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inequality question

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by erdnah » Wed Jul 18, 2007 7:31 am
(A) can't be right, since x could be either positive or negative: If the result in the brackets gets negative (this happens when x > 1), n would be positive; if the results in the brackts remains positive (this happens when x < -1 or 0 < x < 1), n would be negative. i think there are no other ways to fulfill the inequality and thus is never between -1 and 0!

Statement 2 gives you the information, that x^2 < 1; here you know that x is between -1 and 1 excluded.

Taken a and b together, x must be positive (0 < x < 1).
Last edited by erdnah on Wed Jul 18, 2007 10:01 pm, edited 1 time in total.
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by harry_x1 » Fri Jul 20, 2007 12:19 am
I beg to differ on this solution

Statement one tells us that n is positive only if solution of bracket is nagative. and the solution of bracket is negative only when x>1. For all other values of x where x<1, n will always be positive.


the second statement tells us that -1 <x <1.
therefor x is never greater than 1 thus n is always negative.
if n is always neagtive then x solution of bracket has to be positive.
now -.5 and +.5 both result in solution of bracket being positive.
THUS BOTH STATEMENTS ARE INSUFFICIENT AND ANS SHOULD BE E.

can any maths expert please confirm the ans
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by erdnah » Fri Jul 20, 2007 3:41 am
@harry: what tells you that n isn't negative?
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Re: inequality question

by gabriel » Fri Jul 20, 2007 5:13 am
broy wrote:Is x negative?
a. n^3 (1 - x^3) < 0
b. x^2 - 1 < 0

OA is C...(I was thinking A)...how?
i agree with harry over here ...

using the first statement .. we have 2 possiblities .. first n^3<0 and (1- x^3)>0 ... and the other possiblity is n^3>0 and (1-x^3)<0 ... so statement 1 is insufficient ...

using the second statement .. we get the range of value that x can take .. we get x^2< 1 ... so we have -1 < x < 1 .. again insufficient ..

Now combine the 2 statements .. u can see that the expression (1-x^3) is always positive for the range of values - 1 < x < 1 ... and in that case n^3 is always negative .. so the condition n^3(1-x^3)<0 is satisfied for -ve as well as positive values of x .. so we dont have a defnite answer .. so answer is E ..
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by erdnah » Fri Jul 20, 2007 5:43 am
ugh, you're right...
if the results in the brackets remains positive (this happens when x < -1 or 0 < x < 1), n would be negative. i think there are no other ways to fulfill the inequality and thus is never between -1 and 0!
my answer was wrong... the italic part should be x < 1.
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by gabriel » Sat Jul 21, 2007 10:46 pm
broy wrote:Statement 2 gives you the information, that x^2 < 1; here you know that x is between -1 and 1 excluded.--- when x^2 < 1....does that mean -1<x<1? I was thinking that here x < -1 and x < 1 which will lead to x < -1....can someone explain this, pl?

OA is wrong from the above comments then for the solution...I got this question from official 11th edition ...is it?


@ broy .. when u say x< -1 .. it means assumes values to the left of -1 on the number line .. that is numbers like -1.5,-2,-5 ....-100 etc.

A number line would look something like this

________________________________________
..... -5 -4 -3 -2 -1 0 1 2 3 4 5 .......

numbers to the left of any number on the numbers line is lesser than that number ..

.. so if u say x < -1 .. then that would include the numbers to the left of - 1 .. that is -2 , -3, -4 etc. but we have been given that x^2<1 .. if x = -2 .. then x^2 = 4 ... which violates the condition that x^2< 1 ... so the correct answer is x > -1 .... that is the numbers on the right of -1 ..

... always remember whenever we have x^2< a ... then the range of values for x is - root(a) < x < root(a) ....

.. so for x^2<4 .. x wuld vary between -2<x<2 .. hope this clears your doubts ..
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