well since no reply has come for the last sum,i think u guys(i mean the students less than 11th)can have a hint:

take to the RHS

cut off for bits goa Comp Sc-317.. i no a person who got@ tht mark

Chk @ www.bits360.com

consider a function g(x)

such tht d/dx(g(x))=p(x)

so g(x)=x^(n+1) +x^n +...+x^2 +x=x(x^n + x^(n-1)....+x+1)

g(x) = x(x^(n+1)-1)/x-1

so g(x) =0 is when x=0

or x^n+1 -1=0 which gives x=1 or -1...1 is not possible as denom becomes 0

so roots of g(x) are 0,-1..ie 2 roots

so p(x) which is equal to d/dx(g(x)) will have 1 root

consider a function g(x)

such tht d/dx(g(x))=p(x)

so g(x)=x^(n+1) +x^n +...+x^2 +x=x(x^n + x^(n-1)....+x+1)

 p(x) has 1 real root.. this root shud lie between the roots of p(x) as the function g(x) shud @ some pt between the 2 roots(ie @ a maxima or minima)

ig roots of g(x) are p,q then p<a<q

one root of g(x) is obviously 0

now other root is -1

cuz x^n + x^(n-1)....+x+1 becomes 0 @ x=-1 when n is an odd positive integer

so -1<a<0

see the force of compression is basically the tension in the wire thts all...now when u resolve tht(based on my pic) tht tgt components get cancelled

the perp components get added n these 2gether equal in mag but opp dir the mag force

Bhagat singh..

gandhiji was lucky some of his movements paid off..else it wud have been worth nothing..only ppl lik Bhagat singh made the Britsh realize in the tru sense tht we WANTED freedom

consider f(x) = e^x -x-1

f'(x)=e^x-1

f'(x)=0 @ x=0

so 0 is an extrema

f''(x)=e^x always>0 so f''(0)>0

so 0 is a minima

@ x=0

f(x)=f(0)=0.. but this is also the minima..

so f(x)>=0 for all values x..

r u guys/gals able to view the file attached??

im having trouble viewnig it..jus temme if its opening

chk out the dig n proof

in the q it doesn seem to be mentioned wat is the dist of the side parallel to the y axis from the y axis...let us asuume it is 'a'

now for the sides of the loop parallel to x axis,the force will cancel out...cuz @ each  pt on the 2 parallel sides,same force acts but in opp directions

now for 2 sides parallel to y axis:

for side closer to Y axis:

B=Bo(1+a/l)

so force=Bo(1+a/l)*i*l towards neg x..

for the side farther away from Y axis

B=Bo(1+(a+l)/l)

so force=Bo(1+(a+l)/l)*i*l towards pos x..

so net force=Bo(1+(a+l)/l)*i*l-Bo(1+a/l)*i*l= Bo*i*l

express as e^y(f(y) + f'(y))

it is very sad to see wats going on here..

both bhattji n feynmann r senior to me in age n also academically so i dun wanna comment  but it is very saddening to see these arguements time and again..as a student who wants to learn more from this site i request both of u to plz be careful in ur words..u ppl ofc already no this as u r older but i jus thot i shud say this cuz it really takes the shine of this site.. no offense meant..