Let `f:RrarrR` be defined as `f(x)=x^(4)`. Then :

Options

A. f is one-one onto
B. f is many-one onto
C. f is one-one but not onto
D. f is neither one-one nor onto

Answer

To solve the problem, we need to analyze the function \( f: \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = x^4 \). We will determine whether this function is one-to-one (injective) and onto (surjective).

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

To check if \( f \) is one-to-one, we need to verify if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \) for all \( x_1, x_2 \in \mathbb{R} \).

1. Assume \( f(x_1) = f(x_2) \).
\[ x_1^4 = x_2^4 \]

2. Taking the fourth root on both sides gives:
\[ |x_1| = |x_2| \]

3. This implies two cases:
- \( x_1 = x_2 \)
- \( x_1 = -x_2 \)

Since we have found that \( x_1 \) can be equal to \( x_2 \) or \( -x_2 \), the function is not one-to-one. Therefore, \( f \) is many-to-one.

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

To check if \( f \) is onto, we need to see if every element in the codomain \( \mathbb{R} \) has a pre-image in the domain \( \mathbb{R} \).

1. The codomain of \( f \) is \( \mathbb{R} \), which includes all real numbers.
2. The range of \( f(x) = x^4 \) is \( [0, \infty) \) because:
- The minimum value of \( x^4 \) occurs at \( x = 0 \), where \( f(0) = 0 \).
- As \( x \) increases or decreases, \( f(x) \) increases without bound.

3. Since the range \( [0, \infty) \) does not cover all real numbers (it does not include negative numbers), the function is not onto.

### Conclusion

The function \( f(x) = x^4 \) is:
- Not one-to-one (many-to-one).
- Not onto.

Thus, the final answer is that the function is neither one-to-one nor onto.