ISI 2009 subjective MIB

sorry b555 for being so unaware of the term multiplicity :(

i researched on this and found that a root has multiplicity of k if all its derivative till k-1th derivative are 0

now f(x) =x (x+1) (x+2) … (x+2009) - c

f'(x)= (f(x)+c) [ 1/x + 1/x+1 + ..... 1/x+2009 ]

let x be a root with multiplicity >2

now f(x)=0

f'(x)= c[ 1/x + 1/x+1 + ..... 1/x+2009 ]=0 (c≠0 has if c=0 ,clearly 2010 roots possible)

implies

now f"(x) =-c[ 1/x²+1/(x+1)² +....1/(x+2009)²] ≠0

so maximum multiplicity =2

now when multiplicity =2

f'(x)=0

f'(x) is a polynomial of degree 2009 and has only one real root as f"(x)≠0

so only one value of c possible

can u explain it b555 ??

i had a careful look cudnt figure out :(

ok celeestine , see this

u say opposite pairs for paints are chosen in 6C2 . 4C2 = 90 ways

but no, thats wrong. its 90/6 [6=3!]

(because the three pairs are indistniguishable)

thanks b555

"may day!! help!!"

it is 30.ml khanna is wrong. or u r getting the q wrong.

My sincere apologies for posting wrong Question no. 2

Q. 2. This was the correct Question –

f(x) = ax² + bx + c be a quadratic polynomial such that

f(-1), f(0), f(1) belong to [-1, 1].

Prove that

Q. 10. For some fixed real constant c,
Prove that any root of the equation x (x+1) (x+2) … (x+2009) = c can have at most multiplicity 2.
Determine the number of values of 'c' for which the equation has a root of multiplicity 2.

10th one..........

hm this seems interesting. (that doesnt mean its a tough one)

waiting to see if anyone got a better soln

celestine , are u joking or wat ?????????????????????/

a root is ssaid to be of multiplicity `k` if its k times repeated root.

like for x²2+4x+4 , -2 is a root of multiplicity 2

for (x-1)²(x-2) , 1 is a root of multiplicity 2

i hope u get it

abhirup can u pls give exp why its 30 as in #2

this is my method (please dont laugh , as i couldnt find a better one)

begin by writing the polynomial in the form u(x-1)(x+1) + vx + w

then f(-1) = -v + w, f(1) = v + w, f(0) = w - u

there is a formula describing f(x) in terms of f(-1), f(0), f(1)

because a quadratic polynomial is fixed by giving any three of its values (at different points)

google for "lagrange interpolation" if you don't know the formula,

i mean, if f(-1), f(0), f(1) are known then you have a linear system of equations on a,b,c

so it rema+ins to prove that if u,v,w are in [-1,1]

(U = f(-1), V = f(0), W = f(1))
then
|V(1-x^2) + U/2 (x^2-x) + W/2 (x^2+x)| <= 3/2

triangle inequality on the LHS

remains to show that |1-x^2| + |x^2-x|/2 + |x^2+x|/2 <= 3/2,

which is easy to prove

waw terrific sol b555 :)

this was the only Q in ISI above jee level i think .

ur method is so tight u r infact proving modfx <= 3/2 - 1/4

hehe yes celestine

overthinking i suppose :(

and 8th q is a well known question. !!!!

i think most of u know how to do it