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Which of the following is always true?

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New post 24 Jul 2019, 08:00
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Which of the following is always true?


A. If \(\frac{1}{x}\) is greater than \(x\), \(x\) is greater than \(x^2\).

B. If \(x\) is greater than \(\frac{1}{x}\), \(2x\) is greater than \(x\).

C. If \(x\) is greater than \(2x\), \(\frac{1}{x}\) is greater than \(x\).

D. If \(x^2\) is greater than \(x\), \(x^3\) is greater than \(x^2\).

E. If \(x\) is greater than \(\frac{1}{x}\), \(x^2\) is greater than \(\frac{1}{x}\).


 

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New post Updated on: 24 Jul 2019, 08:53
1
Quote:
Which of the following is always true?


A. If 1/x is greater than x, x is greater than x^2.

B. If x is greater than 1/x, 2x is greater than x.

C. If x is greater than 2x, 1/x is greater than x.

D. If x^2 is greater than x, x^3 is greater than x^2.

E. If x is greater than 1/x, x^2 is greater than 1/x.


A- Say x=-2 then 1/x=-1/2 but -2 is not greater than 4.
B- Say x=-1/2 then 1/x=-2 but 2x=-1 is not greater than -1/2
C- Say x=-1 then 2x=-2 but 1/x=-1 is not greater than -1
D- Say x=-2 then x^2=4 but x^3=-8 is not greater than 4

Hence E
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Originally posted by kitipriyanka on 24 Jul 2019, 08:21.
Last edited by kitipriyanka on 24 Jul 2019, 08:53, edited 1 time in total.
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Re: Which of the following is always true?  [#permalink]

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New post 24 Jul 2019, 08:26
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Which of the following is always true?


A. If 1x1x is greater than xx, xx is greater than x2x2.

B. If xx is greater than 1x1x, 2x2x is greater than xx.

C. If xx is greater than 2x2x, 1x1x is greater than xx.

D. If x2x2 is greater than xx, x3x3 is greater than x2x2.

E. If xx is greater than 1x1x, x2x2 is greater than 1x1x.


1 not true at x=-1/2
2. not true at x=-1
3. not true at x=1
4. not true at x=-2
5. valid always IMO E
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New post Updated on: 24 Jul 2019, 09:21
1
Quote:
Which of the following is always true?


A. If 1x1x is greater than xx, xx is greater than x2x2.

B. If xx is greater than 1x1x, 2x2x is greater than xx.

C. If xx is greater than 2x2x, 1x1x is greater than xx.

D. If x2x2 is greater than xx, x3x3 is greater than x2x2.

E. If xx is greater than 1x1x, x2x2 is greater than 1x1x.


POE:
A. If 1x1x is greater than xx, xx is greater than x2x2.
if \(\frac{1}{x}\) > x
for x = 1/2
this eq will always be true. Hence Incorrect


B. If xx is greater than 1x1x, 2x2x is greater than xx.
let x = 2 then this will always be true as 4 > 2.

C. If xx is greater than 2x2x, 1x1x is greater than xx.
let x= -2 then 1/-2 > -2
Hence always true.


D. If x2x2 is greater than xx, x3x3 is greater than x2x2.
for x =-2
then -8 > 4 is not true.
Hence D is the answer.


E. If xx is greater than 1x1x, x2x2 is greater than 1x1x
let x =2 hence 4 > 1/2
Hence this will always be true.


Answer is D

Originally posted by techloverforever on 24 Jul 2019, 08:28.
Last edited by techloverforever on 24 Jul 2019, 09:21, edited 1 time in total.
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New post 24 Jul 2019, 08:28
take cases for positive ,negative ,less than 1 and greater than 1 only E holds true in all
C fails when number is -1.
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New post 24 Jul 2019, 08:31
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Which of the following is always true?

This is an easy question, you could solve it by doing multiple cases to disprove the equation. I solve it by plugging in numbers like -2, 1/-2, 2, 1/2 Hence was able to disprove all but E.


A. If 1x1x is greater than xx, xx is greater than x2x2.
This is an incorrect choice.

B. If xx is greater than 1x1x, 2x2x is greater than xx.
This is an incorrect choice.

C. If xx is greater than 2x2x, 1x1x is greater than xx.
This is an incorrect choice.

D. If x2x2 is greater than xx, x3x3 is greater than x2x2.
This is an incorrect choice.

E. If xx is greater than 1x1x, x2x2 is greater than 1x1x.
Yes, this is the correct choice.

The answer choice is E.
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Which of the following is always true?  [#permalink]

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New post Updated on: 24 Jul 2019, 09:09
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Quote:
Which of the following is always true?

A. If 1/x is greater than x, x is greater than x2.
B. If x is greater than 1/x, 2x is greater than x.
C. If x is greater than 2x, 1/x is greater than x.
D. If x2 is greater than x, x3 is greater than x2.
E. If x is greater than 1x, x2 is greater than 1/x.


x: 2x: 1/x: xˆ2: xˆ3:
[1] 0.5…1…2…0.25…0.12
[2] -0.5…-1…-2…0.25…-0.12
[3] 2…4…0.5…4…8
[4] -2…-4…-0.5…4…-8

(A) 2 cases [1,4] not always true;
(B) 2 cases [2,3] not always true;
(C) 2 cases [2,4] not always true;
(D) 2 cases [2,3,4] not always true;

Answer (E).

Originally posted by exc4libur on 24 Jul 2019, 08:35.
Last edited by exc4libur on 24 Jul 2019, 09:09, edited 1 time in total.
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Re: Which of the following is always true?  [#permalink]

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New post 24 Jul 2019, 08:36
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The Answer is E.

Test different cases to try to prove each false.

A.
Try this with a negative number. Say x is -2. Then, -0.5 is therefore greater than -2, however -2 is not greater than (-2)^2 which is 4. Hence, false.

B.
Which cases x will be greater than 1/x is when x is a positive integer >1, or x is negative integer between 0 and -1.
Hence, test out both cases. e.g. when x is positive 2, then x is greater than 1/2, and 2(2) is greater than 2. If x is -.5, then x is greater than -1/.5 which is -2, but 2(-0.5) which is -1 is LESS than -0.5. Hence false

C. In this case, x will be greater than 2x if x is a negative number. Test out if x is -0.5, then -1/x is -5 which is LESS than x. Hence false


D.
In this case, x^2 will be greater than x if x is either positive integer greater than 1 or negative integer less than 1. However, if it is a negative integer less than 1, then x^3 is LESS than x^2 as x^3 will be negative. Hence false


E. Correct.
Which cases x will be greater than 1/x is when x is a positive integer >1, or x is negative integer between 0 and -1.
Hence, test out both cases. e.g. when x is positive 2, then x^2 is 4 which is greater than 4. If x is -0.1, then x^2 is -.01 which is greater than -0.1
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Re: Which of the following is always true?  [#permalink]

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New post 24 Jul 2019, 08:40
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A. If 1/x is greater than x, x is greater than x^2.
--> Possible Range is x < -1 or 0 < x < 1
--> x is not always greater than x^2 (eg: x = -2, x^2 = 4) -- NO

B. If x is greater than 1/x, 2x is greater than x.
--> Possible Range is -1 < x < 0 or x > 1
--> 2x is not always greater than x (eg: x = -0.5, 2x = -1) -- NO

C. If x is greater than 2x, 1/x is greater than x.
--> Possible Range is x < 0
--> 1/x is not always greater than x (eg: x = -0.5, 1/x = -2) -- NO

D. If x^2 is greater than x, x^3 is greater than x^2.
--> Possible Range is x < -1 or x > 1
--> x^3 is not always greater than x^2 (eg: x = -2, x^3 = -8, x^2 = 4) -- NO

E. If x is greater than 1/x, x^2 is greater than 1/x.
--> Possible Range is -1 < x < 0 or x > 1
--> x^2 is always greater than 1/x (eg: x = -0.5, (x^2, 1/x) = (0.25, -2); eg: x = 2, (x^2, 1/x) = (4, 0.5)) -- YES

IMO Option E

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New post Updated on: 24 Jul 2019, 18:57
1
In this one, is not a bad idea to try values to discard some options, so we can first start trying with negative numbers: -2 and -1/2

A) If 1/x is greater than x, x is greater than x^2. For x = -2 we have that 1/x = -0.5 and x^2=4, so not always true
B) If x is greater than 1/x, 2x is greater than x. For x =-1/2 we have that 1/x = -2 , 2x= -1, so not always true
C) If x is greater than 2x, 1/x is greater than x. For x = -1/2, 2x=-1 1/x = -2, so not always true
D) If x^2 is greater than x, x3 is greater than x^2. Not necessarily, since x could be a negative number, so not always true
E) If x is greater than 1/x, x^2 is greater than 1/x. Always true

(E) is our answer

Originally posted by Mizar18 on 24 Jul 2019, 08:48.
Last edited by Mizar18 on 24 Jul 2019, 18:57, edited 1 time in total.
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New post Updated on: 24 Jul 2019, 21:25
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Let us check option by option:

A. If \(\frac{1}{x}\) is greater than x –> x can be 0.5, -2
x is greater than \(x^2\) -> When x = 0.5 NO

B. If x is greater than \(\frac{1}{x}\)-> x can be -0.5, 5
2x is greater than x -> When x = -0.5 NO

C. If x is greater than 2x -> x can be -1, -0.5
\(\frac{1}{x}\) is greater than x -> When x = -0.5 NO

D. If \(x^2\) is greater than x -> x can be -1, 2
\(x^3\) is greater than \(x^2\) -> When x = -1 NO

E. if x is greater than \(\frac{1}{x}\) -> x can be 2, -0.5
\(x^2\) is greater than \(\frac{1}{x}\) -> Always True

Answer E

Originally posted by Sayon on 24 Jul 2019, 09:00.
Last edited by Sayon on 24 Jul 2019, 21:25, edited 1 time in total.
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New post 24 Jul 2019, 09:02
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For these type of questions we can divide number into 4 zones ie. -infinity to -1, -1 to 0, 0 to 1, 1 to infinity

The question is must be true or always true, for each zone the condition has to satisfy

A. If 1/x is greater than x, means x lies between -infinity to -1 and 0 to 1, then x is not greater than x^2 on one region.

B. not true when x is between -1 to 0

C. not true when x is between -1 to 0

D. not true when x is between -1 to 0

E is correct.
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New post 24 Jul 2019, 09:13
Honestly none of the options work, but I chose A because it looked better.

a) 1/x > x, so -1<x<1. Then, x > x^2. This will work for 0<x<1, but not for -1<x<0. So "not always" true, but found it closest.

b) x>1/x, so x>1 & x<-1. Then, 2x>x so x>0. Again, does not "always" work

c) x>2x, so x<0. Then, 1/x > x, we get -1<x<1. Does not always work.

d) x^2 > x, so x<0 & x>1. Then, x^3>x^2, we get x>1. Does not work for x<0

e) x>1/x, so x>1 & x<-1. Then, x^2>1/x, we get x>1. Does not work for x<-1
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Re: Which of the following is always true?  [#permalink]

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New post 24 Jul 2019, 09:21
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A. If \(1/x\) is greater than \(x\), \(x\) is greater than \(x^2\).

\(1/x>x\)
case 1. \(x\) is positive
\(1>x^2\)
\(x^2-1<0\)
\((x-1)(x+1)<0\)
since x is positive thus \(0<x<1\)
case 2. x is negative
\(1>-x^2\)
\(x^2>-1\)
since \(x\) is negative thus\(-inf<x<-1\)

Clearly, in the second case \(x\) is not greater than \(x^2\).

B. If \(x\) is greater than \(1/x\), \(2x\) is greater than\(x\).

case 1. \(x\) is positive.
\(x^2>1\)
\((x-1)(x+1)>0\)
since is positive thus \(1<x<+inf\)
case 2. \(x\)is negative.
\(-x^2>1\)
\(x^2<-1\)
since \(x\) is negative thus \(-1<x<0\)

Clearly, in the second case \(2x\) is not greater than \(x\).

C. If \(x\) is greater than \(2x\), \(1/x\) is greater than \(x\).

\(x>2x\)
\(x<0\)

Here, if we take \(-1<x<0\) then \(1/x\) is not greater than \(x\)

D. If \(x^2\) is greater than \(x\), \(x^3\) is greater than \(x^2\).

\(x^2>x\)
\(x*(x-1)>0\)
\(-inf<x<0\); \(1<x<+inf\)

Thus \(x\) can be any negative number, but \(x^3\) is not greater than \(x^2\)


E. If \(x\) is greater than \(1/x\), \(x^2\) is greater than\(1/x\)


case 1. \(x\) is positive.
\(x^2>1\)
\((x-1)(x+1)>0\)
since is positive thus \(1<x<+inf\)
case 2. \(x\)is negative.
\(-x^2>1\)
\(x^2<-1\)
since \(x\) is negative thus \(-1<x<0\)

Here, both cases are always true.


The answer is E
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New post 24 Jul 2019, 09:26
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Which of the following is always true?

A. If 1/x is greater than x, x is greater than x^2. --> not always true
case-1: true: 0<x<1, take x = 0.5, 1/0.5>0.5 => 0.5 > 0.5^2
case-2: false: x<-1, take x = -2, 1/-2>-2, but -2 < (-2)^2

B. If x is greater than 1/x, 2x is greater than x.--> not always true
case-1: true: x>1, take x = 2, 2>1/2 => 2*2 > 2
case-2: false: -1<x<0, take x = -0.5, -0.5>1/(-0.5), but 2*(-0.5) < -0.5

C. If x is greater than 2x, 1/x is greater than x.--> not always true
case-1: true: x <-1, take x = -2, -2>2*(-2) => 1/(-2) > -2
case-2: false: -1<x<0, take x = -0.5, -0.5>2*(-0.5), but 1/(-0.5) < -0.5

D. If x^2 is greater than x, x^3 is greater than x^2.--> not always true
case-1: true: x >1, take x = 2, 2^2>2 => 2^3 > 2^2
case-2: false: x<0, take x = -1, (-1)^1>-1, but (-1)^3 < (-1)^2

E. If x is greater than 1/x, x^2 is greater than 1/x.--> correct: always true
case-1: true: x>1, take x = 2, 2>1/2 => 2^2 > 1/2
case-2: true: -1<x<0, take x = -0.5, -0.5>1/(-0.5), but (-0.5)^2 > 1/( -0.5)
case-1: true: if x>0, x >1/x=>x^2>1=> x>1, so x^2>1/x
case-2: true: if x<0, x >1/x=>x^2<1=>-1<x<0, so x <0 & x^2>0 => x^2>1/x
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New post 24 Jul 2019, 09:33
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Lets try to eliminate unfavorable cases here

A. If \(\frac{1}{x}\) is greater than \(x\), \(x\) is greater than \(x^2\).
Consider \(x=-2\),
\(x=-2, x^2=4\)
So \(x<x^2\)

Eliminate (A)


B. If \(x\) is greater than \(\frac{1}{x}\), \(2x\) is greater than \(x\).
Consider \(x=-\frac{1}{2}\)
\(2x=-1, x=-0.5\)
So \(2x<x\)

Eliminate (B)

C. If \(x\) is greater than \(2x\), \(\frac{1}{x}\) is greater than \(x\).
Consider \(x=-1\)
\(\frac{1}{x}=-1, x=-1\)
So, \(\frac{1}{x}=x\)

Eliminate (C)

D. If \(x^2\) is greater than \(x\), \(x^3\) is greater than \(x^2\).
Consider \(x=-\frac{1}{2}\)
\(x^3=-\frac{1}{8}, x^2=\frac{1}{4}\)
So \(x^3<x^2\)

Eliminate (D)

E. If \(x\) is greater than \(\frac{1}{x}\), \(x^2\) is greater than \(\frac{1}{x}\).

We are left with (E)

Answer is (E)
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New post 24 Jul 2019, 09:53
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IMO correct answer is E - Explanation attached
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New post 24 Jul 2019, 09:55
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Option A:- Put x=-0.5 , A out
Option B:- Put x=-2 , B out
Option C:- Put x=-0.5, C out
Option D:- Put x=-1, D out

Option E satisfies all the condition. Hence answer is E.
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New post 24 Jul 2019, 10:01
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tricky question, we can solve by cases:
4 cases -2,1/2,-2,-1/2
A. it fails when we take x is -2
B. it fails when we take x is -1/2
C. it fails when we take x is -1/2
D. it fails when we take x is -2

E will always pass for all cases
hence ans is E
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New post 24 Jul 2019, 10:18
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Which of the following is always true?


A. If \(1/x\) is greater than \(x\), \(x\) is greater than \(x^2\).
\(1/x > x\) when either \(x\) is negative and \(x<-1\) or \(x\) is positive and \(x<1\).
If \(x=1/3\), then \(1/x=1/(1/3)=3 > x=1/3\) => \(x=1/3 > x^2=(1/3)^2=1/9\) OK
If \(x=-5\), then \(1/x=-1/5 > x=-5\) => \(x=-5 > x^2=(-5)^2=25\)
-5 is not greater than 25 Not Correct

B. If \(x\) is greater than \(1/x\), \(2x\) is greater than \(x\).
\(x>1/x\) when either \(x\) is positive and \(x>1\) or \(x\) is negative and \(x>-1\)
If \(x=2\), then \(x=2 > 1/x = 1/2\) => \(2x=2*2=4 > x=2\) OK
If \(x = -1/2\), then \(x=-1/2 > 1/x=1/(-1/2)=-2\) => \(2x=2*(-1/2)=-1 > x=-1/2\)
-1 is not greater than -1/2 Not Correct

C. If \(x\) is greater than \(2x\), \(1/x\) is greater than \(x\).
\(x>2x\) when x is negative
If \(x = -1/3\), then \(x=-1/3 > 2x=-2/3\) => \(1/x=1/(-1/3)=-3 > -1/3\)
-3 is not greater than -1/3 Not Correct

D. If \(x^2\) is greater than \(x\), \(x^3\) is greater than \(x^2\).
\(x^2>x\) when \(x\) is negative or when \(x\) is positive and \(x>1\)
If \(x=2\), then \(x^2=2^2=4 > x=2\) => \(x^3=2^3=8 > x^2=2^2=4\) OK
If \(x=-2\), then \(x^2=(-2)^2=4 > x=-2\) => \(x^3=(-2)^3=-8 > x^2=(-2)^2=4\)
-8 is not greater than 4 Incorrect

E. If \(x\) is greater than \(1/x\), \(x^2\) is greater than \(1/x\).
As all previous statements were incorrect, this statement is correct. But let us still prove it.
If \(x=-1/2\), then \(x=-1/2 > 1/x=1/(-1/2)=-2\) => \(x^2=(-1/2)^2=1/4 > 1/x=1/(-1/2)=-2\). If \(x\) is negative, the second statement is always correct as the square of a negative number is a positive number.
If \(x\) is positive and less than 1, the condition is not adhered to. If \(x\) is positive and more than 1, then the statement is correct.
Correct.

Answer: E
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