Re: ISI 2004 ME - I Answer Key

Posted by AJ on May 09, 2012; 1:58am
URL: http://discussion-forum.276.s1.nabble.com/ISI-2004-ME-I-Answer-Key-tp7515784p7541451.html

ohhh..its asking if limit is continuous.... I was checking continuity of function... :/

and for 11th..

1.1/2 + 1/2.1/3 + 1/3.1/4 + ...........1/n.1/(n+1) .....

each of these terms become... like 1/n-1/(n+1)
(1-1/2)+(1/2-1/3)+(1/3-1/4) .....

so adjacent terms will cancel.. like -1/2 with 1/2 .... leaving only the 1 in beginning..  
so series converges to 1....


and for 21st..
I don't know how to do this minimum distance questions..so, my solution is super clumsy...

so what i did was...I drew this y=x^2 curve ..and the line .. y=2x-4....

and saw all the options... like see the distance between line and these points..
and marked the one that looked minimum ... :p
so.. both (1/2,1/4) and (1,1) .. looks like the answer...

so... cant draw graph here.. but we need the length of perpendicular from given line to these two points..
for this we need the point on line from where the perpendicular starts...

lets take point (1,1) first... our given line had slope of 2..so thats tan theta..
and perpendicular has slope tan(90+theta) = -tan theta = -2

so we have slope of perpendicular..i.e. -2... and a point on perpendicular.. i.e. (1,1) ..
so equation becomes.. y=-2x + 3

and intersection of this perpendicular and given line gives us the point on first line from where the distance has to be measured... the point is.. (7/4, -1/2) .. and distance is 720/16^2...

I did same for point (1/2, 1/4) ... and got the distance as .. (1021/16^2) ... so 1st is less.. so that's the answer .. (1,1)
I tried my best to explain this.. hope u get it..

p.s.: i just did calculations again, so they might be wrong.. .. but that was what I basically did..
and if u find some neat way to do it..do temme!
:)