dse 2007

[seema]

The set : intersection(n=1 to infinity) (-1-n^-1, 1+n^-1) is identical to
a.  (-1,1]
b.  [-1,1)
c.  (-1,1)
d.  [-1,1]
Pls tell me how to solve this.

[neha]

what i am getting is (-1,1)
putting n=1, the interval is (-2,2)
for n=2, the interval is (-1.5,1.5)
.
.
.
as n approaches infinity, interval becomes (-1,1)
so taking the intersection of all these intervals , we get (-1,1)
but in the answer book prepared by Mr. Amit Goyal, it shows [-1,1], wondering why as the interval in the question is open, please help on this,,,

[anon_econ]

neha any such interval will always contain both -1 and 1..no matter how high n is.

[neha]

but the question has an open interval,,, so when n approaches to infinity ,u mean the interval will be [-1,1] ,not (-1,1),, how?

[anon_econ]

say n is very very large. the interval would be something like (-1.000000000000000000000001,1.000000000000000000000001) but it would always include 1 and -1.

[neha]

yeah makes sense to me ,,, i guess the interval will approach to (-1,1) but will not be exactly the same, ,, thanks

[tim]

hey guys cn u help me solve q22 and 23 of the same ppr. (2007)

[Chinni18]

tim
For Q22,
expected wealth w/o house insurance=(0.9)(5,00,000)+(0.1)(3,00,000)=4,80,000
Also his utility fn is concave so risk averse