x>5/2 & x<1/2 shudnt b that tuff
1)FInd area bounded by mod(y)=e-mod(x) with [mod(x)+mod(y)]/2 + [mod(x)-mod(y)]/2 ≤2
2)Find area of region at point P satisfying maximum{PA+PB,PB+PC}<2 where A=(1/2 ,0) ,B=(3/2 , 0) ,C=(5/2 , 0)
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68 Answers
yes tapan.. i din check the nittigritties.. but this is what i was saying :)
Sir, Pl. c my post # 9,10,11
I hav jus considered I case 1< x < 2
Can u chk my workin for dat case as in abov mentioned post nos.
2)Find area of region at point P satisfying maximum{PA+PB,PB+PC}<2 where A=(1/2 ,0) ,B=(3/2 , 0) ,C=(5/2 , 0)
arrey 2nd is the more easier one!!!
I thought you were asking about 1
that is why i was lazying in answering this one...
P=(x,y)
take cases..
x<1/2
1/2<x<3/2
3/2<x<5/2
x>5/2
in each case, it will boil down to
PA+PB<2 or PB+PC<2
which are ellipses!
I think you can take it from this hint
I cant think of a muchsimper/ shorter method.
SIR, u chkd the wrong questn........
QUE1 is pretty easy......
I wanted to know abt QUE2
this is same as the area between |y|=e-|x| and |x|<=2
This is 4 times \int_{0}^{2}{e^{-x}dx}
\int_{0}^{2}{e^{-x}dx}\\ = 1-e^{-2}
check ifthere is any mistake
So shall I consider my ans in post #10 and corresponding workin in #10 as
correct (as a snippet only coz I gotta add more cases)
okzz...........but i never forget to mention that GINT thing whenver i post my questions.......HEre it didnt represent so I didnt wrote........
Sorry Tapan if ur time was wasted.......
it wasnt confusing............whenever [] represent GINT it is always mentioned................
but manu please take care to use symbols that r not confusing!!!!!!!!!1
arre but then mathematician told me it representeed GIF to mene 1/2 hour barbad kar diya iske pichhe........ [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17][17] [17]17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17][17] [17]17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17][17] [17]17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17][17] [17]17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17][17] [17]17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17][17] [17]17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17][17] [17]17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17] [17][17] [17]
arre mani chhod ye sab.
IN THE QUESTIONN [.] represents simple paranthesis, nothin else.... then the que bcums very easy....
Image link...
http://www.artofproblemsolving.com/Forum/images/smiles/rotfl.gif
just use [ im ]
@tapan.. bhai I dont know the final answer..
so i din want to pink it..
but i have had that feeling right from the beginning ki it must be correct..
only that i din have enuf in me to sit down and calculate it ;)
:P
matalb calculate karna hoga... final answer to nahi nikale... bas fig banaye...(actually forgot to solve it yesterday night... so just drew it now...)
Yehe!!!
me 2 same ans!!
priyam can u giv da final value of ur Ans. OR CHK my Calculation in PAGE 2
yeh ques to eureka ne diya tha
he SHOULD be knowing better than anybody
Sir while I was givin this one a 2nd thot I realised the questn says :
2)Find area of region at point P satisfying maximum{PA+PB,PB+PC}<2 where A=(1/2 ,0) ,B=(3/2 , 0) ,C=(5/2 , 0)
So wud not the area of favourable cases on the X-axis be jus wen 1<x<2 coz beyond this bound of X our inequality wud not hold!!
So I guess the answer and working in POST #10,11 wud be my final Ans.
Very SORRY [2] if my analysis is wrong!
*** AND ya if u chk my working pl. note that the origin has been shifted to point B
Thnx for ur relentless support [1]