Messages posted by akki_0410

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Ola gentita pues aki les dejo un vediito .. el trailler de la pelicula EL EFECTO MARIPOSA con el sondtrack del la banda OASIS ... espero les guste x q a mi me fascina .. ah .. ps al final le agregue los 3 finales alternativos .. mas el final original .. aver escojan cual les gusta mas ... :D .. les dejo con la letra de la cancion en ingles ..

STOP CRYING YOUR HEART OUT

Hold up
hold on
don't be scared
you'll never change whats been and gone

may your smile (may your smile)
shine on (shine on)
don't be scared (don't be scared)
your destiny may keep you warm

cos all of the stars
are fading away
just try not to worry
you'll see them some day
take what you need
and be on your way
and stop crying your heart out

get up (get up)
come on (come on)
why're you scared? (i'm not scared)
you'll never change
whats been and gone

cos all of the stars
are fading away
just try not to worry
you'll see them some day
take what you need
and be on your way
and stop crying your heart out

cos all of the stars
are fading away
just try not to worry
you'll see them some day
take what you need
and be on your way
and stop crying your heart out

we're all of us stars
we're fading away
just try not to worry
you'll see us some day
just take what you need
and be on your way
and stop crying your heart out
stop crying your heart out
stop crying your heart out

pie ! ! !

pie ! ! !

$2^9\equiv7\ mod(101)\\ \\ \\ 2^{18}\equiv49\ mod(101)\\ \\ \\ 2^{19} \equiv 98\ mod(101),98 \equiv -3\ mod(101)\\ \\ \\ so,2^{19} \equiv -3\ mod(101)\\ \\ \\ 2^{190} \equiv 3^{10}\ mod(101)\\ \\ \\ 2^{12}.2^{190} \equiv 4.2^{10}.243^{2}\ mod(101)\\ \\ \\ 2^{10} \equiv 14\ mod(101),243^2 \equiv (41)^2 \ mod(101),41^2 \equiv 65\ mod(101)\\ \\ \\ => 2^{202} \equiv 4.2^{10}.3^{10}\ mod(101),4.2^{10}.3^{10} \equiv 4.14.65\ mod(101)\\ \\ \\ 4^{101} \equiv 3640\ mod(101) \equiv 4\ mod(101)\\ \\ \\ 4^{101} \equiv 4\ mod(101)\\ \\ \\ hence\ the\ remainder\ left\ is\ 4$

But, try to remember Fermat's Little Theorem, it's of good help.

yaa p = 2, so p(=2) lies in interval

1.[-16/5,16/5]

2.[0,2]

3.[-2,2]

so 1,2,3 options will be correct . . .

@ avnish . . . it is not given that the 5 no's are in AP . . . .

@ jyotibdha, it must be given in the question that the 5 no's, or atleast the 4 no's 'a,b,c,d' are "positive real" . . . i am taking a,b,c,d, to be positive reals,else the que is unsolvable

$\frac{a+b+c+d+p}{5}=2\\ \\ \\ \frac{a^2+b^2+c^2+d^2+p^2}{5}=4 \\ \\ \\ => a+b+c+d=10-p\ ,\\ \\ a^2+b^2+c^2+d^2=20-p^2\\ \\ \\ now, using\ QM-AM\ ineq.\ to\ a,b,c,d \\ \\ \\ \sqrt{\frac{a^2+b^2+c^2+d^2}{4}} \ge \frac{a+b+c+d}{4}\\ \\ \\ \sqrt{\frac{20-p^2}{4}} \ge \frac{10-p}{4}\\ \\ \\ solving,(p-2)^2 \le 0 \\ \\ =>(p-2)^2=0\\ \\ \\ => p=2$

$I=\int (tan\theta)^\frac{2}{3}d\theta\\ \\ \\ put\ tan\theta=\frac{1}{u^3}\\ \\ \\ I=\int \frac{-3\ du}{(u^6+1)}\\ \\ \\ I=-3 \int \frac{du}{(u^2+1)(u^2+\sqrt{3}u+1)(u^2-\sqrt{3}u+1)}\\ \\ \\ I=\frac{-\sqrt{3}}{2} \int \frac{1}{u(u^2+1)}(\frac{1}{u^2-\sqrt{3}u+1}-\frac{1}{u^2+\sqrt{3}u+1})du\\ \\ \\ I=\frac{1}{2} \int \frac{1}{u^2}(\frac{1}{u^2+1}-\frac{1}{u^2-\sqrt{3}u+1})du+\frac{1}{2} \int \frac{1}{u^2}(\frac{1}{u^2+1}-\frac{1}{u^2+\sqrt{3}u+1})du\\ \\ \\ I=\int \frac{du}{u^2}-\int \frac{du}{u^2+1}-\frac{1}{2} \int (\frac{1}{u^2(u^2-\sqrt{3}u+1)}+\frac{1}{u^2(u^2+\sqrt{3}u+1)})du\\ \\ \\ I=-\int \frac{du}{u^2+1}-\frac{1}{2}\int (\frac{\sqrt{3}u+2}{u^2+\sqrt{3}u+1})du-\frac{1}{2} \int( \frac{-\sqrt{3}u+2}{u^2-\sqrt{3}u+1})du\\ \\ \\ now\ each\ of\ the\ 3\ integrals\ r\ in\ their\ standard\ format\\ &\ hence\ can\ easily\ be\ solved.$

.

$I = \int e^x(\frac{x^4+2}{(1+x^2)^\frac{5}{2}})dx\\ \\ \\ now\ in\ such\ questions,try \ to\ arrange\ the\ polynomial\ terms\\ in\ such\ form\ that\ polynomial\ can\ be\ written\ as\ sum\ diff \ of\\ a\ polynomial\ &\ it$

second que is done in link given by Kabi.

$I=\int_{0}^{1}tan^{-1}\frac{1}{(1-x+x^2)}dx\\ \\ I=\int_{0}^{1}tan^{-1}(\frac{(1-x )+ x}{1-x(1-x)})dx\\ \\ I=\int_{0}^{1}tan^{-1}(1-x)\ dx+\int_{0}^{1}tan^{-1}x\ dx\\ \\ applying\ integration\ b\ parts\ to\ each\\ \\ I=\int_{0}^{1}\frac{1-x}{1+x^2}dx+\frac{\pi}{4}-\int_{0}^{1}\frac{x\ dx}{1+x^2}\\ \\ I=\frac{\pi}{2}-\int_{0}^{1}\frac{2x\ dx}{1+x^2}\\ \\ I=\frac{\pi}{2}-ln2$

.

$put\ x=acos\theta\\ \\ I=a^2\ \int_{0}^{\frac{\pi}{2}}\theta^2sin^2\theta\ d\theta\\ \\ or\ I = a^2\ \int_{0}^{\frac{\pi}{2}}cos^2\theta\ (\theta^2+\frac{\pi^2}{4}-\theta\pi)d\theta\\ \\adding\ both\\ \\ I=\frac{a^2}{2}\int_{0}^{\frac{\pi}{2}}(\theta^2+\frac{\pi^2}{4}cos^2\theta-\pi\theta\ cos^2\theta)d\theta$

now,all 3 integrals are easily seperately integrable . . .

here's another method :

multiply numerator & denominator by (2cosx+1)

$y=\frac{(4cos^2x-1)(2cos2x-1)(2cos4x-1)...(2cos32x-1)}{(2cosx+1)}\\ \\ y=\frac{(2(2cos^2x-1)+1)(2cos2x-1)(2cos4x-1)...(2cos32x-1)}{(2cosx+1)}\\ \\ y=\frac{(2cos2x+1)(2cos2x-1)(2cos4x-1)...(2cos32x-1)}{(2cosx+1)}\\ \\ in\ same\ way\ the\ numerator\ goes\ on\ simplifying\ to\ \\ y=\frac{(4cos^232x-1)}{(2cosx+1)}\\ \\ y=\frac{(2cos64x+1)}{(2cosx+1)}$

$at\ x=\frac{2\pi}{13}\\ cos64x=cos(10\pi-(\frac{2\pi}{13}))=cos\frac{2\pi}{13}\\ y = \frac{(2cos\frac{2\pi}{13}+1)}{(2cos\frac{2\pi}{13}+1)}=1$

we have,

a = b mod (a-b)

aⁿ= bⁿ mod (a-b)

aⁿ - bⁿ= 0 mod(a-b)

hence (a-b) will always be a factor of (aⁿ- bⁿ ) for every real n .

let

a=1+w

b=1+x

c=1+y

d=1+z

so we have,

(1+w+x+wx)+(1+y+z+yz) = 7+x+y+z+w

wx+yz = 5 . . . . . . . (i)

with the constraint 0<=w<=x<=y<=z with all w , x , y , z belongs to integers

we have

wx + yz = 5+0

= 4+1

= 3+2

= 2+3

= 1+4

= 0+5

ignoring first three cases on basis of constraint 0<=w<=x<=y<=z

we have only 3 possible sets of solution,

w x y z

1 1 2 2

1 1 1 4

0 0 1 5

=> (a,b,c,d) = = (2,2,2,5) , (2,2,3,3) and (1,1,2,6) are only possible solution

goiit_user_videos

hope you enjoy it

Name : Akshay Kulkarni

Date of Admission : 11 May 2008 (but i started attending this forum from November 2008 :))

Coll & Branch : IIT - Delhi .B.Tech Chemical Engineering (4yrs)

What ever i am today is because of this place. Thank U . . . Goiit . . .

see,

if u r writing the law in terms of molar conductivity then,

/\_m⁰(BaCl₂) = /\_m⁰Ba⁺² + 2 /\_m⁰Cl ^-

if u r writing the law in terms of equivalent conductivity then,

/\_eq.⁰(BaCl₂) = /\_eq.⁰Ba⁺² + /\_eq.⁰Cl ^-

Hope u got it . . .