Messages posted by parikshit2kool

Matlab bilkul padhai nahi ki hogi

lim x--> 0 (sinx/x)^(1/x²) = [1+(sinx-x)/x]^(1/x²)

Now the above limit is of the type (1 + f)^g, where f---->0 and g----> infinity

Hence, the result is e^f*g = lim x----> 0 e^{(sinx-x)/x^3}= e^-1/6(By using expansions)

AIR 2815, Marks: 246

Being a pure science lover doesn't mean u need to abuse IIT-JEE n tag is as 'THE BIGGEST RAT RACE'........The race seems a 'RAT RACE' for those who know they are unfit to run....

Congrats dude!........Even I have been selected.........But I'll not be going for the interview as I have also been selected for IIT-JEE.....

Anyways, I have very little idea about it

I think Eashwar is right....The expansion of the gas is too evident to ignore

I got AIR-2815 in IIT-JEE

One of my friends was xpecting around 210 by FIITJEE solutions........But he did not even qualify...........Wonder why

Why do you think so?

IMPORTANT INFO: JEE Marks to be uploaded by 12 noon today

Dude.........I had solved this question correctly and got +5

But I don't think I wud b able to explain it to u in goiit

I wud advise u against joining a college......If not JEE, then AIEEE......If not AIEEE, then BIT-SAT.......U would still b in a better situation than u r in now...

If u join a college, then u will neither get time to study for JEE, nor for ur college exams............So at the end of one year, u might flunk ur college exams as well as do bad in JEE.........

Sorry for the pessimistic outlook, but u should b aware of the worst-case scenarios

www.iitg.ac.in/jee

IIT-JEE 2010 Official answer key released on the JEE websites

Yes....This one is in accordance with my actual rank

Marks: 228

AIR: 7147(Overall), 6062(General)

AIR 6062(General)

AIR 7147(Overall)

STATE RANK 23

MARKS: 228

Post ur opinion please......

My rank in JEE is 2815.........Although I've got pretty good chances of getting Mechanical n Chemical in IIT-Guwahati, i still want to clear my doubts about MATHEMATICS N COMPUTINGAs per last year's data, I'm getting B.Tech in M&C in IIT-G as well as MSc in M&C in IIT Kharagpur.......Which wud b a better option?

I was expecting 153 by the official solutions........

Just seen....Somehow............Well my name is not in the main list...but i have a rank of 247 in the additional list

lim x--> 0 (sinx/x)^(1/x2) = [1+(sinx-x)/x]^(1/x2)

Now the above limit is of the type (1 + f)g , where f---->0 and g----> infinity

Hence, the result is ef*g = lim x----> 0 e(sinx-x)/x^3 = e-1/6 (By using expansions)

lim x--> 0 (sinx/x)^(1/x²) = [1+(sinx-x)/x]^(1/x²)

Now the above limit is of the type (1 + f)^g, where f---->0 and g----> infinity

Hence, the result is e^f*g = lim x----> 0 e^{(sinx-x)/x^3}= e^-1/6(By using expansions)