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Algebra

Blazing goIITian

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24 Oct 2008 11:42:17 IST

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2615

Good Questions only for RMO candidates :D

None

1) Prove that the roots of x^5+ax^4+bx^3+cx^2+dx+e=0 cannot be all real if 2a^2<5b

2) Let be a root of x^5-1=0 with $z\ne 1$ Compute the value of $z^{15}+z^{16}+z^{17}+\cdots\cdots+z^{50}$

(Seems like a Wrong Question3) The sum [edited] $\frac{1}{1!9!}+\frac 1{3!7!}+\frac 1{5!5!}+\frac 1{7!5!}+\frac 1{7!3!}$ can be written in the form $\frac {2^a}{b!}$ where a and b are positive integers. Find the ordered pair (a,b) )

4) Prove that if the polynomial $P(x)=a_0x^n+a_1x^{n-1}+\cdots\cdots+a_{n-1}x+a_n$ with integral

coefficients has odd value for $x=0\ and\ x=1$ , then the equation p(x)=0 can't have integral roots.

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Lakshya Bhardwaj

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25 Oct 2008 18:16:50 IST

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Re:Good Questions only for RMO candidates :D

Rajat Sen

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25 Oct 2008 19:39:02 IST

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let S = $\frac{1}{1!.9!}+\frac{1}{3!.7!}+\frac{1}{5!.5!}+\frac{1}{7!.3!}+\frac{1}{9!.1!}$

so, S = $\frac{1}{10!}\left(^{10}C_{9}+^{10}C_{7}+^{10}C_{5}+^{10}C_{3}+^{10}C_{1}\right)$

$^{10}C_{9}+^{10}C_{7}+^{10}C_{5}+^{10}C_{3}+^{10}C_{1}=\frac{2^{10}}{2} = 2^{9}$

so, S = $\frac{2^{9}}{10!}$

rate if you like it! Rolling Eyes

Decoder

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25 Oct 2008 20:12:06 IST

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well rajat...we r jsut pressing on tht thing only for two days. . :D:D

abhishek sinha

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26 Oct 2008 18:06:46 IST

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So the answers are differeing !!

Anyone knows why my method is not giving the right answer ?

pardesi .svk

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26 Oct 2008 20:41:15 IST

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@hamba

wrong notion u need to correct that soon .Diff is maths, anything that is maths is 'allowed' in math oly.

best of luck

Hari Shankar

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27 Oct 2008 09:37:44 IST

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@feynmann: coefficient of x¹⁰ on LHS is $\frac{1}{4} \left(\frac{2^{10}}{10!} + \frac{2^{10}}{10!} \right) = \frac{2^9}{10!}$

Shreya

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27 Oct 2008 10:58:17 IST

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feyman, can u plz give me the proof of
Let , the roots be all reals = q1,q2 ....,q5

then we must have q1^2 +... +q5^2 >=0

or , (q1 +q2 +... +q5)^2 - 2( q1q2 +... +q4q5 )>0

Shreya

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27 Oct 2008 11:09:48 IST

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also, this 1/4 ( e^2x + e^-(2x ) -2 ) = ( x+ x^3/3! + ... ) ( x+ x^3/3! +... )

Now equate the coeff of x^10 from both sides

1/4( 2^5/5! + 2^5/5! ) = 1/1!9! + 1/3!7! +... +1/7!3!

so we get the reqd expression ( RHS ) = 2^4/5!
,my doubt is , how could u equate the coff of x^10 in lhs when it is in terms of e

abhishek sinha

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27 Oct 2008 20:14:22 IST

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So silly of me !!

I can't believe of making such silly mistakes and not even finding it out after several trials !!!!!!!!!!!

Think need to take a V.R. from the forum :( :(

Thanks mate !!

@ Shinee ,

I don't get your doubt for the first one . The steps are so easy ........

For the second one , I have just used series expansion of e^2x and e^(-2x ) , now equated the coefficients .

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