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1 d
2 c 3 a 4 b 5 a 6 c 7 a 8 c 9 a 10 a 11 c 12 b 13 d 14 d 15 d 16 c 17 c 18 a 19 c 20 d 21 c 22 d 23 a 24 c 25 c 26 a 27 b 28 d 29 b 30 d Ref: ISI Sample Exam 2013 
I would be grateful if someone could explain question no:  8, 13 and 28.
Regards 
q8)
total no of tosses =43 no of heads greater than no of tails will be when heads are 22,23,24,25....43 as corresponding to these cases no of tails will be 21,20,...1. so total no of heads will ne 43C22+43C23+43C24+.......43C43= 2^(431)=2^42 
In reply to this post by ISI agent
13)
A.M.>=G.M for a and b ((a+b)/2)^2>= ab a^2+b^2+2ab>=4ab a^2+b^2>=2ab.....(1) similarily b^2+c^2>=2bc.....(2) a^2+c^2>=2ac.....(3) adding 1,2,3, 2(a^2+b^2+c^2)>=2(ab+bc+ac)......(4) as a^2+b^2+c^2=1....(5) 4 and 5 gives ab+bc+ca <= 1.........(6) also (a+b+c)^2>=0 as square of any no is greater than or eqal to zero a^2+b^2+c^2 + 2(ab+bc+ca) >=0 1+ 2(ab+bc+ca) >=0 (ab+bc+ca) >= 1/2 (7) 6 and 7 gives (ab+bc+ca) belongs to [1/2,1] 
Thank you so much. This is really helpful.

In reply to this post by ISI agent
q 28= ans = 2418+61=11

Why '+61'?

actually i forgot which method i used
alternative is 2436+241=11 using priciple of inclusion and exclusion more detail 4!3*(4!/2)+3*(4!/3)1 
hey ,plz some1 explain me 29 n 30. in 29,evn c is correct acc to me. nd in 30 ,i agree vd d, bt shdnt b be correct as sum (sigma) of residual terms is always 0. plz correct me. Thank u :)

Hi Aashna.. :)
30) Thats not the case always. You dont have an intercept term in this model. So, ∑û(i) will not be equal to zero. Hence, option (d) is correct.
:)

In reply to this post by wolverine
Can someone please help me with q9,11, 15 ?

This post was updated on .
9)
let g(x) = x^3 + ax^2 + bx + c ==> f(x) = g(x) ==> the graph of f(x) = graph of g(x) when x>=0 & the graph of f(x)= reflection of f(x) in y axis when x<=0. (since f(x) is even) we know the graph of h(x)=x^3 + 5 (an example). now we can twist g(x) in a way that it becomes h(x) as then it will never touch/cut x axis (when x>=0) ie (for understanding) put a=b=0 and c=5. hence we can construct such g(x) which will never cut x axis when x>=0. for x<=0, it i.e. f(x) is nothing but reflection of g(x) when x>=0. hence minimum no roots possible. (a) 11) f(x)={x[x]} which is nothing but a fractional func. can be rewritten as follows: f(x) = x+1 ; 1 <=x<0 = x ; 0 <=x<1 =x1 ; 1 <=x<2 and so on.... on seeing it closely (and/or seeing the plot), it is clear that its range is from [0,1) and it is periodic with periodicity 1. (just plot it for clarity) So I= Integral (2 to 343) of [{x[x]}^2] = 341* integral (from 0 to 1) of [{x[x]}^2] = 341* integral (from 0 to 1) of [x^2] = 341/3 (c) 15) if we change the order of integration without changing the limits of integration, it becomes fairly simple and ans comes out to be D. however I wasn't able to integrate after changing the order of integration in a proper way. I might be making some mistake. ans comes out to be D but I am not satisfied with my solution.

"You don't have to believe in God, but you should believe in The Book." Paul Erdős 
In reply to this post by duck
Thank u so mch! :). Plz help me vd ques 15. Thank u.

In reply to this post by Sinistral
Heyy
Please explain Q2 and 21. Thank you 
This post was updated on .
for ques no. 2 we will try to use sandwich theorem.
our original series is the middle term: 1.1 +2.2 +3.3 + ... n.n 1.2 +2.3 +3.4 + ... n(n+1) 2.2 + 3.3 + 4.4 + ... (n+1)(n+1)  <  <  n^3 n^3 n^3 left & right hand terms have been made in a way to suit the inequality: now sum up: 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

"You don't have to believe in God, but you should believe in The Book." Paul Erdős 
In reply to this post by Devika
for ques no 21 refer to this:
http://discussionforum.2150183.n2.nabble.com/isi2013URGENTtp7580532p7580533.html

"You don't have to believe in God, but you should believe in The Book." Paul Erdős 
Thank you so much!!!!!!!!!!!

In reply to this post by Sinistral
i wud b grateful if someone can explain ques 20 16 n 24

16)
A1 = {0,2,4,6...} A2 = {0,3,6,9,...} A1 ∩ A2 ={0,6,12,18,24...} = {6k ; k ∈ N} = {(5+1)k ; k ∈ N} = A5 option c 20) if a+b = 2n+1 s.t. a < b < 2n+1 and a,b ∈ N then for the product of ab to be maximum a & b should be as close to the middle value of the series 2n+1 (n=0 to n=n). on close observation it can be inferred that it can happen only if b becomes the middle term and 'a' becomes the term to the left of the middle term (ie b) middle term = (2n+1 + 1)/2 = n+1 =b previous term to the middle term = n+1 1 = n = a so a and b become n,n+1 hence option d 24) t/(t1) = 1/(t1) +1 log y = log x +1 (log denotes log to the base t) log y = log x + log t log y = log (xt) y = xt y/(t1) = xt/(t1) dividing both sides by t1 y logx = x log y x^y = y^x option c

"You don't have to believe in God, but you should believe in The Book." Paul Erdős 
someone please help me out with question no.25!!!thanks:)

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