Question

Consider the following statements
I. The modulus function $f: R \rightarrow R$ given by $f(x)=|x|$ is neither one-one nor onto.
II. The signum function $f: R \rightarrow R$ given by $f(x)=\begin{cases}1, & \text { if } x>0 \\ 0, & \text { if } x=0 \\ -1, & \text { if } x<0\end{cases}$ is bijective.
III. The function $f: N \rightarrow N$ defined by $f(n)=\begin{cases}\frac{n+1}{2}, & \text { if } n \text { is odd } \\ \frac{n}{2}, & \text { if } n \text { is even }\end{cases}$ for all $n \in N$ is not bijective.
Choose the correct option.

• A. I and II are true
• B. II and III are true
• C. I and III are true
• D. All are true

Answer 1

Answer: (C)

I. Here, $f: R \rightarrow R$ is given by $f(x)=|x|=\begin{cases}x, & \text { if } x \geq 0 \\ -x, & \text { if } x < 0\end{cases}$
It is seen that $f(-1)=|-1|=1, f(1)=|1|=1$
Therefore, $ f(-1)=f(1)$ but $-1 \neq 1$
Therefore, $f$ is not one-one.
Now, consider $-1 \in R$
It is known that $f(x)=|x|$ is always non-negative.
Thus, there does not exist any element $x$ in domain $R$ such that
$f(x)=|x|=-1 .$
Therefore, $f$ is not onto.
Hence, the modulus function is neither one-one nor Onto.
II. $f: R \rightarrow R, f(x)=\begin{cases}1, & \text { if } & x>0 \\ 0, & \text { if } & x=0 \\ -1, & \text { if } & x < 0\end{cases}$
It is seen that $f(1)=f(2)=1$ but $1 \neq 2$. Therefore, $f$ is not one-one.
Now, as $f(x)$ takes only three values (1,0 or $-1$ ), therefore for the element $-2$ in codomain $R$, there does not exist any $x$ in domain $R$ such that $f(x)=-2$.
Therefore, $f$ is not onto.
Hence, the Signum function is neither one-one nor onto.
III. Here, $ f: N \rightarrow N$ is defined as $f(n)=\begin{cases}\frac{n+1}{2}, & \text { if } n \text { is odd } \\ \frac{n}{2}, & \text { if } n \text { is even }\end{cases}$ for all $n \in N$. It can be observed that $f(1)=\frac{1+1}{2}=1$ and $f(2)=\frac{2}{2}=1 $ (by definition of $f$ ) $\therefore f(1)=f(2)$, where $1 \neq 2$.
Therefore, $f$ is not one-one. Consider a natural number $n$ in codomain $N$.
Case I When $n$ is odd.
Therefore, $n=2 r+1$ for some $r \in W$.
Then, there exists $4 r+1 \in N$ such that
$f(4 r+1)=\frac{4 r+1+1}{2}=2 r+1$
Therefore, $f$ is onto.
Case II When $n$ is even
Therefore, $n=2 r$ for some $r \in N$
Then, there exists $4 r \in N$ such that $f(4 r)=\frac{4 r}{2}=2 r$
Therefore, $f$ is onto. Hence, $f$ is not a bijective function.