Let f : R - R be defined as f(x) = 2x - 1 and g : R - {1} - R be defined as g(x) = (x-1)/x. Then t

Q: Wait, why do we have to exclude $x=0$ from the domain of $f(g(x))$? It wasn't in the original domain of $g(x)$!

Great catch! Even if $x=0$ wasn't explicitly forbidden in the initial definition of $g(x)$, the composition $f(g(x))$ creates a new expression where $x$ appears in the denominator. If $x=0$, the expression $1 - \frac{2}{x}$ becomes undefined. In composition, the domain must satisfy both the inner function's constraints AND the constraints of the final simplified form.

Q: How do I know if a function is onto without drawing the graph?

Think of 'onto' as a coverage test. A function is onto if its range equals its codomain. Here, the codomain is $\mathbb{R}$. If you can find even one value in $\mathbb{R}$ that the function can never reach, it fails the onto test. By analyzing the horizontal asymptote ($y=1$) and the hole at $x=1$ ($y=-1$), we see the function misses values, so it cannot be onto.

Q: Why is the function one-one? I thought hyperbolas were tricky.

A function is one-one if every $y$-value corresponds to exactly one $x$-value. For $y = 1 - \frac{2}{x}$, if you solve for $x$, you get $x = \frac{2}{1-y}$. For every valid $y$, there is exactly one $x$. Since the function is strictly monotonic in its defined intervals, it never 'turns back' to hit the same $y$ twice.

Q: Does the hole at $x=1$ affect the one-one property?

It doesn't break the one-one property, but it does restrict the range! A hole simply means that a specific point is missing from the graph. As long as the function doesn't map two different $x$-values to the same $y$-value, it remains one-one. The hole just makes the function 'not onto' because that specific $y$-value is never achieved.

Q: Is it possible for a function to be onto but not one-one?

Absolutely! Think of $f(x) = x^2$ if the domain were all real numbers. It hits all positive values (onto), but $f(1) = f(-1) = 1$, so it fails the one-one test. In our current problem, we have the opposite situation: the function is injective (one-one) but fails to cover the entire codomain (not onto).

Question

Let $f : R 	o R$ be defined as $f(x) = 2x - 1$ and $g : R - \{1\} 	o R$ be defined as $g(x) = \frac{x-1}{x}$. Then the composition function $f(g(x))$ is :

Options:

(A) both one-one and onto

(B) onto but not one-one

(C) neither one-one nor onto

(D) one-one but not onto

Sigma Percentile · Accepted Answer

### The Architecture of Composition: A Mathematical Journey

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are dissecting the anatomy of a function.

We are looking at the composition $f(g(x))$, where $f(x) = 2x - 1$ and $g(x) = \frac{x-1}{x}$. Think of this as a two-stage rocket where the inner function $g(x)$ acts as the first stage, processing the input $x$, and its output is then fed into the second stage, $f(x)$.

### Stage 1: The Algebraic Transformation

Before we dive into the properties of the function, we must understand its form. We start by substituting $g(x)$ into $f(x)$:

$$f(g(x)) = 2\left(\frac{x-1}{x}ight) - 1$$

I know, looking at fractions inside functions can feel daunting. But take a breath. Let's expand this carefully.

We distribute the $2$ to get $\frac{2x - 2}{x} - 1$. To combine these, we need a common denominator, which is $x$. So, we rewrite $1$ as $\frac{x}{x}$:

$$f(g(x)) = \frac{2x - 2 - x}{x} = \frac{x - 2}{x}$$

By splitting the fraction, we arrive at the elegant form: $f(g(x)) = 1 - \frac{2}{x}$. This is the soul of our function.

It is a rectangular hyperbola, shifted and scaled. But wait—before we celebrate, we must respect the domain.

The original $g(x)$ was defined for $x 
eq 0$. Furthermore, our new expression $1 - \frac{2}{x}$ is undefined at $x = 0$. Thus, the domain of our composite function is $\mathbb{R} \setminus \{0\}$.

### Stage 2: The One-One Test (Injectivity)

Now, let's test if this function is 'one-one'. A function is one-one if every input produces a unique output.

Mathematically, we assume $f(g(x_1)) = f(g(x_2))$ and see if it forces $x_1 = x_2$:

$$1 - \frac{2}{x_1} = 1 - \frac{2}{x_2}$$

Subtracting $1$ from both sides and multiplying by $-1$, we get $\frac{2}{x_1} = \frac{2}{x_2}$, which immediately implies $x_1 = x_2$.

Geometrically, this means the graph never 'doubles back' on itself. Any horizontal line will intersect the curve at most once. Therefore, our function is **strictly one-one**.

### Stage 3: The Onto Test (Surjectivity)

Finally, we ask: is this function 'onto'? Does it cover the entire codomain $\mathbb{R}$? To be onto, the range must equal the codomain.

Let's look at the behavior of $y = 1 - \frac{2}{x}$. First, consider the horizontal asymptote.

As $x 	o \pm\infty$, the term $\frac{2}{x}$ approaches $0$, meaning $y$ approaches $1$. However, $y$ can never actually equal $1$. That is our first 'missing' value.

Since the range $\mathbb{R} \setminus \{1\}$ is not equal to the codomain $\mathbb{R}$, the function fails to be onto. It misses a critical value!

### The Conclusion

We have walked through the algebra, respected the domain, tested for uniqueness, and verified the range.

We found that our function is **one-one** because it maps distinct inputs to distinct outputs, but it is **not onto** because it leaves a gap in the codomain. You have successfully navigated the complexities of function composition. Keep this rigor in your toolkit—it is the key to mastering **JEE Advanced**.

Animated Solution for Mathematics - Functions: Let $f : R \to R$ be defined as $f (x) = 2 x - 1$ and $g : R - {1} \to R$ be defined as $g (x) = \frac{x - 1}{x}$ . Then the composition function $f (g (x))$ is :

Select Answer:

Visualized Solution

Define $f (x)$ and $g (x)$

Set up Composition $f (g (x))$

Substitute and Simplify

Final Form of $f (g (x))$

Determine Domain of $f (g (x))$

Check One-one Property

Check Onto Property (Asymptote)

Check Onto Property (Hole)

Final Conclusion

The Sigma Insight: Composite Functions

The Architecture of Composition

Stage 1

Stage 2

Stage 3

The Conclusion

Similar Questions

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Select Answer:

Visualized Solution

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

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

Substitute and Simplify

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

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