If R denotes the set of all real numbers, then the function `f : R to R` defined by `f (x) =[x]` is

Options

A. One-one only
B. Onto only
C. Both one-one and onto
D. Neither one-one nor onto

Answer

To determine whether the function \( f: R \to R \) defined by \( f(x) = [x] \) (where \([x]\) denotes the greatest integer less than or equal to \( x \)) is one-to-one (injective) and onto (surjective), we will analyze the properties of this function step by step.

### Step 1: Understand the Function
The function \( f(x) = [x] \) maps any real number \( x \) to the greatest integer less than or equal to \( x \). For example:
- \( f(2.3) = 2 \)
- \( f(3.9) = 3 \)
- \( f(-1.5) = -2 \)

### Step 2: Check if the Function is One-to-One
A function is one-to-one if different inputs produce different outputs.

**Analysis:**
- Consider \( f(0.5) = 0 \) and \( f(0.9) = 0 \).
- Both \( 0.5 \) and \( 0.9 \) map to the same output \( 0 \).

Since we found two different inputs that produce the same output, the function is **not one-to-one**.

### Step 3: Check if the Function is Onto
A function is onto if every element in the co-domain has a pre-image in the domain.

**Analysis:**
- The co-domain of \( f \) is all real numbers \( R \).
- The range of \( f \) consists only of integers (e.g., ..., -2, -1, 0, 1, 2, ...).
- Since there are real numbers (like \( 0.5 \), \( 1.5 \), etc.) that are not integers, these numbers do not have a pre-image in the domain.

Since not every real number has a corresponding input in the domain, the function is **not onto**.

### Conclusion
The function \( f(x) = [x] \) is neither one-to-one nor onto.

### Final Answer
The function \( f: R \to R \) defined by \( f(x) = [x] \) is **not one-to-one** and **not onto**.

---