carcass wrote:
If x is not equal to zero, is \(| x | < 1\) ??
1) \(x^2\) \(< 1\)
2) \(| x |\) < \(\frac{1}{x}\)
I would like to know if I do in the right manner
question: x id non zero, is \(| x | < 1\) ??? now to be < 1 x must be negative, so the question become " is x negative ???
1) for a square to be < 1 the must be for example \(\frac{- 1}{2}\) ----> x is negative . suff
2) for x to be minor of a certain number that is the reciprocal i.e. : \(-2 < - 1/2\) --------> x is negative. suff
is correct or I should rethink the entire part regarding inequalities ?? because this statement should not take you more that 50 seconds to solve. otherwise you get in trouble with this exam.
Thanks
I would suggest you to keep 4 ranges in mind when dealing with squares, reciprocals etc.
............... -1........ 0 ........ 1 ..............
Less than -1, -1 to 0, 0 to 1 and greater than 1
You can do the question logically or by plugging in numbers.
Ques: If x is not equal to zero, is \(| x | < 1\) ??
Logically, | x | implies distance from 0 on the number line. So the question becomes "Is x a distance of less than 1 away from 0?" i.e. "Is -1 < x < 1?" given x is not 0.
Statement 1: \(x^2\) \(< 1\)
Square of a number will be less than 1 only if the absolute value of the number is less than 1. This means \(| x | < 1\). The number needn't be negative. If it is positive, it should be less than 1. If it negative, it should be greater than -1.
Or, do you remember how to solve inequalities using the wave? \(x^2 < 1\) is \(x^2 - 1 < 0\) which is \((x - 1)(x + 1) < 0\). This implies -1 < x < 1 but x is not 0.
Or plug in numbers from the given 4 ranges. YOu will see that x must lie in -1 < x < 1.
Sufficient.
Statement 2: \(| x | < \frac{1}{x}\)
First of all, x cannot be negative since | x | is never negative. Since | x | is less than 1/x, 1/x must be positive. Also, x cannot be greater than 1 since then, | x | will be greater than 1/x.
Or plug in numbers from the given 4 ranges. You will see that x must lie in 0 < x < 1.
Sufficient.
Answer (D)
This is the same reasoning that I followed in the first instance with some errors but the path was correct. Specifically the stimulus evaluation : \(| x | < 1\) ------ in this scenario the first one is \(x < 1\) AND \(-x < 1\) so \(x > -1\) ------> \(