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Its easy.....

multiply the given expression by -e^-a and divide by -e^-a....

now substituting x-a=X....the expression turns out to be in the form of e^X/1+X = p

hence the answer is -p.e^-a

by the by..lemme tell u 1thing...d article was simply outstanding :)

twaz awesome........i remember of my past n present wen i read this.....its not dat m cryin...cuz i ve learnt to live in darkness...n the art of life speaks that u hav to conquer wherever u are....how ever u are..... twaz a really "meaningful" article...this is the first tym ever in goiit i hav ever read such an article.......n pink....if thats d way u really are.....u cn b pretty sure cuz no power abv n below will b able 2 stop u if u know 2 conquer ur fears...ahts off 2 ya :)

Hiee.....

well.....see.....

for integration..u'll get d best explanation in RDSharma's objective maths.....m sure u'll find it useful......

den for more practice..jus go on with arihant.....

n if u hav FIITJEE's RSM, do dat 1 first...

if u still crave for sums (dat chapter is enjoyful

)..u can go with Das Gupta.....

thanx for d tips bro ^_^

NOTE: This book is of a higher level, higher than Irodov sums...but it is highly useful for developing ur basics.....its not a must, but if u have the book with u, then u can surely master mechanics...this book is not specifically for IIT, but u can surely master ur mechanics if u got hro the theory of this book...the entire book is based on calculus aproach...

A1) let the last term be 5x...

.: 5+10+....+5x = 1000000

so, 1+2+...+x= 200000

x(x+1)/2 = 200000

x^2+x- 400000=0

hence solve for x..and the requir no. is 5x....

A2) 1+1/2-1/4+1/8......

= 1+ 1/2 { 1-1/2+1/4..}

= 1+ 1/2 { 1/1-1/2}

= 1+ 1/2*2

= 2

A3) the expression under limit is: (1/2+1+3/2+....+n/2)/n^2/4+n+3

= 1/2{1+2+3+....+n}/n^2/4 +n+3

= {1/2*n/2*(n+1)}/ n^2/4+n+3

= {n^2+n}/{n^2+4n+2}

= {n+1/n}/{n+4/n+12/n^2}

as n tends to infinity, 1/n and 1/n^2 tends to 0...

hence answer is 1...

...Goutham Harsha.K

I find bliss in ignorance [^_^]

well.....

put tanx= t^2

=> sec^2x dx = 2tdt

=> (1+ tan^2 x) dx= 2t dt

=> (1+t^4) dx = 2t dt

=> dx = 2t/(1+t^4) dt

.:

_{[ ]}

^{(tan x)1/2 dx =}_{[ ]}

^{2t^2/(1+t^4) dt}

^{now integrate using standard formulae....}

^{Goutham Harsha (I find bliss in ignorance [^_^] )}

time taken for the first drop to fall= [2h/g]^1/2 = 1 sec

since drops fall uniformly, 5 drops fall in 1 sec...

hence 1 drop falls in 1/5 = 0.2 sec

so, third drop falls wen t= 0.4

.: the third drop shud have fallen for 1-0.4= 0.6 sec

dist covered by 3rd drop= 1/2*g*0.6*0.6

= 1.8 m

.: distance of the third drop from the floor= 5-1.8 = 3.2 m

hope u got it...

thank ya..

Goutham Harsha .K {i find bliss in ignorance ^_^ }

The length of the path is given by square root of (dx)^2+(dy)^2

well...dis is a very simple advice....

see..HCV is preferred all over d country...dis definitely means dat the book has sum stuff...n well..as long as ur concepts r clear, den, the pattern of d exam makes no diference.dis is d truth..but u don need 2 waste ur tym solvin lenghty probs such as sum sums in irodov...so jus get on wid HCV..but remmeber to make notes after evry sum u do..n well...m also in 12th now..i mean..m appearin for IIT in 08....

n so if u want ne help...do mail me to:

gouthamkgh@gmail.com

goutham_kgh@yahoo.com

well...HCVerma is definitely a gud buk..n m also an aspirant for 2008 IIT......

and well..as far as i hav done, HCV doesnt cause much trouble, or better say, it shud not cause ..... but the point is that i take coaching......

so if u were to feel HCV is of a high level, start frm buks dat r of still basic level....say like NCERT...or INTERACTIVE PHYSICS by mtg publications...

Well....it all depends on ur coordinate axis...u can choose it any waay and in any direction.....

it need not be dat the positive direction is up or so on..

but watever b the direction u choose, take care u assign the other quantities their directions acc to coordinate axes...dats all

well...let us assume the vectors to be F1 and 8-F1....now let us assume that F1 is the smaller vector......

for convenience i ll represent F1 by u and 8-F1 by v.....

now,

u^2+v^2+2uvcos

= 16 (formula for the resultant)

so

(u+v)^2+ 2uv+2uv cos

= 16

so, u+v=8 (given)...substituting...

48= 2uv(1-cos

)-------------------------(1)

now, angle made by the resultant is given as:

Tan

= v sin

/u+v cos

...and sice angle is 90 deg, it implies...

tan90 = infinity.....

therefore u+v cos

=0

.: cos

= -u/v-----------------------------(2)

substituting in equation (1)....we get....

48= 2uv(1-u/v)

now, put v=8-u....

u get 24= 8u...

.: u=3, hence V=5...

.: angle btween them = cos^-1(-3/5) (cos inverse of -3/5)

hope m clear..

thank u..

Goutham.K