JEE 2023 — Mathematics PYQ

Let $f(x)=\frac{x}{{\left[1+(x^n)\right]}^{1/n}}$, $x\in \mathbb{P}-\{-1\}$, n \in \mathbb{N}, n > 2

If $f^n(x)=(n{f}_0{f}_0\dots \text{upto }n\text{ times})$ $(x)$

${\lim }_{n\to \infty }{\int }_0^1{x}^{n-2}{f}^n(x)dx=\frac{1}{(1+2x^n{)}^{\frac{1}{n}}}$

$f(f(x))=\frac{x}{(1+3x^n{)}^{\frac{1}{n}}}$

Similarly,

$f^{\prime \prime }(x)=\frac{x}{(1+n{x}^n{)}^{\frac{1}{n}}}$

${\lim }_{n\to \infty }\int \frac{x^{n-2}dx}{(1+n{x}^n{)}^{\frac{1}{n}}}={\lim }_{n\to \infty }\int \frac{x^{n-1}dx}{(1+n{x}^n{)}^{\frac{1}{n}}}$

Now $1+n{x}^n=t$

$x^{n-1}=\frac{dt}{n^2}$

$\Rightarrow {\lim }_{n\to \infty }\frac{1}{n^2}{\left[\frac{t}{1-\frac{1}{n}}\right]}^{1+n}={\lim }_{n\to \infty }\frac{1}{n^2}\left[\frac{t}{t-\frac{1}{n}}\right]$

Let $n=\frac{1}{n}$

${\lim }_{n\to 0}{\left(\frac{1+\frac{1}{n}}{n}\right)}^{-n}-1=0$

$={\lim }_{n\to 0}{\left(\frac{1}{n}+\frac{1}{n^2}\right)}^{-n}-1=0$