Doubt in DSE -2009 paper(option A)

[Niharika Mishra]

Hi

 I am a DSE aspirant and solving 2009 dse option A paper .provided in site.


I have doubt in 3rd,4th and 5th question

3rd question- I think question is wrong because i tried doing it by substitution but exponent has x^2 but we dont have xdx.Please let me know if there is some other way of solving this problem.

3. 1.96∫x2 / 2 dx
¥e
-is approximately
a) 0.025
b) 2p
c) 0.025/ p
d) 0.025 2p

4. Consider maximizing f (x, y) = x2 - y2 subject to the constraint x + y = 1, where x
and y are real numbers. This problem has
a) no solution.
b) a unique solution.
c) 2 solutions.
d) an infinity of solutions.


For this question I found  out partial derivatives.
I took f(x)=x^2-y^2
and g(x)=x+y-1

taking out partial derivatives
^g(x)=(1,1)
^f(x)=(2x,2y)

(2x,2y)=lamb(1,1)

x=lamb/2
y=lamb/2
x+y=1
lamb/2+lamb/2=1
lamb=1
x=1/2
y=1/2

so I am getting unique solution for them x=1/2 and y=1/2
but in answers its given as no solution



5th question I have no clue how u got the solution

[Bhaskar Biswas]

the question number 4 that you have tried, you haven't checked the second order condition. If you would have checked, you could easily see that the solution you derived is the solution for minimisation problem..

f(x,y)= (1/2)^2 - (1/2)^2 = 0
if you take, x=1,y=0, you will get f(x,y)= 1

[Niharika Mishra]

Hi Bhaskar ,

Thanks for ur reply...Please provide explanation for following questions .


5. Suppose P(x) and Q(x) are real polynomials of degree m and k respectively, where
both m and k are less than or equal to the positive integer n. Suppose the equation P(x)
= Q(x) has at least (n + 1) distinct solutions. Which of the following choices best
describes what this situation implies?
a) m = k = n
b) m = k < n
c) P(x) and Q(x) are identical.
d) P(x) and Q(x) are linear

ans c ---but doesnt tht imply tht rhs=lhs and everything will canceled out ,,,moreover even if we assume that coefficients differ than let m=k=5 given n> or = to m and k  
if m and k <n then the polynomials can never have more than n+1 solutions
and if we consider n=m=k ....than option a sud be the answer.

8. In a surprise check in a local bus, 20 passengers were caught without tickets.
The sum of squares and the population standard deviation of the amount in their
pockets were Rs. 2000 and Rs. 6, respectively. If the total fine equals the total amount
discovered on them, and a uniform fine is imposed, then the fine imposed on an
individual is:
a) Rs. 8
b) Rs. 6
c) Rs. 10
d) Rs. 12

For this one I used variance formula and got mean =8  ....is this wat we were suppose to do

12. The demand function for lemonade is Q p d =100 - , and the supply function is
Q p s =10 + 2 , where p is the price in rupees. The government levies a sales tax on
lemonade after which the volume of sales drops to 60. Then the per unit tax on
lemonade is
a) Rs 20
b) Rs 15
c) Rs. 10
d) Rs. 5

My ans option d  which is wrong -----plz find my approach below and plz correct me

For this one I  assumed equilibrium cond supply=demands
100-p=10+2p

p=30

quantity supplied=70

quant demanded= 60 ----after levying sales tax

so total worth of supply = 70*30=2100
but after tax demand is for 60 units ---2100/60=35
therefore  price per unit after tax is 35
therefore tax= 35-30=5

[kiran]

 qno-2 has 2 solutins that are x =0 y =1 or x=1 y=0

[kiran]

sry for mistake its x=1 and y=0 as pointed out by Bhaskar

 i solved it for x2+y2

[duck]

Hello, :)


Q4> its no solution..
@neharika, u have found out wrong partial derivatives. check it out once again.

Q12> its 15
earlier p=30 and q=70

after imposition of tax, q=60
from supplier's point of view, imposition of tax means less price than before
=> q(s)= 10+2p , now becomes q(s)= 10+2(p-t)
=>60= 10+2(p-t)
=>t=5  ( subsituting p=30)

Now, from consumer's side, imposition of tax means he has to pay more price than before,
=> q(d)=100-p , now becomes q(d)=100 -(p+t)
=> 60 = 100-(p+t)
=>t= 10

Therefore, total tax+ 5+10 = 15