Q.Let the function $f : \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = \cos x$, $\forall\, x \in \mathbb{R}$. Show that $f$ is neither one-one nor onto.

Concept: One-One (Injective) and Onto (Surjective) Functions

A function is one-one if distinct inputs give distinct outputs: $x_1 \neq x_2 \Rightarrow f(x_1) \neq f(x_2)$. It is onto if every real number is attained as an output.

Step 1 – Not one-one:

Take $x_1 = 0$ and $x_2 = 2\pi$. Then $x_1 \neq x_2$ but $f(0) = \cos 0 = 1$ and $f(2\pi) = \cos 2\pi = 1$. Since the same output comes from two different inputs, $f$ is not injective.

Step 2 – Not onto: …

Why this approach works

To show a function is not one-one, we need to find two distinct inputs that give the same output. For a periodic function like $\cos x$, this is almost immediate — the function repeats its values every $2\pi$. To show it is not onto, we need to find at least one real number that is never produced as an output. Since $\cos x$ is bounded between $-1$ and $1$, any number outside this interval is impossible to reach.

Let's work through both parts cleanly.

Step-by-step reasoning

1. Showing $f$ is not one-one

A function is one-one (injective) if $f(a) = f(b)$ implies $a = b$. To disprove this, we just need a counterexample.

Take $x = 0$ and $x = 2\pi$. Both are real numbers, and $0 \neq 2\pi$. But:

So $f(0) = f(2\pi)$ while $0 \neq 2\pi$. This directly violates the definition of a one-one function.

2. Showing $f$ is not onto

A function is onto (surjective) if every element of the codomain $\mathbb{R}$ is the image of some input. Here, the codomain is all real numbers. …

Method: Showing a periodic bounded function is neither one-one nor onto

For $f(x)=\cos x$ (or any periodic, bounded function), use periodicity to break one-one and boundedness to break onto.

Steps

Step 1: Break one-one using periodicity

Two inputs differing by a period share the same output. For cosine, $\cos 0=\cos 2\pi=1$ with $0\ne 2\pi$, so it is not one-one.

Step 2: Break onto using the bounded range …

Common Mistakes

Mistake 1: Confusing range with codomain when checking onto

Why it's wrong: the codomain is $\mathbb{R}$ but the range is only $[-1,1]$; the gap is what makes it not onto. Correct approach: pick a codomain value like $2$ that is never an output.

Mistake 2: Trying to prove one-one fails "in general" instead of one clash …