ISI 2012 ME - I Answer Key

[Amit Goyal]

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

[Badmathsboy]

Hey, How are you. Thanks for the answer key but can you please explain me Q.30?
For Q.30, why not option C??

[Amit Goyal]

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.

[archita]

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?

[archita]

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?

[Badmathsboy]

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|..

[lovekesh]

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

[Badmathsboy]

Can you please elaborate? Step by step?

[lovekesh]

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

[Badmathsboy]

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?

[lovekesh]

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

[Badmathsboy]

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

[archita]

plz halp me in q 23.

[chitvan jain]

how to do 16th and 19th quesn?

[Sweta]

Shouldnt the answer to ques 22 be A because integral of 1/x is log|x|. So log|-4| - log|-1| = log4 - log1 =
log4

[Sweta]

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

[archita]

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

[badmathsboy1]

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

[chitvan jain]

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

[lovekesh]

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