@Harsh Shrivastava
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Yes and if the paper is like that we get more scared and I realised I could have got 5-6 marks more had I been not depressed after looking at the paper. I lost all my confidence after looking at the paper.

@Svatejas Shivakumar@Sharky Kesa Can either of you help me verify the phrasing of Q4? Is it "exactly one common language" or "at least one common language"?

It makes a huge difference in appoaching the problem.

WLOG \(a>b>c\) Multiplying both sides by \((a-b)(b-c)(c-a)\), it suffices to prove \(\sum a^3(b-c) \ge 3(a-b)(b-c)(c-a)\).

Expanding both sides and simplifying , this reduces to prove \(\sum (a^3b+3a^2b) \ge \sum( ab^3+3ab^2)\). Using \(abc=1\) this simplifies to show
\[\sum \frac{a^2+3a}{c} \ge \sum \frac{a^2+3a}{b}\]

But this is just rearrangement on \[a^3+3a>b^3+3b>c^3+3c\] and \[1/a < 1/b < 1/c\]

For q2 since a,b,c are distinct WLOG a>b>c.Take a=c+y,b=c+x,and replace a,b with the above values in lhs and on solving you get finally lhs as 3c+x+y which is nothing but a+b+c,now apply am-gm,to prove it.

Apply am -gm for these
a^3/3(a-b)(a-c)+(a-b)/3^(5/2)+(a-c)/3^(5/2)
Which is>=a/3
Now write same expressions for b and c and add all three
We get in rhs-
a+b+c/3>=1(am-gm)
Hence proved :)

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TopNewestsolutions are now uploaded on resonance.

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Have you come to know your result?

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What is your rmo score?

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I registered through FIITJEE so I will receive my marks when I have my next class(tomorrow).

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What is yours? Mine is very bad.

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It was just 5 minutes after the exam was over LOL

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I Did the first one now in 10 min

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It's easy.Which class for you? And did you write RMO?

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Yep, That was easy try this.

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Solution 1.link text

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How many of you all are selected.

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I am selected from Chattisgarh region.

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How much scores pals.

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51

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Yours ???

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Yours?

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@Svatejas Shivakumar @Sharky Kesa Can either of you help me verify the phrasing of Q4? Is it "exactly one common language" or "at least one common language"?

It makes a huge difference in appoaching the problem.

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It is atleast one common language.

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Thanks. I've edited the question accordingly.

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Anyone with 6 th done

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Both of \(6\)'s subdivions are available in Brilliant.

Grasp here and here

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20 is the answer for the 4th one

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Is the following solution correct for P2?

WLOG \(a>b>c\) Multiplying both sides by \((a-b)(b-c)(c-a)\), it suffices to prove \(\sum a^3(b-c) \ge 3(a-b)(b-c)(c-a)\).

Expanding both sides and simplifying , this reduces to prove \(\sum (a^3b+3a^2b) \ge \sum( ab^3+3ab^2)\). Using \(abc=1\) this simplifies to show \[\sum \frac{a^2+3a}{c} \ge \sum \frac{a^2+3a}{b}\]

But this is just rearrangement on \[a^3+3a>b^3+3b>c^3+3c\] and \[1/a < 1/b < 1/c\]

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\[ \sum_cyc (\frac{a^3}{(a-b)(a-c)} = \sum_cyc a = a + b + c \geq 3 \text{ (by AM-GM)} \]

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what is the answer of the 4 th one

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The same question paper was asked in jharkand . Being in eleventh is solving three correctly , enough to qualify ?

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Has anyone perfectly solved Q.4??Log in to reply

Comment deleted 1 year ago

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The Farey sequence relates to question 3.

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For q2 since a,b,c are distinct WLOG a>b>c.Take a=c+y,b=c+x,and replace a,b with the above values in lhs and on solving you get finally lhs as 3c+x+y which is nothing but a+b+c,now apply am-gm,to prove it.

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Terms are [a^3/{3(a-b)(a-c)}]+a-b/(3^{5/2})+a-c/(3^{5/2})>=a/3

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Better use the formatting guide to write the expression.\(\frac{a^{3}}{3(a-b)(a-c)}+\frac{a-b}{3^{5/2}}+\frac{a-c}{3^{5/2}} \geq \frac{a}{3}\)

That looks better and much understandable too!Log in to reply

*a-b is also in bracket.sane with a-c

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Terms can be negative but they are eventually cancelled so we can apply am-gm

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@Svatejas Shivakumar Could you post the solution to problem 6?

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I am from jharkhand.I have solved q1,2,5,6.solution for q4 please

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You are in which class and did you got dps ranchi as exam centre ?

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I am from Jamshedpur,in class 11

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Same paper was for chhattisgarh also!

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Apply am -gm for these a^3/3(a-b)(a-c)+(a-b)/3^(5/2)+(a-c)/3^(5/2) Which is>=a/3 Now write same expressions for b and c and add all three We get in rhs- a+b+c/3>=1(am-gm) Hence proved :)

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You cannot apply AM GM Its not given that a-b,b-c,c-a are +ve

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Comment deleted Oct 16, 2016

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BTW,Rank In Technothlon?

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can u put brackets please

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you can solve the second one by using am gm inequality

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Can you post the solution? Also when does equality occur?

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Which questions did u do?

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Really!! Though it was very easy but don't think it could be an NCERT question.Log in to reply

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Both went good

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It is better to evaluate it to be a+b+c

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It is better if u factorise given expression It is a+b+c

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All these problems except 3rd one is posted in this brilliant with solutions. Please refer the medium rated problems.

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