Is x negative?
1.x^3(1-x^2)<0
2.x^2-1<0
Can anyone offer alternate explanation apart from what is given in the OG?
OA is C...
OG
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I have not seen the expln in OG, but here is my take:
A) assume x is -ve. x^3 is -ve. Divide by x^3 on both sides, you get:
x^2 < 1 => -1<x<0 (we assumed x is -ve)
assume x is +ve. Do the same as above, but this time you get:
x^2 > 1 => x > 1. ( we assumed x is +ve).
So, x could be anywhere between -1 and 0 or > 1. So, not sufficient.
B) x^2-1 < 0 => x lies between -1 and 1.
Combined the results from above, and you get :
x lies between -1 and 0 or x > 1.
x lies between -1 and 1.
So, x must be between -1 and 0.
for confirmation, sub one value for the three ranges and check.
A) assume x is -ve. x^3 is -ve. Divide by x^3 on both sides, you get:
x^2 < 1 => -1<x<0 (we assumed x is -ve)
assume x is +ve. Do the same as above, but this time you get:
x^2 > 1 => x > 1. ( we assumed x is +ve).
So, x could be anywhere between -1 and 0 or > 1. So, not sufficient.
B) x^2-1 < 0 => x lies between -1 and 1.
Combined the results from above, and you get :
x lies between -1 and 0 or x > 1.
x lies between -1 and 1.
So, x must be between -1 and 0.
for confirmation, sub one value for the three ranges and check.