Is |x - 1| < 1? (1) (x - 1)^2 <= 1 (2) x^2 - 1 > 0

Since both parts of the inequality are nonnegative we can take square root and write as you did.

GENERAL RULES FOR THAT:
A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
\(2<4\) --> we can square both sides and write: \(2^2<4^2\);
\(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);

But if either of side is negative then raising to even power doesn't always work.
For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we can not square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
\(-2<-1\) --> we can raise both sides to third power and write: \(-2^3=-8<-1=-1^3\) or \(-5<1\) --> \(-5^2=-125<1=1^3\);
\(x<y\) --> we can raise both sides to third power and write: \(x^3<y^3\).

So for our question we can not square x/|x|< x as we don't know the sign of either of side.

Hope it helps.

Hi,

When dealing with inequalities and absolute values, it helps to know these standard forms.

1. If x^2 < a^2 then the range for x will always be -a < x < a.

2. If x^2 > a^2 then the range for x will always be x > a and x < -a.

3. If |x| < a then the range for x will always be -a < x < a

4. If |x| > a then the range for x will always be x > a and x < -a

Note : The above standard forms hold good even if we have <= or > =. Also the ranges for x^2 < a^2 and x^2 > a^2 are the same for |x| < a and |x| > a. The reason for this is because |x| = \(\sqrt{x^2}\)

So the question here asks us 'Is |x-1| < 1'?

We can break down |x-1|< 1 to -1 < x - 1 < 1. Adding 1 throughout we can rephrase the question to

'Is 0 < x < 2'

Statement 1 : (x - 1)^2<=1

(x - 1)^2 <=1 -----> -1 <= x - 1 <= 1. Adding 1 throughout we get 0 <= x <= 2. This clearly is not sufficient. As x can be equal to 0 and 2.

Statement 2 : x^2 - 1 > 0

x^2 > 1 -----> x > 1 and x < -1. This again is clearly insufficient.

Combining 1 and 2 : From Statement 1 we have 0 <= x <= 2 and from Statement 2 we have x > 1 and x < -1. So to satisfy both the statements the only possible range for x will be 1 < x <= 2, which again is insufficient since the target question is 'Is 0 < x < 2'. x can be 1,5 which gives us a YES and x can also be 2 which gives us a NO.

Answer: E

Hope this helps!

CrackVerbal Academics Team

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