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Find x such that

2001^x+2004^x = 2002^x+2003^x

or you can use that any perfect square is of the form 3k or 3k+1. you get the same result.

Find all n such that n!-1 is a perfect square

Its time you guys devote more time to boards. So this is the last integral:

Evaluate $\int_0^{\frac {\pi}{2}} \frac {x^2}{(\cos x + x\sin x)^2}\ dx$

DAV demolition squad at work again

How does R remain constant?

it does look like a possible JEE prob isnt it, i mean with that 2008 and all that.

nope, sboosy, you dont get to cross 400 with this one.

Hey sorry, I thought you already have the solution.

I have to correct my answer a bit. It should be both m and n should be even.

With an expression like (x-1)^m(x-2)ⁿ, the first derivative vanishes at 3 points, x=1, x=2 and a third point.

We can dispose of the third point easily as you can see that f"(x) is non-zero at that point.

So the critical points to worry about are x=1 and x=2.

You can easily see that f^k(x) will go to zero at both points as long as k<min(m,n) as (x-1) and (x-2) appear in every term.

Each order of derivative reduces the degree of (x-1) and (x-2) by 1. So there is going to be a term that goes like (x-1)^m-k (x-2)ⁿ.

Let's assume that m<n. So, this term becomes (x-2)ⁿ and every other term has (x-1) as a factor. Hence f^m(x) becomes non-zero at x =1 and is zero for all f^k(x) k<m. So from the sufficiency condition for x=1 to be an extremum, m has to be even. (I mean the sufficency condition that if f^k(a) =0, for k<t, and f^t(a) is non zero, then a is an extremum point if t is even and not if t is odd)

Similarly at fⁿ(x), we have non-zero derivative at x =2 and so for extremum condition, n has to be even.

If m and n are not even, the extrema will not occur at these points.

m+n should be even is the requirement you are looking for.

So (b) and (d) are correct

It is sin2008x. not to the power of.

Evaluate $\frac {1}{\displaystyle \int _0^{\frac {\pi}{2}} \cos ^{2006}x \cdot \sin 2008 x\ dx}$

Not quite so simple if you read my posts carefully. I say that the left limits of two equivalent expressions are being equated.

If you still want to argue on whether the limit is correct, just remember you are taking on Riemann, Abel, Cesaro, Liebnitz, Borel and not to speak of our own Ramanujan and a host of advanced calculus students. In short, it is a pretty well defined procedure. Maybe I omitted to say this, but the conditions under which such summations can be performed are well laid out which incidentally cover geometric series with r>1. No magic there!

And I suspect Feynmann himself would have used such a summation at the Shelter Island Conference.

I agree sboosy. In fact before I started the problem I even had this idea that the problem is something like A and B are reporting about an event E.

So, to calculate the prob that their statements match, we have to take into account whether the event happened or not and then we start worrying about in how many ways it could have happened and not happened also.

All I am trying to drive home is that the problem must be defined carefully. And a bit of judgement from our side whether the problem (JEE '75!!) demands that much probing.

No offence meant and I hope no offence taken.

or when they speak the truth they say the same things?