Well I first substitute $x=\frac { 1 }{ t }$. and then using Properties of integration I get :

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

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 calculus That 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
–
Yes this is correct! My teacher also said to substitute $x = \frac {1}{y}$
You should have differentiated the function. $(f'(x) = 0$) would be more correct.

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## Comments

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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.

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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.!!

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Question was integer type and we need to find K...I guess so it is integer. How do got 2?

The answer is 2

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Well I first substitute $x=\frac { 1 }{ t }$. and then using Properties of integration I get :

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

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.

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$x = \frac {1}{y}$ You should have differentiated the function. $(f'(x) = 0$) would be more correct.

Yes this is correct! My teacher also said to substituteThe answer is 2.

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