JEE Main 2024 — Mathematics Algebra

Given,

$$f(x)=\frac{x}{{(1+{x}^{4})}^{1/4}}$$f(x)=(1+x4)1/4x​

Now, finding

$$fof(x)=\frac{f(x)}{{(1+{f}^{4}(x))}^{1/4}}=\frac{\frac{x}{{(1+{x}^{4})}^{1/4}}}{{(1+\frac{{x}^{4}}{1+{x}^{4}})}^{1/4}}=\frac{x}{{(1+2{x}^{4})}^{1/4}}$$fof(x)=(1+f4(x))1/4f(x)​=(1+1+x4x4​)1/4(1+x4)1/4x​​=(1+2x4)1/4x​

And $f(f(f(x)))=\frac{f(x)}{{(1+2{f}^{4}(x))}^{1/4}}=\frac{\frac{x}{{(1+{x}^{4})}^{1/4}}}{{(1+\frac{2{x}^{4}}{1+{x}^{4}})}^{1/4}}=\frac{x}{{(1+3{x}^{4})}^{1/4}}$f(f(f(x)))=(1+2f4(x))1/4f(x)​=(1+1+x42x4​)1/4(1+x4)1/4x​​=(1+3x4)1/4x​

$$\Rightarrow f(f(f(f(x))))=\frac{x}{{(1+4{x}^{4})}^{1/4}}=g(x)$$⇒f(f(f(f(x))))=(1+4x4)1/4x​=g(x)

Now finding,

$$18{\int }_{0}^{\sqrt{2\sqrt{5}}}{x}^{2}g(x)dx$$18∫025​​​x2g(x)dx

$$=18{\int }_{0}^{\sqrt{2\sqrt{5}}}\frac{{x}^{3}}{{(1+4{x}^{4})}^{1/4}}dx$$=18∫025​​​(1+4x4)1/4x3​dx

Let $1+4{x}^{4}={t}^{4}$1+4x4=t4

$$\Rightarrow 16{x}^{3}dx=4{t}^{3}dt$$⇒16x3dx=4t3dt

So, the integral becomes

$$\frac{18}{4}{\int }_{1}^{3}\frac{{t}^{3}\mathrm{dt}}{t}$$418​∫13​tt3dt​

$$=\frac{9}{2}{\int }_{1}^{3}{t}^{2}\mathrm{dt}$$=29​∫13​t2dt

$$=\frac{9}{2}{(\frac{{t}^{3}}{3})}_{1}^{3}$$=29​(3t3​)13​

$$=\frac{3}{2}\cdot 26=39$$=23​⋅26=39