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

celestine it is 30. have a careful look.

can u explain it b555 ??

i had a careful look cudnt figure out :(

thanks b555

@ q 5)... the same q is in MLkhana and the ans was 1...

the solved it as: 6!/6! (since the sides of a cube are similar!!)

whom to trust??

"may day!! help!!"

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

2nd one is easy. u have 3 cnditions so use legrange interpolation to find the polynomial,. then at later stage u would have to solve triangle inequality. done

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

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

waiting to see if anyone got a better soln

Q 10

wats the meaning of multiplicity ????

yup thats the way cele. good work.

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

for Q8

if a_n is nth term its easy to see

a_n = n +[n/10] +[n/100] +.....[n/10^r]..

see #48 , #49 for part 2 ans :)

did u give the ISI b555 ?

and can u solve the Q8 also im not getting the Q at all see #44

no cele , i did not give ISI. actually was thinking my math wasnt of that level.

hehe yes celestine

overthinking i suppose :(

@Mirka
[10][10][10]
thank u for all the ques
ab thoda kaam karte hain [1]

Q1
look according to the question asked the min value of x²+y² should be 0 as we can take (0,0)point
so the answer is 0

but respecting the question and understanding the mistake
it should be mentioned that the point lies on the curves

so

we can see that it cuts the coordinate axis at (0,-2) and (9,0)

we observe that the distance of (0,0) from (-5,12) is 13 so the line continuing shall have the distance of 14 at a distance of 1 from (0,0)

Q5 directly from TMH

MIB is looking easier than SIB...:(

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

i think most of u know how to do it