Finally , I am still a larner and my education is not complete . But I think that u have firm conviction that u have completed all ur mathematics lessons and have nothing to learn anymore   ( I guess this out of ur reaction in pointing out ur mistake ) .

If that is ur attitude let me warn u that " Learning is a continuous process " . So first try to behave urself and then keep refrain of posting useless erroneous solution . That will benefit u and us both .

 x=0 , y= 3^n are still integral solutions  (it is not quite general as  there are vaious n's that don't satisfy it n= 2 for example)

But that is not my point !!

Ok, for the sake of argument let us solve it for integral x,y s . But how on earth do u assume that they are ( 3y)  going to be ONLY of the form 2mx +m^2 .? ( assuming m positive )

It is certainly a sufficient condition but not a necessary one !!

I think u have certain intelligence to get my point .

So ur method can't gurantee that u have found all the integral solns.

take z= r exp ( i@)

Now put this value in the given qty and then u will find a quadratic eqn in r . ( with @ as a parameter)

Solve for r and see for what value of @ r gets max .

Plz  atleast try to read the problem before u start copy-pasting from another book .

The problem asks to find all ordered pairs (x,y) , who has told u that they are going to be integers( unless ur careless assumption ) ?There are other holes in the proof also !!

.

again a wrong one !!

see , u wrote

" Now, with x² of course being a square, adding 3y makes it a perfect square. That means 3y must be of the form  where m is a positive integer ....."

consider x=3, m=5

so y=16 /3 makes it !!

now we have 2mx + m^2 = 2*5*3 +25 = 55 which is not 3y ( i.e. 16 )

So ur whole argument breaksdown !!!

Rewrite the given expn as the normal quadratic form and apply the condition that

thus u will get the answer

Write that expn as

which equals

Now put r= 1, 2, 3 ,,,,,,,,,,,,,,,,,,,,n succesively and add . Note that the cross terms cancel leaving the only first and last term .

So the sum is

got it ?

If it is a definite integral with limits 0 to inf or -inf to inf then it is certainly integrable in closed form.

Do u want that ?

To @ Deedee

Plz note that discontinuity at an isolated point does not make a fn non-integrable , e.g. U have certainly done integration involving box fn ,  is not it ?

Now that box fn ( Greatest integer fn ) is discontinuous corresponding to each integral point .

Secondly, pay attention to the redundant term ( pi/2 ) in the final form of ur obtained simplified form !!!

Can't read anything on screen !!!

Dear , if the bottom most point of the sphere and the plank has got different accelaration then there would be sliding .

and the centre of the sphere and the plank has obviously got different accn .

Because as the eqn has no real roots , p and f(1), f(1/2) ,f(-1/2) must be of the same sign .

( f(x) = the LHS)

and f(1)= p +q+r

f(1/2)= (p+2q +4r)/4

f(-1/2)=(p-2q + 4r)/4 etc

so their product must be >0

Now writ option (B) clearly and chek the answer

otherwise it is (D)

agreed !!!

Thanks .((::

so the ratio is 2:sin (sqrt(2) )

But for the first case ( the actual qn) the ans remains unchanged .

Let f denotes the force of friction . Then for the linear motion of th plank

F-f= M1a--------------(1)

and considering the motion of the sphere standing on the plank ( a non-inertial reference system ), because of no sliding , its angular accn= a1/R ( R = radius of the sphere ). where a1= accn of the  centre of the sphere wrt the plank .Then introducing pseudo force we have : For angular motion of the sphere ::

fR = (2/5 M2 R^2) (a1/R) .........(2)

and for the linear motion of the sphere

M2a- f = M2a1 ........(3)

solving (1) and (2)  and (3)I get the reqd accn .as

a=accn  of the plank =F/( M1 + 2/7 M2)

and a1 = accn of the sphere wrt plank = 5/7 a ( directed opposite to the applied force )

Now accn of the sphere wrt ground = ( a-a1 ) = 2/7 a= 2/7 F/( M1 + 2/7 M2)

Again a sign of stupidity !!

Dear  Phystd,  do u know a little bit about how the conservation of angular moentum comes into the polar eqn ?

It comes from the angular potion of the eqn

as  .....................(1)

since no-force is acting in the angular direction ( as argued by me previously )

so that from (1) anguar momentum turns out to be constant .

This follows entirely from angular eqn with no influence from the radial eqn .

Got it ?

(c) the limit does not exist

put @ = pi/2 -x

so f(x) = - sin {@}/@

Now as @->0 ( as x->pi/2 )we have for @->0+  limit = -1 but as @ ->0-{@}= 1- @ and the limit does not exist

(d) is also possible as x->pi/2- @->0+ so {@}= @ and the RHL is 1