ISI 2017 key

[Asd1995]

Obviously general is worse off. Reservation in earlier years had no impact on seats as reserved candidates held an advantage only in the entrance exam. Now seats are reserved so they have to take minimum amount of reserved students from test AND interview.

Edit: This is assuming reservation policy has changed from this year.

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[Asd1995]

1. Vectors are independent so dimension of space is not 1, probably 4 as I recall

3. Function is not concave, I remember it was quasi-something.

4.Homogenous of degree 2

2. Onto functions are 2^k - 2

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[Asd1995]

Sure. To be more descriptive for others:

Onto functions are total functions minus into functions. Into functions are 2 i.e map all domain to output 1 or to output 2. Hence.

Homogenity is f(tx,ty)=t^k*f(x,y). For this problem k=2/

You can google f=xy, it is quasiconcave. I just used the mathematical expression that had double order derivatives.

[Abhitesh]

getting same.

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[Abhitesh]

Q6. c Covar by 100 rest same.
Q7. c x=1/sqrt(p)
Q8. d A non zero solution.
Q9. c n^k * (k)!   {not sure}
Q10. d none
Q11. c >= 1 - alpha - beta.

[Asd1995]

9. I got n(n-1)(n-2)....(n-k plus 1)

A simple way to do it was to see when n=k, total ways should be n factorial as you have one person per row, and they can be allotted in n! ways.

11. I marked d, unsure. You're probably correct.

Rest all are same.

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[Akshay Pamnani]

Ah Shit,11 will be greater than or equal to 1-alpha-Beta.
Man I messed that one up in exam.

[Akshay Pamnani]

8-A is not necessarily true as it can be a line also.
   
10)Just consider disjoint sets.Prob will be 0

[Akshay Pamnani]

9)That reasoning is incorrect,since they said atmost n queues.

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[Akshay Pamnani]

8th will be (D)
As ther determinant is 0,hence non trivial solutions
You can't be sure of line or plane,both are possible

9th one,has to be one of c or d,but I am able to disprove both for some values of k and n.
I marked C,but can't be sure

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[Abhitesh]

Q9. Any person can be assigned to any one of the n queues which can be done in n ways. So for k persons it's n^k.
Now k persons can deboard in k! ways. In any one queue where P1 is ahead of P2 or P2 is ahead of P1 are different arrangements.

Q11 P(A int B) = P(A) + P(B) - P(A union B). P(A union B)<=1