 2013, 07:57, edited 2 times in total.



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Re: Is x > x ^3 ? [#permalink]
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08 Mar 2013, 06:43
vinaymimani wrote: rxs0005 wrote: Is x > x ^3 ?
(1) x < 0 (2) x^2  x^3 > 2 Nothing to add after Bunuel's explanation. But just writing down how to get the inequality x<1 from F.S 2 for those who couldn't get it. F.S 1 clearly not sufficient. Take x= 1,you get a NO for stem, but for 0.5, you get a YES. F.S 2, states \(x^2x^32>\)0 or\(( x^21)(x^31)>0\) or\((x1)[(x+1)  (x^2+1+x)]>0\) or\((x1)(x^2)\)>0. Thus, (x1) has to be negative or x1<0 or x<1. Sufficient. B. +1 Kudos. Thanks for showing how to solve this inequality! There is a minor change though... x1 >0 gives x<1 and not x <1. Bunuel seems to have considered 1 and 1 as the roots. Any reasons for that?



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Re: Is x > x ^3 ? [#permalink]
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08 Mar 2013, 07:56
Quote: +1 Kudos. Thanks for showing how to solve this inequality! There is a minor change though... x1 >0 gives x<1 and not x <1. Bunuel seems to have considered 1 and 1 as the roots. Any reasons for that?
thanks for pointing out the mistake. It was a splendidly foolish mistake. Apologies.
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Re: Is x > x ^3 ? [#permalink]
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Re: Is x > x ^3 ? [#permalink]
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09 Oct 2014, 21:50
rxs0005 wrote: Is x > x ^3 ?
(1) x < 0 (2) x^2  x^3 > 2 You can either use inequalities here or the big picture approach. When is x greater than x^3? It is when x < 1 or 0 < x <1. (Recall that we should know the relations of x, x^2 and x^3 in the ranges 'less than 1', '1 to 0', '0 to 1' and 'greater than 1') (1) x < 0 When x is between 1 and 0, x^3 is greater than x. When x < 1, then x is greater than x^3. So this statement alone is not sufficient. (2) x^2  x^3 > 2 Now, this is not very easy to handle using inequalities. Without the cube, we would have just taken 2 to the other side and solved the quadratic. But this will be more easily solved using the big picture. Think of a number line. What does x^2  x^3 > 2 imply? It means x^2 is greater than x^3 and is 2 units to the right of x^3 on the number line. x^2 is never negative so it must be to the right of 0. Now there are two cases: Either x^2 is between 0 and 1 or it is greater than 1. If x^2 were between 0 and 1, x would be between 1 and 1 and x^3 would be between 0 and 1. In this case, difference between x^3 and x^2 cannot be greater than 2. Hence this is not possible. So x^2 must be to the right of 1 and hence, x would be greater than 1 or less than 1. If x were greater than 1, x^3 would be greater than x^2 so x cannot be greater than 1. Hence x must be less than 1. When x is less than 1, then x IS greater than x^3. Sufficient alone. Answer (B) Again, spend some time checking out the relations of x, x^2 and x^3 on the number line.
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Re: Is x > x ^3 ? [#permalink]
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10 Oct 2014, 05:37
BunuelIs x> x^3? Is x>x^3? > is x^3x<0? > is (x+1)x(x1)<0? is x<1 or 0<x<1 when (x+1)x(x1)<0,, doesnt it gives us the rage as x>1 or x<1 ... because ..as you explained in an another question that Intersection points are the roots of the equation x^24x+3=0, which are x_1=1 and x_2=3. "<" sign means in which range of x the graph is below xaxis. Answer is 1<x<3 (between the roots). From this i understood that "<" sign means roots to the RIGHT of the smaller root and to the LEFT of the bigger root). I am getting it wrong?? please explain



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Re: Is x > x ^3 ? [#permalink]
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Re: Is x > x ^3 ? [#permalink]
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04 Feb 2016, 10:08
rxs0005 wrote: Is x > x ^3 ?
(1) x < 0 (2) x^2  x^3 > 2 Question : Is x > x ^3 ?for x to be greater than x^3 Case 1: Either x < 1 or Case 2: 0 < x < 1 Statement 1: x < 0 x may be 0.5 or x may be 2 hence NOT SUFFICIENT Statement 2: x^2  x^3 > 2 For this statement to be true x < 1, (Just try some values to substitute in expression to check acceptable values of x), Hence SUFFICIENT Answer: Option B
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Re: Is x > x ^3 ? [#permalink]
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04 Feb 2016, 18:41
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 > x ^3 ? (1) x < 0 (2) x^2  x^3 > 2 When it comes to inequality questions, it is crucial that if range of que includes range of con, that con is sufficient. When you modify the original condition and the question, they become x^3x<0?, x(x1)(x+1)<0? > x<1, 0<x<1?. Then, there is 1 variable(x), which should match with the number of equations. So you need 1 more equation. For 1) 1 equation, for 2) 1 equation, which is likey to make D the answer. For 1), if x<0, the range of que doesn’t include the range of con, which is not sufficient. For 2), in x^3x+2<0, (x+1)(x^22x+2)<0, x^22x+2=(x1)^2+1 is derived, which is always bigger than 0. Then, x+1<0 > x<1, in which the range of que includes the range of con. Thus, it is sufficient and the answer is B. > 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.
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Re: Is x > x ^3 ? [#permalink]
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02 Mar 2016, 11:16
Hi Bunuel, For x^2x^3>2 how did you directly arrive at the intervals <1 between 1 and 1 and >1. When we deal with (1x) >2 we get x <1. But how did you chose the other point 1?[/quote] Check here: x24x94661.html#p731476, inequalitiestrick91482.html, everythingislessthanzero108884.html?hilit=extreme#p868863, xyplane71492.html?hilit=solving%20quadratic#p841486[/quote][/quote] i could not understand with those links... as x^2x^3>2 we can write as X^2(1X)>2 so roots are 0,1 So range will be x>1 or x<0 or in number line if i plot i get +ve values within 0<x<1(i m plotting with roots 0 and 1 only in consideration) Plz help......... secondly plz clear me also that the for selecting different roots is same for different inequality like x^2x^3> 2 and x^2x^3> 0Thanks











