The function `f:R rarr R` defined as `f(x) = x^3` is:

Options

A. One-one but not onto
B. Not one-one but onto
C. Neither one-one nor onto
D. One-one and onto

Answer

To determine whether the function \( f: \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = x^3 \) is one-to-one (injective) and onto (surjective), we will analyze the definitions of these properties step by step.

### Step 1: Check if the function is one-to-one (injective)

A function is one-to-one if for every pair of distinct inputs, the outputs are also distinct. In mathematical terms, \( f \) is one-to-one if:
\[ f(x_1) = f(x_2) \implies x_1 = x_2 \]
for all \( x_1, x_2 \in \mathbb{R} \).

**Proof:**
Assume \( f(x_1) = f(x_2) \). This means:
\[ x_1^3 = x_2^3 \]
Taking the cube root of both sides, we get:
\[ x_1 = x_2 \]
Since we have shown that \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \), the function \( f(x) = x^3 \) is one-to-one.

### Step 2: Check if the function is onto (surjective)

A function is onto if for every element \( y \) in the codomain \( \mathbb{R} \), there exists at least one element \( x \) in the domain \( \mathbb{R} \) such that:
\[ f(x) = y \]

**Proof:**
Let \( y \) be any arbitrary element in \( \mathbb{R} \). We need to find \( x \) such that:
\[ f(x) = y \implies x^3 = y \]
To solve for \( x \), we take the cube root:
\[ x = y^{1/3} \]
Since \( y^{1/3} \) is a real number for any real \( y \), we can find an \( x \) in \( \mathbb{R} \) for every \( y \) in \( \mathbb{R} \). Therefore, the function \( f(x) = x^3 \) is onto.

### Conclusion

Since the function \( f(x) = x^3 \) is both one-to-one and onto, we conclude that:

**The function is one-to-one and onto.**

Thus, the correct option is:
**Option 4: One-to-one and onto.**

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