Jnu maths doubt

[Spiti]

Thanks

[Ashima]

c

[Spiti]

How?

[Ashima]

Sorry its a!

[Ashima]

|a| < 3
= -3<a<3
|b| < 5
= -5<b<5
add them

-8<a+b< 8
= |a+b|<8

so a

[Dreyfus]

I m getting a....

[Spiti]

This answer key says c
http://discussion-forum.2150183.n2.nabble.com/JNU-2011-Answer-Key-td7563304.html

[Ashima]

C is also possible. It just has an extra equals to!

It has to be a.

See, -3<a<3
-5<b<5

-2<a-b<2
|a-b|< 2

and not |a-b|<=2

[Spiti]

Ya. I reckon option a is correct. Thanks

[ViV]

|a+b|<8
=> -8< (a+b) < 8
=> -8-a < b < 8-a
Now, its given that |a|<3, => -3< a <3
If i pick 2.5 as a,
then, i would get -10.5<b< 5.5
And we need |b|< 5, i.e -5<b<5
So we are getting a contradiction here.
I chose |a-b| <,= 2 because
-2 <,= (a-b) <,= 2
-2-a < -b < 2-a ( not strict inequality )
a+2 > b > a-2
a-2 < b <  a+2
Now, if I choose any -3 < a < 3
I will always get |b| < 5. ( you can check by taking numbers)

[Ashima]

Oh yes!
How'd i miss that!

Also, if I take |a-b|<2

Case 1            
(a-b)<2

Case 2
(a-b)>2

Then two sub cases of each of these two will come, by putting a as -3 and 3.
This give me for case 1: b>1 and b > -5, union of which is b> -5.

for case 2: b<-1 and b <5 union again is b< 5

So |b| < 5

So yeah its C. Thanks Viv!

[The Villain]

I didnt get case 2..Pls explain again..

[Ashima]

You understood the two main cases right? Opening |a-b| wala?

Then there will be two sub cases, where in you put -3 and 3 as a's values. Just do it. You will find 4 values for b.

2 for each case.

Within each case take the union. Till this understood?

[The Villain]

Okk i got it nw...got confused..thanxx