1) find the no of right angled triangle with integer sides and inradius 2008
2) evalute this limit
\lim_{x\rightarrow \infty }((x-1)(x-2)(x+3)(x+5)(x+10))^{1/5}-x
341another way of looking at it is that in terms of AM-GM.
You know that AM=GM when the numbers are equal.
Here as x→∞, each of the terms tends to each other.
So
\sqrt [5] {(x-1)(x-2)(x+3)(x+5)(x+10)} ≈
\frac{(x-1)+(x-2)+(x+3)+(x+5)+(x+10)}{5} = x+3
Hence the limit is 3
1actually i don hav the answers for these q[2]....so cant tell wats the right ans
62i read it long time back...
probably in DM Berton (I dont even remember the spelling of the author :P)
62for the first question use the fact that
pythagorean triplets are of the form
2x^2+2x, 2x+1 \text{ and } 2x^2+2x+1
and constant multiples of these ...
also you can show that this x used above is equal to inradius :)
so you are simply looking at the number of factors of 2008
:)
1thankz nishant bhaiyya & hari sir.............................now i got it............
1well neva thought of applying am=gm here.......that jus solve the prob in 2 steps...
but nishant sir where u think der is prob applying that[1]
62Prophet sir, I Have my reservations in saying that the numbers are equal near infinity..
I think a better way to say that would be to take x common and then use the same logic on the numbers
1-1/x, 1-2/x, 1+3/x, 1+5/x and 1+10/x
am I being uselessly cautious?
1oh yeah we hav to only find coeeficient of x4 in that.....actully that thing was trubling since i thouoght multiplying a hole lot of five things will be quiet a big task....
thnx a lot [1]
62yup that is correct
so this expression will reduce to
\left[x^5+15x^4+ax^3+bx^2+cx+d \right]^{1/5}-x
Here, a, b and c are things we are not concerned about and hence are not finding out...
Now take x common
and use the approximation (1+x)1/n = 1+x/n+... higher terms of x..
\\=x\left[ \left[1+15/x+a/x^2+b/x^3+c/x^4+d/x^5 \right]^{1/5}-1\right] \\=x\left[ 1+\left( 15/x+a/x^2+b/x^3+c/x^4+d/x^5\right)/5-1\right] \\=x\left[ \left( 3/x+a/5x^2+b/5x^3+c/5x^4+d/5x^5\right)\right]
which tends to 3 when x goes to infinity
1i got ur hint but cudnt get how to apply in this q[2]....i mean do we hav to then multiply all (x-1)(x-2)..... to get a form of ax5 + bx4.........to get a form of
5√(ax5 + bx4.........) -x....
62see as a hint
what is the limit when n tends to infinity of
\\\sqrt{x^2+ax+b}-x \\=x\left[\sqrt{1+a/x+b/x^2}-1 \right] \\=x\left[(1+a/2x+b/2x^2)-1 \right]
= a/2
The above holds because √1+t ≈ 1+t/2 when t is very small...
Or otherwise rationalize it...
1yup that was my doubt :)
if i wud hav given it as a challenge...i wud hav metioned that its not my doubt[1]
62oh it was your doubt?!!
I thought you gave this one for the others..
we have discussed 2 questions on this one..
one was for 4 terms instead of 5 and one was for general n
For general n, i think the proof was given by Kaymant sir.. anyways the idea remains the same Let me try and post it
62no madhu..
this one was solved long back..
the answer is (-1-2+3+5+10)/5 = 3