Maths doubt !

[Anjali]

[Anjali]

[Anjali]

[RajEco]

1. a chisquare  

2. c x<a  
m not sure.

[Akshay Jain]

Oprion A-The sum of square of N independent random variables dat follow standaed normal dist den dat sum follows a chi2 dist vd N degrees of freedom...its a theorem

[Akshay Jain]

Its option B-x not equal to a....x is in neighborhood so it can b less den or greater dn a

[Anjali]

Akshay , answer to 2nd is option D ( x=a )

[Noel]

For question 2
V€(a)= ]a-€,a+€[ for all €>0
if x is in V€(a) then
a-€ < x< a+€
|x-a|<€ for all €>0
since this condition is satisfied by all €>0,let € tend to 0
then |x-a| tends to 0
hence x=a
although this answer is contradictory as every €-neighbourhood of a contains infinitely many points distinct from a

[Granpa Simpson]

I dont know how answer to question 2 is x=a, because V(a)= deleted neighborhood of a i.e. (a-e,a)U(a,a+e)...so the answer should be x not equal to a because a excluded from the set V.

[Homer Simpson]

i think its 2(b) as well. it cant be x = a

[Anjali]

Yeah I know b seems correct , however answer is d !
Can anyone a good reference for sets ( like definition based ques ) ? Will RD sharma help ?

[Anjali]

What about 3rd ques ?

[Granpa Simpson]

For the 3rd question the answer will be d i guess..!!!!

[Anjali]

No its A !

[Noel]

[Anjali]

I did not get the step where x doesn't belong to a-€ , a+€ , when x is not equal to a , so it will belong to a-€,a+€

[Granpa Simpson]

Please upload the complete answers with explanations..dat will be helpful for all of us...!!!

[Noel]

I hope you got it till |x-a|>€
now take two cases as follows
case 1. (x-a)>€ which implies that x>a+€
case 2. -(x-a)>€ which implies that x<a-€
combining both cases we see that x does not lie in the €-neighbourhood of a..which is a contradiction to the given hypothesis..so when x is not equal to a,it cant belong to
]a-€,a+€[ for all €>0 hence x=a..hope it's clear now :)

[Granpa Simpson]

Thanx Noel..

[Anjali]

Great Noel ! Got it ! Thanks !
And subhayu I just know the answers , not the explanations
People what about 3rd ques ?