Let f, g: N rightarrow N such that f(n+1)=f(n | vedclass

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(D) Given $f(n+1) = f(n) + f(1)$. This is a linear recurrence relation.For $n=1$,$f(2) = f(1) + f(1) = 2f(1)$.For $n=2$,$f(3) = f(2) + f(1) = 3f(1)$.B

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If $g$ is onto,then $fog$ is one-one

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(D) Given $f(n+1) = f(n) + f(1)$. This is a linear recurrence relation.For $n=1$,$f(2) = f(1) + f(1) = 2f(1)$.For $n=2$,$f(3) = f(2) + f(1) = 3f(1)$.By induction,$f(n) = n f(1)$.Since $f: N \rightarrow N$,$f(1)$ must be a positive integer $k \in N$.Thus,$f(n) = kn$. Since $k \geq 1$,$f(n_1) = f(n_2) \Rightarrow kn_1 = kn_2 \Rightarrow n_1 = n_2$,so $f$ is one-one. Statement $C$ is true.If $fog$ is one-one,then $f(g(x_1)) = f(g(x_2)) \Rightarrow g(x_1) = g(x_2)$ because $f$ is one-one. Since $fog$ is one-one,$x_1 = x_2$. Thus $g$ is one-one. Statement $A$ is true.If $f$ is onto,then for any $y \in N$,there exists $n \in N$ such that $f(n) = y$,i.e.,$kn = y$. This is only possible for all $y$ if $k=1$,so $f(n) = n$. Statement $B$ is true.If $g$ is onto,$fog$ is not necessarily one-one. For example,if $g(n) = 1$ for all $n$ (not onto) or any non-injective function,$fog$ will not be one-one. Even if $g$ is onto,if $g$ is not one-one,$fog$ cannot be one-one. Thus,statement $D$ is $NOT$ true.

Column-$I$	Column-$II$
$A$. $f$ is one-one and onto,if	$1$. $A = R^{+}, B = R$
$B$. $f$ is one-one but not onto,if	$2$. $A = B = R$
$C$. $f$ is onto but not one-one,if	$3$. $A = R, B = R^{+}$
$D$. $f$ is neither one-one nor onto,if	$4$. $A = B = R^{+}$

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

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