Let [tex]\displaystyle f(x)= \lim_{n\to \infty}\left( \dfrac{n^n (x+n)\left(x+\dfrac{n}{2}\right) \dots \left(x+\dfrac{n}{n}\right)}{n! (x^2+n^2)\left(x^2+\dfrac{n^2}{4}\right)\dots \left( x^2+\dfrac{n^2}{n^2}\right)}\right)^{\dfrac{x}{n}}[/tex] for all x > 0 , then;

[tex] A ) \ f\bigg(\dfrac{1}{2}\bigg) \geq f(1) \\\\ B)\ f\bigg(\dfrac{1}{3}\bigg) \leq f\bigg(\dfrac{2}{3}\bigg) \\\\ C) \ f'(2)\leq 0 \\\\ D) \ \dfrac{f'3(3)}{f(3)}\geq \dfrac{f'(2)}{f(2)} [/tex]​

The solution is Option C.

The equation f'( 2 ) ≤ 0 is true

What is an Equation?

Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side.

It demonstrates the equality of the relationship between the expressions printed on the left and right sides.

Coefficients, variables, operators, constants, terms, expressions, and the equal to sign are some of the components of an equation. The "=" sign and terms on both sides must always be present when writing an equation.

Given data ,

Let the equation be represented as A

Now , the value of A is

[tex]f ( x ) = \lim_{n \to \infty} [ \frac{( n^{n} ( x + n ) ( x + \frac{n}{2} ) ....( x + \frac{n}{n} )}{n! (x^{2}+n^{2}) ( x^{2} + \frac{n^{2}}{4} ) ....( x^{2} +\frac{n^{2}}{n^{2} } ) } ]^{\frac{x}{n} }[/tex]

Taking logarithm on both sides of the equation , we get

[tex]ln f ( x ) = \lim_{n \to \infty} \frac{x}{n} [\sum _{r=1}^nln\frac{1}{r/n} + \sum _{r=1}^nln\frac{(x + \frac{1}{r/n} )}{(x^{2} + (\frac{1}{r/n})^{2} ) }[/tex]

On simplifying the derivatives of the equation , we get

[tex]ln f ( x ) = x [\int\limits^1_0 {ln ( 1/t )} \, dt + \int\limits^1_0 {ln \frac{( x+1/t )}{(x^{2} + 1/t^{2} ) } } \, dt][/tex]

Now , substituting the limits in the equation , we get

[tex]ln f ( x ) = x [ \int\limits^1_0 {( - ln t+ ln (tx+1)-lnt-ln(t^{2} x^{2}+1)+lnt^{2}] } \, dt[/tex]

Substitute the value of tx = z in the equation , we get

[tex]ln f ( x ) = \int\limits^1_0 {ln\frac{(1+z)}{1+z^{2} } } \, dz[/tex]

Taking exponential on both sides of the equation , we get

[tex]f ( x ) = e^{\int\limits^x_0 {{ln\frac{(1+z)}{1+z^{2} } } \, dz} \, }[/tex]

Differentiating with respect to x , we get

[tex]\frac{f' (x ) }{f ( x ) } = ln\frac{1+x}{1+x^{2} }[/tex]

Now , on further simplification of the equation , we get

f ( x ) > 0 for x > 0

And , f' ( x ) > 0

[tex]ln\frac{1+x}{1+x^{2} } > 0[/tex]

And ,

x > x²

0 < x < 1

Therefore , f ( x ) is increasing ∀ x ∈ ( 0 , 1 )

Hence , the equation f' ( 2 ) ≤ 0 is true

To learn more about equations click :

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Explanation:

[tex] \rm f(x) = \lim \limits_{n \to \infty } \left( \dfrac{ \displaystyle \rm \prod \limits_{r = 1}^n \left(1 + \dfrac{rx}n \right ) }{\displaystyle \rm\prod \limits_{r = 1}^n \left(1 + \bigg(\dfrac{rx}n \bigg)^{2}\right )} \right)^{ \dfrac{x}{n} }[/tex]

[tex] \large \rm = {e}^{ x \int \limits_{0}^{1} ln(1 + xy)dy - ln(1 + (xy {)}^{2} ) dy }[/tex]

[tex]\rm f'(x)=f(x) \ln \bigg( \dfrac{1 + x}{1 + {x}^{2} } \bigg)[/tex]

[tex]\rm For \: x \in(0,1) \: it \: i s \: increasing \: function[/tex]

[tex]\rm f'(2) = f(2) \ln( \frac{3}{5} ) < 0[/tex]

[tex]\rm \dfrac{f'(3)}{f(3)} =ln \bigg( \dfrac{2}{5} \bigg), \dfrac{f'(2)}{f(2)} = ln \bigg( \dfrac{3}5 \bigg) [/tex]

X^2(x+5)
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