Let the function `f:R to R` be defined by `f(x)=cos x, AA x in R.` Show that `f` is neither one-one nor onto.

Explanation

To show that the function \( f: \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = \cos x \) is neither one-one nor onto, we will analyze both properties step by step.

### Step 1: Show that \( f \) is not one-one

A function is one-one (or injective) if different inputs produce different outputs. In other words, if \( f(a) = f(b) \) implies that \( a = b \).

1. Consider \( f(0) \) and \( f(2\pi) \):
\[ f(0) = \cos(0) = 1 \]
\[ f(2\pi) = \cos(2\pi) = 1 \]
Here, \( f(0) = f(2\pi) \) but \( 0 \neq 2\pi \).

2. Since we found two different inputs (0 and \( 2\pi \)) that yield the same output (1), we conclude that \( f \) is not one-one.

### Step 2: Show that \( f \) is not onto

A function is onto (or surjective) if every element in the codomain has a pre-image in the domain. In this case, we need to check if every real number can be expressed as \( \cos x \) for some \( x \in \mathbb{R} \).

1. The range of the cosine function is limited to the interval \([-1, 1]\). This means:
\[ \text{Range of } f = \{ y \in \mathbb{R} : -1 \leq y \leq 1 \} \]

2. The codomain of the function \( f \) is \( \mathbb{R} \), which includes all real numbers.

3. Since there are real numbers (for example, \( 2 \) or \( -3 \)) that are not in the range of \( f \), we conclude that \( f \) is not onto.

### Conclusion

Since we have shown that the function \( f(x) = \cos x \) is neither one-one nor onto, we can summarize our findings as follows:
- \( f \) is not one-one because there exist distinct \( x \) values that yield the same \( f(x) \).
- \( f \) is not onto because the range of \( f \) is not equal to the codomain \( \mathbb{R} \).