ISI 2014 PEA discussion

[shiva]

MSQE-MQEK-MQED-PEA-2014.pdfISI
I have uploaded the question paper of ISI 2014 for your reference and also the solutions on solving them. I do have certain doubts with some of the questions as mentioned in my solutions, if any of you know the answer, please answer them. If there is anything wrong with my answers(provided below), please correct me.
1)B
2)D
3)C
4)B
5)A
6)D
7)A
8)C
9)A
10)C
11)D
12)C
13)B
14)C(Doubt)
15)C
16)C
17)B
18)A
19)C
20)A
21)D
22)B
23)B
24)B
25)A
26)D
27)C
28)B
29)Doubt
30)D

[Seecha]

this is the  sample questions for 2015 if you see 10 years papar
1. how you have done question no. 2 ,11,18,20,26,30 ?
2. for answers" http://discussion-forum.2150183.n2.nabble.com/ISI-2015-PEA-Answer-Key-tt7596538.html" .... check your answers here .

[shiva]

Thank you for the link, Solutions to 30 and 26. will send the remaining shortly.

On 12 Feb 2017 10:55, "Seecha [via Discussion forum]" <[hidden email]> wrote:

this is the  sample questions for 2015 if you see 10 years papar
1. how you have done question no. 2 ,11,18,20,26,30 ?
2. for answers" http://discussion-forum.2150183.n2.nabble.com/ISI-2015-PEA-Answer-Key-tt7596538.html" .... check your answers here .

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[shiva]

11. The minimum value of a function in the form of (x+1/x) is always 2, so by this the minimum value of this function is 7.

20. Take three conditions as x<a1, a1<x<a2, x<a2 and form the functions and plot the graph of it and then you can observe the answer.

2. For any function[0,1]---[0,1], let function be convex or concave or a straight line, only d option satisfies. I have answered this based on options and intuition. NO specific procedure followed by me.
 
If you know the solutions to Q14, 24, 29. Please send the solutions.

[Seecha]

Thank you for the solutions  
Q. 24 ... ans ....
 Here as P(X=2)=1
E(X)=2
so, E(X-E(X))^2n= E(2-2)^2n=0

[Abhitesh]

Q14. This is from mean value theorem. We can always find k in (1,2) such that f'(k) = 3a+b. This is not necessarily k.
Answer should be (D).

Q29. Probability distribution table for X
X      :  -2         -1             0           1              2
P(X) : 1/5       1/5            1/5        1/5           1/5
E(X)=0  SD(X) = \sqrt(2)
Probability distribution table for Y=|X|
Y      :  2         1             0           1              2
P(Y) : 1/5       1/5            1/5        1/5           1/5
E(Y)= 6/5  SD(V) = \sqrt(2)
Probability distribution table for XY
XY      :  -4         -1             0           1              4
P(XY) : 1/5       1/5            1/5        1/5           1/5
E(XY) = 0
Now, P(XY) \not= P(X)P(Y). Therefore X and Y are dependent.
r= [E(XY)-E(X)E(Y)]/SD(X).SD(Y)=0

[Divyanshu]

Hi Shiva,

Could you please explain how the answer of 6 is D. I am trying and am getting it to be C.