ISI 2012 ME - I Answer Key

[deepak]

 the answer is the 1 - CDF of a binomial random variable with parameters n and p
ie option B nC100 p^100 (1-p)^100

[surbhi sharma]

shouldnt the answer of q3 A6..???

[Preet]

Well  lets see..
For each n€N; define An = {(n + 1)k : k€N}

Say i chose an n= 1 then
A1= {2k} for k€N => A1= {2,4,6,8,10,12,..,18....}
A2= {3k} for k€N => A2= {3,6,9,12,15,18}

Now,  A1∩A2= (6,12,18....)
i.e. A1∩A2 = A5= {(n+1)k for k€N and n=5}

surbhi sharma wrote

shouldnt the answer of q3 A6..???

[sonal]

hey plz help m wd question 27 9 n 3

[Sinistral]

3)
A1 = {0,2,4,6...}
A2 = {0,3,6,9,...}
A1 ∩ A2 ={0,6,12,18,24...} = {6k ; k ∈ N} = {(5+1)k ; k ∈ N} = A5
option c

9)
take for eg f(x)=x
               [f(x)]^3 = x^3  (ie n=3)
watch the behaviour of the two functions x and x^3. we know the graph of the two functions. we also know that x^3 increases for all x ∈ R (evident from its graph)
so option A

i dont want to go into the formal proof (for the generalised case) right now.

27)
let 1<n<n+1 be 2 such integers ie a=n & b=n+1
so a^2 +b^2 = n^2 +(n+1)^2 = 2(n^2 +n) +1
lets see option a: it is always odd, so not true.
option b: <b>Always prime no ?
            well it took Dr manindra agrawal a couple of decades to find an algorithm for generating prime numbers. it cant be that simple :P One can also verify by taking n=6. so false.
option c): it is prime for n=2 . so not true.
hence option d

[sonal]

hey plz help m out wd ques 7 in me 2...

[taanya]

could you explain q23 ??

[ISI agent]

Try substituting different values that maximizes A = X+3Y+5Z s.t. X>=Y>=Z and X+Y+Z=9.
E.G.: - Taking X=4, Y=3, Z=2 will give A = 23 and so on.

The only value that maximizes A is when X=Y=Z=3 where A = 27.

Hope this helps.

[soumyab_sl]

please help me with q)14

we are gating 1) for root x             - x should be greater than equal to zero
                     2) for roor 3-x         - x should be x less than equal to three
                     3) fir root x^2 - 4x   - x equal to zero and x greater than equal to zero

so combining 1,2 and 3 we get f(x)  is real for x belonging to  [0,3]u[4,infinity)...............but the answer is only zero...............can u explain why???????????????????                

[Sinistral]

soumyab_sl wrote

so combining 1,2 and 3 we get f(x)  is real for x belonging to  [0,3]u[4,infinity)...............but the answer is only zero...............can u explain why???????????????????

So basically, u r trying to make the values under the root sign > 0 (all at the same time)
So, u need to take the intersection (NOT union) in the above case. but  it is evident that the values under the root sign cant be greater than zero at the same time (for a fixed value of x).

So, it is observed that x and x^2-4x become zero at x=0 and 3-x also doesn't become negative. So, {0} is a part of the domain. Now we need to check if anything else is also a part of the domain apart from 0.


After checking at various intervals,like for x<0 , x> 3, etc, which make the expression under the root sign negative; we find that not all 3 expression (the expressions under the root sign) can become < 0 at the same time. and for the remaining two (which are becoming negative for some value of x) it can never happen that the imaginary part of the two expressions get cancelled on addition for a fixed value of x. So basically we are only left with {0}.

Apologies for not coming out with a pure mathematical solution..

[soumyab_sl]

Sinistral  - thank you so much - it helped

[soumyab_sl]

is there any age limit of doing msc in economics in top INDIAN institutes......kindly reply, it would be a great help..

[Ridhika]

how is (11+6root 2)=(3+root2)^2 ?!

[ojaswini]

just apply the (a+b)^2 = a^2+b^2+2ab formula.

On Wed, Feb 26, 2014 at 7:01 PM, Ridhika [via Discussion forum] <[hidden email]> wrote:

how is (11+6root 2)=(3+root2)^2 ?!

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[neha 1]

hi can you please explain me ques 28
why not option a)2

[duck]

Hi n1.. :)

Question is asking you to find "non-negative" roots.

[neha 1]

but we are suppose to solve 2 equations right?
xsq-3x-10=0 and xsq +3x-10=0
so we get 4 roots 2 are non negative
please explain

[duck]

xsq-3x-10=0 for x>0
⇒ x= 5, -2
But "-2" is not a valid root because x>0

xsq+3x-10=0 for x<0
⇒ x = -5, 2
But "2" is not a valid root because x<0.

Therefore, only two roots: {5,-5}
Hence, only one non-negative root: 5

[neha 1]

Thanks :)

[shraddha.bhatia11]

Please help me with question 11 and 12 of 2012