ISI 2011 SAMPLE PAPER ME-I Answer key

[Amit Goyal]

1. (d)
2. (d)
3. (b)
4. (d)
5. (c)
6. (c)
7. (d)
8. (a)
9. (c)
10. (b)
11. (d)
12. (c)
13. (a)
14. (d)
15. (c)
16. (a)
17. (d)
18. (b)
19. (d)
20. (b)
21. (c)
22. (c)
23. (a)
24. (a)
25. (b)
26. (a)
27. (b)
28. (c)
29. (c)
30. (b)

[Badmathsboy]

Hi. Can you please help me with Q.7. I take log and then try to equate using Lopital's rule? What am I doing wrong? Maybe I cannot apply the rule when f(x) is not equal to 0, while g(x) = 0. However, I get stuck after that. A little help please?
Can you also discuss Q.19. What two functions do you subtract? Is it {p,r} and {r,P}? Thank you in advance.

[Shatakshi]

For Q7, you can take an example and do it. For eg, f(x)=3x^3 satisfies the given conditions and once you apply ln, you get 0/0 form and applying L-hospital's will be very simple now.

[Mainak]

can u pls xplain why the answer of the question no 25 is 65?? it should be 55 i think

[abc]

hey can u post 2011 isi sample paper here?? or any link to it ?

[taanya]

here's the link!

http://economicsentrance.weebly.com/uploads/1/1/0/5/1105777/msqe2011.pdf

[riyaf]

can some one plz explain questions 7,11, 17 and 18.

[Ridhika]

q7 - take f(x) = 3x^3 (or any function that satisfies the equation) and solve

q11. because the terms are in AP x2 = x1 +d ; x3= x1 + 2d and so on till xn = x1 +(n-1)d; No subsitute this ino the equation and solve.. the answer comes different from the 3 equations that are witten.. so ans is (d)

q18) It is clear by looking that the equation will be satisfied at (0,1). So we know |x|+|y| will be 1 at least. So we can eleminate options a and c immediately.
Now if we can show |x|+|y| = 2 (d) is our answer else (b) must be answer.

if |x|+|y| = 2; then to satisfy the eq of the circle, we need x^2 + (2-x)^2 = 1 => we find this has no real solution. So (b) must be the answer.

Hope this helps.

Somebody, please help with questions 16 and 17? and 21, 24 and 25

In q1 I get 6^1/2 .. which means option b.. how is it d?

[SoniaKapoor]

Someone pls help me with quest 17

[Bankelal]

Hi Sonia,

As f(x)= px + qx^2,
thus f(y)= py + qy^2. Replace it in integral.

Hope it will work fine!

Regards,

Aditya

On Sat, Apr 5, 2014 at 2:39 AM, SoniaKapoor [via Discussion forum] <[hidden email]> wrote:

Someone pls help me with quest 17

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NAML

[SINGHAM]

Radhika,

25. |log x*x1|+|log x*x2|+ |log x/x1|+|log x/x2|

Now we rearrange terms as |log x*x1|+|log x/x1|+|log x*x2|+|log x/x2|

Let we focus on first two terms |log x*x1|+|log x/x1|
|log x*x1|+|log x/x1|=|log x+ log x1|+|log x- log x1|=|log x1+ log x|+|log x1- log x|, because of absolute value.

Now,
|log x1+ log x|+|log x1- log x|>=|log x1+ log x+log x1- log x|=|log x1+ log x1|=|2 log x1|=2|log x1|

Similarly,
|log x*x2|+|log x/x2|>=2|log x2|

So, |log x*x1|+|log x/x1|+|log x*x2|+|log x/x2|>=2|log x1|+2|log x2|>=2|log x1+log x2|

So, |log x*x1|+|log x*x2|+ |log x/x1|+|log x/x2|=|log x1+log x2| only in case when each term is zero. But, that can happen only when (x,x1,x2)=(1,1,1)


Will you please help me with 22 ?

[neha 1]

pls explain ques 21

[Bankelal]

Hi Neha 1,

Idea is to use det(A^n) = (det(A))^n.

If there's any doubt please let me know.

Aditya

On Sun, Apr 13, 2014 at 12:37 AM, neha 1 [via Discussion forum] <[hidden email]> wrote:

pls explain ques 21

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[The Villain]

Pls help me with quest 19..

[Bankelal]

Hi Ron,

For an onto function, every element in b in B must have an element a in A such that f(a)=b. condition (i)

thus total # of functions possible= 2^4=16.

this includes those functions as well where one or more elements from set B are excluded 

say for all a in A, f(a)=p or f(a)=r, thus violating condition (i)  for onto fn.

thus we need to take of these 2 cases.

thus total # of onto fns = 16-2=14.

On Sun, Apr 13, 2014 at 3:21 AM, Ron [via Discussion forum] <[hidden email]> wrote:

Pls help me with quest 19..

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[The Villain]

Thanxx man!!

[manisha]

can someone please explain Q-3 and Q-6

[Ridhika]

Thanks a lot Singham.

For 22, I solved by elimination of options:

a) if 2A is an ineger => A is an integer . If A+B is an integer and we know A is an integer, then B must also be an ineger. In this case for f(x) to be an integer, C must be an integer. However that is not true.

b) Now if C is an integer we need x(Ax + B) [let this funtion be g] to be an integer as well. x is an integer. so for g to be an integer, Ax +B must be an integer. we know 2A is not an integer, therefore A is not an integer. if A+B is an integer, it is not necessary for Ax+B to be an integer. Therfeore this is not sufficient.

c) If 2A is an integer => A is an integer. if A is an integer and A=B in an integer, B must also be an integer. And finally C must be an integer. And if A , B , C are all integers, f(x) must be an integer. Therefore this condition is sufficient.

There may be a more amthematical way of approaching this, but Im not sure how.. this seems to make sense though and it was quick.. ;)

[Ridhika]

Aditya I dont follow what u mean by this?  please help!

Someone please help with 16, 20 and 25 !?! Please please!!

[Dreyfus]

For 20th, the coefficient of rth term in expansion of (1+x)^n is given by nC(r-1)......now substitute the values....
Plz help me wid the concept used in solving q24