21wat approach did u use 4 this one?
repeated lebnitz??? but then vo ∫ kaise gaya?????/
Today's Question of Open test(FIITJEE AITS)
Let f(x) be a polynomial function of degree 4 satisfying..
(1∫xA(t)B(t)dt)(1∫xC(t)D(t)dt)-(1∫xA(t)C(t)dt)(1∫xB(t)D(t)dt)=f(x)
where A(x) B(x) C(x) D(x) are non constant continuous and differentiable function. Its is given that the leading coefficient(coefficient of x4) of f(x) is 1..
Q. Find f(x)
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25 Answers
21
9thanks prophet sir
that q turned out to be a easy q made to appear difficult in the aits exam conditions ;)
341and putnam is an a college level competition in the US. Its quite a prestigious one and this qn was asked in one of them. I will search for the solution and post it if its more illumining
341Celestine asked how many numbers between 1 and 2000 are such that the sum of the digits of their square is 21.
The sum of the digits of the square is 21 means that we have a square that is divisible by 3 and not by 9. which is absurd. So, no such integers exist
341Sorry, I wasnt watching this thread.
But celestine has already given the solution.
The idea is this: suppose you want to prove that (x-a)k divides f(x), then one possible approach is to prove that a is a root of f(x) repeated k times. in other words f(a) = f'(a) = f"(a) =..= fk-1(a) = 0.
Here if the integrals are p,q,r and s , then we have to prove that
for f(x) = ps-qr, we have f(1) = f'(1) = f"(1) = f"'(1) = 0
We have to note that p(1) = q(1) = r(1) = s(1) = 0
So f(1) = 0. f'(1) = (p's + ps' - qr' - q'r) at x = 1 and this again zero.
f'(1) = p's' - q'r' = 2( ABCD-ACBD) = 0
and so on ...(I havent gone further. But it will all turn out well :D)
1getting f'(x)=0.........anyone clarify how 4 degree polynomial ...prophet sir...
21cele,
didnt get ur questn dude ??? [7]
can u post it in a diff post with the entire questn....
jus chk i gues thers sumthin missin in da abov que as the 2 limits r only not given [2]...
9guys did ne one get how to do the sum of digits of a sqr q ?
pls help
Q how many nos btw 1,2000 exist such that the sum of the digits of its sqr is 21.
1[11][11]what a question !!
plz give a bit of ffree air to - it will suffocate to non existence [4]
i had to glare to find it
21hi nihit......
so did u get da ans in dat optics que???
did u get my message??????
9if u diff it repeatedly ull get
f1,f'1,f''1,f'''1 =0 so f is x-1^4
1i didn't see he was online that time
well now here we are in wait of a solution from prophet sir
"i am quite sure, prophet will have a solution... "
if he hasn't he will find one and it will be an interesting one too..
21an old putnam problem rehashed iska kya matlab hai???????????/
1philip, i am quite sure, prophet will have a solution...
i am very sure [1]
33are soln nahi hai bas isitarah hai..
"given soln:
Since f(x) is divisible by (x-1)4 so f(x) is (x-1)4
"
1you can rest assured no one here (at least the likes of me) are going to try it
so you can post the soln right away abhi