Question:

<p>one-one: Let f(m) = f(n)<br /> Three cases arise</p> <ol start="1" style="list-style-type: lower-roman;"> <li>m and n are both odd<br /> clearly, f(m) = f(n)&nbsp;<span class="math-tex">$\Rightarrow \frac{1}{2}(m+1)=\frac{1}{2}(m+1)$</span><br /> = m + n</li> <li>m and n are both even<br /> in this case f(m) = f(n)&nbsp;<span class="math-tex">$\Rightarrow \frac{m}{2}+\frac{n}{2} \Rightarrow m=n$</span></li> <li>m is even n is odd<br /> in this case f(m) = f(n)<br /> <span class="math-tex">$\Rightarrow \quad \frac{n}{2}=\frac{1}{2}(n+1)$</span><br /> <span class="math-tex">$\Rightarrow x \neq 3$</span><br /> Hence f is not one-one.</li> </ol> <p>Onto: let n&nbsp;<span class="math-tex">$\in$</span>&nbsp;N (co-domai Now f(m)) = n<br /> i.e.&nbsp;<span class="math-tex">$m=\left\{\begin{array}{l} \frac{1}{2}(m+1), \quad \text { in is } odd \\ \frac{m}{2}, \quad m \text { is even } \end{array}\right.$</span><br /> So, for every n is co-domain. These exists m is domain r.t<br /> n = f(m)<br /> So, f is onto<br /> Hence f is many-one onto function.<br /> f: N&nbsp;<span class="math-tex">$\rightarrow$</span>&nbsp;N: f(x) =&nbsp;<span class="math-tex">$\left\{\begin{array}{l} \frac{1}{2}(n+1), \text { when } n \text { is odd } \\ \frac{n}{2}, \text { when } n \text { is even. } \end{array}\right.$</span>&nbsp;<br /> One-One function</p> <table border="1" cellpadding="3" cellspacing="0" style="width:100%;"> <tbody> <tr> <td>When n is odd</td> <td>When n is even</td> </tr> <tr> <td>f(1) = 1</td> <td>f(2) = 1</td> </tr> <tr> <td>f(3) = 2</td> <td>f(4) = 2</td> </tr> </tbody> </table> <p>It is clear from the above that the function is many-one&nbsp; and Onto function<br /> &nbsp;</p>