if a is the root of the eqn. f(x)=f^--1(x)....then f(x) =x @ x=a....

so we have x + sin (pix) /pi= x...(actually im not being able to make out wat uve written clearly)

or sin(pix)/pi =0....and using this u can get to the soln....but the main funda is as explained in the first line...

jus. given the brief method...

rate if useful...

do one thing bring the mod to the other side and square both sides up...to get

(/Z1+Z2/)² = (/Z1-Z2/)² and then u can use the fact that /Z/ ² = Z {(Z)bar}.....to get the reqd. result...

rate if useful...

u cud do it usng dis way also.....put z= R (cos@ + i sin@)

u have the reqd. expression as / R (cos @ + isin@ ) + (1/R)(cos @ - isin@ ) / = 2 ...u cud dus proceed and solve for max value of R....

rate if useful...

look me really not getting ne way out of dis one....but neways ill post wat i was trying....

let us concider a 5rd degree eqn. of  x⁵+ax⁴+bx³+cx²+dx+e.....having integral roots....say x1 x2 x3...x5....

E == summation (formula edittor not working)

den E (x1) = -a......E (x1.x2) =b......E (x1.x2.x3)=-c.....E (x1.x2.x3.x4)=d.....x1.x2.x3.x4.x5=-e

we have -e=-a or e=a.....nd we also have { E (x1.x2.x3) / (-e)} = E (1/(x1.x2)) =c/e

using am >= hm for x1.x2,  x2.x3, .......

we have the result b/(⁵C₂) >= (⁵C₂)/(c/e)--------------1)

nd going in same lines we also have E (x1.x2)/ (-e) = E (1/x1x2x3) = (-b/e)

and applying am >= hm for x1.x2.x3, x2.x3.x4,........

we have -c/(⁵C₃) >= (⁵C₃)/(-b/e)--------------------------2)

using dese 2 eqn.s and also the fact dat a=e.....we cud probably get sumwhere....

sorry....dis im not able to get beyond dis...

i may be totally wrong...

please help me out...

frnz....dis yr my jee rank wasnt negood.....dinget ne branch nd thus as a fallout have decided to give JEE 2009....i have decided to drop an yr.....and i am in a fix now as i donno where to start from....and what shud be my flow....so i wud really accpeciate sumone comng ahead and giving me a rough flow....or a schematic way of studies....(i mean what shud be the order in which shud go in each subject)....help needed....plzzz

will shift towards the other carbon....coz benzly carbocation is acuired by doing so and it is very stable....

arre yaar...a²-b²=(a+b)(a-b) so (a+b) and (a-b) are two factors.....for a prime number the factor shud be either itself or 1

we cannot have (a+b)=1 as a and b are positive integers...

so we can have a-b=1

multiplying both sides by (a+b) we have (a²-b²) =(a+b) 

hence the problem....

and btw vatika i guess u got a so called "AIR 21"...na.....so go ahead and tell us wat branch u got....cmon'... 

name:pramod...

AIR: 4322....

branch alloted : no branch of my choice alloted...

mood : hungry for revenge...(jokes apart interestd in jee 09)

congo to all those who have got branches alloted in any of the IITs or ISMU or ITBHU....congo even to those who didnt get a branch of choice this yr....coz picture abhi baaki hai mere dost....congo to u ppl as well....

~~~~~~~ALL THE BEST~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

a fiitjee favourite.....

dekho the net force in the horizontal direction on the rod=0 so the cm will follow a straight line motion in a down ward dirn....if given a velocity in a horizontally...it wud have assumed the path of a projectile....but as V-->0 it will assume a vertically down wards path....now to get a relation between the vel and ang vel abt the pt of contact with the smooth floor....let the cm have gone down by a distance 'x' so wat we have is (l/2 - x) = (l/2)sin@ ....@ being the inclination with the horizontal at ne instant of time 't'... do diff. we have -v =wLcos@ /2-------1)

so now u can conserve energy summing up the problem thus...

mgL/2 = 1/2m {v(@)}² + 1/2 (mL²/3) {W(@)}² + mgH(@) ------------2)

but frm 1) we have w(@) =(-2vsec@ /L)------3)

and H(@) = (L/2)*sin@ -------------4)

so using 3) and 4) in 2) one can solve for v(@) .....

pardon ne silly mistakes from my side....im in the habbit of making of many.....

hope it helps...

rate if useful....

dere are 2 ways of doing this one...1st the technical way...

concider the thin wire at a distance 'x' from the centre of width 'dx'...

the magnetic field at the pt due to it will be...

-00S⁺⁰⁰(mu) Kdx / (2(pi) (x²+L²)^1/2) sin@ = -00S⁺⁰⁰(mu)L Kdx / (2(pi) (x²+L²)) = (mu) K/2  (assume L is the distance btween the pt and the plane)

the 2nd method is a shorter one...

using gauss' law for magnetic field....

SB. dl  = (mu) I_(total)    

= B  (2L) = (mu) KL        assume L is the length (a large length lasting frm -00 to +00)

or B= (mu) K/2

hope its helpful..

rate if useful....

T.B i guess its a 5th degree eqn. yaar....

look for 1 rational root this eqn shud bare minimum have 1 root....or it shud touch the x-axis one time.....

so if a is the soln. of F'(x) =0....the F(a)=0....

or x=(17/5)^1/4and we have F((17/5)^1/4) =0 solving this u shud be able to get the bare minimum value of a i think so....

rate if useful...

dude...but AM > GM is only valid for Positive numbers naa....how do u assure that all the roots are positive yaar....cant u do it this way....we haveF' =0 at 2 pts...+(17/5)^1/4 and ...-(17/5)^1/4

so we realise that we can at max have just 3 distinct roots and the 2 are repeated... so for all roots real...the basic criteria wud probably be F(+(17/5)^1/4 ) . F(-(17/5)^1/4 ) < 0

solving which we may et the reqd. condition on a....

evn i am not sure of my method...but bhai...plz think of this...(AM>GM)

a,b,c,d and e are four consequtive natural numbers....such that :

a+b+c+d+e is a perfect cube.....nd b+c+d is a perfect square....then find the minimum value of c.....

try it out...

nice Q.....does this really happen....mite be happening....dude....btw...it is not just the sun thats pulling the plannets.....even the plannets are pulling the sun..(evn they have atteractive forces)...toh...evn tho' not really significant.....it does move a bit...so a movement of sun cud also be affective and cud nullify the change in centre of mass(concidering that the mass of the sun to be pretty high compared to the masses of the plannets...think on it....this is a debatable Q....evn as said by varun...there are other sources that exert external gravitational forces on the solar system.....