Let f : R → R be a function defined f(x) = x/(1+x⁴)^(1/4) and g(x) = f(f(f(f(x)))) then 18 ∫[0 to √2

Q: Why can't we just expand the function composition algebraically? It seems like a lot of work.

You *could*, but you would be walking into a trap. JEE Advanced problems are designed to test your ability to spot patterns, not your stamina for tedious algebra. If you expand blindly, you will get lost in a sea of exponents and likely make a sign error. Always look for the recursive structure first—it is the hallmark of a true problem solver.

Q: Why did we choose $t^4$ as our substitution instead of just $t$?

Great question! If you choose $t = 1+4x^4$, you are left with a $t^{1/4}$ term, which is still messy to integrate. By choosing $t^4 = 1+4x^4$, the fourth root $(t^4)^{1/4}$ becomes simply $t$. It is a strategic choice to make the algebra collapse elegantly. Always choose a substitution that kills the most complex part of the integrand immediately.

Q: I keep getting the limits wrong. How do I handle the upper limit $\sqrt{2}\sqrt[4]{5}$?

Take it slow. When $x = \sqrt{2}\sqrt[4]{5}$, then $x^4 = (\sqrt{2})^4 \cdot (\sqrt[4]{5})^4 = 4 \cdot 5 = 20$. Then $1+4(20) = 81$. Since $t^4 = 81$, $t=3$. Always simplify the power of the limit separately before plugging it into the substitution equation. Precision is your best friend here.

Question

Let $f : \mathbb{R} 	o \mathbb{R}$ be a function defined $f(x) = \frac{x}{(1+x^4)^{1/4}}$ and $g(x) = f(f(f(f(x))))$ then $18 \int_0^{\sqrt{2}\sqrt[4]{5}} x^2 g(x) dx$ is equal to

Options:

(A) 33

(B) 36

(C) 42

(D) 39

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare of nested functions. You see $f(f(f(f(x))))$ and your instinct might be to panic. But in the world of JEE Advanced, panic is the enemy. Pattern recognition is your greatest weapon.

We are given the function:
$$f(x) = \frac{x}{(1+x^4)^{1/4}}$$

Our goal is to evaluate $18 \int_0^{\sqrt{2}\sqrt[4]{5}} x^2 g(x) \, dx$, where $g(x)$ is the four-fold composition of $f$. The secret to this problem is not brute force; it is observation. Let us calculate $f(f(x))$ and see if a structure emerges.

When we substitute $f(x)$ into itself, we get:
$$f(f(x)) = \frac{f(x)}{(1 + (f(x))^4)^{1/4}}$$

Now, look closely at the denominator. When we raise $f(x)$ to the power of $4$, the fourth root vanishes:
$$(f(x))^4 = \frac{x^4}{1+x^4}$$

Adding the $1$ inside the denominator's root gives us:
$$1 + \frac{x^4}{1+x^4} = \frac{1+x^4+x^4}{1+x^4} = \frac{1+2x^4}{1+x^4}$$

When we place this back into the expression, the $(1+x^4)^{1/4}$ terms cancel out perfectly. We are left with:
$$f(f(x)) = \frac{x}{(1+2x^4)^{1/4}}$$

### The Power of Generalization

Do you see it? When we applied the function once, the coefficient of $x^4$ was $1$. When we applied it twice, the coefficient became $2$. It is almost as if the function is 'counting' the number of compositions.

By the principle of mathematical induction, we can confidently generalize:
$$f_n(x) = \frac{x}{(1+nx^4)^{1/4}}$$

This is the 'Aha!' moment. We don't need to compute $f(f(f(f(x))))$ manually. We simply set $n=4$. Thus:
$$g(x) = f_4(x) = \frac{x}{(1+4x^4)^{1/4}}$$

The complexity has evaporated, replaced by a clean, elegant expression. This is the beauty of mathematics—what seems chaotic is often just a simple pattern waiting to be revealed.

### The Calculus of Elegance

Now, we turn our attention to the integral:
$$I = \int_0^{\sqrt{2}\sqrt[4]{5}} x^2 \cdot \frac{x}{(1+4x^4)^{1/4}} \, dx = \int_0^{\sqrt{2}\sqrt[4]{5}} \frac{x^3}{(1+4x^4)^{1/4}} \, dx$$

This is a classic substitution problem. We have an $x^3$ in the numerator and an $x^4$ term in the denominator. This is a match made in heaven for the substitution $1+4x^4 = t^4$.

Differentiating both sides, we get $16x^3 \, dx = 4t^3 \, dt$, which simplifies to:
$$x^3 \, dx = \frac{1}{4} t^3 \, dt$$

Now, we must be careful with the limits. When $x=0$, $t^4 = 1+0 = 1$, so $t=1$. When $x = \sqrt{2}\sqrt[4]{5}$, we calculate $x^4 = 2^2 \cdot 5 = 20$. Then $t^4 = 1 + 4(20) = 81$, so $t=3$.

Our integral transforms into:
$$I = \int_1^3 \frac{1}{4} \frac{t^3}{t} \, dt = \frac{1}{4} \int_1^3 t^2 \, dt$$

### The Final Victory

Look at how far we have come. We started with a terrifying composition of functions and a complex integral, and we have reduced it to the integral of $t^2$. The integral of $t^2$ is $\frac{t^3}{3}$.

Evaluating this from $1$ to $3$:
$$I = \frac{1}{4} \left[ \frac{t^3}{3} \right]_1^3 = \frac{1}{4} \left( \frac{27}{3} - \frac{1}{3} \right) = \frac{1}{4} \cdot \frac{26}{3} = \frac{13}{6}$$

Finally, the question asks for $18I$. We multiply our result by $18$:
$$18I = 18 \cdot \frac{13}{6} = 3 \cdot 13 = 39$$

There it is. The **final answer is 39**. You didn't just solve a problem; you navigated a maze of complexity and found the path of least resistance.

Animated Solution for Mathematics - Definite Integration: Let $f : R \to R$ be a function defined $f (x) = \frac{x}{( 1 + x ^{4} ) ^{1/4}}$ and $g (x) = f (f (f (f (x))))$ then $18 \int_{0}^{245} x^{2} g (x) d x$ is equal to

Select Answer:

Visualized Solution

Defining the Function $f (x)$

Setting up $f (f (x))$

Substituting $f (x)$ into $f (f (x))$

Simplifying $f (f (x))$

Identifying the Pattern

Defining $g (x)$ for $n = 4$

Setting up the Integral $I$

Applying Substitution Method

Changing the Limits of Integration

Substituting into the Integral

Evaluating the Definite Integral

Final Calculation for $18 I$

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Power of Generalization

The Calculus of Elegance

The Final Victory

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Visualized Solution

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