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Is x > 1? (1) (x+1)(x  1) > 0 (2) x < 5
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18 Jun 2012, 15:46
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Is x > 1? (1) \((x+1)(x  1) > 0\) (2) \(x < 5\)
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Re: Is x > 1? (1) (x+1)(x  1) > 0 (2) x < 5
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18 Jun 2012, 16:15
Good question. +1. Is x> 1?(1) (x+1) (x  1) > 0. Consider two cases: If \(x>0\) then \(x=x\) so \((x+1) (x  1) > 0\) becomes \((x+1) (x  1) > 0\) > \(x^21>0\) > \(x^2>1\) > \(x<1\) or \(x>1\). Since we consider range when \(x>0\) then we have \(x>1\) for this case; If \(x\leq{0}\) then \(x=x\) so \((x+1) (x  1) > 0\) becomes \((x+1) (x  1) > 0\) > \((x+1) (x+1) > 0\) > \((x+1)^2>0\) > \((x+1)^2<0\). Now, since the square of a number cannot be negative then for this range given equation has no solution. So, we have that \((x+1) (x  1) > 0\) holds true only when \(x>1\). Sufficient. (2) x < 5 > \(5<x<5\). Not sufficient. Answer: A. Hope it's clear.
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Re: Is x > 1? (1) (x+1)(x  1) > 0 (2) x < 5
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18 Jun 2012, 16:26
Thanks Bunuel. Can you please explain how did you get this? If \(x>0\) then \(x=x\)
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Re: Is x > 1? (1) (x+1)(x  1) > 0 (2) x < 5
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18 Jun 2012, 16:30
enigma123 wrote: Thanks Bunuel. Can you please explain how did you get this?
If \(x>0\) then \(x=x\) Check this: mathabsolutevaluemodulus86462.htmlAbsolute value properties:When \(x\leq{0}\) then \(x=x\), or more generally when \(some \ expression\leq{0}\) then \(some \ expression={(some \ expression)}\). For example: \(5=5=(5)\). When \(x\geq{0}\) then \(x=x\), or more generally when \(some \ expression\geq{0}\) then \(some \ expression={some \ expression}\). For example: \(5=5\). Hope it helps.
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Re: Is x > 1? (1) (x+1)(x  1) > 0 (2) x < 5
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19 Jun 2012, 12:22
Bunuel wrote: Good question. +1.
Is x> 1?
(1) (x+1) (x  1) > 0. Consider two cases:
If \(x>0\) then \(x=x\) so \((x+1) (x  1) > 0\) becomes \((x+1) (x  1) > 0\) > \(x^21>0\) > \(x^2>1\) > \(x<1\) or \(x>1\). Since we consider range when \(x>0\) then we have \(x>1\) for this case;
If \(x\leq{0}\) then \(x=x\) so \((x+1) (x  1) > 0\) becomes \((x+1) (x  1) > 0\) > \((x+1) (x+1) > 0\) > \((x+1)^2>0\) > \((x+1)^2<0\). Now, since the square of a number cannot be negative then for this range given equation has no solution.
So, we have that \((x+1) (x  1) > 0\) holds true only when \(x>1\). Sufficient.
(2) x < 5 > \(5<x<5\). Not sufficient.
Answer: A.
Hope it's clear. Dear Bunuel, i got (x+1)^2<0 . and further solved to x+1<0 giving x <1 and ended up in wrong answer E. square of a number cannot be negative. but just confused for (x+1)^2 where x is a variable?



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Re: Is x > 1? (1) (x+1)(x  1) > 0 (2) x < 5
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20 Jun 2012, 01:07
kashishh wrote: Bunuel wrote: Good question. +1.
Is x> 1?
(1) (x+1) (x  1) > 0. Consider two cases:
If \(x>0\) then \(x=x\) so \((x+1) (x  1) > 0\) becomes \((x+1) (x  1) > 0\) > \(x^21>0\) > \(x^2>1\) > \(x<1\) or \(x>1\). Since we consider range when \(x>0\) then we have \(x>1\) for this case;
If \(x\leq{0}\) then \(x=x\) so \((x+1) (x  1) > 0\) becomes \((x+1) (x  1) > 0\) > \((x+1) (x+1) > 0\) > \((x+1)^2>0\) > \((x+1)^2<0\). Now, since the square of a number cannot be negative then for this range given equation has no solution.
So, we have that \((x+1) (x  1) > 0\) holds true only when \(x>1\). Sufficient.
(2) x < 5 > \(5<x<5\). Not sufficient.
Answer: A.
Hope it's clear. Dear Bunuel, i got (x+1)^2<0 . and further solved to x+1<0 giving x <1 and ended up in wrong answer E. square of a number cannot be negative. but just confused for (x+1)^2 where x is a variable? It doesn't matter that \(x\) is a variable, it's still some number and so is \(x+1\). So, \((x+1)^2\) is a square of that number and it cannot be negative. Also your way of solving is not correct: \((x+1)^2<0\) does not mean \(x+1<0\) it means that \(x+1<0\). From that you could deduce the same: since absolute value cannot be negative then this equation has no solution. Hope it's clear.
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Re: Is x > 1? (1) (x+1)(x  1) > 0 (2) x < 5
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21 Jun 2012, 01:04
Hi Bunuel, I understood your solution but can you please tell me where am I going wrong?
When I see the expression, (x+1)(x1)>0, I immediately think that these two brackets must be either positive or negative.
Hence, if that take both of them are positive, then x>1 and and x>1 & x<1
For these two ranges we realize when we start plugging in number that only for x>1 the equation hold true.(1)
Similarly, if we take both of them to be negative, then x<1 and x<1 & x>1
for x<1 for example, 2, one of the brackets turn out to be positive, which does not satisfy our initial assumption that both must be negative.
for x<1 ex 0, one bracket is +ve and one is ve <Not Desired>; for 0, the inequality is not > 0 <not desired>
if we go still less it follows the first case
for x>1 pretty much follows like the above. Hence nothing desired.(2)
Hence, x>1
A is sufficient.



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Re: Is x > 1? (1) (x+1)(x  1) > 0 (2) x < 5
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21 Jun 2012, 02:41
pavanpuneet wrote: Hi Bunuel, I understood your solution but can you please tell me where am I going wrong?
When I see the expression, (x+1)(x1)>0, I immediately think that these two brackets must be either positive or negative.
Hence, if that take both of them are positive, then x>1 and and x>1 & x<1
For these two ranges we realize when we start plugging in number that only for x>1 the equation hold true.(1)
Similarly, if we take both of them to be negative, then x<1 and x<1 & x>1
for x<1 for example, 2, one of the brackets turn out to be positive, which does not satisfy our initial assumption that both must be negative.
for x<1 ex 0, one bracket is +ve and one is ve <Not Desired>; for 0, the inequality is not > 0 <not desired>
if we go still less it follows the first case
for x>1 pretty much follows like the above. Hence nothing desired.(2)
Hence, x>1
A is sufficient. First of all I wouldn't recommend to solve this question the way you are doing. Next, when you consider both multiples to be negative and get \(x<1\) from the first one, then the second multiple automatically transformes to \((x1)\), since if \(x<1<0\) then \(x=x\). So, we have that \(x1<0\) must also be true or \(x>1\), which contradicts the case for the first multiple (\(x<1\)). So, both \(x+1\) and \(x1\) can not be negative. Also I think you got x<1 & x>1 from x<1, and if yes, then it's not correct: \(x<1\) means that \(1<x<1\). So, again \(x<1\) (for the first multiple to be negative) and \(1<x<1\) (for the second multiple to be negative) cannot simultaneously be true. Hope it's clear.
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Re: Is x > 1? (1) (x+1)(x  1) > 0 (2) x < 5
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27 Jan 2013, 23:17
fozzzy wrote: Is x>1
1) (x+1)(lxl1) > 0 2) lxl < 5 Statement 1: For (x+1)(lxl1) > 0, we should have either (x+1)>0 and (lxl1) > 0 or (x+1)<0 and (lxl1) < 0 when (x+1)>0 and (lxl1) > 0 (x+1)>0 => x>1 (lxl1) > 0 => x>1 or x <1 From above two, possible solution is x>1 when (x+1)>0 and (lxl1) < 0 (x+1)<0 => x<1 (lxl1) < 0 => 1<x<1 Both of these can not be satisfied by any value of x. Hence we get only 1 solution, x>1. which is what we wanted to ascertain. Sufficient. Statement 2: x <5 => 5<x<5 Clearly not sufficient to tell whether x>1 or not. Ans A it is.



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Re: Is x > 1? (1) (x+1)(x  1) > 0 (2) x < 5
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Updated on: 28 Jan 2013, 00:25
so you just have to find the common region if its done on a number line and we ignore one of the cases, since there isn't a common region?
Originally posted by fozzzy on 28 Jan 2013, 00:21.
Last edited by fozzzy on 28 Jan 2013, 00:25, edited 1 time in total.



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Re: Is x > 1? (1) (x+1)(x  1) > 0 (2) x < 5
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06 Feb 2013, 03:21
andrew40 wrote: Bunuel wrote: Good question. +1.
Is x> 1?
(1) (x+1) (x  1) > 0. Consider two cases:
If \(x>0\) then \(x=x\) so \((x+1) (x  1) > 0\) becomes \((x+1) (x  1) > 0\) > \(x^21>0\) > \(x^2>1\) > \(x<1\) or \(x>1\). Since we consider range when \(x>0\) then we have \(x>1\) for this case;
If \(x\leq{0}\) then \(x=x\) so \((x+1) (x  1) > 0\) becomes \((x+1) (x  1) > 0\) > \((x+1) (x+1) > 0\) > \((x+1)^2>0\) > \((x+1)^2<0\). Now, since the square of a number cannot be negative then for this range given equation has no solution.
So, we have that \((x+1) (x  1) > 0\) holds true only when \(x>1\). Sufficient.
(2) x < 5 > \(5<x<5\). Not sufficient.
Answer: A.
Hope it's clear. Sorry, I don't understand why we should consider the range where x>0. because of the absolute value? We need to get rid of the modulus in the expression to solve it and this is the way to do that. Check here: isx134652.html#p1097668Hope it helps.
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Re: Is x > 1? (1) (x+1)(x  1) > 0 (2) x < 5
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18 May 2013, 05:56
Bunuel wrote: Good question. +1.
Is x> 1?
(1) (x+1) (x  1) > 0. Consider two cases:
If \(x>0\) then \(x=x\) so \((x+1) (x  1) > 0\) becomes \((x+1) (x  1) > 0\) > \(x^21>0\) > \(x^2>1\) > \(x<1\) or \(x>1\). Since we consider range when \(x>0\) then we have \(x>1\) for this case;
If \(x\leq{0}\) then \(x=x\) so \((x+1) (x  1) > 0\) becomes \((x+1) (x  1) > 0\) > \((x+1) (x+1) > 0\) > \((x+1)^2>0\) > \((x+1)^2<0\). Now, since the square of a number cannot be negative then for this range given equation has no solution.
So, we have that \((x+1) (x  1) > 0\) holds true only when \(x>1\). Sufficient.
(2) x < 5 > \(5<x<5\). Not sufficient.
Answer: A.
Hope it's clear. Hi, This is my first post so was little conscious to ask my doubt. In the above question, we took the roots as 1<x>1. However, in this question as mentioned below (unable to post the link as per new member rule) is ((X3)^2)^1/2 = 3X ? 1) X does not = 3 2) XX > 0 the roots are 3<x>3. Can you please explain the difference?



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Re: Is x > 1? (1) (x+1)(x  1) > 0 (2) x < 5
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22 May 2013, 16:45
Please help,
When I try this question for statement 1: (x+1) (x  1) > 0 (x+1)x(x+1)>0 (x+1)x >(x+1) x > 1
x>1 ; x<1



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Re: Is x > 1? (1) (x+1)(x  1) > 0 (2) x < 5
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23 May 2013, 02:14
smartyman wrote: Please help,
When I try this question for statement 1: (x+1) (x  1) > 0 (x+1)x(x+1)>0 (x+1)x >(x+1) x > 1
x>1 ; x<1 Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know its sign.So you cannot reduce both parts of inequality (x+1)x>(x+1) by x+1 as you don't know the sign of x+1: if x+1>0 you should write x>1 BUT if x+1<0 you should write x<1 (flip the sign). Hope it helps.
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Re: Is x > 1? (1) (x+1)(x  1) > 0 (2) x < 5
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02 Jun 2013, 10:59
Hi guys,
I just want to present an easier way to prove that I is sufficient.
For I to be positive what are the conditions?
1. x+1 > 0 this means x >1 2.x 1 > 0 this means x > 1 or x < 1
then you must draw the number line [b]1[b]01> then draw this inequalities on this line and look for any unity. then you will fine x > 1
for the other side, I mean, x+1 <0 and x  1 <0 you wont find any unity. so I is sufficient. Hope it clears. cheers



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Re: Is x > 1? (1) (x+1)(x  1) > 0 (2) x < 5
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23 Jun 2013, 17:59
So here is my question:
For #1 we have two cases: positive and negative
For x≥0 (x+1)(x1)>0 (x+1)(x1)>0 x^21>0 x^2>1 Of course, x could be 2 or negative two for all we know, so it seems like this is insufficient as we are testing for the range of x>1
For x≤0 (x+1)(x1)>0 (x+1)(x1)>0 x^21>0 x^2>1
I am a bit unsure where I went wrong here.



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Re: Is x > 1? (1) (x+1)(x  1) > 0 (2) x < 5
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24 Jun 2013, 02:11
WholeLottaLove wrote: So here is my question:
For #1 we have two cases: positive and negative
For x≥0 (x+1)(x1)>0 (x+1)(x1)>0 x^21>0 x^2>1
Of course, x could be 2 or negative two for all we know, so it seems like this is insufficient as we are testing for the range of x>1 x can never be 2, as because you have assumed that \(x\geq0\). Also, from the inequality you have correctly arrived at, i.e. \(x^2>1\) \(\to\) x>1 OR x<1. As assumption was \(x\geq0\). thus only x>1 condition is valid. Also,as x>1 automatically makes \(x\geq0\), thus the correct range is x>1. Sufficient. Quote: For x≤0 No need to include equality with zero twice. (x+1)(x1)>0 (x+1)(x1)>0 This leads to \((x+1)^2>0\) and this is not possible for any real value of x. So,there is no solution for this. x^21>0 x^2>1
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Re: Is x > 1? (1) (x+1)(x  1) > 0 (2) x < 5
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28 Aug 2013, 13:08
I mean to say, we define X as +x when x is greater than equal to zero, and x when x is less than zero. But I have noticed you at times have taken x as x based on x less than equal to zero ? I mean equality cant be on both sides right ? Infact I saw you explained definition of mod function keeping equality on both +x and  x, I hope you are able to make out what doubt I have.



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Re: Is x > 1? (1) (x+1)(x  1) > 0 (2) x < 5
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29 Aug 2013, 04:34
ygdrasil24 wrote: I mean to say, we define X as +x when x is greater than equal to zero, and x when x is less than zero. But I have noticed you at times have taken x as x based on x less than equal to zero ? I mean equality cant be on both sides right ? Infact I saw you explained definition of mod function keeping equality on both +x and  x, I hope you are able to make out what doubt I have. The point is that 0=0=0. So, in that definition we can include = sign in both cases. Hope it helps.
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Re: Is x > 1? (1) (x+1)(x  1) > 0 (2) x < 5
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04 Apr 2014, 22:15
enigma123 wrote: Is x > 1?
(1) (x+1) (x  1) > 0
(2) x < 5 Sol: (x+1) (x  1) > 0 The RHS can only be positive if both are positive or both are negative lets take 5 as value of x===> ve *+ve ===>not possible let's take 0 as value of x ==> 1* (1) ===>not possible let's take 1 as value (+ve) ===> 2* (0)===> not possible let's take 1 as value (ve) ====>0 *0 ====>not possible so x != ve, x != 0, x != 1 conditions apply. only remaining option is x>1 ....sufficient 2> (5)@@@@@@(5) 5<x<5 ...not sufficient since x is less than 1 here final sol is A




Re: Is x > 1? (1) (x+1)(x  1) > 0 (2) x < 5
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