JEE MAIN - Mathematics (2024 - 30th January Evening Shift - No. 1)
Let $$f: \mathbb{R} \rightarrow \mathbb{R}$$ be a function defined by $$f(x)=\frac{x}{\left(1+x^4\right)^{1 / 4}}$$, and $$g(x)=f(f(f(f(x))))$$. Then, $$18 \int_0^{\sqrt{2 \sqrt{5}}} x^2 g(x) d x$$ is equal to
36
33
39
42
Explanation
$$\begin{aligned} &f(x)=\frac{x}{\left(1+x^4\right)^{1 / 4}}\\ &f \circ f(x)=\frac{f(x)}{\left(1+f(x)^4\right)^{1 / 4}}=\frac{\frac{x}{\left(1+x^4\right)^{1 / 4}}}{\left(1+\frac{x^4}{1+x^4}\right)^{1 / 4}}=\frac{x}{\left(1+2 x^4\right)^{1 / 4}}\\ &f(f(f(f(x))))=\frac{x}{\left(1+4 x^4\right)^{1 / 4}} \end{aligned}$$
$$18 \int_\limits0^{\sqrt{2 \sqrt{5}}} \frac{x^3}{\left(1+4 x^4\right)^{1 / 4}} d x$$
$$\begin{aligned} & \text { Let } 1+4 \mathrm{x}^4=\mathrm{t}^4 \\ & 16 \mathrm{x}^3 \mathrm{dx}=4 \mathrm{t}^3 \mathrm{dt} \\ & \frac{18}{4} \int_\limits1^3 \frac{\mathrm{t}^3 \mathrm{dt}}{\mathrm{t}} \\ & =\frac{9}{2}\left(\frac{\mathrm{t}^3}{3}\right)_1^3 \\ & =\frac{3}{2}[26]=39 \end{aligned}$$
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