Question

➟ Don't spam
➟ Explaination needed​

Answer 1

$\large\underline{\sf{Solution-}}$

$\sf \: f(x) = \tt \:\lim_{n\to \infty } {\bigg(\dfrac{ {n}^{n}(x + n)\bigg(x + \dfrac{n}{2} \bigg) - - \bigg(x + \dfrac{n}{n} \bigg)}{n!( {x}^{2} + {n}^{2})\bigg( {x}^{2}+\dfrac{ {n}^{2} }{4} \bigg) - - \bigg( {x}^{2}+\dfrac{ {n}^{2} }{ {n}^{2} } \bigg)} \bigg) }^{\dfrac{x}{n} }$

$\sf = \tt \:\lim_{n\to \infty } {\bigg(\dfrac{ {n}^{n}n(1 + \frac{x}{n} ) \frac{2}{n} \bigg(1 + \dfrac{2x}{n} \bigg) - - \frac{n}{n}\bigg(1 + \dfrac{nx}{n} \bigg)}{n! {n}^{2} ( \frac{ {x}^{2} }{ {n}^{2} } +1) \frac{4}{ {n}^{2} } \bigg(1+\dfrac{ {4x}^{2} }{ {n}^{2} } \bigg) - - \frac{ {n}^{2} }{ {n}^{2} } \bigg(1+\dfrac{ {n}^{2} {x}^{2} }{ {n}^{2} } \bigg)} \bigg) }^{\dfrac{x}{n} }$

$\sf = \tt \:\lim_{n\to \infty } {\bigg(\dfrac{(1 + \dfrac{x}{n} ) \bigg(1 + \dfrac{2x}{n} \bigg) - - \bigg(1 + \dfrac{nx}{n} \bigg)}{( \dfrac{ {x}^{2} }{ {n}^{2} } +1) \bigg(1+\dfrac{ {4x}^{2} }{ {n}^{2} } \bigg) - - \bigg(1+\dfrac{ {n}^{2} {x}^{2} }{ {n}^{2} } \bigg)} \bigg) }^{\dfrac{x}{n} }$

$\rm \: = \displaystyle \: \tt \:\lim_{n\to \: \infty } \prod_{r=1}^n {\bigg(\dfrac{1 + \dfrac{rx}{n} }{1 + \dfrac{ {r}^{2} {x}^{2} }{ {n}^{2} } } \bigg) }^{ \dfrac{x}{n} }$

Now, taking log on both sides, we get

$\rm \:log \: f(x) =log \bigg \{ \displaystyle \: \tt \:\lim_{n\to \: \infty } \prod_{r=1}^n {\bigg(\dfrac{1 + \dfrac{rx}{n} }{1 + \dfrac{ {r}^{2} {x}^{2} }{ {n}^{2} } } \bigg) }^{ \dfrac{x}{n} } \bigg \}$

$\rm \:log \: f(x) = \bigg \{ \displaystyle \: \tt \:\lim_{n\to \: \infty } \dfrac{x}{n} \sum_{r=1}^n log{\bigg(\dfrac{1 + \dfrac{rx}{n} }{1 + \dfrac{ {r}^{2} {x}^{2} }{ {n}^{2} } } \bigg) }\bigg \}$

Now, using limit as a sum, we get

$\rm :\longmapsto\:log \: f(x) = x \displaystyle \int_{0}^1 \: log\bigg( \dfrac{1 + xy}{1 + {x}^{2} {y}^{2} } \bigg)dy$

To evaluate this integral, we use method of Substitution

$\rm :\longmapsto\:Put \: xy = t$

$\rm :\longmapsto\:x\dfrac{dy}{dt} = 1$

$\rm :\longmapsto\:xdy = dt$

Now, we have to change the limits too

when y = 0, t = 0

and

when y = 1, t = x

So above integral can be reduced to

$\rm :\longmapsto\:log \: f(x) = \displaystyle \int_{0}^x \: log\bigg( \dfrac{1 + t}{1 + {t}^{2}} \bigg)dt$

Now, Differentiate using Leibniz Rule, we get

$\rm :\longmapsto\:\dfrac{f'(x)}{f(x)} = log\bigg( \dfrac{1 + x}{1 + {x}^{2} } \bigg)$

$\rm :\longmapsto\:f'(x) =f(x) \: log\bigg( \dfrac{1 + x}{1 + {x}^{2} } \bigg)$

As x > 0, it implies f(x) > 0.

Now at x = 2

$\rm :\longmapsto\:f'(2) =f(2) \: log\bigg( \dfrac{1 + 2}{1 + {2}^{2} } \bigg)$

$\rm :\longmapsto\:f'(2) =f(2) \: log\bigg( \dfrac{3}{5 } \bigg)$

$\bf\implies \:f'(2) < 0 \: \: \: \: as \: log \dfrac{3}{5} < 0$

It implies f(x) is an increasing function.

$\bf\implies \:f\bigg( \dfrac{1}{3} \bigg) \leqslant f\bigg(\dfrac{2}{3} \bigg)$

Hence,

Option (b) is correct.