JEE Main 2023 — Definite Integration Question with Solution

$$\begin{array}{rl}& \text{ We have, }f(x)=\frac{x}{{\left(1+x^n\right)}^{1/n}}\\ & \\ & \therefore f(f(x))=\frac{f(x)}{(1+{\left[f(x{)}^n\right]}^{1/n}}\\ & \\ & =\frac{\frac{x}{{\left(1+x^n\right)}^{1/n}}}{{\left(1+\frac{x^n}{1+x^n}\right)}^{1/n}}\\ & \\ & =\frac{x}{{\left(1+2x^n\right)}^{1/n}}\\ & \\ & f(f(f(x)))=\frac{x}{{\left(1+3x^n\right)}^{1/n}}\\ & \\ & \text{ Similarly, }{f}^n(x)=\frac{x}{{\left(1+n{x}^n\right)}^{1/n}}\end{array}$$

Now, $\underset{n\to \infty }{\lim }{\int }_0^1{x}^{n-2}\left(f^n(x)\right)dx$

$$\begin{array}{rl}& =\underset{n\to \infty }{\lim }\underset{0}{\overset{1}{\int }}{x}^{n-2}\frac{x}{{\left(1+n{x}^n\right)}^{1/n}}dx\\ & \\ & =\underset{n\to \infty }{\lim }\underset{0}{\overset{1}{\int }}\frac{x^{n-1}}{{\left(1+n{x}^n\right)}^{1/n}}dx\end{array}$$

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

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

When, $x=0$, then $t=1$

When, $x=1$, then $t=1+n$

$\begin{aligned} & =\lim _{n \rightarrow \infty} \int_1^{1+n} \frac{d t}{n^2(t)^{1 / n}} =\lim _{n \rightarrow \infty} \frac{1}{n^2} \int_1^{1+n} \frac{d t}{(t)^{1 / n}} \\\\ & =\lim _{n \rightarrow \infty} \frac{1}{n^2}\left(\frac{t^{1-\frac{1}{n}}}{1-\frac{1}{n}}\right)_1^{1+n} \\\\ & =\lim _{n \rightarrow \infty} \frac{1}{n(n-1)}\left[(1+n)^{1-\frac{1}{n}}-1\right]\end{aligned}$

Put $n=\frac{1}{h}$

When, $n\to \infty$, then $h\to 0$

$$\begin{array}{rl}& =\underset{h\to \infty }{\lim }\frac{1}{\frac{1}{h}\left(\frac{1}{h}-1\right)}\left[{\left(1+\frac{1}{h}\right)}^{1-h}-1\right]\\ & \\ & =\underset{h\to \infty }{\lim }\frac{1}{\frac{1}{h}\left(\frac{1}{h}-1\right)}\left[1-(1-h)\left(1+\frac{1}{h}\right)+\dots -1\right]\\ & \\ & =\underset{h\to 0}{\lim }\frac{h^2}{1-h}\left[-(1-h)\left(1+\frac{1}{h}\right)+\dots \right]\\ & \\ & =\underset{h\to 0}{\lim }-h[h+1+\dots ..]=0\end{array}$$