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finally today i got the time to reply here frnds

every bdy listen , i also read saame thing from hall and knight whatever u have read, but the fact is ur logic isnt clear , let me explain it..................................

Suppose A and B are two boolean expression and it is provable that
A => B ( A implies B) now mathematics says that from here it cannot be proved that (~A) = > (~B) where ~X means not of statement X, the only thing that can be proved from here is

if ( A => B) then (~B => ~A) or writing mathematically (A=>B)=> (~B=>~A)
suppose the following statement for a polynomial f(x)
A = whether polynomial function has n equal root or not.
A = T if it has & A = F if it doesnt.
B = derivative of Polynomial has (n-1) equal roots or not

from here A => B as Hall and Knight suggest ( alright!!!)

~A = f(x) do not have n equal roots
~B = f``(x) do not have (n-1) equal roots

and you people are trying to say that (~A) => (~B)
which mathematics say cannot hold only on the basis of A=> B.
So i think , its clear now.
i hope everyone must understood this now.

again yaar , solution is wrong to an extent , because u have canceled (1-s)
that means s= 1 , that is theta = (2pie*n + pie/2) is missing from ur answer

well akriti , what i can say i also cleared JEE few years ago, i never requires this , even at that time paper was more tougher than now ............................neways this can be solved by converting it to standard form of biquadratic equation.

and whosoever ask this question , please understand that this is not in the JEE syllabus ( biquadratic equation) , and if it will come then it will be some special case.

If u wanna know solution of this question i can write that here , but nothin will be happened by that , it willl be very good for u if u give ur time to topics which are in syllabus

Thnxs

rahul_c i told u several times taht u r giving wrong answers y dunt u understand this, please do not give wrong answer untill and unless u understand the concept carefully , ur concepts are wrong , please see this example

f(x) = (x-3)^2 + 4
this function has no solution but , f``(x) = 2(x-3) has real solution , i am telling u again , u only know answers , why & how terms are immaterial to u, figure it out , otherwise u know very will what will happen

Upto the level i think , there shdnt be any solution to this equation ,but i am not able to figure out a standard proof , let me check one more time

I think it need more explanation , the above lines arent enough :-

First important think to note about that it is the carbonyl compound , so either it will proceed through their acidic hydrogen or necleophilic centre cabonyl carbon, in the present case , NaOH acts as a base to extract the acidic hydrogen of ------CH3
and this will continue until no hydrogen is left and these hydrogen will be replaced by halogen , as soon as all hydrogen are consumed. Now -OH will attack carbonyl carbon as it is now more electron deficient due to -I effect of halogens , after nucleophilic attack , instead of additon , -CX3 grop will leave , leadin to final product.

Few special points:-

1) Knowing importance of leaving group is very important , it is this ability that decide the fate of the reaction. I suppose u must have notice here a carbanion act as a leaving group!!!!!!, one of the few rare case where carbanion act as a leaving group( obviuosuly becuase of three halogens)
2) second , rate of the reaction is decide by this step , we can have various exception cases where haloform reaction did not occur , u shd be able to able figure that using this step.
3) One important fact also, ethyl alchol also gives Iodoform test , because they are oxidised to aldehyde in NaOH and X2.

for getting under 100 in JEE 2006 , it was around 340 , i dunt know exactly , but i was +- 10 from here.

In JEE 2005 ,for getting into 100 it was 126
in JEE 2004 , for getting into 100 it was 112 +- 2 .

hi everyone ,
I think Prakhyath is absolutely correct , i dont know whether others are correct or not but this man is correct.i personally think that arhinat books arent good for mathematics , they are just complete collection of questions from different sources.
What u need is understanding of mathematics , which i am very sorry to say hardly any indian writer offer. And as far as rankers saying is concerned , most of them follow their coaching teacher ( for any sort of new questions) arihant books are used mainly for the purpose of "practice" . For learning mathematics , i am very sorry to say but arihant isnt good( if u have source for learning u may use it for solving questionns), if u want to understand mathematics , take my words , do only optimum amount of problem preferably from some good foriegn authors as mentioned , but let me tell you analyse those problem properly , once u solve those problem try to find out why this question was asked , what kind of mistake u can make in these questions , further from here twist the problem , and once u r able to twist the problem , i spps u got the hold of mathematical concept. Last but not the least , evrybdy get the answer in mathematics , but there are few who know they are exactly correct(and after making mistake u even dont know that u have commited the mistake!!!!). mathematics is not abt solving , its about learning how you can make mistake ????

i dunt know why , puneet has cancel out the ( 1 + 2yp^3) from both side, i think one more possible solution is
shifting all the terms on the sam side , taking (1 + 2yp^3) as common
=> either the above solution or ( 1 + 2yp^3) = 0
=> p^3 = -1/2y
=> p = -1/(2y)^(1/3)

putting it in the equation that x = y/2p - y^2p^2/2 , we willget another possible solution.

i think u never need to remember any formula in Ionic equilibrium , this is 1 field in chemistry where u really dont need anything to remember except few basic rules .But the important factor is getting hold on those basic facts , and it will be more better for you if u start solving ionic problem from derivation rather than from knowing formula , otherwise u will be always in the shades of commiting mistakes

i do not have book , right here the steps of the reaction

brush up ur concept , i agree there will be hydrogen bonding after enol is formed , but whats so special about it , it should have been special if acetone would not been able to form H-bond,H-bonding was present with acetone as well as with enol form , so from where extra stabilisation came???????????? nobody knows ???

If u still think that u r correct provide me stepwise mechanism with reasoning at each step why that step is happening , i am giving u explanation why enol form will not be dominant, it will be more good if u can provide me some source from where i can read that it is feasible , because i strictly feel that it is unfeasible , there may be 5-10% yield but it can never be major product , it will be also a new thing for me as i never think abt anything like this.

rahul , ur solution is wrong

Do u have any idea what u r doing in the answer

In question it is given that , points are non-coplanar and u r trying to find cases when points are non-coplanar, well dude there are infinite number of ways in which these points can be non-coplanar(think over it why), and next on what basis u r finding ways , how u r arranging these points ........................................................................................
And last fact , u r trying to say that given 4 points whenever u will make them non-coplanar they will give u a plane in between them which is equidistant from all four points ..............................i.e , completely irrelevant

Last but not the least question says , find the planes equidistant from 4 non-coplanar points that means points are fixed in space , and only information we are having about them is that they are coplanar , nothing more than that

Friends let me write the solution of this problem and also the mistake in the solution provided by rashmi

What rashmi has done is that she has taken 3 points and then she draws a plane passing through all those points and then taken 4th point and she says we will find a plane which is equidistant from the plane we have drawn and from the last point ...........................this is correct , but is it exhaustive ......................no definitely not
What about those cases in which plane will separate four points , two above it and two below it...................................definitely that case isnt covered , and there is
4C2/2 ways to do this(divided by two because planes will be same in half of the case) , i.e , 3 cases.
Total number of possible planes are 7

This is correct because it is exhaustive , means we cannot choose 4 points on same , plane as that will violate their non-coplanarity and 3,1 and 1,3 case are same , and (2,2) case is now covered.