# ISI 2013 ME - I Answer Key

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## Re: ISI 2013 ME - I Answer Key

 i think this ques was solved in this forum a few weeks back: take f(x)=exp(x+2) note that it satisfies all the 3 given conditions. integrate to get option c ---  "You don't have to believe in God, but you should believe in The Book." -Paul ErdÅ‘s
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## Re: ISI 2013 ME - I Answer Key

 In reply to this post by taanya For Ques 25, u need to find that f(x0 which satisfies all the three conditions: 1. f''(x)/f'(x)=1 2. f(0) =e^2 3. f(1) = e^3 U can derive the function f(x) from these. (1) implies that f(x) might be such a function whose first order and second order derviative are same making its division equal to one. In Differentiation, e^x is such a function whose both order derivatives are the same. If we put f(x) = e^(x+2) it satisfies (2) => f(0) = e^(0+2) = e^2                (3) => f(1) = e^(1+2) = e^3 and its satisfies (1) also... f'(x) = e^(x+2)                                     f''(x) = e^(x+2) Finally u get your f(x) = e^(x+2) Integrating it => {e^(x+2)} from -2 to 2 => e^(2+2)  -  e^(-2+2) => e^4  -  e^0 (we know e^0 = 1) Ans. (c) e^4 - 1
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## Re: ISI 2013 ME - I Answer Key

 In reply to this post by Amit Goyal can anybody please explain how to solve Q.26 the objective function problem
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## Re: ISI 2013 ME - I Answer Key

 How many maths questions does one need to look at getting right to crack the exam? About 23 right?
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## Re: ISI 2013 ME - I Answer Key

 Q.26 is a simple Linear programming Problem..what you need to do is just plot the constraints on the X-Y plane s.t. x, y >= 0. Now shade the region that satisfy all of the constraints and find the corner points of that region. In this case these corner points turn out to be (0,2), (0,4), (3,0), (5,0), (1,5) and (6,3).    Now we put these points in Z= 5x + 7y and it turns out to be that for (0,2), Z is minimum and is equal to 14, which is the answer.
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## Re: ISI 2013 ME - I Answer Key

 pls explain ques 28 and 18
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 In reply to this post by Sinistral can someone please explain Q-3,4 and 18
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## Re: ISI 2013 ME - I Answer Key

 In reply to this post by Sinistral Hi, In ques no 9 lets say |x|=z then z^3+az^2+bz+c=0 minimum real root=1 of the above equation as complex root always appear in conjugate. i.e. Z=alpha(lets say) be the real root then x=+ alpha and x= - alpha 2 real roots minimum Can you please tell what's wrong in this approach? Thanks
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## Re: ISI 2013 ME - I Answer Key

 In reply to this post by Amit Goyal can someone explain ques no 2 please ? I am still doubtful about the exact procedure to be used.
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## Re: ISI 2013 ME - I Answer Key

 In reply to this post by Amit Goyal In q15 I have tried to integrate first w.r.t.to y.and then X but I am not getting a constant but a fn of X. Can someone tell me where I am gng wrong?
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## Re: ISI 2013 ME - I Answer Key

 why will the answer to 4th question be b) 1/n+1. It shud be c) n/n+1. Solving both p(x=0)&p(x=1), we'll get c only. Someone plz check. Amir sir plz check..
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## Re: ISI 2013 ME - I Answer Key

 Hi Bhavya.. :) you'll get the following equation: 1-p-np=0 after solving this.. p = 1/n+1 So, option(b) :)
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## Re: ISI 2013 ME - I Answer Key

 In reply to this post by sam92 @ Sam - I am having the same issue.. :( @ Manisha q3--> a1.a2..an=1 so all ai = 1 so the function is 2^n q4) nC0 (1-p)^n = nC1 . p. (1-p)^(n-1) => 1-p = np(1-p) solve this and u get ans (b) Can somebody please please explain how to go about 24, 16, 28 and 9 ( I do not understand the explanation given in the post previously for 28 and 9 .. what is the logic for solving?)
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## Re: ISI 2013 ME - I Answer Key

 In reply to this post by Sinistral @ Sinistral.. in your explanation for 2.. n(n+1)(2n+1)       1.2 +2.3 +3.4 + ... n(n+1)        (n+1)(n+2)(2n+3) -6  -------------  <  -------------------------   <   ---------------------       6n^3                      n^3                                     6n^3       ratio of coefficients of highest order of n in the numerator and denominator gives the limit of the fraction as n tends to infinity. which is nothing but 2/6 = 1/3 for both the LHS & RHS terms. hence option c What is the condition for using this rule (the part I have italicized?) Do we have to always express as an inequality on both sides to use this?
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## Re: ISI 2013 ME - I Answer Key

 Also where did the -6 (in bold) appear from?
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## Re: ISI 2013 ME - I Answer Key

 Can some one help me with detailed answers for Q1. 10 outcomes of a random variable were recorded. The sample mean is 0 and sample variance is 4. It is discovered that two outcomes were recorded incorrectly: one outcome was -5 instead of -6 and another was 11 instead of 12. What is correct variance? a) 4            b) 3.6              c) 7.4                 d) 5.2 Q2. Amit has a box with 6 red and 3 green balls. Amita has a box with 4 red and 5 green balls. Amit randomly draw a ball and put it into Amita's box and Amita too randomly pics up a ball from her box. What 's the probability that the ball drawn by the two were of diff colors. a) 1/3         b) 2/15       c) 4/15          d) 7/15 Q3. A traffic light on the way to the University is red 40% of the time. What's the probability of getting red  light (i) two days in a row, (ii) two out if three days.   a) 0.16 & 0.20     b) 0.16 & 0.29     c) 0.24 & 0.29      d) none of the above Which book should I go for cost functions, production function and market sums? Please help me. Tanushree
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## Re: ISI 2013 ME - I Answer Key

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## Re: ISI 2013 ME - I Answer Key

 @mittar.chardikala Take (x + 1/x) = t. After squaring it and proper substitution you will get, 2t^2 - 3t - 5 = 0 after solving, t = 2.5, -1 since min(x + 1/x) = 2, t = -1 not possible therefore, x = 2 or 1/2, hence, their product = 1
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## Re: ISI 2013 ME - I Answer Key

 In reply to this post by Amit Goyal Economics 6 & 7. PLease have a look. My answer in 6 Employment decreases , Output same for both. For 7 I FEEL its true. But how? Please let me know.