JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Definite Integrals

Write $f(x)=\frac{x}{(1+x^4{)}^{1/4}}$. For any $k$, define $h_k(x)=\frac{x}{(1+k{x}^4{)}^{1/4}}$; note $f=h_1$.

Compute $f(h_k(x))$: since $h_k(x{)}^4=\frac{x^4}{1+k{x}^4}$,
$1+h_k(x{)}^4=\frac{1+(k+1)x^4}{1+k{x}^4}$,
so
$f(h_k(x))=\frac{h_k(x)}{{\left(1+h_k(x{)}^4\right)}^{1/4}}=\frac{x}{(1+k{x}^4{)}^{1/4}}\cdot \frac{(1+k{x}^4{)}^{1/4}}{(1+(k+1)x^4{)}^{1/4}}=\frac{x}{(1+(k+1)x^4{)}^{1/4}}=h_{k+1}(x)$.

So each application of $f$ raises the coefficient by 1: $f(x)=h_1(x)$, $f(f(x))=h_2(x)$, and continuing, $g(x)=f(f(f(f(x))))=h_4(x)=\frac{x}{(1+4x^4{)}^{1/4}}$.

Now
$18{\int }_0^{\sqrt{2\sqrt{5}}}{x}^{2}g(x)dx=18{\int }_0^{\sqrt{2\sqrt{5}}}\frac{x^3}{(1+4x^4{)}^{1/4}}dx$.

Substitute $t^4=1+4x^4$, so $4t^{3}dt=16x^{3}dx$, i.e. $x^{3}dx=\frac{t^3}{4}dt$. At $x=0$, $t=1$; since ${\left(\sqrt{2\sqrt{5}}\right)}^4=(2\sqrt{5}{)}^2=20$, at the upper limit $t^4=81\Rightarrow t=3$.

The integral becomes
$18{\int }_1^3\frac{t^3/4}{t}dt=\frac{18}{4}{\int }_1^3{t}^{2}dt=\frac{9}{2}{\left[\frac{t^3}{3}\right]}_1^3=\frac{9}{2}\cdot \frac{26}{3}=39$.