Let the function f:R→R be defined by f(x)=cosx,∀x∈R. Show that f is neither one-one nor onto

Options

A. into
B. one-one
C. onto
D. none of these

Answer

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 the properties of the cosine function.

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

A function is one-one (or injective) if different inputs map to different outputs. To show that \( f \) is not one-one, we need to find at least two different values of \( x \) such that \( f(x_1) = f(x_2) \) but \( x_1 \neq x_2 \).

Let's take \( x_1 = \frac{\pi}{2} \) and \( x_2 = -\frac{\pi}{2} \):

\[ f\left(\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0 \]
\[ f\left(-\frac{\pi}{2}\right) = \cos\left(-\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0 \]

Here, we see that:

\[ f\left(\frac{\pi}{2}\right) = f\left(-\frac{\pi}{2}\right) = 0 \]

However, \( \frac{\pi}{2} \neq -\frac{\pi}{2} \). Therefore, since two different inputs give the same output, \( f \) is not one-one.

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

A function is onto (or surjective) if every element in the codomain (in this case, \( \mathbb{R} \)) has a pre-image in the domain \( \mathbb{R} \). The range of the cosine function is limited to the interval \([-1, 1]\).

To show that \( f \) is not onto, we need to demonstrate that there are values in \( \mathbb{R} \) that are not achieved by \( f(x) = \cos x \).

For example, consider \( y = 2 \). There is no \( x \in \mathbb{R} \) such that \( \cos x = 2 \) because the cosine function only takes values in the range \([-1, 1]\).

Thus, since there are values in \( \mathbb{R} \) (like 2) that are not in the range of \( f \), we conclude that \( f \) is not onto.

### Conclusion

We have shown that the function \( f(x) = \cos x \) is neither one-one nor onto.

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