Let `RR ` be the set of all real numbers and `ƒ : RR to RR ` be a continuous function. Suppose `|ƒ(x) – ƒ(y)| ge |x – y| `for all real numbers x and y. Then

Options

A. ƒ is one-one, but need not be onto
B. ƒ is onto, but need not be one-one
C. ƒ need not be either one-one or onto
D. ƒ is one-one and onto

Answer

To solve the problem, we need to analyze the given condition for the function \( f: \mathbb{R} \to \mathbb{R} \) which states that:

\[ |f(x) - f(y)| \geq |x - y| \quad \text{for all } x, y \in \mathbb{R}. \]

### Step 1: Show that \( f \) is one-one (injective)

Assume \( f(x) = f(y) \) for some \( x, y \in \mathbb{R} \).

Using the given condition, we have:

\[ |f(x) - f(y)| \geq |x - y|. \]

Since \( f(x) = f(y) \), we can substitute this into the inequality:

\[ |0| \geq |x - y|. \]

This simplifies to:

\[ 0 \geq |x - y|. \]

The absolute value \( |x - y| \) is always non-negative, which implies that:

\[ |x - y| = 0 \implies x = y. \]

Thus, we have shown that \( f \) is one-one (injective).

### Step 2: Show that \( f \) is onto (surjective)

To show that \( f \) is onto, we need to demonstrate that for every \( b \in \mathbb{R} \), there exists an \( a \in \mathbb{R} \) such that \( f(a) = b \).

Since \( f \) is continuous and we have already established that \( f \) is one-one, we can analyze the behavior of \( f \) as \( x \) approaches \( \infty \) and \( -\infty \).

1. **Consider the limit as \( x \to \infty \)**:

From the given condition, we can derive that:

\[ |f(x) - f(0)| \geq |x - 0| = |x|. \]

Hence, as \( x \to \infty \):

\[ |f(x) - f(0)| \to \infty \implies f(x) \to \infty. \]

2. **Consider the limit as \( x \to -\infty \)**:

Similarly, we have:

\[ |f(x) - f(0)| \geq |x| \implies f(x) \to -\infty \text{ as } x \to -\infty. \]

Since \( f \) is continuous and takes values from \( -\infty \) to \( \infty \), by the Intermediate Value Theorem, \( f \) must take every value in \( \mathbb{R} \). Thus, \( f \) is onto (surjective).

### Conclusion

Since \( f \) is both one-one and onto, we conclude that:

\[ \text{The function } f \text{ is a bijection (one-one and onto).} \]

### Final Answer

Thus, the final answer is:

**\( f \) is one-one and onto.**

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