# ISI 2012 ME - I Answer Key

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## ISI 2012 ME - I Answer Key

 Administrator This post was updated on . 1 C  2 C  3 C  4 B  5 C  6 A  7 D  8 A  9 A 10 C 11 A 12 B 13 A 14 C 15 A 16 A 17 D 18 B 19 C 20 D 21 C 22 D 23 A 24 D 25 C 26 A 27 D 28 B 29 B 30 A
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## Re: ISI 2012 ME - I Answer Key

 This post was updated on . Hey, How are you. Thanks for the answer key but can you please explain me Q.30? For Q.30, why not option C??
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## Re: ISI 2012 ME - I Answer Key

 when x < 0, the function is -x/(1-x). At x = -2, the value is 2/3 and at x = -1, the value is 1/2. And you can see that the function is not increasing.
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## Re: ISI 2012 ME - I Answer Key

 In reply to this post by Amit Goyal sir i have a doubt in question 22 i.e integral of 1/xdx,LL=-4 AND UL=-1.sir the integral is log x and log x is not defined for x<0 and therefore area under -4 and -1 is not defined.so shouldnt be the answer be undefined? nd sir i am not able to approach question17,19,23,all of them are based on maximizing over all possible values,can u please solve any one of these questions. q.17What is the maximum value of a(1-a)b(1-b)c(1-c), where a, b, c vary over all positive fractional values?
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## Re: ISI 2012 ME - I Answer Key

 In reply to this post by Amit Goyal and one more general question about ISI,is the level of question paper same as sample paper?how many question do v need to solve in order to qualify isi maths section?
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## Re: ISI 2012 ME - I Answer Key

 This post was updated on . In reply to this post by archita Hey Archita, you can approach 23 with reasoning. Z is multiplied by 5, so try to maximize z first, you cannot increase it when you reach 3 (due to x>=y>=z). Then you maximize y and then the remaining is x. And for the log question, definite integral of 1/x is log|x|..
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## Re: ISI 2012 ME - I Answer Key

 For q.17 just apply A.M>= G.M u will get the answer
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## Re: ISI 2012 ME - I Answer Key

 Can you please elaborate? Step by step?
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## Re: ISI 2012 ME - I Answer Key

 Let diff numbers be a, 1-a, b, 1-b, c, 1-c A.M will be (a+1-a+b+1-b+c+1-c)/6 U get 1/2 G.M is {a.(1-a)b(1-b)c(1-c)}^1/6 Since A.M>= G.M 1/2>= G.M Just take power 6 on both sides 1/64 >= a.(1-a)b(1-b)c(1-c) Therefore maximum value it can attain is 1/64
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## Re: ISI 2012 ME - I Answer Key

 Thanks. How do you solve for 19? And what is the right explanation for Q.20? I got to know the answer by assuming a=b=c=1, but is there any maths principle i am overlooking?
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## Re: ISI 2012 ME - I Answer Key

 for q.19 since all terms of second bracket are in squares and if i want to minimize them, one of em i took them to be equal. let c=a then i can write (2a+b)(a-b)^2 again, minimizing them...the minimum value (a-b)^2 with strictly positive value can be 1 therefore, b=2, a=c=1 put into 2a+b u get 4 u can try in a more noble way, which is pure mathematics, but rite now neither i have that much time nor i would be having during xam so u knoe what will be our best strategy :P for q.20 with no condition on a, b, c try a=1, b=c=0 nothing showed up bingo D :D
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## Re: ISI 2012 ME - I Answer Key

 Yo! Thanks man. Keep checking 2011 answers too. I might just ask something there. :P
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## Re: ISI 2012 ME - I Answer Key

 plz halp me in q 23.
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## Re: ISI 2012 ME - I Answer Key

 how to do 16th and 19th quesn?
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## Re: ISI 2012 ME - I Answer Key

 In reply to this post by Badmathsboy Shouldnt the answer to ques 22 be A because integral of 1/x is log|x|. So log|-4| - log|-1| = log4 - log1 = log4
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## Re: ISI 2012 ME - I Answer Key

 oh..sorry! It'll be -log4. My bad!
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## Re: ISI 2012 ME - I Answer Key

 in question 16 -(11+6root 2)=(3+root2)^2 hence anwer is 3+root2+3-root2=6 which is positive nd even...
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## Re: ISI 2012 ME - I Answer Key

 In reply to this post by Badmathsboy Sorry, i meant a=1 and b=c=0
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## Re: ISI 2012 ME - I Answer Key

 In reply to this post by archita thnx archita....plz help me wid d 19th one also.