Mathematics - Functions Question with Solution | TestHub

Given,

$f\left(x\right)=\frac{x}{{\left(1+x^4\right)}^{1/4}}$

Now, finding

$fof\left(x\right)=\frac{f\left(x\right)}{{\left(1+f^4\left(x\right)\right)}^{1/4}}=\frac{\frac{x}{{\left(1+x^4\right)}^{1/4}}}{{\left(1+\frac{x^4}{1+x^4}\right)}^{1/4}}=\frac{x}{{\left(1+2x^4\right)}^{1/4}}$

And $f\left(f\left(f\left(x\right)\right)\right)=\frac{f\left(x\right)}{{\left(1+2f^4\left(x\right)\right)}^{1/4}}=\frac{\frac{x}{{\left(1+x^4\right)}^{1/4}}}{{\left(1+\frac{2x^4}{1+x^4}\right)}^{1/4}}=\frac{x}{{\left(1+3x^4\right)}^{1/4}}$

$\Rightarrow f\left(f\left(f\left(f\left(x\right)\right)\right)\right)=\frac{x}{{\left(1+4x^4\right)}^{1/4}}=g\left(x\right)$

Now finding,

$18{\int }_0^{\sqrt{2\sqrt{5}}}{x}^{2}g\left(x\right)dx$

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

Let $1+4x^4=t^4$

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

So, the integral becomes

$\frac{18}{4}{\int }_1^3\frac{t^3\mathrm{dt}}{\mathrm{t}}$

$=\frac{9}{2}{\int }_1^3{t}^2\mathrm{dt}$

$=\frac{9}{2}{\left(\frac{t^3}{3}\right)}_1^3$

$=\frac{3}{2}·26=39$