JEE Main 2023 — Mathematics Calculus

Given,

$f(x)=\frac{x}{{(1+{x}^{n})}^{\frac{1}{n}}},x\in \mathbb{R}-{-1},n\in \mathbb{N},n>2$f(x)=(1+xn)n1​x​,x∈R−−1,n∈N,n>2,

${f}^{n}(x)=(fofof....$fn(x)=(fofof.... upto $n$n times) $(x)$(x),

Now finding $f(f(x))=\frac{f(x)}{{[1+{(f(x))}^{n}]}^{\frac{1}{n}}}$f(f(x))=[1+(f(x))n]n1​f(x)​

$$\Rightarrow f(f(x))=\frac{\frac{x}{{(1+{x}^{n})}^{\frac{1}{n}}}}{{[1+{(\frac{x}{{(1+{x}^{n})}^{\frac{1}{n}}})}^{n}]}^{\frac{1}{n}}}$$⇒f(f(x))=[1+((1+xn)n1​x​)n]n1​(1+xn)n1​x​​

$$\Rightarrow f(f(x))=\frac{x}{{[1+2{x}^{n}]}^{\frac{1}{n}}}$$⇒f(f(x))=[1+2xn]n1​x​

Similarly, ${f}^{n}(x)=\frac{x}{{(1+n{x}^{n})}^{\frac{1}{n}}}$fn(x)=(1+nxn)n1​x​

Now solving the integral we get,

$$I={\int }_{0}^{1}\frac{{x}^{n-1}}{{(1+n{x}^{n})}^{\frac{1}{n}}}dx$$I=∫01​(1+nxn)n1​xn−1​dx

Now let, $1+n{x}^{n}={t}^{n}$1+nxn=tn

$$\Rightarrow {n}^{2}{x}^{n-1}dx=n{t}^{n-1}dt$$⇒n2xn−1dx=ntn−1dt

$$I={\int }_{1}^{{(1+n)}^{\frac{1}{n}}}\frac{\frac{1}{n}{t}^{n-1}dt}{t}$$I=∫1(1+n)n1​​tn1​tn−1dt​

$$\Rightarrow I={[\frac{1}{n}\frac{{t}^{n-1}}{n-1}]}_{1}^{{(1+n)}^{\frac{1}{n}}}$$⇒I=[n1​n−1tn−1​]1(1+n)n1​​

$$\Rightarrow I=\frac{1}{n(n-1)}({(1+n)}^{1-\frac{1}{n}}-1)$$⇒I=n(n−1)1​((1+n)1−n1​−1)

Now solving the limit we get,

$$\underset{n\rightarrow \infty }{\mathrm{lim}}\frac{{(1+n)}^{1-\frac{1}{n}}-1}{n(n-1)}$$n→∞lim​n(n−1)(1+n)1−n1​−1​

Now let $n=\frac{1}{h}$n=h1​ so the limit changes to,

$$\underset{h\rightarrow 0}{\mathrm{lim}}\frac{{(1+\frac{1}{h})}^{1-h}-1}{\frac{1}{h}(\frac{1}{h}-1)}$$h→0lim​h1​(h1​−1)(1+h1​)1−h−1​

$$=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{(1+\frac{1}{h}(1-h))-1}{\frac{1}{h}(\frac{1}{h}-1)}$$=h→0lim​h1​(h1​−1)(1+h1​(1−h))−1​

$$=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{(1-h)}{(\frac{1}{h}-1)}$$=h→0lim​(h1​−1)(1−h)​

$$=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{h(1-h)}{(1-h)}$$=h→0lim​(1−h)h(1−h)​

$$=\underset{h\rightarrow 0}{\mathrm{lim}}h=0$$=h→0lim​h=0