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(A-IV, B-III, C-II, D-I) $f: R \rightarrow R$, $f(x) = \cos(112x - 37)$. Since $f(x)$ is a periodic function, it is many-one, hence not injective. The

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(A-IV, B-III, C-II, D-I) $f: R \rightarrow R$, $f(x) = \cos(112x - 37)$. Since $f(x)$ is a periodic function, it is many-one, hence not injective. The range is $[-1, 1]$, which is a proper subset of the codomain $R$, so it is not surjective. Thus, $A \rightarrow IV$.$(B)$ $f: [-2, 2] \rightarrow [-4, 4]$, $f(x) = x|x|$. This can be written as $f(x) = \begin{cases} -x^2 & -2 \leq x $(C)$ $f: R \rightarrow R$, $f(x) = (x-2)(x-3)(x-5)$. Since $f(2) = f(3) = f(5) = 0$, it is not injective. As it is a cubic polynomial, its range is $R$, so it is surjective. Thus, $C \rightarrow II$.$(D)$ $f: N \rightarrow N$, $f(n) = n+1$. Since $f(n_1) = f(n_2) \implies n_1+1 = n_2+1 \implies n_1 = n_2$, it is injective. The range is ${2, 3, 4, \dots}$, which is not equal to the codomain $N$ (as $1$ is not in the range), so it is not surjective. Thus, $D \rightarrow I$.

$A$. $f: R \rightarrow R$ defined by $f(x) = \cos(112x - 37)$	$I$. Injection but not surjection
$B$. $f: A \rightarrow B$ defined by $f(x) = x\|x\|$ when $A = [-2, 2]$ and $B = [-4, 4]$	$II$. Surjection but not injection
$C$. $f: R \rightarrow R$ defined by $f(x) = (x-2)(x-3)(x-5)$	$III$. Bijection
$D$. $f: N \rightarrow N$ defined by $f(n) = n+1$	$IV$. Neither injection nor surjection
	$V$. Composite function

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