Consider the function $F; N →N $ given by $f(x)=\left\{\begin{matrix} x+1, \text{if x is odd}\\x-1, \text{if x is even}\end{matrix}\right.$

f is both one-one and onto

The correct answer is Option (4) → f is both one-one and onto

$f(x)=\left\{\begin{matrix} x+1\\x-1\end{matrix}\right.$

let $x_1,x_2$ be

(i) both odd

then $f(x_1)=f(x_2)⇒x_1+1=x_2+1⇒x_1=x_2$

(ii) both even

then $f(x_1)=f(x_2)⇒x_1-1=x_2-1⇒x_1=x_2$

(iii) one odd, one even

$f(x_1)=f(x_2)⇒x_1-1=x_2+1⇒x_1-x_2=2$ Not possible

(difference between even and odd number can't be even)

⇒ f is one one

also $f^{-1}(x)=\left\{\begin{matrix} x+1&even\,x\\x-1&odd\,x\end{matrix}\right.$

for every y there exists atleast one x ⇒ onto

f is both one one, onto