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I'm new to this forum. I'm working through some OG DS questions and I find that some of the explanations to ans are not clear or take too long to solve in 2 mins.

Q:DS154 Is x negative? 1) x^3(1-x^2) < 0 2) x^2-1 < 0

Thanks

(1) \(x^3(1-x^2)<0\), two cases (one of the multiples negative, another positive):

1, is equivalent to x(x-1)(x+1) >0 ( x^2 is always>0 and i have inverted the sign)

this gives x>1 and -1<x<0 , thus not sufficient

2, gives -1< x<1 not sufficient.

if you combine the result we get -1<x<0 , thus x is negative and thus C.

Another way, since we know both are not sufficient individually.

2nd equation gives x^2 -1 <0 => 1- x^2 >0 now in first equation, if its -ve then x^3 must be -ve....thus both taken together sufficient.
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Thanks guys for the quick solution..Heres a question..

For st1, if I take a +ve number that is \(>=2\) the inequality holds true, then why do we need to consider st2?

Look forward to the replies..

Thats wrong approach towards DS question.

x>=2 satisfy the equation 1 , what about x = -1/2, that is also satisfying the equation.Then how we can be sure whether x is + or -... sufficiency comes when we are sure this statement as a whole ans the question and there is no other possibility which will be against it,.

Before starting another DS question do get this concept, else no use of DS questions. letme know if you got it else i will try to explain in other way.
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Alternative approach by taking numbers. Statement1: \(x^3(1-x^2) < 0\) From this we can say \(x <> 0\) and \(x <>+1 or -1\).

Consider x=2 Then the equation results in \(8*-3 <0\). Satisfy the equation. Consider x= 1/2 then the equation results in \(1/8*(3/4) >0\). Does not satisfy the equation.

Consider x=-2 Then the equation results in \(-8*-3 >0\). Does not satisfy the equation. Consider x= -1/2 then the equation results in \(-1/8*(3/4) <0\).Satisfy the equation.

This statement is not sufficient to conclude.

Statement2: \(x^2-1 <0\) From the statement we can say that \(x <>=1 or -1\) Consider x=2 then the equation results in \(3 >0\). Does not satisfy the equation. Consider x=1/2 then the equation results in \(-3/4 <0\). Satisfy the equation.

Consider x=-2 then the equation results in \(3 >0\). Does not satisfy the equation. Consider x=-1/2 then the equation results in \(-3/4 <0\). Satisfy the equation.

This statement alone is not sufficient.

Combining both the statements when x is -ve that is -1/2 , it satisfies both the equations.

(1) \(x^3(1-x^2) < 0\) Two cases: A. \(x>0\) and \(1-x^2<0\), which means \(x<-1\) or \(x>1\) --> \(x>1\); B. \(x<0\) and \(1-x^2>0\), which means \(-1<x<1\) --> \(-1<x<0\).

So \(x^3(1-x^2) < 0\) holds true for two ranges of \(x\): \(x>1\) and \(-1<x<0\). If \(x\) is in the first range answer to the question is NO, but if \(x\) is in the second range answer to the question is YES. Two different answers. Not sufficient.

(2) \(x^2-1<0\) --> \(-1<x<1\). x can be positive as well as negative. Not sufficient.

(1)+(2) \(x>1\) or \(-1<x<0\) AND \(-1<x<1\) --> intersection of the ranges from (1) and (2) is \(-1<x<0\). \(x\) is negative. Sufficient.

Answer: C.

aiyedh wrote:

1. x^3(1-x^2) <0 x^3- x^5<0 x^3 < x^5 , which mean that x is positive no negative. ( sufficinet).

2. x^2<1

-1 < x < 1

which mean that we do not know exactally wheather it is positive or negative. ( not sufficient).

the answer is A

Red part is not correct. \(x^3 < x^5\) does not hold true for all positive \(x-es\). For example \(x=0.5\) --> \(x^3>x^5\). Also \(x\) can be negative in the range \(-1<x<0\) and \(x^3 < x^5\) will hold true.
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here, is small note: as soon as we see one number can be either positive or negative we know the statement is not sufficient. so statement 1) \(x^3(1-x^2) < 0\) , here \(x^3\) can +ve or -ve => not suff. similarly; \(x^2-1 < 0\) => (x-1)(x+1)<0 ; here also x-1 or x+1 can +ve or -ve not suff.

combining both : since \(x^2-1 < 0\) or 1- \(x^2 > 0\) hence from statement 1 \(x^3 < 0\) => x< 0 => suff.

while solving stmt 1 X^3(1-X^2)<0 X^3(X+1)(1-X)<0 Then i took the roots as -1,0,+1 on the number line to find the range as per vertias prep graph approach but am getting ranges as 0>X<1 and x<-1 which is wrong .please do correct me
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