# ISI 2011 SAMPLE PAPER ME-I Answer key

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## ISI 2011 SAMPLE PAPER ME-I Answer key

 Administrator 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)
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## Re: ISI 2011 SAMPLE PAPER ME-I Answer key

 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.
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## Re: ISI 2011 SAMPLE PAPER ME-I Answer key

 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.
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## Re: ISI 2011 SAMPLE PAPER ME-I Answer key

 In reply to this post by Amit Goyal can u pls xplain why the answer of the question no 25 is 65?? it should be 55 i think
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## Re: ISI 2011 SAMPLE PAPER ME-I Answer key

 hey can u post 2011 isi sample paper here?? or any link to it ?
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## Re: ISI 2011 SAMPLE PAPER ME-I Answer key

 In reply to this post by Amit Goyal can some one plz explain questions 7,11, 17 and 18.
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## Re: ISI 2011 SAMPLE PAPER ME-I Answer key

 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 25In q1 I get 6^1/2 .. which means option b.. how is it d?
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## Re: ISI 2011 SAMPLE PAPER ME-I Answer key

 In reply to this post by Amit Goyal Someone pls help me with quest 17 MA Economics DSE 2014-16
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## Re: ISI 2011 SAMPLE PAPER ME-I Answer key

 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] wrote: Someone pls help me with quest 17 If you reply to this email, your message will be added to the discussion below: http://discussion-forum.2150183.n2.nabble.com/ISI-2011-SAMPLE-PAPER-ME-I-Answer-key-tp6329651p7586122.html To start a new topic under General Discussions, email [hidden email] To unsubscribe from General Discussions, click here. NAML "Woh mara papad wale ko!"
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## Re: ISI 2011 SAMPLE PAPER ME-I Answer key

 In reply to this post by SoniaKapoor 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 ?
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## Re: ISI 2011 SAMPLE PAPER ME-I Answer key

 pls explain ques 21
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## Re: ISI 2011 SAMPLE PAPER ME-I Answer key

 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] wrote: pls explain ques 21 If you reply to this email, your message will be added to the discussion below: http://discussion-forum.2150183.n2.nabble.com/ISI-2011-SAMPLE-PAPER-ME-I-Answer-key-tp6329651p7586730.html To start a new topic under General Discussions, email [hidden email] To unsubscribe from General Discussions, click here. NAML "Woh mara papad wale ko!"
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## Re: ISI 2011 SAMPLE PAPER ME-I Answer key

 In reply to this post by Amit Goyal Pls help me with quest 19..
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## Re: ISI 2011 SAMPLE PAPER ME-I Answer key

 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] wrote: Pls help me with quest 19.. If you reply to this email, your message will be added to the discussion below: http://discussion-forum.2150183.n2.nabble.com/ISI-2011-SAMPLE-PAPER-ME-I-Answer-key-tp6329651p7586759.html To start a new topic under General Discussions, email [hidden email] To unsubscribe from General Discussions, click here. NAML "Woh mara papad wale ko!"
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## Re: ISI 2011 SAMPLE PAPER ME-I Answer key

 In reply to this post by Amit Goyal Thanxx man!!
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## Re: ISI 2011 SAMPLE PAPER ME-I Answer key

 In reply to this post by Amit Goyal can someone please explain Q-3 and Q-6
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## Re: ISI 2011 SAMPLE PAPER ME-I Answer key

 In reply to this post by SINGHAM 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.. ;)
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