ISI 2015 PEA Answer Key - Discuss

[L]

Amit Sir usually provides the answer key for maths section. But, somehow this year he hasn't updated.
I have got following answers, check if any is wrong. (updated after discussion)
 1. (b)
 2. (d)
 3. (c)
 4. (b)
 5. (a)
 6. (c)
 7. (a)
 8. (c)
 9. (a)
10. (c)
11. (d)
12. (c)
13. (b)
14. (d)
15. (c)
16. (c)
17. (b)
18. (a)
19. (a)
20. (d)
21. (d)
22. (c) -- notation wrong (2nd option)
23. (b)  
24. (b)
25. (a)
26. (d)
27. (a)
28. (b)
29. (b)
30. (d)

[soumen08]

Im working through the paper, and the first anomaly I noticed is that 8 should be (a)
Here is some evidence to support my claim -
http://www.wolframalpha.com/input/?i=domain+of+sqrt%28x%5E2+-+1%29+-+log%28sqrt%281-x%29%29

Another, simpler way of thinking about this is to simply consider the value of the function at x = -2

[soumen08]

Also, what you seem to have done for question 14 is f(2) - f(1)

By the stationary point theorem (some lemma I forget), We can say that at some point, f'(x) = (f(2) - f(1) )/1 . This is also fairly intuitive.

Consider our function as being 100*x^2 + 100*x + 1
So that, a = 100, b = 100

Note our function is defined on [1,2] as required by the question.

Now, your answer says that k = 3a+b = 400 which isnt between 1 and 2 as required by the question.

Actually, the whole question is wrong. The answer options should have been f'(k) = 3a+b etc. In that case, option c would be right.

What do you think?

[onionknight]

The answer would be A if the question didn't mention x>=0.

[onionknight]

In question 5, the square matrix is |a2+bc     ab|     . To make this a null matrix, a=0 and either of b and
                                                 |ac           bc|             c=0 should be sufficient. How'd you get a?

[onionknight]

I'm getting A for 18th, why do you think it is C?

[onionknight]

How'd you do 26?

[L]

Q.5) Does "square matrix of matrix" means square of the matrix? I didn't read in that way. I was reading it as square matrix whose rows are equal to columns. I am not sure exactly what it means over here?

Q.8) As onionknight said, if x>= was not given, then it would have been (A)

Q.14) I think you are correct. Answer should be (d)

Q.18) I just did it by logic. If g1 and g2 are straight lines in third quadrant, then, |f(x*, y*)| > |f(x',y')|
Tell me If I am wrong.

Q.26)Search Gambler's Ruin on Google, you will get the solution. The solution is a bit tedious to write over here. This kind of problem can only be done if you have done similar problems beforehand. During exam, I can't even think of any solution like this.

[onionknight]

Hmm, I'm not so sure about 5 either, but it'd be a little too straight forward if what you think is correct.

In 18, when one constraint is applied, the solution to the optimization problem would occur at a point in some subset of R2 (the subset would be defined by the constraint). Now when we apply another constraint, we will find the maximum within the subset defined by the intersection of the two sets given by the two constraints. Now this point was there in the set in the first case also but it was definitely less than the first maximum and that's why it wasn't a solution to the maximisation problem. I don't see any role for the modulus here.

[Rajat]

How do you do the do the 16th ?
In the exam, I would have marked C but only because i know the answer is between 18 and 36 . How to get exact 27 ?

[onionknight]

Let the scores of the three people be X1 , X2 and X3. X1, X2 and X3 can take values between 1 and 6. Now you need to find all cases where the sum of these values is 10. X1+X2+X3=10 . So essentially you need to find in how many ways you can express 10 as a sum of 3 integers.

To do these sort of problems, there is a nifty trick. Imagine 10 blocks and 2 dividers, now you have 10+2=12 spaces for these 12 things. When you select any two spaces for the two dividers by 12C2, you essentially divide the 10 blocks into 3 different groups which are your 3 different integers that sum up to 10. But here since X1 and X2 and X3 can't be 0, define Yi=Xi+1 so your problem would reduce to Y1+Y2+Y3=7 which gives (7+2)C2=9C2=36. Now subtract the cases where one of the Yi's is 6 and 7 (since this would make Xi 7 or 8). There will be 6 cases for one of the Yi's=6 and 3 for one of the Yi's=7.
So finally 36-6-3= 27.

[Rajat]

Hi Onionknight,

You are right. I am aware of the blocks-and-dividers technique and that is how I have solved it too but i am not able to arrive at 27. I get 33 as the answer. Please tell me where i am going wrong.

There are 10 blocks and two dividers are to be placed in the 9 spaces in between. For the first divider, we have 9 options. For the second divider we have 8 options. So total number of ways is 9x8/2 = 36 unique ways. (Dividing by 2 because there will be duplicate counting). But this 36 number of ways will also include those configurations where one divider is placed in the 7th and 8th space.

Case 1: Now, if one divider is in the 7th space (i.e after the 7th block), the other divider can either be in the 8th space or the 9th space. This means two configurations.
Case 2: If one divider is in the 8th space, the other divider can only be place in the 9th space.This implies one configuration.

So we should subtract these 2+1 impossible configurations from 36 to give 33 .
I know i am wrong somewhere, not sure where !

[Cermank]

Can you share the link of the question paper itself? I can't find it online :/

[onionknight]

The group of blocks which contains 7 or 8 could be in the middle and end as well (apart from it being the first block in the case you've mentioned)

[Rajat]

@onionknight, right. Thanks..

[onionknight]

http://economicsentrance.weebly.com/lessons.html

[L]

Q. 18) Say the max value lies in 3rd quadrant (value is -ve) under first constraint. If the second constraint lies under first constraint, then, the max value will be greater than the first's, as it smaller -ve value, but, in absolute terms it will be lesser.

[onionknight]

If the value under the second and first constraint together is larger (smaller -ve) than the value under the first constraint, then the solution under the first constraint isn't the solution at all since we might as well choose the value obtained after adding the other constraint as we are maximising f(x,y) and not |f(x,y)|

[L]

I got it now. You are correct.

[kk]

are u sure the answer of question 1 is B?? applying L hopsital rule we get lim x tends to 0+ cos{ sqrt x}
that is cos of  decimal part..