Messages posted by abhujabal

thanks guys ........but i want a sort of comparision between cse or eee of nit,trichy(which i think is best nit) and chemical of bits,pilani.......

Hey guys i am really in a fix........i dont wanna loose the iit tag but i dont seem to b getting anything good in iit........i can get definitely better options in bits,pilani or goa or hyderabad and in NITs........

well at this score what can i get in bits,pilani........as far as i think i may get chemical......and in nit,trichy i may get cse or eee........which is better ..........chemical in bits,pilani or eee in nit,trichy?????

also i wanna tell something informal........i want a good campus wid gud facilities.......so plz consider that too.......

plz help me out guys........i hav always been bailed out of my problems by u all.......plz help me this time too......

yeah exactly!!!!!! I am really too worried......and i donno what to do wid revision....i seem to hav no time.....boards hav started and i donno how to revise for iit and how to rejunevate my problem solving skills which i once had in plenty in me.......
plz help me phoenix,rydbyrd,waterdemon,himanshu and other elite goiitians.......i know only u all can solve my problem.....that's why i hav come to u all for help........

Well frnds, I wanna share something wid u all......I hav got 20 percentile in Fiitjee Open Test.....I am really shocked.....I donno what to do??? ......I am very scared to c my parents disappointed from me......hence i haven't told them about this.....I don't know whether I shall qualify IIT or not.......though I am working very hard for that.....I am really feeling demoralised.......I hav tried my level best to improve but to no avail......two to three months ago I used to get 4th or 5th rank in Fiitjee phase tests( I am in regular classroom program).....But I donno what has happened to me......Now I don't remeber what I studied in inorganic chem.....can't make many mechanisms in organic......trignometry and coordinate geometry has got evaporated from my mind......and suddenly i ain't able to attempt electricity and magnetism problems which i used to do in class 12 few months ago.......I am really demoralised......I donno how to revise???

I am working hard but require a good planning and approach.......

Can anyone help this demoralised chap to get back on the right path???

I wud be grateful to u all....

well the comments of so many goiitians itself narrates the greatness of this article......

It's gr8 .....all at one place......

really inspiring........Well I knew this story as I hav read books of shiv Khera......but felt nice reading this once again as I had read it two yrs ago.......

A real image of a distant object is formed by a plano-convex lens on its principal axis. Spherical abberation is smaller if curved side faces the object......

This is because , in case of plane surface facing the object, the total deviation occurs at a single surface but in case of curved sueface facing the object, the total deviation is distributed at both the surfaces......

I know only this much about sperical abberation .......but I don't think more is required as far as IIT JEE is concerned........

I think the answer is 1

6 = 2^x.3^x+y

2.3 = 2^x.3^x+y

Comparing we get, x+y=1

I hopw my answer is correct....

If yes then plz rate ....

But frnds I also want to know to know from which books to revise......as I don't have classroom notes.....
So plz suggest some book to revise text from......

Frnds .......what is wattless current.......

In Brilliant package of AC, a question has been asked......

If a AC circuit has 0.7H inductor in series with a resistance of 200ohm

An AC source of 220V , 50 Hz is given...

Find the wattless component of current through the circuit.......

PLz explain what is wattless current.........

Is the answer 30.6cm???

Here consider liquid lens having one side plane and other side convex of radius R(which is to be found)
Then there glass lens which has one side plane and other side concave of radius R
Then use lens makers formula for both to find the individual focal length of each.....

1/f₁ = (

_l-1)/R (for liquid lens)

1/f₂ = -(

_g-1)/R (for glass lens)

Then since there is plane mirror......it's focal length is infinity

And since the ray of light is reflected we can write,

1/F = 2/f₁ + 2/f₂+ 0 = 0.34/R

Now this whole arrangement behaves as a mirror.....

Using mirror formula,

-1/180 + (-1/180) = -0.34/R

R= 30.4 cm

PLz rate if satisfied........and nudge me if any doubts r there.......

Hey guys I have almost completed my syllabus........

How shud I start revision.......I mean from which books shud I start revision.......

I shall soon get Fiitjee GMP and BT YG File.......And I don't think it is possible to do both.....So how shud I use these materials to the best......

And also I don't have classroom notes.......

So how shud I revise the topics.......if from books then from which books.......

Plz help me frnds......

Well I am getting an odd 105.....got much less in maths......

Fantabulous......

f(x)=xlnx

f'(x)=1+ lnx =0 for min or max

at lnx= -1, x=1/e, f(x) is min as f''(x)=e>0

f(x)= -1/e at x=1/e

f(x) ^_>

at x^_>

Hence the range is ( -1/e,

)

I hope I am correct.......cheers....

Plz rate if it helps u.....

Ramya was going the right till she made a small mistake.....and went the wrong way....

f(x+y) +f(x-y) =2f(x).f(y)

Differenciating both sides, w.r.t x, we get

f '(x+y) (1+y') +f '(x-y)(1-y') = 2 f '(x).f(y) +2f(x).f '(y) .y'

y is independent of x at any value, hence dy/dx will be zero.

Hence the above expression becomes,

f '(x+y) +f'(x-y) = 2f '(x).f(y)

now put x=0, and y=0 in the eqn

f '(0) + f '(0) = 2f '(0).f(0)

2f '(0) =2f '(0) .K

f(0)=K

hence , f'(0)=0

therefore,2f'(0)[K-1] =0

K=1 (This was where she made mistake)

Now in

f(x+y) +f(x-y) =2f(x).f(y)

put, x=0, and u get

f(y) +f(-y) = 2f(0).f(y)

f(y) +f(-y) = 2 K f(y)

f(y) +f(-y) = 2f(y)

Now putting -y instead we have

f(-y) +f(y) = 2f(-y)

Hence f(y)=f(-y)

Hence the given function is always even for all real nos.

I hope now I am correct!!!!

IN such questions u shud use this:-

_{[0 ]}

^{[ a]} f(x)dx = _{[0 ]}

^{[ a]} f(a-x)dx

Here , I = _{[0 ]}

^{[ ]} xdx/(1+sinx) = _{[ ]}

^{[ ]} (

- x)dx/(1+sinx)

[ as sin(

-x)=sinx]

2I = _{[0 ]}

^{[ ]}

dx/(1+sinx) = 2

_{[0 ]}

^{[/2 ]} dx/(1+sinx)

Now rationalise it by multiplying both numerator and denominator with 1-sinx

2I = 2

_{[0 ]}

^{[/2 ]} (1-sinx)dx/(1-sin²x)

= 2

_{[0 ]}

^{[/2 ]} (1-sinx)dx/cos²x

= 2

_{[ 0]}

^{[/2 ]} (sec²x-tanxsecx)dx

= 2

[ tanx - secx ]₀^/2

= 4

I = 2

I hope my explanation is correct......

plz rate if it helps u.......

_{[x ]}

_{[0 ]} [ f(1+x)/f(1) ]^1/x

= _{[x ]}

_{[0 ]} [ f(1+x)/3 ]^1/x

=_e^{_{[x ][0 ]}[ ((f(1+x)/3) - 1 )/x ]}

because when _{[x ]}

_{[0 ]} f(x)^g(x) ,where _{[x ]}

_{[0 ]} f(x)=1, _{[ x]}

_{[0 ]} g(x) ^_>

there _{[x ][0 ]} f(x)^g(x)= _e^{_{[x ][0 ]} (f(x) -1)/g(x)}

= _e^{_{[x ][0 ]}[ f'(1+x) /3 ]} (using l hospital rule.....)

= e²

I hope the answer is correct....plz rate....I haven't got one since long.....

I TMH AIEEE is good for AIEEE........
U mainly have to concentrate on speed for AIEEE as the questions asked r not very tough.......on the contrary tougher questions r asked in IIT......
BEST of luck!!!

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