maths

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maths

ritu
1..cosider a twice differentiable functn...f:R-R and a,b belong to R such that a<b.f(a)=f(b)=0 and f''(x)+f'(x)-1=0 for every x belong to [a,b].then,
a.f has a maximum but not a minimum over the open interval (a,b)
b.f has a minimum bt nt a max over (a,b)
c.f has neither max nor min over (a,b)
d.f has a max and a min over (a,b)
(dse 2010)

m not getting anything pls explain the logic behind this questn..


2..in ques like ---which of the following is greater..?
5^44 or 4^55....can we solve this questions without using log???




3.consider a firm producing a single good with cost functn----
C(x)=5 if x=0
        10+10x,if x>0


firm's sunk and fixed costs are respectively---
a.10 and 10
b.10 and 0
c.0 and 10
d.5 and 10


4.suppose we have a chair with "n" legs and it stands with all its legs touching the floor,regardless of floor quality.then n is
a.2
b.3
c.4
d.5
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Re: maths

ritu
5.How to find demand in this case where min fun has 3 values?
a) u(x,y)= min {2x+1,x+y,2y+1}
               px=py=1
               total income=2


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Re: maths

ritu
In reply to this post by ritu
5.How to find demand in this case where min fun has 3 values?
a) u(x,y)= min {2x+1,x+y,2y+1}
               px=py=1
               total income=2


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Re: maths

anon_econ
In reply to this post by ritu
1) First of all since the function is continuous & has the same value at 2 points, it must have at least 1 max or  min b/w the 2 points. Now, at a max/min, f'(x) has to be equal to 0. In that case f"(x) would be 1, which is positive. According to the 2nd order condition, if f"(x) is positive, f always reaches a minimum.
3) Sunk cost is what cannot be recovered. Fixed cost is what u have to pay once u decide to produce and is independent of the production level. So the ans is d.
Don't know about the other 2. Question 4 is extremely strange.
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Re: maths

anon_econ
In reply to this post by ritu
U just have to see which function has the lowest value in which region.
To begin with, when x>y, 2x+1>2y+1. Within this region (x>y) see where x+y would be less than 2y+1.
Finally what I got was:
For U>1,
When x>y, x-y<1, U=x+y
When x>y, x-y>1, U=2y=1
When x<y, x-y>-1, U=x+y
When x<y, x-y<-1, U+2x+1.

Plot one of the ICs to see the shape. U get 2 kinks.
The optimum is any point on the b.c b/w (3/2,1/2) & (1/2,3/2). It's clear once you plot it.
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Re: maths

duck
In reply to this post by ritu
Q2) It can be done in the following way>

4^3> 5^2
=> 4^6 > 5^5
=> 4^54 > 5^45
(multiplying powers by 9 on both the sides)
=> 4^55 > 5^45
Therefore, we have  4^55 > 5^44
:)
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Re: maths

ritu
@vasudha...
i guess 1st questn was an application of intermediate value theorem right????????thank u i understood 1st questn completely...god bless u :)

@duck...
thanx...can u post a similar questn like i posted...i"ll get to kno then if i m able to solve.:)..