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Thanks to my friend Rohith , he has done beautifully in BITSAT 2008 and he has uploaded these for you guys.....

Hope you make full use of it Here goes the link This time esnips.com

http://www.esnips.com/nsdoc/1cae3ef6-b498-46f2-ad12-979266b5ba7f

Enjoy and do well

All the best.

Edit : Looks like some technichal errors ; However here goes the mediafire link as well :

http://www.mediafire.com/?r9xojo9wujw

Sorry for the inconvenience

$** Put \;up\;for\;general\;benefit\;not\;for\;rates **$

$Nothing\;is \;mentioned \;whether\;repetition \;is\;allowed\;or\;not$

$If \;repetition\;is\;allowed\;answer \;will\;be\;different\;$

$In\;that\;case\;answer\;will\;be\;different\;$

$It \;should\;be ; 8.{9}^{3} + 7.{8}^{3}+6.{7}^{3} +.....+1.{2}^{3}$

$Otherwise \;answer \;should \;be\;the\;1\;given\;by\;allamraju\;$

$Edit : Thanks\; allamraju\;I \;didnt\;notice\;it..Sorry.$

$Your\;answer\;is\;correct.$

$Its\;clearly\;given\;different\;digits.\;Sorry\;for\;the\;inconvenience$

Click here ,

You will find 2 solutions.

I am sorry that this thing comes so late . I know , many of you have already had the exam , but still so that users who have not yet taken the test maybe benifitted, I have put up 5 BITSAT mock test papers from Brilliant tutorials. I know many of you maybe having it , still ....... for those who don't have it.....

Check this rapidshare link :

http://rapidshare.com/files/114798938/BITSAT_Mock_test_papers_from_Brilliant_tutorials.zip.html

All the best to all of you. May all of you do very well.

~ Greatdreams. The papers are in pdf format , for which you need Adobe reader

You will get the software here in any case if you don't have that. You will get it here.

Edit : For people who are finding error please check this mediafire link :

http://www.mediafire.com/?bdeiwlttu2z

http://www.srmist.org/srmweb/admissionsReport/onlineRank.jsp

This might be of some help.

Yes vulgar parts of the post has been edited.

Thank you Sir , My fellow moderators and me hope to remove the trouble makers who have unfortunately crept in. Thank you once again.

$Good \;work ; however\;it\;can\; be\;written \;as\;:$

$5^{99} = 5^{3}.5^{96} = 125 . (625)^{24}$

$= [13.9 +8](1 + 48.13)^{24}$

$Now \;expand\;binomially\;;$

$Hence \; it\;is=(13.9+8)[1+^{24}C_{1}.48.13 + ^{24}C_2(48.13)^{2}+..................]$

$Hence\; 8\;is\;remainder\;!$

$w.r.t \;the\;previous\;post ;$

$a \;=\;b(mod)c \rightarrow when\;a\;is\;divided\;by\;c\;;b\;is\;the\;remainder$

$P.S-36\;seconds\;differnce\;between\;rabiit\;and\;me.$

$Answer \;has\; been\;edited\;a\;bit!$

$For\;these \;sort\;of\;problems\;you\;have\;to\;take :$

$x = a{cos}^{2}t + b{sin}^{2}t$

$Hence\;x-a\;=\;b{sin}^{2}t - a(1-{cos}^{2}t)$

$= (b - a){sin}^{2}t$

$Similarly ;\;b-x\;=(b-a){cos}^{2}t$

$Hence \;dx =\;2(b-a)sint\;cost\;dt$

$x = a ; \;sint = 0 \;and\; x = b ;cost=0$

$Thus I = \int_{0}^{\pi/2} 2(b-a)sint costdt / (b-a)sintcost$

$= 2\;[ t ]^{\pi/2}_0$

$= \pi$

$Aliter;\; take\;x-a\;=t^{2}$

$adjust\; limits\;and\;thus; b-x = b-a-t^{2}$

$Hence; I =\int_{0}^{\sqrt{b-a}} 2tdt/(t)\sqrt{b-a-t^{2}}$

$= 2 {sin}^{-1}[ t/\sqrt{b-a}]$

$= 2[ sin^{-1}1 -sin^{-1}0]$

$= 2.\pi/2$

$= \pi$

Whether you are a Hindu or a Muslim or a Christian..............in the end you are a human being and we should get this thing clear in our heads .

If I go back to my History Civics days I can't remember exactly the number (as I used to draft the constitution according to my requirements in my answer sheet !!)

but I am sure the Constitution provides you the right to practise your religion as a Fundamental right.Indian society is certainly incomplete without Islam ~ all walks of society!!.(I love Mughal dishesand I go to a friend's house right after Id to enjoy the delicaices. )

I can't figure out what Pragun said in his initial post but better keep away from these topics .......discussions and arguements on this topic could go on for ever .You seemed to have incited a few people!

However Ash I think instead of religion if we consider every human being to be an entity instead and that God exists in all of us .....will certainly make the world a much better place to live.[please don't mind ~ totally personal opinion]

As regards terrorism; people in general don't take to such ways until of course they are brain washed or its a well known thing that these things originate at places where people are the poorest.Look at Afghanistan or the tribal areas of Pakistan or Bangladesh or perhaps the northern parts of Sri Lanka ........as for Sri Lanka if we look into it carefully we will find that the people have been denied of their basic fundamental needs for years; hence the hostility.[ economic factors after all take centre stage ]

Quoting Rishipratim : " mr. pragun....

i find dat u r makin highly misleading comments....

first of all...

u r makin comments dat r against the ethics of this site..."

Don't you think the whole discussion is against the principles of the site??Hence I request admin to block this thread for further discussion.

Please help me guys by requesting admin to do so.

Good work Ramkumar for finding such an excellent problem ,

Sometimes at Goiit also we also have a few discussions that does bamboozle tried and tested mathematicians . Isn't it?

Hope Bhattji comes up with his questions again and Koni , Nadeem , Sboosy ,Elasti , Akhil etc ......... again participate in solving and making things beautiful up here!

${x}^{12}+{x}^{9}+{x}^{4}-{x} + 1 \; is\; defined\; for > 0$

$Hence \; {x}^{4}({x}^{8}\; + 1) - x ({x}^8 + 1) + 1 > 0$

$So , ({x}^{8} +1)x \; ({x}^{3} - 1) + 1 > 0$

$If \;x\ge1 \;or \le -1 , given\; expression\; assumes \;positive\; values$

$If \;-1<x\le0 \; then \;again \;expression \;is > 0$

$Considering \; 0<x<1\; inequality \;still \;remains \;true$

$So\;domain \;(-\infty , \infty)$

$sorry \;a \;little\; bit \;edition :$ $f(x) \; is\; defined \;for \;the \;expression\; being \;\ge and\; not \;> 0$

1 method of solving these type of problems :

Given straight lines : y + 2x = 3 and y + 2x - 5 = 0

They pass through the point , (2,3)

r = - [ ax1 + by1 + c / (acos $\theta$ + b sin $\theta$ ]

where (x1 , y1) = 2,3

Hence r = - [ 2(2) + 1(3) - 3 / 2 cos $\theta$ + sin $\theta$ ] = - ( 4 / 2 cos $\theta$ + sin $\theta$ )

Now its given that it makes an intercept of length 2 units , hence

r+2 = - [ 2(2) + 1(3) - 5 / 2 cos $\theta$ + sin $\theta$ ] = - ( 2 / 2 cos $\theta$ + sin $\theta$ )

Hence subtracting the two we get ,

r+2 - r = 2 = 2 / (2cos $\theta$ + sin $\theta$ )

Hence 2 cos $\theta$ + sin $\theta$ = 1

or 4 cos² $\theta$ = ( 1 - sin $\theta$ ) ²

So 4(1 - sin $\theta$ ) (1 + sin $\theta$ ) - (1 - sin $\theta$ ) ² = 0

or (1 - sin $\theta$ ) [ 4 (1 + sin $\theta$ ) - [ 1 - sin $\theta$ ] = 0

Hence sin $\theta$ = - 3/5

Thus cos $\theta$ = 4/5

or sin $\theta$ = 1 and cos $\theta$ = 0

So two cases are ,

1.) x-2 / 4/5 = y-3 / (-3/5)

Hence , -3(x-2) = 4(y - 3)

or 4y + 3x = 18

2.) x-2/ 0 = y-3/1

or x-2/y-3 = 0

Hence x-2 = 0

Thus the 2 equations are ,

x = 2 and 4y + 3x = 18 ................ans