DSE 2010

[SoniaKapoor]

CONTENTS DELETED

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[Arushi :))]

ques 52 ans is a
because we see the limit of newton quotient as h tends to zero.
We have to take the value of a function at a point , then check the value of the function at that particular point plus some 'h', then subtract the initial from the latter.
Then we say that as h->o the newton quotient tends to the the tangent (derivative) at that point x.
So, in equation 1, the initial point is x-h , new point is x. And we see what newton quotient approaches as
h->o. and this gives the derivative . So this statement is true.
But in the second equation, initial point is x+h , new point is x+2h , but the newton quotient is not written correctly, Its 2h in the denominator. so it wont give us the derivative of the function.
So this statement is incorrect .

[Arushi :))]

Ques 53,
ii statement is false,
consider if x<0 and y>0
Then LHS gives 4 , RHS gives 14.
For statement i, consider all possible cases of x>o , y>0, x<0 , y<0,
it would hold in all the cases. Hence , its true.
So, answer would be a

[Arushi :))]

Ques 54:
We can rewrite the function here using the expansion of x^y
g= f(x,y)-1/2*f(y,y) , when x>y (its given x^y =y , when x>y)
    f(x,y)- 1/2*f(x,x) , when x=y ( x^y = x, when x=y)
    1/2*f(x,x)           , when x<y ( x^y =x, when x<y)
Now its given f is increasing in x & y.
We will consider three cases here.
<b>CASE I x>y
Consider 3>1,
g(3,1) = f(3,1)-1/2*f(1,1)
If I increase x here and take 4>1, g(4,1) =f(4,1)-1/2*f(1,1)
g(4,1)- g(3,1) = f(4,1)- f(3,1) And this term is positive because f is increasing in x.
So g is increasing in x when x>y
If I increase y here and take 3>2
g(3,2)=f(3,2)-1/2*f(2,2)
g(3,2)-g(3,1)= [f(3,2)-f(3,1)]-1/2*(f(2,2)-f(1,1))
The sign of this term is uncertain , because the first term is positive and second is negative
So we cant say whether g is increasing/ decreasing in y, when x>y
CASE II x=y
g(x,y)= f(x,x)-1/2*f(x,x)= 1/2*f(x,x)
Here x is always equal to y.
So increasing either of them would lead to a higher f value.
If x or y is increased 1/2*f(x',y')>1/2*f(x,y) , implies
g is increasing in both x & y
CASE III when x<y
Consider 1<3
g(1,3)= 1/2*f(1,1)
If x is increased here, 2<3
g(2,3) = 1/2*f (2,2) => g(2,3)> g(1,3)
Hence, the function is increasing in x
If y is increased here, 1<4
g(1,4)= 1/2*f(1,1)
g(1,4) = g(1,3) So the function is not strictly increasing, value is not changing on increasing y
Therefore, we can conclude that g is always increasing in x, however, it may or may not be increasing in y.
Hence, part c is the answer

[SoniaKapoor]

Thanxx arushi...bt i asked the wrong quest..
please tell me abt quedt 31-34 on exchange

[Arushi :))]

OMG ! Not now but ! m sleeping

[SoniaKapoor]

its oky no problem...Do it tomorrow...
Thankyou..goodnight :-))

[econ14]

question on exchange-
we are given the following endowments; e1(0,1) and e2(2,0).
agent 1 budget px.x + py.y = m1
                      px.0 + py.1 = m1
                      so m1= py

agent 2 budget px.x + py.y = m2
                     px.2 + py.0 + m2
                     so m2= 2px

let assume px= 1
so m1=py and m2= 2.
now CASE 1;  
 If py< 1
then agent 1 will consume all x and agent 2 will consume none of x but only y.
total demand for x must equal total supply therefore- dd by 1 + dd by 2 = total supply
                                                                         m1/px +  0        =   2
                                                                          py/1                  =    2
                                                                         but we have py < 1 therefore this case cannot happen.
 

CASE 2 ;
Py>1
Then by same process we equate dd = ss
here  agent 1 will consume only x and agent 2 will also consume only x.
                                     dd by 1 + dd by 2 = ss
                                     m1/px + m2/px  = 2
                                     py      +  2       = 2
                                        py= 0                                                                                                                         as py> 1 this is not possible.  

CASE 3;
PX=PY = 1
 dd by 1 + dd by 2 = 2
m1/px +    ( any thing between 0 to 2) = 2
1        +    ( fill it )                            = 2
we can put 1 which satisfies the equation
therefore our competitive equilibrium allocation will be
x1y1 = ( 1,0) & x2y2 = ( 1,1) option d.
also our equilibrium price ratio is 1/1 = 1.option a.