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by rohan_vus » Fri Jan 08, 2010 8:21 pm
papgust wrote:Is x between -1 and 1?

(1) x^3 is less than x.
(2) x^3 is less than x^2.

Why it isn't A? OA is E
Consider x = -2 ..x^3<x...So answer to original question that if x is between -1 and 1 is 'No'.
Consider x = .5 then x^3<x..So answer to oroginal question that if x is between -1 and 1 is 'Yes'.
Thus A is ruled out

Algebrically A is reduced to x^3-X<0 --> x(x^2 - 1) < 0..Now situation varies depending on x is _ve or +ve..
If x is +ve then x^2 - 1 < 0 ..this means |x| < 1..which satisfies the original questionas 'yes'.
But of x = -ve then x^2 - 1 >0 ...which means |x| > 1.. so x could be any integer greater than 1 or any integer less than -1..This condition comes as 'No' to the original question asked in the question stem
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by pandeyvineet24 » Fri Jan 08, 2010 8:23 pm
for stmt 1,
put x = -5, x^ 3 = -125, x = -5 therefore x^3 < x and x not between -1 and 1
put x = 1/2, x^3 =1/8, x =1/2 , therefore x^3 < x and x between -1 and 1
clearly insuff.

for stmt 2
put x = 1/2, x^2 =1/4, x^3 = 1/8, --> x^3 < x ^2 and x between -1 and 1
put x = -5 , x^3 = -125, x^2 = 25 --> x^3 < x ^2 and x not between -1 and 1
insuff.

combine 2, use same 2 values as above. together insuff
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by papgust » Sat Jan 09, 2010 3:20 am
rohan_vus wrote: Algebrically A is reduced to x^3-X<0 --> x(x^2 - 1) < 0..Now situation varies depending on x is _ve or +ve..
If x is +ve then x^2 - 1 < 0 ..this means |x| < 1..which satisfies the original questionas 'yes'.
But of x = -ve then x^2 - 1 >0 ...which means |x| > 1.. so x could be any integer greater than 1 or any integer less than -1..This condition comes as 'No' to the original question asked in the question stem
Hey Rohan,

Thanks for replying. I have a doubt here.

In (1), simplify the eqn to x(x^2-1) < 0 -> from which you get x^2 < 1 and x < 0. As x < 0, we only get a single scenario and we don't have to consider x as +ve.

What is wrong in this approach?
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by rohan_vus » Sat Jan 09, 2010 4:33 am
Hey papgust

You are right !.. Its errata on my part as i wrote it my midnight time just before sleeping:) ..

First inequlaity we agree its reduced to x(x^2-1) < 0... So this equantion is satisfied in following ways
a) x < 0 and (x^2-1) > 0
b) x > 0 and ( x^2-1) < 0

We need to consider 2 possibilities , if x would have been considered to be INTEGER then possibility (b) we wont get and then in that case only possibility (a) exists and which is what agrees to your solution

But since x could be non integer so we must consider both the possibilities

Now if x < 0 then |x| > 1..which means x < -1 as x is -ve..So this value ranges DONT fit into the ranges ( -1<x<1)---- So its 'NO' in this case

Now when x is > 0 , then x^2-1 < 0 ..==>|x| < 1..Here the ranges for 0<x<1 ( as is +ve so ignoring the -ve possibilities)..These values ranges fit into the ranges which is asked ( -1<x<1)---- -- So 'Yes' in this case ..

But still situation is same , i.e, A is not sufficient to give a definite answer
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by papgust » Sat Jan 09, 2010 7:54 am
Now it makes sense. I failed to consider 2 possibilities here with opposite signs. Thank you very much!
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