If -1

Question

If -1 x^4
III. x^3 > x^5

A. I only 
B. I and II 
C. II only 
D. II​ and III 
E. I, II, and III

vivekkumar987 · Answer

I. x^3 < x^4 fails when x = 1/2
II. x^2 > x^4 100 % true in all case
III. x^3 > x^5 fails when x=-1/2

So in my IMO : C

ravisinghal · Answer

C I think. Take for any example any fraction and plug that in :- for stmt 1 :- take x = -1/2 - 1/8 < 1/16 which is true x = 1/2 1/8 > 1/16 hence not true stmt 2 : - both the quantitienties x^ 2 and x^4 will be positive and 1/4 > 1/16 which will be true for any fraction stmt 3 :- x3 > x^5 will not be true for negative fractions -1/8 < -1/32

babuvgmat · Answer

Ans:-

Given :- -1<x<0 and 0<x<1 means all fractions in between .

Must be true.

I. if x = -1/2 (true) but if x =1/2 (false) eliminate it.
II. Generally irrespective of signs all even powers will be positive and if power is greater value also will be greater for integers but for fractions it is reverse.
ex:- test for x =1/2 or -2/3 and so on.
 This option Holds good

III. odd powers this statement is false for all positive numbers and true for all negatives.

for fractions if x= 1/2 (true) but for x=-1/2 (false) eliminate.

SO only II must be true . Ans is  C

Abhishek009 · Answer

Plug in some values of x Say x = 1/2 Now check the options I. x^3 < x^4 {1/2}^3 < {1/2}^4 => \frac{1}{8} > \frac{1}{16} II. x^2 > x^4 {1/2}^2 > {1/2}^4 => \frac{1}{4} > \frac{1}{16} III. x^3 > x^5 {1/2}^3 < {1/2}^5 => \frac{1}{8} < \frac{1}{32} Hence we find that only option C. II satisfies the given condition... :lol:

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If -1 < x < 1, and x 0, which of the following inequalities must be

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