Is │x + 1│ < 2 ?

goyalsau wrote:Is │x + 1│ < 2 ?

1) (x-1)2 < 1
2) x2 - 2 < 0

Is |x+1|<2?
means is -2 < x+1 < 2?

1) (x-1)2 < 1
this means -1 < x-1 < 1
0<x<2
1<x+1<3
From this we cant say whether x+1 is within the range -2 < x+1 < 2
hence INSUFF

2) x2 - 2 < 0
-root(2) < x < root(2)
1-root(2) < x+1 <1+root(2)
-0.414 < x+1 < 2.414
From this we cant say whether x+1 is within the range -2 < x+1 < 2
Hence INSUFF

Combining:

1 < x+1 < 2.414
From this we cant say whether x+1 is within the range -2 < x+1 < 2
hence INSUFF

pick E.

First i would like to say Awesome Work,
Thanks a lot, Can you please explain one more thing.

kvcpk wrote: Is |x+1|<2?

means is -2 < x+1 < 2

In Modes we have to take 2 cases one is +ve and -ve
While taking +ve

+ve X + 1 < 2

-ve X + 1 < - 2 ( IS it because of this -ve sign you have flipped the sign )
and then it become X + 1 > 2


Please Correct if i am wrong any where in this.

goyalsau wrote:First i would like to say Awesome Work,
Thanks a lot, Can you please explain one more thing.

kvcpk wrote: Is |x+1|<2?

means is -2 < x+1 < 2

In Modes we have to take 2 cases one is +ve and -ve
While taking +ve

+ve X + 1 < 2

-ve X + 1 < - 2 ( IS it because of this -ve sign you have flipped the sign )
and then it become X + 1 > 2


Please Correct if i am wrong any where in this.

Instead of getting confused with these signs, I would suggest you this:

There are 3 cases:

When mod(something) = some value
|x| = z
then x = z or x=-z

When mod(something) < some value
|x| < z
then -z < x <z
Simply to remember: x lies between negative and positive values of z

When mod(something) > some value
|x| > z
then x<-z OR x>z
Simply to remember: x does not lie between negative and positive values of z

Hope this helps!!

Let me know if you have any questions.

kvcpk wrote: Is |x+1|<2?

means is -2 < x+1 < 2

Hello guys,

Let me show you a DIFFERENT way (quicker and safer) to answer this question in the affirmative!

If a and b are any two real numbers, the DISTANCE between them (in the real line) is ALWAYS equal to |a-b| , to be honest, it is DEFINED this way (check some values for a and b to see this is very reasonable!

That understood, once you realize that |x+1| = |x-(-1)| = distance (x, -1), the question is:

Is the distance between x and the number -1 less than 2 (units of length)? And, of course, (draw a line with the number -1 as a "reference", then -2 and +2 to the left and right.... and verify that) this is equivalent to the question is x between -1 -2 = -3 and -1 +2 = 1 (both extremities excluded)? Please note that -3 < x < 1 (as a question) is equivalent to the question you put, that is, -2 < x+1 < 2 ?

I called this the "geometric representation of modulus". Very useful, no doubt! I hope you like (and use) it!!

Regards,
Fabio.