If x is not equal to zero, is |x| < 1 ?

Perfect Explanation, Evajager, @Carcass this is a very common mistake which I am also prone to make.
Thanks @Evajager - Your explanation for {B}

I remember skipping a Question just like this in one of my exams because I gt stuck!!! in solving - | x | < 1/x

Thanks guys you rock!!!

Correction: in (2) and we are again in the situation from (1) above. - not exactly, but similar as we have \(x^2<1\) but in addition \(x>0.\) In (1) \(x\) could also be negative.
The conclusion is still correct, as now \(|x|=x<1.\)

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 \(x > -1\) ------> \(- 1 < x < 1\)

Thanks Mod

The general rule of thumb with DS + ineq questions is if you can avoid plug numbers DO IT!

If x is not equal to zero, is |x| < 1 ?

Unraveling the question we have case1 where x < 1 OR case 2 where x > -1

REMEMBER that with ineq. DS questions A suffiecient asnswer is the answer that verifies case1 OR case2 OR BOTH

(1) x^2 < 1
"ROOTING" both sides (i.e. raised both sides in 1/2 power) we have sqrt(x^2) = sqrt(1) = 1 hence |x| < 1 which is SUFFICIENT as verifies case1 and case2 (effectivly if you solve it you get exactly the 2 cases from the question stem). Is important to know that "ROOTING" is allowed here because BOTH sides of the enequality are NOT NEGATIVE.


(2) |x| < 1/x
Solving the ineq. above gives two cases:

caseA: x < 1 / x and caseB x > - 1/x

CaseA: Since x>0 we can multiply both sides and get x^2 < 1. Rooting both sides again (as shown in (1) above) you get |x| < 1 this verifies again case1 and case2 from the question stem. At this point YOU DONT need to go and check caseB.

D is the correct answer

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