ISI 2013 ME - I Answer Key

[Amit Goyal]

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

[ISI agent]

I would be grateful if someone could explain question no: - 8, 13 and 28.

Regards

[wolverine]

q-8)
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^(43-1)=2^42

[wolverine]

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]

[ISI agent]

Thank you so much. This is really helpful.

[srawan]

q 28-------------=      ans   = 24-18+6-1=11

[ISI agent]

Why '+6-1'?

[srawan]

actually i forgot  which method i used

alternative is
24-36+24-1=11
using priciple of inclusion and exclusion
more detail
4!-3*(4!/2)+3*(4!/3)-1

[Aashna]

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 :)

[duck]

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.

[monica]

Can someone please help me with q-9,11, 15 ?

[Sinistral]

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
      =x-1  ;  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.

[Aashna]

Thank u so mch! :).  Plz help me vd ques 15. Thank u.

[Devika]

Heyy

Please explain Q2 and 21.

Thank you

[Sinistral]

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

[Sinistral]

for ques no 21 refer to this:
http://discussion-forum.2150183.n2.nabble.com/isi-2013-URGENT-tp7580532p7580533.html

[Devika]

Thank you so much!!!!!!!!!!!

[sonal]

i wud b grateful if someone can explain ques 20 16 n 24

[Sinistral]

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/(t-1) = 1/(t-1) +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/(t-1) = xt/(t-1)     dividing both sides by t-1
    y logx   = x log y
       x^y   =  y^x
 option c    

[taanya]

someone please help  me out with question no.25!!!thanks:)