DSE 2012-Q 57

[kangkan]

Hi...

I am getting option a as the answer while the givn answer is c..can some one please upload their working?

Please let me know the formula of e that you are using

http://economicsentrance.weebly.com/uploads/1/1/0/5/1105777/2012-option-a.pdf

[The Villain]

@Kangkan
assume ...e(i'm not able to type the other e)=eP/P*
Now try..u'll get it

[kangkan]

hI..i used this definition but i am getting the expression for option C,but insted of dividing by  P*,i ma getting multiplying by P*..can you post your working please?

[The Villain]

Y=C+I+G+X-IM/e    (WHERE e IS REAL EXC ONE)
Y=Co+c1Y+d1Y-d2r+G+x1Y*-x2e-m1Y/e-m2
Y(1-c1-d1+m1/e)=Co+G-d2r-x2e-m2
let Co+G-d2r-m2=A

Y(1-c1-d1+m1P*/pE)=A-x2EP/P*
now diff wrt P*
u'll get it :)))

[kangkan]

got it..was making a stupid mistake

[Shefali]

Q58-59 anyone?

[Akshay Jain]

http://discussion-forum.2150183.n2.nabble.com/DSE-2012-paper-discussion-Please-join-td7590218i20.html
hi shefali...plz refer to the above link....its discussed here

[kangkan]

@ Akshay..can you please help with 55 of 2008

http://economicsentrance.weebly.com/uploads/1/1/0/5/1105777/28_june_2008_option_a.pdf

I understand that f(x)= n for all values 0<x<1/n
                             = 0 otherwise

But as n tends to zero the set A will get smaller and smaller ,therefore g(x)=0 for all X when n->0..but f(x)=n*g(x)..how to evaluate this limit?

Thanks :)

[Akshay Jain]

Hi Kangkan...m really not sure about my approach to this question...I will call it a jugaad instead of what it shud be (I really dont no)
but good thing is dat the ans is coming.....

[kangkan]

A null set is finite....such a simple idea ,and i kept missing it for one whole day :(

Thanks a lot

[Aranya]

Can anybody help me out for qsn no 46? plz!

[Akshay Jain]

no kangkan a set {0} is not a null set, it has an element....a null set is {} this
a finite set is a set which has coutable no. of elements

[Akshay Jain]

of wich ppr???

[Aranya]

dse 2012

[Akshay Jain]

Hi aranya..
the function (x+2)^3 in interval (-3,-1] has a maximum at x=-1 and not at x=-3 coz of the open interval at this point
the function x^(2/3) has a minimum at x=0 in interval (-1,2) as the function is not defined for values < 0.
so there are 2 points.

[Granpa Simpson]

Hiii Akshay can it be proceeded in this manner?
Since the first function does not pass the even order derivative test so it cannot have a global maxima or a minima however in the interval (-3,-1] it achieves the locally maximum value at x=-1.

[Akshay Jain]

Yes...thats wat the ques asks....about local extrema....and not global extrema.....

[Granpa Simpson]

Thanx a lot..!!!!!

[Aranya]

thanks @akshay