If the value of

\(\displaystyle f(k) = \int_0^{\infty} \frac{x^{k}}{2x^{6} + 4x^{5} + 3x^{4} + 5x^{3} + 3x^{2} + 4x + 2} dx \)

Is minimum

Then find

\(101k\), (k is an integer)

If the value of

\(\displaystyle f(k) = \int_0^{\infty} \frac{x^{k}}{2x^{6} + 4x^{5} + 3x^{4} + 5x^{3} + 3x^{2} + 4x + 2} dx \)

Is minimum

Then find

\(101k\), (k is an integer)

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TopNewestI don't know whether I am right or not but I am getting \(k=2\), here's a solution.

Differentiating both sides with respect to k we get :

\(\large { f }^{ ' }(k)=\int _{ 0 }^{ \infty }{ \frac { ln(x){ .x }^{ k } }{ (2x^{ 6 }+4x^{ 5 }+3x^{ 4 }+5x^{ 3 }+3x^{ 2 }+4x+2) } dx }=I \)

I differentiated here using newton leibnitz rule.

Splitting the integral into two parts we get :

\(\large I={ f }^{ ' }(k)=\int _{ 0 }^{ 1 }{ \frac { ln(x){ .x }^{ k } }{ (2x^{ 6 }+4x^{ 5 }+3x^{ 4 }+5x^{ 3 }+3x^{ 2 }+4x+2) } dx } +\int _{ 1 }^{ \infty }{ \frac { ln(x){ .x }^{ k } }{ (2x^{ 6 }+4x^{ 5 }+3x^{ 4 }+5x^{ 3 }+3x^{ 2 }+4x+2) } dx }\)

In the second part put \(x=\frac{1}{y}\) to get :

\(\large { f }^{ ' }(k)=\int _{ 0 }^{ 1 }{ \frac { ln(x){ .x }^{ k } }{ (2x^{ 6 }+4x^{ 5 }+3x^{ 4 }+5x^{ 3 }+3x^{ 2 }+4x+2) } dx } +\int _{ 0 }^{ 1 }{ \frac { -ln(y).y^{ 4-k } }{ (2y^{ 6 }+4y^{ 5 }+3y^{ 4 }+5y^{ 3 }+3y^{ 2 }+4y+2) } dy } \)

Combining both integrals we get :

\(\large { f }^{ ' }(k)=\int _{ 0 }^{ 1 }{ \frac { ln(x){ .(x }^{ k }-{ x }^{ 4-k }) }{ (2x^{ 6 }+4x^{ 5 }+3x^{ 4 }+5x^{ 3 }+3x^{ 2 }+4x+2) } dx }\)

Equating the derivative equal to \(0\) we get :

\(k=4-k\)

\(\boxed{\Rightarrow k=2}\)

I won't do the second derivative test as I know this has the minimum possible value. – Ronak Agarwal · 2 years ago

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Is The answer is 202 means K=2 ??

( I have feeling That it might not be correct)

Since You Didn't Specify That 'K' is any real number or an integer So I have Assumed That

K is integer.. If it is real then please mention it.!! – Deepanshu Gupta · 2 years ago

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The answer is 2 – Krishna Sharma · 2 years ago

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\(f(k)=\int _{ 0 }^{ \infty }{ \frac { { x }^{ k }+{ \frac { { x }^{ 4 } }{ { x }^{ k } } } }{ (2x^{ 6 }+4x^{ 5 }+3x^{ 4 }+5x^{ 3 }+3x^{ 2 }+4x+2) } dx } \).

Since Denominator is an \(Palindrome\) Then it must be Products of Perfect squares. So Denominator is always Positive.

And To Minimize integral then We have To minimize The numerator. (

For a Particular\(x\))So Apply

\(AM\quad =\quad GM\). in numerator Keeping \(x\) constant .

Or

Also thinks that in Numerator as 'k' increases First term is Increases and Second term is decreases

So Minimum Value of Numerator occurs when \(first\) \(term\)=\(Second\) \(Term\). in Numerator

Or

You can also Prove it by using

calculusThat numerator is minimum when \(K=2\). Hence The Integral also minimum.This Gives \(k=2\).

But I'am not sure that this method is correct or not. – Deepanshu Gupta · 2 years ago

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– Megh Choksi · 1 year, 10 months ago

Very beautiful approach , really very niceLog in to reply

– Deepanshu Gupta · 1 year, 10 months ago

thank you [ feeling blessed :) ]Log in to reply

The answer is 2. – Krishna Sharma · 2 years ago

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@Michael Mendrin – Krishna Sharma · 2 years ago

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@Calvin Lin @Daniel Liu @Pi Han Goh

Or anyone with good integration skills plz post a solution – Krishna Sharma · 2 years ago

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