Let f : R → R be a function defined by f(x) = [2(1 - x²⁵/2)(2 + x²⁵)]^(1/50). If the function g(x) =

Q: Wait, why do we get $|x|$ instead of just $x$ when simplifying $f(f(x))$?

That is a fantastic question! In algebra, the $n^{th}$ root of $x^n$ is $x$ only if $n$ is odd. When $n$ is even, like our $50^{th}$ root, the result must be non-negative. Therefore, $(x^{50})^{\frac{1}{50}} = |x|$. This is a classic JEE trap designed to test your attention to domain and range constraints.

Q: Why can't we just expand the brackets in the original function $f(x)$ without using the difference of squares identity?

You absolutely could! However, the difference of squares identity $(a-b)(a+b) = a^2 - b^2$ is a powerful tool that collapses complex expressions instantly. Using it saves time and reduces the chance of arithmetic errors, which is vital in a high-pressure exam environment.

Q: How do I know for sure that $3^{\frac{1}{50}}$ is between 1 and 2?

Think about the powers of the boundaries. We know $1^{50} = 1$ and $2^{50}$ is a massive number (much larger than 3). Since the function $h(x) = x^{50}$ is strictly increasing for $x > 0$, if $1^{50} < 3 < 2^{50}$, then taking the $50^{th}$ root of all parts preserves the inequality: $1 < 3^{\frac{1}{50}} < 2$. It is a simple but rigorous way to bound the value.

Question

Let $f : \mathbb{R} 	o \mathbb{R}$ be a function defined by $f(x) = \left[2\left(1 - \frac{x^{25}}{2}ight)(2 + x^{25})ight]^{\frac{1}{50}}$. If the function $g(x) = f(f(f(x))) + f(f(x))$, then the greatest integer less than or equal to $g(1)$ is ______.

Sigma Percentile · Accepted Answer

### The Illusion of Complexity: Unmasking the Function $f(x)$

Imagine you are staring at a problem that looks like a tangled knot of exponents and roots. The function

$$f(x) = \left[2\left(1 - \frac{x^{25}}{2}\right)(2 + x^{25})\right]^{\frac{1}{50}}$$

might seem like a nightmare, but in the world of JEE Advanced, complexity is often just a mask for elegance. Let us peel back the layers together.

### Phase 1: The Algebraic Collapse

First, let us focus on the heart of the beast: the expression inside the brackets. We have $2(1 - \frac{x^{25}}{2})(2 + x^{25})$.

If we distribute that leading $2$ into the first parenthesis, the expression transforms into $(2 - x^{25})(2 + x^{25})$.

Do you see it? This is the classic difference of squares identity: $(a-b)(a+b) = a^2 - b^2$.

Here, $a = 2$ and $b = x^{25}$. Squaring these gives us $4 - (x^{25})^2$, which is simply $4 - x^{50}$.

Suddenly, the intimidating expression has collapsed into something beautiful and manageable:

$$f(x) = (4 - x^{50})^{\frac{1}{50}}$$

### Phase 2: The Nested Trap

Now, we tackle the composition $f(f(x))$. We are essentially feeding the output of $f(x)$ back into the function.

So, $f(f(x)) = (4 - (f(x))^{50})^{\frac{1}{50}}$. Substituting our simplified $f(x)$, we get:

$$f(f(x)) = \left( 4 - \left[ (4 - x^{50})^{\frac{1}{50}} \right]^{50} \right)^{\frac{1}{50}}$$

Watch closely as the powers of $50$ and $\frac{1}{50}$ cancel out. We are left with $(4 - (4 - x^{50}))^{\frac{1}{50}}$, which simplifies to $(x^{50})^{\frac{1}{50}}$.

Here is where the JEE examiner tests your precision. Because $50$ is an even power, the $50^{th}$ root of $x^{50}$ is not just $x$; it is the absolute value, $|x|$.

Thus, $f(f(x)) = |x|$.

### Phase 3: The Final Composition

With $f(f(x)) = |x|$, finding $f(f(f(x)))$ becomes a breeze. We simply apply $f$ to $|x|$:

$$f(f(f(x))) = f(|x|) = (4 - |x|^{50})^{\frac{1}{50}}$$

Since $|x|^{50}$ is identical to $x^{50}$ (because any even power swallows the negative sign), this simplifies right back to $(4 - x^{50})^{\frac{1}{50}}$, which is just $f(x)$.

### Phase 4: The Grand Finale

We are asked to find the greatest integer less than or equal to $g(1)$, where $g(x) = f(f(f(x))) + f(f(x))$. Substituting our findings, we get $g(x) = f(x) + |x|$.

Evaluating at $x = 1$:

$$g(1) = f(1) + |1| = (4 - 1^{50})^{\frac{1}{50}} + 1 = 3^{\frac{1}{50}} + 1$$

We know that $1^{50} = 1$ and $2^{50}$ is far greater than $3$. Therefore, $1 < 3^{\frac{1}{50}} < 2$.

Adding $1$ to the entire inequality gives us $2 < 3^{\frac{1}{50}} + 1 < 3$.

Since $g(1)$ is strictly between $2$ and $3$, the greatest integer less than or equal to $g(1)$ is **2**.

Animated Solution for Mathematics - Functions: Let $f : R \to R$ be a function defined by $f (x) = [2 (1 - \frac{x ^{25}}{2}) (2 + x^{25})]^{\frac{1}{50}}$ . If the function $g (x) = f (f (f (x))) + f (f (x))$ , then the greatest integer less than or equal to $g (1)$ is ______.

Enter Numerical Value:

Visualized Solution

The Function $f (x)$

Simplifying $f (x)$

Difference of Squares

Simplified $f (x)$

Finding $f (f (x))$

Substitution for $f (f (x))$

Simplifying the Power

The Absolute Value Trap

Finding $f (f (f (x)))$

$f (f (f (x))) = f (x)$

Constructing $g (x)$

Evaluating $g (1)$

Estimating $3^{\frac{1}{50}}$

Greatest Integer $⌊ g (1)⌋$

The Sigma Insight: Composite Functions

The Illusion of Complexity

Phase 1

Phase 2

Phase 3

Phase 4

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Visualized Solution

The Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Difference of Squares

Simplified .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(f(x))

Substitution for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(f(x))

Simplifying the Power

The Absolute Value Trap

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(f(f(x)))

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(f(f(x)))=f(x)

Constructing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(1)

Estimating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3501​

Greatest Integer .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ⌊g(1)⌋