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Answer should be C. iii is incorrect because when x > x2, which is when 0
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tommannanchery
How many of the following equalities are possible for at least one value of X?

(i) \(x^2\) > X > \(x^3\)
(ii) \(x^2\) > \(x^3\) > X
(iii) \(x^3\) > X > \(x^2\)

A. i only
B. ii only
C. i and ii only
D. i, ii, and iii only
E. none

Answer is D.

For (i) - Consider x=1/3. => We can have X= 2/27 which is between 1/9 and 1/27

For (ii)- Consider x= 1/2= > We can have X=0 which is less than 1/4 and 1/8.

For (iii) - Consider x= 2 => We can have X = 5 which is between 8 and 4.

Hence, we can have each of the above statements valid for different vales of x and X.

NOTE : I have solved the above question based on the assumption that x and X are two different variables.
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tommannanchery
How many of the following equalities are possible for at least one value of X?

(i) \(x^2\) > X > \(x^3\)
(ii) \(x^2\) > \(x^3\) > X
(iii) \(x^3\) > X > \(x^2\)

A. i only
B. ii only
C. i and ii only
D. i, ii, and iii only
E. none

Answer is D.

For (i) - Consider x=1/3. => We can have X= 2/27 which is between 1/9 and 1/27

For (ii)- Consider x= 1/2= > We can have X=0 which is less than 1/4 and 1/8.

For (iii) - Consider x= 2 => We can have X = 5 which is between 8 and 4.

Hence, we can have each of the above statements valid for different vales of x and X.

NOTE : I have solved the above question based on the assumption that x and X are two different variables.

if the highlighted part is correct then there is no need of any option there always exist a number in between

Karishma

Please give your 2 cents also because source mentioned is veritas... :roll:
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tommannanchery
How many of the following equalities are possible for at least one value of X?

(i) \(x^2\) > X > \(x^3\)
(ii) \(x^2\) > \(x^3\) > X
(iii) \(x^3\) > X > \(x^2\)

A. i only
B. ii only
C. i and ii only
D. i, ii, and iii only
E. none

Answer is D.

For (i) - Consider x=1/3. => We can have X= 2/27 which is between 1/9 and 1/27

For (ii)- Consider x= 1/2= > We can have X=0 which is less than 1/4 and 1/8.

For (iii) - Consider x= 2 => We can have X = 5 which is between 8 and 4.

Hence, we can have each of the above statements valid for different vales of x and X.

NOTE : I have solved the above question based on the assumption that x and X are two different variables.

if the highlighted part is correct then there is no need of any option there always exist a number in between

Karishma

Please give your 2 cents also because source mentioned is veritas... :roll:

Correct bro. I agree to what you said. :)

Can anyone please help here?
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tommannanchery
How many of the following equalities are possible for at least one value of X?

(i) \(x^2\) > X > \(x^3\)
(ii) \(x^2\) > \(x^3\) > X
(iii) \(x^3\) > X > \(x^2\)

A. i only
B. ii only
C. i and ii only
D. i, ii, and iii only
E. none

The question is based on your understanding of how x, x^2 and x^3 behave in different regions of the number line.

x^2 > x for all negative x and when x > 1
x^3 > x for x > 1 and -1 < x < 0.


(i) \(x^2\) > X > \(x^3\)
True for all negative x.

(ii) \(x^2\) > \(x^3\) > X
True for -1 < x< 0

(iii) \(x^3\) > X > \(x^2\)
x is greater than x^2 between 0 and 1 only. But in that region x^3 is not greater than x. So this will not hold.

Answer (C)
Responding to a pm:
Quote:
I have a doubt for the second one?

-0.5 (-1/2) lies in the range -1<x<0

so if we consider x=-1/2; then x2 = -1/4 and x3 = -1/8

in this case x3>x2>x -_. so second case does not hold

Note that square of a real number is never negative.

If

\(x = -1/2,\)

\(x^2 = (-1/2)^2 = (-1/2)*(-1/2) = 1/4\)

\(x^3 = (-1/2)*(-1/2)*(-1/2) = -1/8\)

So
\(x^2 > x^3 > x\)
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tommannanchery
How many of the following equalities are possible for at least one value of X?

(i) \(x^2\) > X > \(x^3\)
(ii) \(x^2\) > \(x^3\) > X
(iii) \(x^3\) > X > \(x^2\)

A. i only
B. ii only
C. i and ii only
D. i, ii, and iii only
E. none


Bunuel: Can you please check this question? I did not find any values for which (iii) holds. Shouldn't OA be C?
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tommannanchery
How many of the following equalities are possible for at least one value of X?

(i) \(x^2\) > X > \(x^3\)
(ii) \(x^2\) > \(x^3\) > X
(iii) \(x^3\) > X > \(x^2\)

A. i only
B. ii only
C. i and ii only
D. i, ii, and iii only
E. none


Bunuel: Can you please check this question? I did not find any values for which (iii) holds. Shouldn't OA be C?

Yes. The correct answer is C, not D. Edited. Thank you.
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saichandm
Answer should be C. iii is incorrect because when x > x2, which is when 0<x<1), x3 is less than x2!!

Posted from my mobile device
For a problem that hinges on a conceptual understanding of properties of numbers with exponents, plugging in is a natural first approach. So try values in meaningfully different ranges. Say x is 2: that makes
\(x^2\)=4
and \(x^3\)=8
\(So, x^3 > x^2 > x\)
which doesn't work for any of the inequalities.
Now say x is -2: that makes \(x^2\)
\(x^3\)=-8

\(So, x^2 > x > x^3\)

Having tried both negative and positive integers, consider what other ranges are left: fractions between 0 and 1 and fractions between 0 and 1. Inequality (ii) will be true for any fraction between 0 and -1, so it's possible, but inequality (iii) never works: the only numbers for which x > \(x^2\) are fractions between 0 and 1, but for all such numbers \(x^2\) is also greater than \(x^3\)

Answer C
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i think for these kind of questions knowing how the graph looks and intersects helps a lot
x,x^2 and x^3 should be something that must be remembered

even graphs for 1/x and 1/(x^2) should be remembered
and we can easily see all of them intersect at 1 or -1 depending on the graph and thats where the > order changes.

once we know the above graphs. graphical transformation helps a lot
x-1 is nothing but x shifted by 1 unit down. so instead of cutting at x=0 it cuts at x=1
|x+5| is x shifted by 5 unit up and then taking a mirror image along x axis
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