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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


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??


when x < 0, the function is x/(1x). 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.


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(1a)b(1b)c(1c), where a, b, c
vary over all positive fractional values?


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?


This post was updated on .
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 logx..


For q.17
just apply A.M>= G.M
u will get the answer


Can you please elaborate? Step by step?


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


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?


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)(ab)^2
again, minimizing them...the minimum value (ab)^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


Yo! Thanks man. Keep checking 2011 answers too. I might just ask something there. :P


how to do 16th and 19th quesn?


Shouldnt the answer to ques 22 be A because integral of 1/x is logx. So log4  log1 = log4  log1 =
log4


oh..sorry! It'll be log4. My bad!


in question 16 (11+6root 2)=(3+root2)^2
hence anwer is 3+root2+3root2=6 which is positive nd even...


Sorry, i meant a=1 and b=c=0


thnx archita....plz help me wid d 19th one also.


guys
what answer you are getting for last question ME I of 2011?

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