Considering the positions on the number line above, which of the follo

Very true.

I had taken both the conditions into account before I cancelled the options A and B

Also , when x < x^3

x can either lie between -1 and 0
Or
x can be any integer above 1

lets see the values x can take..

1) \(x^3<x^2.................x^3-x^2<0...............x^2(x-1)<0...................\)
this means \(x<1\) as \(x^2\) cannot be negative...

2) \(x<x^3............. x^3-x>0...........x(x^2-1)>0.........\)...
so both x and x^2-1 will be of the same sign..

a) if x is +ive, \(x^2-1>0...............x^2>1............ x>1... OR ...x<-1.......\) BUT x is +ive so x>1..
But we have seen above x<1.. so this is not possible

b) if x is -ive, \(x^2-1<0...............x^2<1............ x<1... OR ...x>-1.......\) BUT x is -ive so x lies between -1 and 0.....

From above 2 points, x lies between -1 and 0..... so x is -3/5
C

We are looking for a number where x < x^3 < x^2 .

This particular question can be easily solved through options.

A) 5/3.. this number is >1, so its square would be lesser than its cube. Eliminated.

B) 3/5.. this number lies between 0 and 1. So with increasing powers, value will keep on decreasing. So its cube will be lesser than the number itself . Eliminated.

C) -2/5 or -0.4. Its cube = -0.064, which is closer to zero hence greater than the number itself. Its square is positive and hence will be greater than both number and its cube. This is the one.

D) -5/2 or -2.5. This number is less than 1, its cube will be lesser than the number, not greater. Eliminated.

Hence C answer

+1 for option C. It is clear that this combination is possible only for negative numbers. Try out with negative options available. Only option C fits the bill.

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