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in order for x < 1/x .. x must be a positive fraction and can not be equal zero.

let x = 1/2 or 1/3

then x^3 = 1/8
x^2 = 1/4
x = 1/2

X > X^2 > X^3

Therefore I choose (D) both II and III
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in order for x < 1/x .. x must be a positive fraction and can not be equal zero.

let x = 1/2 or 1/3

then x^3 = 1/8
x^2 = 1/4
x = 1/2

X > X^2 > X^3

Therefore I choose (D) both II and III
..

Your response helped me work out once again.. Thanks for the clue..

x < 1/x. This is possible in two scenarios (positive fraction) 0 < x < 1 and negative number which is x < -1.

If x = 1/2, x^2 = 1/4, x^3 = 1/8
so x > x^2 > x^3
If x = -2, x^2 = 4, x^3=-8
so x^2 > x > x^3.

Now looking into options only Options B is true in both the cases.
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Emdad
If \(x<\frac{1}{x}\), then which of the following must be true?
I. \(x^3> x^2\)
II.\(x^2> x^3\)
III. \(x > x^2\)

(A) only I
(B) only II
(C) only III
(D) both II and III
(E) none of these

Posted from my mobile device


\(x<\frac{1}{x}.......x-\frac{1}{x}<0........\frac{x^2-1}{x}<0\)
Two cases
1) If x>0, then \(x^2-1<0.....x^2<1........0<x<1\)
2) If x<0, then \(x^2-1>0.....x^2>1........x<-1\)

I. \(x^3> x^2\).......When 0<x<1, this is not true.
II.\(x^2> x^3\).....This is always true in the ranges x<-1 and 0<x<1
III. \(x > x^2\) .......When x<-1, this is not true.

B

Hi

Can you please explain this
1) If x>0, then \(x^2-1<0.....x^2<1........0<x<1\). (If x^2 - 1<0 then x^2<1, should it be -1<x<1 then?)
2) If x<0, then \(x^2-1>0.....x^2>1........x<-1\). (If x2-1>1 then x^2>1, should it be x<-1 and x>1 then?)
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chetan2u
Emdad
If \(x<\frac{1}{x}\), then which of the following must be true?
I. \(x^3> x^2\)
II.\(x^2> x^3\)
III. \(x > x^2\)

(A) only I
(B) only II
(C) only III
(D) both II and III
(E) none of these

Posted from my mobile device


\(x<\frac{1}{x}.......x-\frac{1}{x}<0........\frac{x^2-1}{x}<0\)
Two cases
1) If x>0, then \(x^2-1<0.....x^2<1........0<x<1\)
2) If x<0, then \(x^2-1>0.....x^2>1........x<-1\)

I. \(x^3> x^2\).......When 0<x<1, this is not true.
II.\(x^2> x^3\).....This is always true in the ranges x<-1 and 0<x<1
III. \(x > x^2\) .......When x<-1, this is not true.

B

Hi

Can you please explain this
1) If x>0, then \(x^2-1<0.....x^2<1........0<x<1\). (If x^2 - 1<0 then x^2<1, should it be -1<x<1 then?)
2) If x<0, then \(x^2-1>0.....x^2>1........x<-1\). (If x2-1>1 then x^2>1, should it be x<-1 and x>1 then?)

Yes, the range would be as you have mentioned.
But we are taking cases
1) x>0....-1<x<1, but x>0, so common range is 0<x<1
2) x<0....Here too answer will take the range such that x<0, so only x<-1 fits in, and we can discard x>1.
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