Functions Question 11
Question 11 - 2024 (30 Jan Shift 2)
Let $f: R \rightarrow R$ be a function defined $f(x)=\frac{x}{\left(1+x^{4}\right)^{1 / 4}}$ and $g(x)=f(f(f(f(x))))$ then $18 \int_{0}^{\sqrt{2 \sqrt{5}}} x^{2} g(x) d x$
(1) 33
(2) 36
(3) 42
(4) 39
Show Answer
Answer (4)
Solution
$f(x)=\frac{x}{\left(1+x^{4}\right)^{1 / 4}}$
$f \circ f(x)=\frac{f(x)}{\left(1+f(x)^{4}\right)^{1 / 4}}=\frac{\frac{x}{\left(1+x^{4}\right)^{1 / 4}}}{\left(1+\frac{x^{4}}{1+x^{4}}\right)^{1 / 4}}=\frac{x}{\left(1+2 x^{4}\right)^{1 / 4}}$
$f(f(f(f(x))))=\frac{x}{\left(1+4 x^{4}\right)^{1 / 4}}$
$18 \int_{0}^{\sqrt{2 \sqrt{5}}} \frac{x^{3}}{\left(1+4 x^{4}\right)^{1 / 4}} d x$
Let $1+4 \mathrm{x}^{4}=\mathrm{t}^{4}$
$16 \mathrm{x}^{3} \mathrm{~d} \mathrm{x}=4 \mathrm{t}^{3} \mathrm{dt}$
$\frac{18}{4} \int_{1}^{3} \frac{\mathrm{t}^{3} \mathrm{dt}}{\mathrm{t}}$
$=\frac{9}{2}\left(\frac{t^{3}}{3}\right)_{1}^{3}$
$=\frac{3}{2}[26]=39$
