# Let the function $f\;:\;R \rightarrow R$ be defined by$f(x)=\cos x,\forall \;x\;\in R$.Show that f is neither one-one nor onto

Toolbox:
• A function $f: A \rightarrow B$ where for every $x1, x2 \in X, f(x1) = f(x2) \Rightarrow x1 = x2$ is called a one-one or injective function
• A function$f : X \rightarrow Y$ is said to be onto or surjective, if every element of Y is the image of some element of X under f, i.e., for every $y \in Y$, there exists an element x in X such that $f(x) = y$.
Given $f(x)=\cos x ,\forall x,\in R$
Let $f(x)=f(y)=1$
Step1: Injective or One-One function:
but $0 \neq 2\pi$
Hence f is not a one-one function
Step 2: Surjective or On-to function:
Consider $y=2 \in R$
There does not exist any $x \in R$ such that $f(x)=\cos x=y=2$
$\cos x$ can take values from -1 to 1 only
Hence f is not a onto function
Solution: Therefore f is neither one-one nor onto
edited Mar 25, 2013 by meena.p