For x ∈ R - {0, 1}, let f₁(x) = 1/x, f₂(x) = 1 - x and f₃(x) = 1/(1 - x) be three given functions. I

Q: Wait, why do we use a dummy variable like $t$ here? Can't we just solve for $J(x)$ directly?

That is a fantastic question! When we reach the step $J(\frac{1}{x}) = \frac{x}{x - 1}$, we have an expression for the function evaluated at $\frac{1}{x}$, not at $x$. The dummy variable $t = \frac{1}{x}$ is a powerful tool that allows us to 're-label' our input. By setting $t = \frac{1}{x}$, we are essentially saying, 'Let's look at this function from the perspective of its input $t$.' It transforms the expression into a standard $J(t)$ format, which we can then easily map back to $J(x)$.

Q: Is the order of composition $(f_2 \circ J \circ f_1)(x)$ always evaluated from right to left?

Yes, absolutely! In mathematics, the composition $(f \circ g)(x)$ is defined as $f(g(x))$. This means $g$ acts on $x$ first, and then $f$ acts on the result of $g(x)$. When you have a chain like $(f_2 \circ J \circ f_1)(x)$, you must work from the inside out: $f_1$ acts on $x$, then $J$ acts on that result, and finally $f_2$ acts on the result of $J$. Think of it as an assembly line where the item moves from right to left.

Q: What happens if $x=0$ or $x=1$? The domain says $R - \{0, 1\}$. Does that change our algebra?

The domain restriction $R - \{0, 1\}$ is crucial for the functions to be well-defined. For instance, $f_1(x) = \frac{1}{x}$ is undefined at $x=0$, and $f_3(x) = \frac{1}{1-x}$ is undefined at $x=1$. When we perform our algebraic manipulations, we assume $x$ is within the valid domain. These restrictions ensure that we never divide by zero during our steps. It is a sign of a rigorous mathematical mind to notice these constraints!

Question

For $x \in R - \{0, 1\}$, let $f_1(x) = \frac{1}{x}$, $f_2(x) = 1 - x$ and $f_3(x) = \frac{1}{1 - x}$ be three given functions. If a function, J(x) satisfies $(f_2 \circ J \circ f_1)(x) = f_3(x)$ then J(x) is equal to :-

Options:

(A) $f_3(x)$

(B) $f_1(x)$

(C) $f_2(x)$

(D) $\frac{1}{x}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in front of a complex, multi-stage factory assembly line. Each machine in this line is a function, and the raw material is our variable $x$.

We are given three machines: 
$f_1(x) = \frac{1}{x}$, 
$f_2(x) = 1 - x$, and 
$f_3(x) = \frac{1}{1 - x}$.

Our mission is to identify the hidden machine $J(x)$ that, when placed between $f_1$ and $f_2$, produces the same output as $f_3$.

### Peeling the Onion

The equation $(f_2 \circ J \circ f_1)(x) = f_3(x)$ looks intimidating, but let's apply the golden rule of composition: work from the inside out. The notation $(f_2 \circ J \circ f_1)(x)$ is shorthand for $f_2(J(f_1(x)))$.

Think of it as an onion; we need to peel away the layers one by one. First, we replace $f_1(x)$ with its definition, $\frac{1}{x}$.

Now, our equation becomes:
$$f_2\left(J\left(\frac{1}{x}\right)\right) = f_3(x)$$

### The Algebraic Dance

Now, let's look at the outermost function, $f_2$. We know $f_2(X) = 1 - X$. If we treat the entire block $J(\frac{1}{x})$ as our input $X$, then $f_2(J(\frac{1}{x}))$ becomes $1 - J(\frac{1}{x})$.

On the right side, we substitute the definition of $f_3(x)$, which is $\frac{1}{1 - x}$. Equating these, we get:
$$1 - J\left(\frac{1}{x}\right) = \frac{1}{1 - x}$$

This is the moment of truth. With a bit of careful transposition, we move $J(\frac{1}{x})$ to one side and the fraction to the other:
$$J\left(\frac{1}{x}\right) = 1 - \frac{1}{1 - x}$$

Simplifying the right side:
$$J\left(\frac{1}{x}\right) = \frac{1 - x - 1}{1 - x} = \frac{-x}{1 - x} = \frac{x}{x - 1}$$

### The Dummy Variable Trick

We are almost there, but we have $J(\frac{1}{x})$ instead of $J(x)$. We use a 'dummy variable' substitution. Let $t = \frac{1}{x}$, which implies $x = \frac{1}{t}$.

By substituting $x = \frac{1}{t}$ into our expression, we transform the equation into:
$$J(t) = \frac{\frac{1}{t}}{\frac{1}{t} - 1}$$

Simplifying the denominator, $\frac{1}{t} - 1$ becomes $\frac{1 - t}{t}$. Thus:
$$J(t) = \frac{\frac{1}{t}}{\frac{1 - t}{t}}$$

The $t$ in the denominators cancels out beautifully, leaving us with:
$$J(t) = \frac{1}{1 - t}$$

### The Grand Finale

Since $t$ is just a placeholder, we can replace it with $x$ to find the general form of our function:
**$J(x) = \frac{1}{1 - x}$**

Looking back at our original list of functions, we see that this is exactly $f_3(x)$. We have discovered that **$J(x) = f_3(x)$**.

It is a beautiful, symmetric result. You have navigated the layers of composition, performed the algebraic dance, and used the dummy variable to reveal the truth. This is the essence of JEE mathematics: not just calculating, but understanding the structure of the problem.

Animated Solution for Mathematics - Functions: For $x \in R - {0, 1}$ , let $f_{1} (x) = \frac{1}{x}$ , $f_{2} (x) = 1 - x$ and $f_{3} (x) = \frac{1}{1 - x}$ be three given functions. If a function, J(x) satisfies $(f_{2} \circ J \circ f_{1}) (x) = f_{3} (x)$ then J(x) is equal to :-

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Visualized Solution

Given Functions and Equation

Expanding the Composition

Substitute $f_{1} (x)$

Apply $f_{2}$ Definition

Substitute $f_{3} (x)$

Isolate $J (\frac{1}{x})$

Take Common Denominator

Simplify the Expression

Change of Variable

Substitute $x = \frac{1}{t}$

Simplify the Complex Fraction

Final Conclusion

Key Takeaway

The Sigma Insight: Composite Functions

Analyzing the Setup

Peeling the Onion

The Algebraic Dance

The Dummy Variable Trick

The Grand Finale

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Select Answer:

Visualized Solution

Given Functions and Equation

Expanding the Composition

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f1​(x)

Apply .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f2​ Definition

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f3​(x)

Isolate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } J(x1​)