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..

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TopNewest@Sandeep Bhardwaj – Anandhu Raj · 1 year, 10 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, 10 months agoLog in to reply

– Anandhu Raj · 1 year, 10 months ago

Thanks sir :)Log in to reply

\(n = 6 \), \(m = 2\) 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, 10 months ago

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– Sandeep Bhardwaj · 1 year, 10 months ago

You're right. I made a mistake by neglecting n=6. Thanks !Log in to reply

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– Anandhu Raj · 1 year, 10 months ago

I got this question from an old question paper. Going by the key code the answer is 4(but no steps given)Log in to reply

– Sandeep Bhardwaj · 1 year, 10 months ago

Check I've posted a brief solution too.Log in to reply