# ISI 2007 ME-I - Doubts

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## ISI 2007 ME-I - Doubts

 Question paper - http://economicsentrance.weebly.com/uploads/1/1/0/5/1105777/isi2007.pdfQ8) if P= log(base x)xy and Q = log(base y)xy then P+Q = A)PQ B) P/Q C) Q/P D) (PQ)/2 Q12) - I got (A) exactly but Amit sirs solution says B -- someone please explain how? http://economicsentrance.weebly.com/uploads/1/1/0/5/1105777/isi2007.pdfQ14) What does notation f(x;1) and f(x;0) mean? Q21) Again I got (A) = 1 .. please explain how the answer is B? 23) How to approach this question?!25) Again I got 240 possibilities.. please explain how the answer is 168 27) M=n/2 I get that.. but how to calculate variance? 28) ?? They all seem correct to me!! What am I missing? Please respond!! Thank you!
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## Re: ISI 2007 ME-I - Doubts

 For 12, take log on both sides and then differentiate wrt x. put x = 0 , you will get (B) “Operator! Give me the number for 911!”
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## Re: ISI 2007 ME-I - Doubts

 Thanks a lot Tsuki.. For 12 I differentiated directly and got (A).. why is it wrong to differentiate directly?
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## Re: ISI 2007 ME-I - Doubts

 In reply to this post by Ridhika Q14) Here, f(x;θ)=θ*f(x;1)+(1-θ)*f(x;0), where both f(x;1) and f(x;0) are pdf’s, Now, ∫ f(x;θ)dx= θ*∫ f(x;1)dx+ (1-θ)*∫ f(x;0)dx. Since ∫ f(x;1)dx=∫ f(x;0)dx=1. ∫ f(x;θ)dx=θ+(1-θ)=1. Hence option d, f(x;θ) is a pdf for every θQ21) Integrate the function over (0,1), then (1,sqrt(2)), then (sqrt(2),3/2)) and you will get b as answer.  "I don't ride side-saddle. I'm as straight as a submarine"
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## Re: ISI 2007 ME-I - Doubts

 In reply to this post by Ridhika  "I don't ride side-saddle. I'm as straight as a submarine"
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## Re: ISI 2007 ME-I - Doubts

 In reply to this post by Ridhika Q23) Just use normal algebraic equation and then compare the options, dats the easiest way to get the answer, u will get a relationship in the form of this identity (p1+2*p2) = 750*p1*p2, compare the options and u will get that both p1=p2 and p2=(3/4)p1 satisfies the identity.  "I don't ride side-saddle. I'm as straight as a submarine"
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## Re: ISI 2007 ME-I - Doubts

 In reply to this post by Ridhika  "I don't ride side-saddle. I'm as straight as a submarine"
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## Re: ISI 2007 ME-I - Doubts

 Thanks a lot Subhayu.. got these!!Also I think you have posted the ans to 29 instead of 28..
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## Re: ISI 2007 ME-I - Doubts

 for Q28..all of the options seems to go with the problem given..  "I don't ride side-saddle. I'm as straight as a submarine"
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## Re: ISI 2007 ME-I - Doubts

 Question 21, could someone share the workings for integral of fractional function. Thanks
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## Re: ISI 2007 ME-I - Doubts

 The working has been shared..u just check the previous posts..!!  "I don't ride side-saddle. I'm as straight as a submarine"
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## Re: ISI 2007 ME-I - Doubts

 In reply to this post by rongmon Just noticed above workings. thanks.
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## Re: ISI 2007 ME-I - Doubts

 In reply to this post by Granpa Simpson Subhayu, could you briefly elaborate on how you derived the second step in the solution
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## Re: ISI 2007 ME-I - Doubts

 "I don't ride side-saddle. I'm as straight as a submarine"
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## Re: ISI 2007 ME-I - Doubts

 Right, how does that lead to the next equation exactly? Sorry for the bother!
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## Re: ISI 2007 ME-I - Doubts

 In the interval [0,1], [x] =[x^2]=0, in the interval [1, sqrt(2)], [x]=[x^2]=1 and in the interval [sqrt(2), 3/2], [x]=1 and [x^2]=2.  "I don't ride side-saddle. I'm as straight as a submarine"
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## Re: ISI 2007 ME-I - Doubts

 In reply to this post by rongmon Its not at all bothering brother..absolutely ok..!  "I don't ride side-saddle. I'm as straight as a submarine"
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## Re: ISI 2007 ME-I - Doubts

 Grateful!
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## Re: ISI 2007 ME-I - Doubts

 I got 25 number 3 can only be in 3rd , 4th or 5th positions as it must be preceeded and suceeded by 2 numbers. Now if it is in 3rd place -> it can be preceeded in 2! ways and succeded in 4! ways. So total = 48. Exactly the reverse argument holds when 3 is in 5th place. So again there are 48 cases. When 3 is in 4th place .. there are 3 places before and 3 after. in the 3 places before 1 and 2 must be there.. so we need to do 3P2 = 6. similarly, in the 3 places after 3 we can do 3P2 for finding correct places of 6 and 7.. so we have 6*6=36 ways so far and in addition the 2 empty places can be filled by 4 or 5.. so 36*2=72 now 48 + 48 + 72=168 I hope this helps someone :D