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(1) By itself is not sufficient. (2) tells us x is a +ve or -ve fraction. Not sufficient.

(Together) If x is +ve fraction it does not satisfy st. (1) (plug in x = 1/2 for testing) If x is -ve fraction it satifies (1). so x is a -ve fraction and hence x < 0. SUFFICIENT.
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How does 1) X^3(1-X^2)<0 led to x<0 or x>1 eg: X=-2 -2^3 (1-(-2)^2) = -8*-3 = 24 eg: X=-1 , result 0<0 , not true. eg:X=0 result 0<0, not true X=1 , 0<0, not true.

How does 1) X^3(1-X^2)<0 led to x<0 or x>1 eg: X=-2 -2^3 (1-(-2)^2) = -8*-3 = 24 eg: X=-1 , result 0<0 , not true. eg:X=0 result 0<0, not true X=1 , 0<0, not true.

i feel it is only when X>1 , X^3(1-X^2)<0

Actually just re-write with factors:

x^3(1-x^2) < 0 So: (x^3)(1-x)(1+x) < 0

so either all three or negative or any one can be negative for the expression to be true.

If (x^3) is negative, x < 0 If (1-x) is negative, x > 1 If (1+x) is negative, x < -1

When you combine statement 1 and 2, the only overlap is x < -1 so we know for sure that x is negative

Hope this answers your question.
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"Nowadays, people know the price of everything, and the value of nothing."Oscar Wilde

i think we need to consider the three cases in parallel eg: if X<0 what will happen to other two expressions

If (x^3) is negative, x < 0 say X is -0.5, X^3 is -ive , 1-X is +ive and 1+X too is positive. expression holds. Now say X=-2, X^3 is -ive , 1-X is +ive but 1+X is -ive . expression doesnt hold . so we cannot say that if X<0 , X^3 (1-X) (1+X) < 0

1+X will be -ive only when X <-1 , in that case 1-X will be +ive and X^3 will be -ive , over all expression will be +ive so this set of values of X also ruled out.

now 1-X will be -ive for all values of X greater than 1 and the other two X^3 and 1+X will be positive in this case. we can safely say that for all X>1 the expression will hold.

may be i am wrong though in my thought process , pls advise.

Statement-1 If X is -ve: x^3 will be -ve, then (1-x^2) has to be positive, can't say If X is +ve: x^3 will be +ve then (1-x^2) has to be negative, can't say Insufficient

Statement-2 X^2-1<0, this is possible only when x is fraction. But fraction can be negative or positive, Insufficient

Both statements together

we get when x is -ve, x^3 is negative, and 1-x^2 is positive ( as x is a fraction now eg. 1/2) when x is +ve, not valid Hence x is -ve, when both statements combined together.

Hence C

Hope I am clear.
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(1) x^3(1-x^2)<0 --> x^3(1-x)(1+x)<0 --> the roots are -1, 0, and 1 (equate the expressions to zero to get the roots and list them in ascending order), this gives us 4 ranges: x<-1, -1<x<0, 0<x<1, and x>1.

Now, test some extreme value: for example if x is very large number then x and 1+x are positive, while 1-x is negative, which gives negative product for the whole expression, so when x>1 the expression is negative. Now the trick: as in the 4th range expression is negative then in 3rd it'll be positive, in 2nd it'll be negative again and finally in 1st it'll be positive: + - + -. So, the ranges when the expression is negative are: -1<x<0 and x>1.

So, x could be negative as well as positive. Not sufficient.

Or: x^3(1-x^2)<0 --> x(1-x^2)<0 --> x<x^3 --> -1<x<0 or x>1. Not sufficient.

(2) x^2-1<0 --> x^2<1 --> -1<x<1. Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is -1<x<0, hence the answer to the question whether x<0 is YES. Sufficient.