Is 0<x<1? (I) x^2 < x^{1/3} (II) x^3<x^{1/3 : DS Archive
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# Is 0<x<1? (I) x^2 < x^{1/3} (II) x^3<x^{1/3

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SVP
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Is 0<x<1? (I) x^2 < x^{1/3} (II) x^3<x^{1/3 [#permalink]

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24 Jan 2005, 21:39
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Is 0<x<1?

(I)$$x^2 < x^{1/3}$$
(II)$$x^3<x^{1/3}$$
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24 Jan 2005, 22:07
i think A.
statement i is suff.......... and statement ii is not suff...
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25 Jan 2005, 06:37
Is it (D)?

Is 0<x<1?

(I)x^2 < x^(1/3) - raise both sides to the power 3: x^2/3 < x

(II)x^3<x^(1/3) - same: raise both sides to the power of 3: x^9 < x
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25 Jan 2005, 07:27
I go for D

that is if negative numbers are allowed! in that case we have an imaginary numbers, then in that case its A
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25 Jan 2005, 07:44
"A" for me.

state 1: x = 1/8....satisfies.....x = -1/8 doesn't ....so x has to be between 0 and 1...suff.

state 2: x = 1/8....satisfies....x = -8...satisfies....insuff.
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25 Jan 2005, 18:18
Is 0<x<1?

(I)x^2 < x^(1/3)
(II)x^3<x^(1/3)

for 0<x<1 , in either cases cubing the inequations will help.
on cubing, (I) becomes x^6 < x which is true for all 0< x< 1
(II) becomes x^9 < x which agian is true but almost a repetition of (I) . Therefore (I) alone is sufficient .
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25 Jan 2005, 21:37
In this case don't you think D should be the answer??

vsom wrote:
Is 0<x<1?

(I)x^2 < x^(1/3)
(II)x^3<x^(1/3)

for 0<x<1 , in either cases cubing the inequations will help.
on cubing, (I) becomes x^6 < x which is true for all 0< x< 1
(II) becomes x^9 < x which agian is true but almost a repetition of (I) . Therefore (I) alone is sufficient .

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25 Jan 2005, 23:51
A
x^9<x is true not only for 0<x<1, but also for x<-1 that's why 2) Insuff
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26 Jan 2005, 06:24
I guess the question of -ve numbers doesnt arise. We have to assume real numbers only. And in that case, statement 1 is sufficient. No need for statement 2. But isnt statement 2 sufficient by itself too? Any comments?
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26 Jan 2005, 08:17
II is not sufficient, x could be positive or negative.
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31 Jan 2005, 04:31
Banerjee,

Can you let me know how you chose the numbers?What are your considerations for choosing the numbers?Thank you.Rgds

Anna

banerjeea_98 wrote:
"A" for me.

state 1: x = 1/8....satisfies.....x = -1/8 doesn't ....so x has to be between 0 and 1...suff.

state 2: x = 1/8....satisfies....x = -8...satisfies....insuff.

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31 Jan 2005, 04:31
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# Is 0<x<1? (I) x^2 < x^{1/3} (II) x^3<x^{1/3

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