Question

Let $f , g : N \rightarrow N$ such that $f ( n +1)= f ( n )+ f (1)$ $\forall n \in N$ and $g$ be any arbitrary function. Which of the following statements is $NOT$ true?

• A. If fog is one-one, then $g$ is one-one
• B. If $f$ is onto, then $f ( n )= n \forall n \in N$
• C. $f$ is one-one
• D. If $g$ is onto, then fog is one-one

Answer 1

Answer: (D)

$f(n+1)-f(n)=f(1)$
$\Rightarrow f(n)=n f(1)$
$\Rightarrow f$ is one-one
Now, Let $f \left( g \left( x _{2}\right)\right)= f \left( g \left( x _{1}\right)\right)$
$\Rightarrow g \left( x _{2}\right)= g \left( x _{1}\right)$ (as $f$ is one-one )
$\Rightarrow x _{1}= x _{2}$ ( as fog is one-one )
$\Rightarrow g$ is one-one
Now, $f(g(n))=g(n) f(1)$
may be many-one if
$g ( n )$ is many-one