Mapping `f : R to R` which is defined as `f (x) = cos x,x in R` will be

Options

A. Neither one-one nor onto
B. One-one
C. Onto
D. `One-one onto

Answer

To determine the properties of the mapping \( f: \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = \cos x \), we need to analyze whether this function is one-to-one (injective) and/or onto (surjective).

### Step 1: Check if the function is one-to-one (injective)
A function is one-to-one if different inputs produce different outputs. To check this, we can analyze the behavior of the cosine function.

1. **Consider two values \( x_1 \) and \( x_2 \) such that \( f(x_1) = f(x_2) \)**:
\[ \cos x_1 = \cos x_2 \]
This equality implies that:
\[ x_1 = 2n\pi \pm x_2 \quad \text{for some integer } n \]
Hence, there can be multiple values of \( x \) that yield the same \( f(x) \). For example, \( \cos(0) = \cos(2\pi) = 1 \).

**Conclusion**: The function \( f(x) = \cos x \) is not one-to-one.

### Step 2: Check if the function is onto (surjective)
A function is onto if every element in the codomain has a pre-image in the domain.

1. **Identify the range of \( f(x) = \cos x \)**:
The cosine function oscillates between -1 and 1 for all real numbers \( x \). Therefore, the range of \( f(x) \) is:
\[ \text{Range}(f) = [-1, 1] \]

2. **Compare the range with the codomain \( \mathbb{R} \)**:
The codomain is all real numbers \( \mathbb{R} \), but the range of \( f \) only covers the interval from -1 to 1. Thus, there are many real numbers (e.g., 2, -2) that are not covered by the function.

**Conclusion**: The function \( f(x) = \cos x \) is not onto.

### Final Conclusion
Since the function \( f(x) = \cos x \) is neither one-to-one nor onto, the correct answer is that the mapping is **neither one-one nor onto**.

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