Let $A = \{1, 2, 3, 4, 5\}$ and $B = \{1, 2, 3, 4, 5, 6\}$. Then the number of functions $f: A \rightarrow B$ satisfying $f(1) + f(2) = f(4) - 1$ is equal to

Answer the following by appropriately matching the lists based on the information given in the paragraph.
Let $f(x) = \sin(\pi \cos x)$ and $g(x) = \cos(2\pi \sin x)$ be two functions defined for $x > 0$. Define the following sets whose elements are written in increasing order:
$X = \{x : f(x) = 0\}, Y = \{x : f'(x) = 0\}$
$Z = \{x : g(x) = 0\}, W = \{x : g'(x) = 0\}$
$List-I$ contains the sets $X, Y, Z$ and $W$. $List-II$ contains some information regarding these sets.

$List-I$	$List-II$
$(I) X$	$(P) \supseteq \{\frac{\pi}{2}, \frac{3\pi}{2}, 4\pi, 7\pi\}$
$(II) Y$	$(Q) \text{ an arithmetic progression}$
$(III) Z$	$(R) \text{ NOT an arithmetic progression}$
$(IV) W$	$(S) \supseteq \{\frac{\pi}{6}, \frac{7\pi}{6}, \frac{13\pi}{6}\}$
	$(T) \supseteq \{\frac{\pi}{3}, \frac{2\pi}{3}, \pi\}$
	$(U) \supseteq \{\frac{\pi}{6}, \frac{3\pi}{4}\}$

$(1)$ Which of the following is the only $CORRECT$ combination?
$(1) (II), (R), (S)$ $(2) (I), (P), (R)$ $(3) (II), (Q), (T)$ $(4) (I), (Q), (U)$
$(2)$ Which of the following is the only $CORRECT$ combination?
$(1) (IV), (Q), (T)$ $(2) (IV), (P), (R), (S)$ $(3) (III), (R), (U)$ $(4) (III), (P), (Q), (U)$

Let $N$ be the set of positive integers. For all $n \in N$,let $f_n = (n+1)^{1/3} - n^{1/3}$ and $A = \{n \in N : f_{n+1} < \frac{1}{3(n+1)^{2/3}} < f_n\}$. Then,

The number of non-constant functions $f: X \to Y$ where $X = \{0, 1, 2\}$ and $Y = \{1, 2, 3, 4, 5, 6, 7, 8\}$ such that $f(i) \leq f(j)$ whenever $i < j$ is:

If $f(x)=\frac{2^{x}+2^{-x}}{2}$,then $f(x+y) \cdot f(x-y)$ is

The function $f(x) = [|x|] - |[x]|$ where $[x]$ denotes the greatest integer function: