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If the \({ r }^{ th }\) term in the expansion of \({ \left( 2x+\frac { 1 }{ { x }^{ m } } \right) }^{ n },(n\epsilon N)\) is independent term of \(x\) and \({ r }^{ th }\) term in expansion of \({ \left( \frac { 1 }{ { x }^{ 4 } } -2{ x }^{ n-6 } \right) }^{ m },(m\epsilon N)\) is independent term of \(x\), then minimum possible value of \(\left[ \frac { { m }^{ 2 }+{ n }^{ 2 } }{ 9 } \right] \) is? (where \(\left[ . \right]\) represents greatest integer function)

What I have done is write down the general term in both expansions and equated the powers of \(x\) to zero in both cases and tried to connect them...But it goes nowhere..

Note by Anandhu Raj
1 year, 7 months ago

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@Sandeep Bhardwaj Anandhu Raj · 1 year, 7 months ago

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When you try to connect, you got two equations in \(m,n \ and \ r\) then substitute the value of \(r\) in terms of \(m,n\) from one equation in the second equation and simplify.

On simplifying you will get :

\(4m^2+4m=n^2-2n\)

\(4m^2-n^2=-2(2m+n)\)

\((2m+n)(2m-n)=-2(2m+n)\)

So, either \(2m+n=0\) or \(2m-n=-2\).

The first one is not possible because \(m,n \in \mathbb{N}\) .

So \(2m-n=-2\)

\(2m=n-2\).

Now, we'll focus on the value of \(n\) :

From the second equation, it is sure that \(n \geq 6\).

If \(n=6 , \implies m=2 \) (Min. Permissible values)

Hence the answer comes to be \(4 \).

edited !

Hope you'll enjoy it ! Sandeep Bhardwaj · 1 year, 7 months ago

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@Sandeep Bhardwaj Thanks sir :) Anandhu Raj · 1 year, 7 months ago

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@Sandeep Bhardwaj Why \(n > 6? \)

\(n = 6 \), \(m = 2\) also satisfies the condition And also gives the minimum value of 4.

Clarification:- check the last term of second equation with n= 6( power of \(1/x^4\) equals 0). Krishna Sharma · 1 year, 7 months ago

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@Krishna Sharma You're right. I made a mistake by neglecting n=6. Thanks ! Sandeep Bhardwaj · 1 year, 7 months ago

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Comment deleted May 31, 2015

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@Sandeep Bhardwaj I got this question from an old question paper. Going by the key code the answer is 4(but no steps given) Anandhu Raj · 1 year, 7 months ago

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@Anandhu Raj Check I've posted a brief solution too. Sandeep Bhardwaj · 1 year, 7 months ago

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