(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).
inequality question
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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
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
- gabriel
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i agree with harry over here ...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?
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 ..
ugh, you're right...
my answer was wrong... the italic part should be x < 1.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!
- gabriel
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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 ..