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 \$h0cK3r's messages in the community 1 2 3 ... 27 28 29 GO Go to Page...
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 Discussion Forums -> This Post 2 points    (0    in 1 votes )   [?]
yaar Q is theta

a=2
b=1

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
yaar do one thing
put a=rcosQ b=rSinQ
and then solve the problems!
 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
ppl say it does better than others(career point in AIEEE)
but this year only rank<1000 can be correctly predicted

above 1000 the prediction can fail!
(im not saying this ppl are)
 Discussion Forums -> This Post 10 points    (2    in 2 votes )   [?]
method of subsitution is simple!!!
dx/(x4[ 2](1+x2))
now see it like this
dx/(x5[ 2](1+1/x2))
(1/x2 )(1/x3 )dx/[ 2](1+1/x2))
now just put 1/x2=t

bye!!!
 Discussion Forums -> This Post 7 points    (1    in 2 votes )   [?]
hey shoreline even then magciklo,s answer is right

f(x)= x2   +   0 x e-t f(x-t) dt
here f(0)=0+ 0 0 e-t f(-t) dt =0
we differenciate
f`(x)=2x+e-x *f(0)
f`(x)=2x
so answer will still be same

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
i have a different way to solve this question

here * is the cross product
b*(a*c) =(b.c)a-(b.a)c=0
hence we can say that
either a and c are parellel or anti parallel
or A*C and B are parellel or anti parallel
 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]
a4y2=(2a-x)x5
you see if we put y as -y the equation will not change
so the funcition will be symmetric about x axis
y=0 at x=0 and
y=0 at x=2a

so the required area will be

A=2[0 ][2a ]x5(2a-x)/a4
A=2[0 ][2a ]   (x2/a2)x(2a-x)
now put x=2at
A=2[0 ][1 ] 4t2.2at(1-t)    .2a.dt
A=32a2[0 ][1 ] t2t(1-t)dt
now put t=|sin Q|2
for t>0 and Q>0
t=sin2Q
dt=2cosQ.sinQ  t=0 Q=0
t=1 Q=pi/2

A=64a2[0 ][pi/2 ]  Sin6 Q Cos2 Q

A=64a2   [[0 ][pi/2 ]  Sin6 Q - 0 [pi/2 ]  Sin8 Q

using    the property      In=(n-1)/n In-2

The area of circle=pi*a2

 Discussion Forums -> This Post 7 points    (1    in 2 votes )   [?]
well if an equation
ax^2+by^2+2hxy+2gx+2fy+c=0 is given
then the equation of its angle bisector
is
(x^2-y^2)/(a-b)=xy/h
 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
i think that this should be double integral.
 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]
well  general method i can give u a very good idea
here

suppose a function be xm
then the nth derevative will be like this see

y=xm
y`=mxm-1
y``=m*(m-1)xm-2
for the nth derevative
dny/dxn=m*(m-1)(m-2)(m-3)(m-4)............(m-n)xm-n
we multiply and divide by n!
dny/dxn=m!/n!xm-n
dny/dxn=mPn xm-n
this is the method
using this we can calculate for

general formula of nth derivative of (ax+b)-r   is
(-1) (n+r-1)! (a)n / (ax+b)n+r (r-1)!

for such problems we need to find a relation between the derevatives!!
as a function of the number of times differenciated!!

these things dont come in JEE!!

all this successive differenciation and leibnitz theorum
i.e sucessive differenciation of product of two functions!!

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
hey man if it shd have one real solution then the point at which the curve touches the x axis shd be an extremum point
f`(x)=0
and also
f(x)=0

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
hey man
insted of mugging formulas try to understand from where do they come!!
try doing the derevations then ull not have problems the problems which come in JEE on this chapter are easy and scoring!
 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
the question is wrong it should be
nth derivative of 1/(1-x2)
let y= 1/(1-x2)
resolving the above expression in partial fractions we get
y=1/2[1/(1+x)+1/(1-x)]

now differentiating n times we get
y=n! /2( 1/(1-x)n+1) + (-1)n /(1+x)n+1 )

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
i think rama prediction for top ranks will hold OKAY!!
otherwise if we go lower down the growth will be exponential type!!

i mean

if there are 50 students between 350 -400
then there will be 150 students between 300-310
 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
man im telling u make amity ur last option.
It has good facilities
Wireless Internet,Centrally air-conditioned campus
but they require a good sum of money
60,000 per semester and a semester is 6 months.
If u get 85% abv in boards u get selected directly by only giving interveiew.85%- 50% fee waiver,90% abv-100% fee waiver
gud placements fr CS students.
If u are a thinking student u migh find the level of their question papers slightly easy than the conventionel engineering level.
Any ordinary student not knowing much about 11th-12th concepts can get better marks than a thinking intelligent student depending on the hard work and the cramming ability which gives u an advantage of about 20-25 marks in any of their major exams.
PLEASE TAKE ONLY AS LAST OPTION AND THAT TOO ONLY IF U GET CS.
 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
make amity ur last option
lemme tell u it has good placements for only computer science students!!
for others very less opportunities
 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
heh nups is right even im frm amity school
and i know very well how the college functions!!!

money is only what they have

good placements

more than 90% in boards
100% fee waiver

more than 85%
50% fee waiver

--------------

and its a state university

good placements!!!(only in computer science)
perot systems,HCL,HUTCH, and others

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
well paper waseasy
you just have to keep composure

there were no sitters

compared to questions which came last year!!
 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
well guyz even i gave bitsat today mine was complete disaster!!
nothin went right well
im got 312 only!!!

i messed up all those tenses and passive forms and arranging words
 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]
akku actually this is a question frm electrochem series!!

we will come 2 now dat way!!
which will get oxidised or reduced!!
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