Mix Examples of Relation and Function Questions in English

(A) Given $S = (0,1) \cup (1,2) \cup (3,4)$ and $T = \{0, 1, 2, 3\}$.
$(A)$ Since $S$ is an infinite set and $T$ has $4$ elements,there are $4^{|S|}$ functions from $S$ to $T$. Since $|S|$ is infinite,there are infinitely many functions. Thus,$(A)$ is true.
$(B)$ $A$ strictly increasing function $f: S \to T$ must map each connected component of $S$ to a single value because the image of a connected set under a continuous function is connected. However,$T$ is a discrete set. For $f$ to be strictly increasing,it must be constant on each component. Since $T$ is finite,there are only finitely many such functions. Thus,$(B)$ is false.
$(C)$ $A$ continuous function $f: S \to T$ must be constant on each connected component of $S$ because the image of a connected set $(0,1)$,$(1,2)$,and $(3,4)$ under $f$ must be a connected subset of $T$. The only connected subsets of $T$ are singleton sets. Thus,$f$ must take a constant value on each of the $3$ components. There are $4$ choices for each component,so there are $4 \times 4 \times 4 = 64$ such functions. Since $64 \le 120$,$(C)$ is true.
$(D)$ Any continuous function $f: S \to T$ is a constant function on each connected component of $S$. $A$ constant function is differentiable everywhere on its domain. Thus,$(D)$ is true.
Therefore,the correct statements are $(A)$,$(C)$,and $(D)$.

(A) For option $[A]$,let $g(x) = e^x - \int_0^x f(t) \sin t \, dt$. Since $f(t) \in (0,1)$ and $\sin t \in [0,1]$ for $t \in (0,1)$,the integral $\int_0^x f(t) \sin t \, dt < \int_0^x 1 \, dt = x$. Thus $g(x) > e^x - x$. Since $e^x - x > 0$ for all $x$,$g(x)$ cannot be zero.
For option $[B]$,let $h(x) = x^9 - f(x)$. Since $f(x) \in (0,1)$,$h(0) = 0^9 - f(0) = -f(0) < 0$ and $h(1) = 1^9 - f(1) = 1 - f(1) > 0$. By the Intermediate Value Theorem $(IVT)$,there exists $c \in (0,1)$ such that $h(c) = 0$.
For option $[C]$,let $k(x) = f(x) + \int_0^{\pi/2} f(t) \sin t \, dt$. Since $f(x) > 0$ and the integral is positive,$k(x) > 0$ for all $x \in (0,1)$. Thus,it cannot be zero.
For option $[D]$,let $m(x) = x - \int_0^{\pi/2 - x} f(t) \cos t \, dt$. Then $m(0) = 0 - \int_0^{\pi/2} f(t) \cos t \, dt < 0$ and $m(1) = 1 - \int_0^{\pi/2 - 1} f(t) \cos t \, dt$. Since $\int_0^{\pi/2 - 1} f(t) \cos t \, dt < \int_0^{\pi/2 - 1} 1 \, dt = \pi/2 - 1 \approx 0.57 < 1$,$m(1) > 0$. By $IVT$,there exists $c \in (0,1)$ such that $m(c) = 0$.

(C, B) For $f(x) = \sin(\pi \cos x)$,$f(x) = 0 \implies \pi \cos x = n\pi \implies \cos x = n$. Since $n \in \mathbb{Z}$ and $|\cos x| \le 1$,$n \in \{-1, 0, 1\}$. Thus $x = n\pi$ or $x = n\pi \pm \frac{\pi}{2}$,which simplifies to $x = \frac{k\pi}{2}$ for $k \in \mathbb{N}$. $X = \{\frac{k\pi}{2} : k \in \mathbb{N}\}$. This is an arithmetic progression with common difference $\frac{\pi}{2}$. Thus $(I) \to (Q)$.
For $f'(x) = \cos(\pi \cos x) \cdot (-\pi \sin x) = 0$. Either $\sin x = 0 \implies x = n\pi$ or $\cos(\pi \cos x) = 0 \implies \pi \cos x = (2n+1)\frac{\pi}{2} \implies \cos x = \pm \frac{1}{2}$. $Y = \{n\pi, n\pi \pm \frac{\pi}{3}\}$. This is not an $AP$. $Y$ contains $\{\frac{\pi}{3}, \frac{2\pi}{3}, \pi\}$. Thus $(II) \to (R), (T)$.
For $g(x) = \cos(2\pi \sin x) = 0 \implies 2\pi \sin x = (2n+1)\frac{\pi}{2} \implies \sin x = \pm \frac{1}{4}, \pm \frac{3}{4}$. This is not an $AP$. Thus $(III) \to (R)$.
For $g'(x) = -\sin(2\pi \sin x) \cdot (2\pi \cos x) = 0$. Either $\cos x = 0 \implies x = (2n+1)\frac{\pi}{2}$ or $\sin(2\pi \sin x) = 0 \implies 2\pi \sin x = n\pi \implies \sin x = \frac{n}{2}$. Thus $\sin x \in \{0, \pm \frac{1}{2}, \pm 1\}$. $W$ contains $\{\frac{\pi}{6}, \frac{7\pi}{6}, \frac{13\pi}{6}\}$. Thus $(IV) \to (S)$.
Matching: $(1)$ $(II), (R), (T)$ is not an option,but $(II), (R), (T)$ is correct. Re-evaluating options: $(1)$ $(II), (R), (S)$ is false. $(2)$ $(I), (P), (R)$ is false. $(3)$ $(II), (Q), (T)$ is false. $(4)$ $(I), (Q), (U)$ is correct as $X$ is an $AP$ and $U$ is a subset of $X$. For $(2)$,$(III), (R), (U)$ is correct as $Z$ is not an $AP$ and $U$ is a subset of $Z$ (since $\sin(\pi/6)=1/2$ and $\sin(3\pi/4)=1/\sqrt{2}$,wait,$Z$ requires $\sin x = \pm 1/4, \pm 3/4$. Re-checking $Z$: $g(x)=0 \implies \sin x = \pm 1/4, \pm 3/4$. $U$ is not a subset of $Z$. Correcting: $(IV), (P), (R), (S)$ is correct for $(2)$.

(B) Let $\theta = \pi x - \frac{\pi}{4}$. Since $x \in [0, 2]$,$\theta \in [-\frac{\pi}{4}, \frac{7\pi}{4}]$.
Then $2\pi x = 2\theta + \frac{\pi}{2}$,so $\sin(2\pi x) = \cos(2\theta)$.
Also $3\pi x + \frac{\pi}{4} = 3(\theta + \frac{\pi}{4}) + \frac{\pi}{4} = 3\theta + \pi$,so $\sin(3\pi x + \frac{\pi}{4}) = \sin(3\theta + \pi) = -\sin(3\theta)$.
The inequality $f(x) \geq 0$ becomes $(3 - \cos(2\theta)) \sin \theta - (-\sin(3\theta)) \geq 0$.
$(3 - (1 - 2\sin^2 \theta)) \sin \theta + \sin(3\theta) \geq 0$.
$(2 + 2\sin^2 \theta) \sin \theta + (3\sin \theta - 4\sin^3 \theta) \geq 0$.
$2\sin \theta + 2\sin^3 \theta + 3\sin \theta - 4\sin^3 \theta \geq 0$.
$5\sin \theta - 2\sin^3 \theta \geq 0 \Rightarrow \sin \theta (5 - 2\sin^2 \theta) \geq 0$.
Since $5 - 2\sin^2 \theta > 0$ for all $\theta$,we must have $\sin \theta \geq 0$.
Thus,$\theta \in [0, \pi]$.
$0 \leq \pi x - \frac{\pi}{4} \leq \pi$ $\Rightarrow \frac{\pi}{4} \leq \pi x \leq \frac{5\pi}{4}$ $\Rightarrow x \in [\frac{1}{4}, \frac{5}{4}]$.
Therefore,$\alpha = \frac{1}{4}$ and $\beta = \frac{5}{4}$,so $\beta - \alpha = 1$.

(A) Consider the property $f(x) + f(1-x) = \frac{4^x}{4^x+2} + \frac{4^{1-x}}{4^{1-x}+2}$.
$= \frac{4^x}{4^x+2} + \frac{4/4^x}{4/4^x + 2} = \frac{4^x}{4^x+2} + \frac{4}{4 + 2 \cdot 4^x} = \frac{4^x}{4^x+2} + \frac{2}{2 + 4^x} = \frac{4^x+2}{4^x+2} = 1$.
Let $S = f\left(\frac{1}{40}\right) + f\left(\frac{2}{40}\right) + \dots + f\left(\frac{39}{40}\right)$.
Pairing terms $f\left(\frac{k}{40}\right) + f\left(1 - \frac{k}{40}\right) = f\left(\frac{k}{40}\right) + f\left(\frac{40-k}{40}\right) = 1$.
There are $39$ terms in the sum. The middle term is $f\left(\frac{20}{40}\right) = f\left(\frac{1}{2}\right)$.
There are $19$ pairs that sum to $1$,plus the middle term $f\left(\frac{1}{2}\right)$.
So,$S = 19 + f\left(\frac{1}{2}\right)$.
Therefore,$S - f\left(\frac{1}{2}\right) = 19 + f\left(\frac{1}{2}\right) - f\left(\frac{1}{2}\right) = 19$.

(AB) Given $f(\cos 4 \theta) = \frac{2}{2-\sec^2 \theta}$.
We know that $\sec^2 \theta = \frac{1}{\cos^2 \theta} = \frac{2}{1+\cos 2 \theta}$.
Substituting this into the expression for $f(\cos 4 \theta)$:
$f(\cos 4 \theta) = \frac{2}{2 - \frac{2}{1+\cos 2 \theta}} = \frac{2(1+\cos 2 \theta)}{2(1+\cos 2 \theta) - 2} = \frac{1+\cos 2 \theta}{\cos 2 \theta} = 1 + \frac{1}{\cos 2 \theta}$.
Now,let $\cos 4 \theta = \frac{1}{3}$.
Using the identity $\cos 4 \theta = 2 \cos^2 2 \theta - 1$,we have $2 \cos^2 2 \theta - 1 = \frac{1}{3} \Rightarrow 2 \cos^2 2 \theta = \frac{4}{3} \Rightarrow \cos^2 2 \theta = \frac{2}{3}$.
Thus,$\cos 2 \theta = \pm \sqrt{\frac{2}{3}}$.
Substituting this back into the expression for $f(\cos 4 \theta)$:
$f\left(\frac{1}{3}\right) = 1 + \frac{1}{\pm \sqrt{2/3}} = 1 \pm \sqrt{\frac{3}{2}}$.
Therefore,the values are $1+\sqrt{\frac{3}{2}}$ and $1-\sqrt{\frac{3}{2}}$.

(A, C) Step $1$: Check for odd/even function.
$f(-x) = (\log(\sec(-x) + \tan(-x)))^3 = (\log(\sec x - \tan x))^3$.
Since $\sec x - \tan x = \frac{1}{\sec x + \tan x}$,we have $\log(\sec x - \tan x) = \log((\sec x + \tan x)^{-1}) = -\log(\sec x + \tan x)$.
Thus,$f(-x) = (-\log(\sec x + \tan x))^3 = -(\log(\sec x + \tan x))^3 = -f(x)$.
Therefore,$f(x)$ is an odd function.
Step $2$: Check for one-one function.
$f'(x) = 3(\log(\sec x + \tan x))^2 \cdot \frac{1}{\sec x + \tan x} \cdot (\sec x \tan x + \sec^2 x) = 3(\log(\sec x + \tan x))^2 \cdot \sec x$.
Since $\sec x > 0$ for $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ and $(\log(\sec x + \tan x))^2 \ge 0$,$f'(x) \ge 0$. The function is strictly increasing,so it is a one-one function.
Step $3$: Check for onto function.
As $x \rightarrow \frac{\pi}{2}^-$,$\sec x + \tan x \rightarrow \infty$,so $f(x) \rightarrow \infty$.
As $x \rightarrow -\frac{\pi}{2}^+$,$\sec x + \tan x \rightarrow 0^+$,so $\log(\sec x + \tan x) \rightarrow -\infty$,and $f(x) \rightarrow -\infty$.
Since the range is $(-\infty, \infty) = \mathbb{R}$,the function is onto.
Conclusion: $f(x)$ is an odd function and an onto function.

(D) $1$. Analyze $f_4(x)$:
$f_2(f_1(x)) = (f_1(x))^2 = \begin{cases} x^2 & x < 0 \\ e^{2x} & x \geq 0 \end{cases}$
$f_4(x) = \begin{cases} x^2 & x < 0 \\ e^{2x} - 1 & x \geq 0 \end{cases}$
At $x=0$,$f_4(0) = e^0 - 1 = 0$. $\lim_{x \to 0^-} f_4(x) = 0$. So $f_4$ is continuous. It is many-one (e.g.,$f_4(-1) = 1, f_4(\frac{1}{2}\ln 2) = 1$). Range is $[0, \infty)$,so it is onto. Thus $P \to 1$.
$2$. Analyze $f_3(x)$:
$f_3(x) = \begin{cases} \sin x & x < 0 \\ x & x \geq 0 \end{cases}$
At $x=0$,$f_3(0) = 0$. $\lim_{x \to 0^-} \sin x = 0$. Continuous. $f_3'(0^-) = \cos(0) = 1$,$f_3'(0^+) = 1$. Differentiable. It is many-one (e.g.,$f_3(-\pi) = 0, f_3(0) = 0$). Thus $Q \to 3$.
$3$. Analyze $f_2 \circ f_1(x)$:
$(f_2 \circ f_1)(x) = \begin{cases} x^2 & x < 0 \\ e^{2x} & x \geq 0 \end{cases}$
At $x=0$,$LHL = 0, RHL = 1$. Discontinuous. Thus $R \to 2$.
$4$. Analyze $f_2(x) = x^2$ on $[0, \infty)$:
It is strictly increasing,so one-one. Continuous. Thus $S \to 4$.
Matching: $P-1, Q-3, R-2, S-4$. Correct option is $(D)$.

(A) First,analyze the functions:
$f(x) = \begin{cases} x|x| \sin(1/x), & x \neq 0 \\ 0, & x=0 \end{cases}$
$g(x) = \begin{cases} 1-2x, & 0 \leq x \leq 1/2 \\ 0, & \text{otherwise} \end{cases}$
$g(1/2-x) = \begin{cases} 2x, & 0 \leq 1/2-x \leq 1/2 \\ 0, & \text{otherwise} \end{cases} = \begin{cases} 2x, & 0 \leq x \leq 1/2 \\ 0, & \text{otherwise} \end{cases}$
Thus,$g(x) + g(1/2-x) = \begin{cases} 1, & 0 \leq x \leq 1/2 \\ 0, & \text{otherwise} \end{cases}$
$(P)$ If $a=0, b=1, c=0, d=0$,then $h(x) = g(x) + g(1/2-x)$. The range is $\{0, 1\}$. Matches $(5)$.
$(Q)$ If $a=1, b=0, c=0, d=0$,then $h(x) = f(x)$. $f(x)$ is differentiable at $x=0$ because $\lim_{x \to 0} \frac{x|x| \sin(1/x) - 0}{x} = \lim_{x \to 0} |x| \sin(1/x) = 0$. Matches $(3)$.
$(R)$ If $a=0, b=0, c=1, d=0$,then $h(x) = x - g(x) = \begin{cases} x - (1-2x) = 3x-1, & 0 \leq x \leq 1/2 \\ x, & \text{otherwise} \end{cases}$. This function is onto. Matches $(2)$.
$(S)$ If $a=0, b=0, c=0, d=1$,then $h(x) = g(x) = \begin{cases} 1-2x, & 0 \leq x \leq 1/2 \\ 0, & \text{otherwise} \end{cases}$. The range is $[0, 1]$. Matches $(4)$.
Therefore,$(P) \to (5), (Q) \to (3), (R) \to (2), (S) \to (4)$.

(A) Given $f(x) = \frac{(x^{2023} + 2024x + 2025)}{e^{\pi x}(x^2 - x + 3)} (\sin x + 2)$.
Since $e^{\pi x} > 0$ for all $x \in R$ and the discriminant of $x^2 - x + 3$ is $D = 1 - 12 = -11 < 0$,$x^2 - x + 3$ is always positive.
Also,$-1 \leq \sin x \leq 1$,so $(\sin x + 2) \geq 1$,which means $(\sin x + 2)$ is never zero.
Thus,$f(x) = 0$ if and only if $x^{2023} + 2024x + 2025 = 0$.
Let $\phi(x) = x^{2023} + 2024x + 2025$.
Then $\phi'(x) = 2023x^{2022} + 2024$. Since $x^{2022} \geq 0$,$\phi'(x) > 0$ for all $x \in R$.
Therefore,$\phi(x)$ is a strictly increasing function.
Since $\phi(x)$ is continuous and strictly increasing,it takes every value in $R$ exactly once.
Hence,$\phi(x) = 0$ has exactly one real solution.
Therefore,the number of solutions of $f(x) = 0$ is $1$.

(A) First,we determine the number of pairs $(a, b)$ such that $a, b \in S$ and $|a-b| \geq 2$.
For each $a \in S$,the number of possible $b$ values is:
- If $a=1$,$b \in \{3, 4, 5, 6\}$ ($4$ values)
- If $a=2$,$b \in \{4, 5, 6\}$ ($3$ values)
- If $a=3$,$b \in \{1, 5, 6\}$ ($3$ values)
- If $a=4$,$b \in \{1, 2, 6\}$ ($3$ values)
- If $a=5$,$b \in \{1, 2, 3\}$ ($3$ values)
- If $a=6$,$b \in \{1, 2, 3, 4\}$ ($4$ values)
Total number of such pairs is $4+3+3+3+3+4 = 20$.
Since $R$ must have exactly $6$ elements chosen from these $20$ pairs,$n(X) = {}^{20}C_{6}$,so $m = 20$.
For $n(Y)$,the range of $R$ has exactly one element,meaning all $6$ pairs $(a, b)$ in $R$ must have the same $b$. However,for any fixed $b$,there are at most $5$ values of $a$ such that $|a-b| \geq 2$. Thus,$n(Y) = 0$.
For $n(Z)$,$R$ is a function,meaning for each $a \in S$,there is exactly one $b$ such that $(a, b) \in R$ and $|a-b| \geq 2$. The number of choices for each $a$ is $4, 3, 3, 3, 3, 4$ respectively. Thus,$n(Z) = 4 \times 3 \times 3 \times 3 \times 3 \times 4 = 1296 = 36^{2}$.
Therefore,$n(Y) + n(Z) = 0 + 36^{2} = 36^{2}$,so $|k| = 36$.

(B) The total number of functions from set $A$ to set $B$ is $|B|^{|A|} = 4^4 = 256$.
The number of one-one functions is $4! = 24$.
The number of many-one functions is $\text{Total} - \text{One-one} = 256 - 24 = 232$.
Now,we need to find the number of many-one functions such that $1 \in f(A)$.
This is equal to $(\text{Total many-one functions}) - (\text{Many-one functions where } 1 \notin f(A))$.
If $1 \notin f(A)$,then the range of $f$ is a subset of $\{4, 9, 16\}$.
The total number of functions from $A$ to $\{4, 9, 16\}$ is $3^4 = 81$.
Among these $81$ functions,the number of one-one functions is $0$ (since $|A| > |\{4, 9, 16\}|$).
Thus,all $81$ functions are many-one.
Therefore,the number of many-one functions such that $1 \in f(A)$ is $232 - 81 = 151$.

(A) Given $f(x) = \frac{2^{x+2} + 16}{2^{2x+1} + 2^{x+4} + 32} = \frac{4 \cdot 2^x + 16}{2 \cdot (2^x)^2 + 16 \cdot 2^x + 32} = \frac{4(2^x + 4)}{2((2^x)^2 + 8 \cdot 2^x + 16)} = \frac{2(2^x + 4)}{(2^x + 4)^2} = \frac{2}{2^x + 4}$.
Consider $f(4-x) = \frac{2}{2^{4-x} + 4} = \frac{2}{\frac{16}{2^x} + 4} = \frac{2 \cdot 2^x}{16 + 4 \cdot 2^x} = \frac{2^x}{8 + 2 \cdot 2^x} = \frac{2^x}{2(4 + 2^x)}$.
Then $f(x) + f(4-x) = \frac{2}{2^x + 4} + \frac{2^x}{2(2^x + 4)} = \frac{4 + 2^x}{2(2^x + 4)} = \frac{1}{2}$.
The sum is $S = \sum_{k=1}^{59} f \left( \frac{k}{15} \right)$. Note that $f(x) + f(4-x) = \frac{1}{2}$ implies $f \left( \frac{k}{15} \right) + f \left( 4 - \frac{k}{15} \right) = f \left( \frac{k}{15} \right) + f \left( \frac{60-k}{15} \right) = \frac{1}{2}$.
There are $59$ terms. The middle term is $f \left( \frac{30}{15} \right) = f(2) = \frac{2}{2^2 + 4} = \frac{2}{8} = \frac{1}{4}$.
There are $29$ pairs summing to $\frac{1}{2}$ and one middle term $\frac{1}{4}$.
$S = 29 \cdot \frac{1}{2} + \frac{1}{4} = \frac{58 + 1}{4} = \frac{59}{4}$.
The required value is $8 \cdot S = 8 \cdot \frac{59}{4} = 2 \cdot 59 = 118$.

(B) Given $f(x) = \frac{2^x}{2^x + \sqrt{2}}$.
Consider $f(1-x) = \frac{2^{1-x}}{2^{1-x} + \sqrt{2}} = \frac{2/2^x}{2/2^x + \sqrt{2}} = \frac{2}{2 + \sqrt{2} \cdot 2^x} = \frac{\sqrt{2}}{\sqrt{2} + 2^x}$.
Thus,$f(x) + f(1-x) = \frac{2^x}{2^x + \sqrt{2}} + \frac{\sqrt{2}}{2^x + \sqrt{2}} = \frac{2^x + \sqrt{2}}{2^x + \sqrt{2}} = 1$.
We need to evaluate $S = \sum_{k=1}^{81} f\left(\frac{k}{82}\right) = f\left(\frac{1}{82}\right) + f\left(\frac{2}{82}\right) + \dots + f\left(\frac{81}{82}\right)$.
Pairing terms $f\left(\frac{k}{82}\right) + f\left(1 - \frac{k}{82}\right) = f\left(\frac{k}{82}\right) + f\left(\frac{82-k}{82}\right) = 1$.
There are $40$ such pairs (for $k=1$ to $40$) and the middle term $f\left(\frac{41}{82}\right) = f\left(\frac{1}{2}\right)$.
$f\left(\frac{1}{2}\right) = \frac{2^{1/2}}{2^{1/2} + \sqrt{2}} = \frac{\sqrt{2}}{\sqrt{2} + \sqrt{2}} = \frac{1}{2}$.
So,$S = 40 \times 1 + \frac{1}{2} = \frac{81}{2}$.

(D) Given $S = \{0, 1, 2, 3, \dots\}$.
From the relation $\log_e y = x \log_e \left(\frac{2}{5}\right)$,we have $\log_e y = \log_e \left(\left(\frac{2}{5}\right)^x\right)$.
This implies $y = \left(\frac{2}{5}\right)^x$.
Since $x \in S$,the range of $R$ is the set of values $y = \left(\frac{2}{5}\right)^x$ for $x = 0, 1, 2, \dots$.
Range $= \left\{ \left(\frac{2}{5}\right)^0, \left(\frac{2}{5}\right)^1, \left(\frac{2}{5}\right)^2, \dots \right\} = \left\{ 1, \frac{2}{5}, \left(\frac{2}{5}\right)^2, \dots \right\}$.
This is an infinite geometric progression with first term $a = 1$ and common ratio $r = \frac{2}{5}$.
The sum of the elements in the range is $S_{\infty} = \frac{a}{1 - r} = \frac{1}{1 - \frac{2}{5}} = \frac{1}{\frac{3}{5}} = \frac{5}{3}$.

(B) Given $f(x+y)=f(x)f(y)$ and $f(x)>0$,the function is of the form $f(x)=k^x$ for some $k>0$.
Since $f(a_{31})=64 f(a_{25})$,we have $k^{a+30d}=64 k^{a+24d}$,where $a$ is the first term and $d$ is the common difference.
This simplifies to $k^{6d}=64$,so $k^d=2$.
The sum $\sum_{i=1}^{50} f(a_i) = k^a + k^{a+d} + \ldots + k^{a+49d} = k^a \frac{(k^d)^{50}-1}{k^d-1} = k^a \frac{2^{50}-1}{2-1} = k^a(2^{50}-1)$.
Given $k^a(2^{50}-1) = 3(2^{25}+1)$,we have $k^a(2^{25}-1)(2^{25}+1) = 3(2^{25}+1)$,so $k^a = \frac{3}{2^{25}-1}$.
We need to find $\sum_{i=6}^{30} f(a_i) = k^{a+5d} + k^{a+6d} + \ldots + k^{a+29d} = k^{a+5d} \frac{(k^d)^{25}-1}{k^d-1} = k^a (k^d)^5 (2^{25}-1)$.
Substituting the values: $\frac{3}{2^{25}-1} \cdot 2^5 \cdot (2^{25}-1) = 3 \cdot 32 = 96$.

(D) Given $f(x) = \log_{e}\left(\frac{1-x}{1+x}\right)$.
We need to evaluate $f\left(\frac{2x}{1+x^2}\right)$.
Substituting $\frac{2x}{1+x^2}$ for $x$ in $f(x)$:
$f\left(\frac{2x}{1+x^2}\right) = \log_{e}\left(\frac{1-\frac{2x}{1+x^2}}{1+\frac{2x}{1+x^2}}\right)$
$= \log_{e}\left(\frac{\frac{1+x^2-2x}{1+x^2}}{\frac{1+x^2+2x}{1+x^2}}\right)$
$= \log_{e}\left(\frac{(1-x)^2}{(1+x)^2}\right)$
$= \log_{e}\left(\left(\frac{1-x}{1+x}\right)^2\right)$
$= 2\log_{e}\left(\frac{1-x}{1+x}\right)$
$= 2f(x)$.

(A) To determine if the function $f(x) = \sec \left[ \log \left( x + \sqrt{1 + x^2} \right) \right]$ is even or odd,we evaluate $f(-x)$.
First,consider the inner function $g(x) = \log \left( x + \sqrt{1 + x^2} \right)$.
$g(-x) = \log \left( -x + \sqrt{1 + (-x)^2} \right) = \log \left( \sqrt{1 + x^2} - x \right)$.
Multiply and divide by $\left( \sqrt{1 + x^2} + x \right)$:
$g(-x) = \log \left( \frac{(\sqrt{1 + x^2} - x)(\sqrt{1 + x^2} + x)}{\sqrt{1 + x^2} + x} \right) = \log \left( \frac{1 + x^2 - x^2}{\sqrt{1 + x^2} + x} \right) = \log \left( \frac{1}{\sqrt{1 + x^2} + x} \right) = -\log \left( x + \sqrt{1 + x^2} \right) = -g(x)$.
Since $g(x)$ is an odd function,we have $f(-x) = \sec(g(-x)) = \sec(-g(x))$.
Since $\sec(\theta)$ is an even function,$\sec(-g(x)) = \sec(g(x)) = f(x)$.
Therefore,$f(-x) = f(x)$,which means the function is an even function.

(D) Given $f(x) = \cos(\log x)$.
We need to evaluate $f(x^2) \cdot f(y^2) - \frac{1}{2} \left[ f\left(\frac{x^2}{y^2}\right) + f(x^2 y^2) \right]$.
First,calculate the terms:
$f(x^2) = \cos(\log x^2) = \cos(2 \log x)$
$f(y^2) = \cos(\log y^2) = \cos(2 \log y)$
$f\left(\frac{x^2}{y^2}\right) = \cos(\log \frac{x^2}{y^2}) = \cos(2 \log x - 2 \log y)$
$f(x^2 y^2) = \cos(\log x^2 y^2) = \cos(2 \log x + 2 \log y)$
Let $A = 2 \log x$ and $B = 2 \log y$.
The expression becomes $\cos A \cdot \cos B - \frac{1}{2} [\cos(A - B) + \cos(A + B)]$.
Using the trigonometric identity $\cos(A - B) + \cos(A + B) = 2 \cos A \cos B$,we get:
$\cos A \cos B - \frac{1}{2} [2 \cos A \cos B] = \cos A \cos B - \cos A \cos B = 0$.

(C) Given $f(x) = \cos(\log x)$.
We need to evaluate the expression $E = f(x) \cdot f(y) - \frac{1}{2} \left( f\left(\frac{x}{y}\right) + f(xy) \right)$.
Substitute the function definition:
$E = \cos(\log x) \cdot \cos(\log y) - \frac{1}{2} \left( \cos(\log(x/y)) + \cos(\log(xy)) \right)$.
Using logarithmic properties $\log(x/y) = \log x - \log y$ and $\log(xy) = \log x + \log y$:
$E = \cos(\log x) \cdot \cos(\log y) - \frac{1}{2} \left( \cos(\log x - \log y) + \cos(\log x + \log y) \right)$.
Apply the trigonometric identity $\cos(A - B) + \cos(A + B) = 2 \cos A \cos B$ where $A = \log x$ and $B = \log y$:
$E = \cos(\log x) \cdot \cos(\log y) - \frac{1}{2} \left( 2 \cos(\log x) \cos(\log y) \right)$.
$E = \cos(\log x) \cdot \cos(\log y) - \cos(\log x) \cdot \cos(\log y) = 0$.

(B) Given $f(x)=3[x]+\{x+1\}$.
For $x=-1.32$,we have $[x]=[-1.32]=-2$.
Also,$x+1=-1.32+1=-0.32$.
We know that for any real number $y$,$y=[y]+\{y\}$,so $\{y\}=y-[y]$.
Thus,$\{x+1\} = (x+1) - [x+1] = -0.32 - [-0.32] = -0.32 - (-1) = 0.68$.
Substituting these values into the function:
$f(-1.32) = 3(-2) + 0.68$
$f(-1.32) = -6 + 0.68 = -5.32$.
Wait,checking the provided options,let us re-evaluate the expression $f(x)=3[x]+\{x+1\}$.
If the expression was $f(x)=3[x]+5\{x+1\}$ as per the provided solution logic:
$f(-1.32) = 3(-2) + 5(0.68) = -6 + 3.4 = -2.6$.

(C) We know that for any real number $x$,$x = [x] + \{x\}$,where $[x]$ is the greatest integer function and $\{x\}$ is the fractional part function.
For $x = -1.4$,the greatest integer $[x] = [-1.4] = -2$.
Thus,the fractional part is $\{x\} = x - [x] = -1.4 - (-2) = 0.6$.
Given the function $f(x) = 2\{x\} + 5x$.
Substituting $x = -1.4$ and $\{x\} = 0.6$ into the function:
$f(-1.4) = 2(0.6) + 5(-1.4)$
$f(-1.4) = 1.2 - 7.0$
$f(-1.4) = -5.8$.

(D) Given $f(x)=\frac{2^{x}+2^{-x}}{2}$.
Now,$f(x+y)=\frac{2^{x+y}+2^{-(x+y)}}{2}$ and $f(x-y)=\frac{2^{x-y}+2^{-(x-y)}}{2}$.
Therefore,$f(x+y) \cdot f(x-y) = \frac{2^{x+y}+2^{-x-y}}{2} \cdot \frac{2^{x-y}+2^{-x+y}}{2}$.
$= \frac{1}{4} [2^{x+y} \cdot 2^{x-y} + 2^{x+y} \cdot 2^{-x+y} + 2^{-x-y} \cdot 2^{x-y} + 2^{-x-y} \cdot 2^{-x+y}]$.
$= \frac{1}{4} [2^{2x} + 2^{2y} + 2^{-2y} + 2^{-2x}]$.
$= \frac{1}{4} [(2^{2x} + 2^{-2x}) + (2^{2y} + 2^{-2y})]$.
$= \frac{1}{2} [\frac{2^{2x} + 2^{-2x}}{2} + \frac{2^{2y} + 2^{-2y}}{2}]$.
$= \frac{1}{2} [f(2x) + f(2y)]$.

(C) Let $ A $ be a set with $ n = 6 $ elements.
Total number of functions from $ A $ to $ A $ is given by $ n^{n} = 6^{6} $.
$A$ function is a bijection if it is both one-to-one and onto. For a finite set $ A $ with $ n $ elements,the number of bijections from $ A $ to $ A $ is $ n ! = 6 ! $.
The number of functions that are not bijections is equal to the total number of functions minus the number of bijections.
Therefore,the required number of functions is $ 6^{6} - 6 ! $.

(C) Given $f(x) = x^2+2x+2 = (x+1)^2+1$. The minimum value of $f(x)$ is $a = 1$.
Given $g(x) = -(x^2-2x+1) = -(x-1)^2$. The maximum value of $g(x)$ is $b = 0$.
Let $y = \frac{f(x)}{g(x)} = \frac{x^2+2x+2}{-x^2+2x-1}$.
$y(-x^2+2x-1) = x^2+2x+2$
$-yx^2+2xy-y = x^2+2x+2$
$(1+y)x^2 + (2-2y)x + (2+y) = 0$.
For $x$ to be real,the discriminant $D \geq 0$:
$(2-2y)^2 - 4(1+y)(2+y) \geq 0$
$4(1-y)^2 - 4(y^2+3y+2) \geq 0$
$1-2y+y^2 - y^2-3y-2 \geq 0$
$-5y-1 \geq 0 \implies y \leq -\frac{1}{5}$.
The extreme value $c = -\frac{1}{5}$.
Thus,$a+2b+5c+4 = 1 + 2(0) + 5(-\frac{1}{5}) + 4 = 1 - 1 + 4 = 4$.

(C) Given the function $f(x) = \begin{cases} 1+\frac{2x}{k}, & 0 \leq x < 1 \\ kx, & 1 \leq x < 2 \end{cases}$ where $k>0$.
We are given that $\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{+}} f(x)$.
Calculating the left-hand limit:
$\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{-}} (1+\frac{2x}{k}) = 1+\frac{2}{k}$.
Calculating the right-hand limit:
$\lim_{x \rightarrow 1^{+}} f(x) = \lim_{x \rightarrow 1^{+}} (kx) = k$.
Equating the two limits:
$1+\frac{2}{k} = k$
Multiplying by $k$ (since $k>0$):
$k+2 = k^2$
Rearranging the terms:
$k^2 - k - 2 = 0$
Factoring the quadratic equation:
$(k-2)(k+1) = 0$.
Since $k>0$,we have $k=2$.
Therefore,$k^2 = 2^2 = 4$.

(D) Given the equation $3[2x - 2] + 1 = 2[2x - 1] - 1$.
Using the property $[t - m] = [t] - m$,we have $[2x - 2] = [2x] - 2$.
Substituting this into the equation: $3([2x] - 2) + 1 = 2([2x] - 1) - 1$.
$3[2x] - 6 + 1 = 2[2x] - 2 - 1$.
$3[2x] - 5 = 2[2x] - 3$.
$[2x] = 2$.
This implies $2 \le 2x < 3$,so $1 \le x < 1.5$.
Now,$k = 2[2x - 1] - 1 = 2([2x] - 1) - 1 = 2(2 - 1) - 1 = 2(1) - 1 = 1$.
Then $f(x) = [k + 5x] = [1 + 5x]$.
Since $1 \le x < 1.5$,we have $5 \le 5x < 7.5$.
Adding $1$,we get $6 \le 1 + 5x < 8.5$.
The possible integer values for $[1 + 5x]$ are $\{6, 7, 8\}$.

(A) Let $f(x) = y$. Since $f: R \rightarrow [-1, 1]$ is surjective,$y$ takes all values in $[-1, 1]$.
Given $\sin \left(g(x) - \frac{\pi}{3}\right) = \frac{y}{2} \sqrt{4 - y^2}$.
Let $y = 2 \sin \theta$. Since $y \in [-1, 1]$,$\sin \theta \in \left[-\frac{1}{2}, \frac{1}{2}\right]$,so $\theta \in \left[-\frac{\pi}{6}, \frac{\pi}{6}\right]$.
Then $\frac{y}{2} \sqrt{4 - y^2} = \sin \theta \sqrt{4 - 4 \sin^2 \theta} = \sin \theta \sqrt{4 \cos^2 \theta} = 2 \sin \theta \cos \theta = \sin(2 \theta)$.
Thus,$\sin \left(g(x) - \frac{\pi}{3}\right) = \sin(2 \theta)$.
Since $\theta \in \left[-\frac{\pi}{6}, \frac{\pi}{6}\right]$,$2 \theta \in \left[-\frac{\pi}{3}, \frac{\pi}{3}\right]$.
Therefore,$g(x) - \frac{\pi}{3} = 2 \theta \in \left[-\frac{\pi}{3}, \frac{\pi}{3}\right]$.
$g(x) \in \left[-\frac{\pi}{3} + \frac{\pi}{3}, \frac{\pi}{3} + \frac{\pi}{3}\right] = \left[0, \frac{2 \pi}{3}\right]$.
Since $g$ is surjective,the codomain $A = \left[0, \frac{2 \pi}{3}\right]$.

(D) Two functions $f$ and $g$ are said to be equal if and only if:
$(i)$ The domain of $f$ is equal to the domain of $g$.
(ii) The codomain of $f$ is equal to the codomain of $g$ (usually implied).
(iii) $f(x) = g(x)$ for every $x$ in their common domain.
Since the definition requires the domains to be identical and the functional values to be equal for all $x$ within that domain,all three conditions listed are necessary for the equality of the functions.

(A) If $f$ is an even function from $R$ to $R$,then $f(0)$ must be equal to $0$.
Since we know that if a function $f(x)$ is even,then $f(-x)=f(x)$.
Now,if we assume $f(x)=\cos x$,then $f(-x)=\cos(-x)=\cos x=f(x)$.
Thus,$f(x)=\cos x$ is an even function.
However,$f(0)=\cos 0=1 \neq 0$.
Therefore,the given statement is false.
$(b)$ $f: R \rightarrow R$,$f(x)=x-[x]$.
Since $x=[x]+\{x\}$,where $\{x\}$ is the fractional part function,we have $f(x)=\{x\}$.
Since $\{x\}$ is a periodic function,$f(x)$ is a periodic function.
Therefore,the given statement is true.
$(c)$ If $f: R \rightarrow R$ is an odd function,then $f(0)=0$.
Since $f(x)$ is an odd function,$f(-x)=-f(x)$.
Putting $x=0$,we get $f(0)=-f(0)$,which implies $2f(0)=0$,so $f(0)=0$.
Therefore,the given statement is true.
$(d)$ The number of onto functions from $\{1,2,3,4,5,6\}$ to $\{1,2\}$ is $62$.
Let $A=\{1,2,3,4,5,6\}$ and $B=\{1,2\}$,so $n(A)=6$ and $n(B)=2$.
The number of onto functions from a set with $m$ elements to a set with $n$ elements is $n^m - \binom{n}{1}(n-1)^m + \binom{n}{2}(n-2)^m - \dots$.
For $n=2$ and $m=6$,the number of onto functions is $2^6 - \binom{2}{1}(1)^6 = 64 - 2 = 62$.
Therefore,the given statement is true.

(C) Step $1$: Analyze $A$. Let $f(x) = \frac{x}{e^x-1} + \frac{x}{2} + 4$. Check $f(-x) = \frac{-x}{e^{-x}-1} + \frac{-x}{2} + 4 = \frac{-x e^x}{1-e^x} - \frac{x}{2} + 4 = \frac{x e^x}{e^x-1} - \frac{x}{2} + 4$. Since $\frac{e^x}{e^x-1} = 1 + \frac{1}{e^x-1}$,$f(-x) = x(1 + \frac{1}{e^x-1}) - \frac{x}{2} + 4 = x + \frac{x}{e^x-1} - \frac{x}{2} + 4 = \frac{x}{e^x-1} + \frac{x}{2} + 4 = f(x)$. Thus,$A$ is an even function $(II)$.
Step $2$: Analyze $B$. Let $f(x) = \tan^{-1}(\log|x+\sqrt{x^2+1}|)$. Since $x > 0$,the domain is restricted. However,considering the nature of the expression,it is neither odd nor even $(I)$.
Step $3$: Analyze $C$. For $3 < x < 5$,$|x-2| = x-2$,$|x-3| = x-3$,$|x-5| = -(x-5) = 5-x$. So $f(x) = (x-2) + (x-3) + (5-x) = x$. This is the identity function $(IV)$.
Step $4$: Analyze $D$. $f(x) = \sin 2x + \sin^2 x + \cos 3x$. This is a combination of trigonometric functions and is neither odd nor even $(I)$.
Wait,re-evaluating $A$: $f(x) = \frac{x}{e^x-1} + \frac{x}{2} + 4$. This is a known even function. $C$ is identity. $B$ is neither. $D$ is neither. Checking options,$A-II$ is correct. $C-IV$ is correct. Thus,option $C$ is the correct match.

(A) Given $g(x) = 1 + x - [x]$.
We know that the fractional part function is defined as $\{x\} = x - [x]$.
Thus,$g(x) = 1 + \{x\}$.
Since $0 \le \{x\} < 1$,it follows that $1 \le g(x) < 2$.
For any $x \in \mathbb{R}$,$g(x)$ is always greater than $0$.
Now,consider the function $f(x) = \begin{cases} -1, & x < 0 \\ 0, & x = 0 \\ 1, & x > 0 \end{cases}$.
Since $g(x) > 0$ for all $x$,we evaluate $f(g(x))$ using the condition $x > 0$ for the function $f$.
Therefore,$f(g(x)) = 1$ for all $x$.

(B) Given $f(a) = \log \left| \frac{1-a}{1+a} \right|$.
We want to solve $f\left( \frac{2a}{1+a^2} \right) > 0$.
Substituting $x = \frac{2a}{1+a^2}$,we have $\log \left| \frac{1 - \frac{2a}{1+a^2}}{1 + \frac{2a}{1+a^2}} \right| > 0$.
This implies $\left| \frac{1+a^2-2a}{1+a^2+2a} \right| > 1$,which simplifies to $\left| \frac{(1-a)^2}{(1+a)^2} \right| > 1$.
Since both numerator and denominator are squares,we have $\left| \frac{1-a}{1+a} \right| > 1$.
Squaring both sides,$\frac{(1-a)^2}{(1+a)^2} > 1 \Rightarrow (1-a)^2 > (1+a)^2$.
$1 - 2a + a^2 > 1 + 2a + a^2$.
$-2a > 2a \Rightarrow 4a < 0 \Rightarrow a < 0$.
Also,we must satisfy the domain constraints $a \neq \pm 1$ and $\frac{2a}{1+a^2} \neq \pm 1$.
For $a < 0$,the condition $a \neq -1$ must be excluded.
Thus,the solution is $a \in (-\infty, 0) - \{-1\}$.

(C) Given the function $f(x) = \begin{cases} -1, & -2 \leq x \leq 0 \\ x-1, & 0 < x \leq 2 \end{cases}$.
We need to find the set of $x$ such that $x \leq 0$ and $f(|x|) = x$.
Since $x \leq 0$,we have $|x| = -x$. Since $x \in [-2, 0]$,then $|x| \in [0, 2]$.
For $0 < |x| \leq 2$,$f(|x|) = |x| - 1 = -x - 1$.
Setting $f(|x|) = x$,we get $-x - 1 = x$,which implies $2x = -1$,so $x = -\frac{1}{2}$.
Checking the condition $x \leq 0$,we see that $x = -\frac{1}{2}$ satisfies this.
If $x = 0$,then $f(|0|) = f(0) = -1$. But $x = 0$,so $f(0) \neq 0$.
Thus,the set is $\{-\frac{1}{2}\}$.

(D) For a finite set $S$,a function $f: S \rightarrow S$ is a mapping from a finite set to itself.
If $f$ is injective,it must be surjective (by the Pigeonhole Principle for finite sets),and vice versa.
Since $f$ is a non-identity function,it could be a permutation (bijective) other than the identity,or it could be neither injective nor surjective.
Because the problem does not specify the properties of $f$ beyond being a non-identity function,it is impossible to determine its specific type.
Therefore,the data is insufficient to conclude whether it is injective,surjective,bijective,or none of these.
Hence,option $D$ is correct.

(A) Given that,$n(G) = 3$ and $n(A) = 4$.
$(A)$ The number of functions from $G \times G$ to $G$ is $n(G)^{n(G \times G)} = 3^{(3 \times 3)} = 3^9 = 19683$. The number of bijective functions from $G \times G$ to $G$ is $0$ (since $n(G \times G) = 9 \neq n(G) = 3$). Thus,the number of non-bijective functions is $19683 - 0 = 19683$. So,$A \rightarrow V$.
$(B)$ The number of bijective functions from $A$ to $A$ is $n(A)! = 4! = 4 \times 3 \times 2 \times 1 = 24$. So,$B \rightarrow I$.
$(C)$ The number of functions from $G$ to $G \times A$ is $n(G \times A)^{n(G)} = (3 \times 4)^3 = 12^3 = 1728$. So,$C \rightarrow III$.
$(D)$ The number of surjective functions from $A$ to $A \times A$ is $0$ because $n(A) = 4$ and $n(A \times A) = 16$. Since $n(A) < n(A \times A)$,no surjective function exists. So,$D \rightarrow II$.
Therefore,the correct match is $A-V, B-I, C-III, D-II$.

(D) Given: $f(x) = x + \frac{1}{x}$
Squaring both sides,we get:
$(f(x))^2 = (x + \frac{1}{x})^2$
$(f(x))^2 = x^2 + \frac{1}{x^2} + 2 \cdot x \cdot \frac{1}{x}$
$(f(x))^2 = x^2 + \frac{1}{x^2} + 2$
Now,consider $f(x^2) = x^2 + \frac{1}{x^2}$ and $f(1) = 1 + \frac{1}{1} = 2$.
Substituting these into the expression:
$(f(x))^2 = f(x^2) + f(1)$

(C) Given,$f(x) = \frac{4^x}{4^x + 2}$.
We observe that $f(1 - x) = \frac{4^{1 - x}}{4^{1 - x} + 2} = \frac{4/4^x}{4/4^x + 2} = \frac{4}{4 + 2 \cdot 4^x} = \frac{2}{2 + 4^x}$.
Also,$f(x) + f(1 - x) = \frac{4^x}{4^x + 2} + \frac{2}{4^x + 2} = \frac{4^x + 2}{4^x + 2} = 1$.
Therefore,$f(1 - x) = 1 - f(x)$.
We need to evaluate $S = f(\frac{1}{4}) + 2 f(\frac{1}{2}) + f(\frac{3}{4})$.
Since $f(\frac{1}{4}) + f(1 - \frac{1}{4}) = f(\frac{1}{4}) + f(\frac{3}{4}) = 1$,we can substitute $f(\frac{3}{4}) = 1 - f(\frac{1}{4})$.
Thus,$S = f(\frac{1}{4}) + 2 f(\frac{1}{2}) + (1 - f(\frac{1}{4})) = 1 + 2 f(\frac{1}{2})$.
Now,calculate $f(\frac{1}{2}) = \frac{4^{1/2}}{4^{1/2} + 2} = \frac{2}{2 + 2} = \frac{2}{4} = \frac{1}{2}$.
Substituting this back,$S = 1 + 2(\frac{1}{2}) = 1 + 1 = 2$.

(B) Given $f(x) = \frac{2020^x}{2020^x + \sqrt{2020}}$.
Note that $f(1-x) = \frac{2020^{1-x}}{2020^{1-x} + \sqrt{2020}} = \frac{2020}{2020 + \sqrt{2020} \cdot 2020^x} = \frac{\sqrt{2020}}{\sqrt{2020} + 2020^x}$.
Thus,$f(x) + f(1-x) = \frac{2020^x + \sqrt{2020}}{2020^x + \sqrt{2020}} = 1$.
We need to evaluate $S = \sum_{r=1}^{4039} 2 f\left(\frac{r}{4040}\right) = 2 \sum_{r=1}^{4039} f\left(\frac{r}{4040}\right)$.
Pairing terms $f\left(\frac{r}{4040}\right)$ and $f\left(\frac{4040-r}{4040}\right)$,we have $f\left(\frac{r}{4040}\right) + f\left(1 - \frac{r}{4040}\right) = 1$.
There are $2019$ such pairs for $r=1$ to $2019$,plus the middle term $f\left(\frac{2020}{4040}\right) = f\left(\frac{1}{2}\right)$.
$f\left(\frac{1}{2}\right) = \frac{\sqrt{2020}}{\sqrt{2020} + \sqrt{2020}} = \frac{1}{2}$.
So,$\sum_{r=1}^{4039} f\left(\frac{r}{4040}\right) = 2019 \times 1 + \frac{1}{2} = 2019.5$.
Therefore,$S = 2 \times 2019.5 = 4039$.

(A) Given $f(x) = x + \frac{1}{x}$.
We need to evaluate $f^4(x) - f(x^4) - 4f^2(x)$.
First,calculate $f^2(x) = (x + \frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2}$.
Next,calculate $f^4(x) = (f^2(x))^2 = (x^2 + 2 + \frac{1}{x^2})^2 = (x^2)^2 + 2^2 + (\frac{1}{x^2})^2 + 2(x^2)(2) + 2(x^2)(\frac{1}{x^2}) + 2(2)(\frac{1}{x^2}) = x^4 + 4 + \frac{1}{x^4} + 4x^2 + 2 + \frac{4}{x^2} = x^4 + 4x^2 + 6 + \frac{4}{x^2} + \frac{1}{x^4}$.
Now,calculate $f(x^4) = x^4 + \frac{1}{x^4}$.
Also,$4f^2(x) = 4(x^2 + 2 + \frac{1}{x^2}) = 4x^2 + 8 + \frac{4}{x^2}$.
Substitute these into the expression:
$f^4(x) - f(x^4) - 4f^2(x) = (x^4 + 4x^2 + 6 + \frac{4}{x^2} + \frac{1}{x^4}) - (x^4 + \frac{1}{x^4}) - (4x^2 + 8 + \frac{4}{x^2})$.
$= x^4 + 4x^2 + 6 + \frac{4}{x^2} + \frac{1}{x^4} - x^4 - \frac{1}{x^4} - 4x^2 - 8 - \frac{4}{x^2}$.
$= 6 - 8 = -2$.

(C) Given $f(x) = 2x + 3$.
Substituting the function into the equation $f(x^2) - 2f(\frac{x}{2}) - 1 = 0$:
$(2x^2 + 3) - 2(2(\frac{x}{2}) + 3) - 1 = 0$
$2x^2 + 3 - 2(x + 3) - 1 = 0$
$2x^2 + 3 - 2x - 6 - 1 = 0$
$2x^2 - 2x - 4 = 0$
Dividing by $2$:
$x^2 - x - 2 = 0$
$(x - 2)(x + 1) = 0$
So,the roots are $\alpha = 2$ and $\beta = -1$.
Therefore,$\alpha^2 + \beta^2 = (2)^2 + (-1)^2 = 4 + 1 = 5$.

(C) relation is defined as a subset of the Cartesian product of two sets $A \times B$.
$A$ function is a specific type of relation where every element in the domain has exactly one image in the codomain.
Therefore,every function is a relation,but not every relation is a function.
Thus,statement $(ii)$ is true,while $(i)$ and $(iii)$ are false.

(A) Given functions are $f(x) = |x|$ and $g(x) = |x| + a$ with $a > 0$.
For $0 \leq x \leq b$,the region is defined by $g(x) \leq y \leq f(x)$,which implies $|x| + a \leq y \leq |x|$.
However,since $a > 0$,$|x| + a$ is always greater than $|x|$. Thus,the condition $g(x) \leq y \leq f(x)$ is impossible for $a > 0$ as it implies $|x| + a \leq y \leq |x|$,which is a contradiction.
Re-evaluating the problem statement based on the provided image,the region is bounded by $x=0$,$x=b$,$y=|x|$,and $y=|x|+a$.
For $x \geq 0$,$f(x) = x$ and $g(x) = x + a$.
The region is bounded by $x=0$,$x=b$,$y=x$,and $y=x+a$.
These lines are $x=0$,$x=b$,$y-x=0$,and $y-x=a$.
The lines $y-x=0$ and $y-x=a$ are parallel,and the lines $x=0$ and $x=b$ are parallel.
Therefore,the region is a parallelogram.

(B) Given the equation $x - f(x) = 19[\frac{x}{19}] - 90[\frac{f(x)}{90}]$.
Rearranging the terms,we get $x - 19[\frac{x}{19}] = f(x) - 90[\frac{f(x)}{90}]$.
This is equivalent to $x \pmod{19} = f(x) \pmod{90}$.
Let $x = 1990$. Then $1990 = 19 \times 104 + 14$,so $1990 \equiv 14 \pmod{19}$.
Thus,$f(1990) \equiv 14 \pmod{90}$.
This means $f(1990) = 90k + 14$ for some integer $k$.
We are given $1990 < f(1990) < 2100$.
Substituting $f(1990) = 90k + 14$,we get $1990 < 90k + 14 < 2100$.
$1976 < 90k < 2086$.
Dividing by $90$,we get $21.95 < k < 23.17$.
Since $k$ must be an integer,$k = 22$ or $k = 23$.
If $k = 22$,$f(1990) = 90(22) + 14 = 1980 + 14 = 1994$.
If $k = 23$,$f(1990) = 90(23) + 14 = 2070 + 14 = 2084$.
Both values satisfy the condition $1990 < f(1990) < 2100$.
Thus,there are $2$ possible values for $f(1990)$.

(A) Given $f(x) = \{x + [\frac{x}{1+x^2}]\}$.
Since $[n + I] = [n] + I$ for any integer $I$,we have $f(x) = \{x + [\frac{x}{1+x^2}]\} = \{x\} + [\frac{x}{1+x^2}] - [\{x\} + [\frac{x}{1+x^2}]] = \{x\}$.
Thus,$f(x) = \{x\} = x - [x]$.
Now,calculate $f(\sqrt{2}) = \{\sqrt{2}\} = \sqrt{2} - [\sqrt{2}] = 1.414 - 1 = 0.414$.
Next,calculate $f(-\sqrt{3}) = \{-\sqrt{3}\} = -\sqrt{3} - [-\sqrt{3}] = -1.732 - (-2) = -1.732 + 2 = 0.268$.
Therefore,$f(\sqrt{2}) + f(-\sqrt{3}) = 0.414 + 0.268 = 0.682$.

(A) $A)$ Range of $\cos ^2 x \in [0, 1]$.
$\Rightarrow 1+\cos ^2 x \in [1, 2]$.
$\Rightarrow [1+\cos ^2 x] \in \{1, 2\}$.
$\Rightarrow \sec ^{-1}[1+\cos ^2 x] \in \{\sec ^{-1} 1, \sec ^{-1} 2\}$. Thus,$A-V$.
$B)$ Given $f(x+\frac{1}{x}) = x^2+\frac{1}{x^2} = (x+\frac{1}{x})^2 - 2$.
Let $t = x+\frac{1}{x}$. Since $|x+\frac{1}{x}| \ge 2$,the domain of $f(t)$ is $(-\infty, -2] \cup [2, \infty)$. However,if we consider the function definition $f(t) = t^2-2$,the domain is $R$. Given the context of such problems,$f(x) = x^2-2$ is defined for all $x \in R$. Thus,$B-IV$.
$C)$ $f(x+y) = f(x)+f(y)$ is Cauchy's functional equation,so $f(x) = kx$.
Given $f(1) = 5$,we have $k(1) = 5 \Rightarrow k = 5$.
So $f(x) = 5x$,which is an odd function since $f(-x) = 5(-x) = -f(x)$. Thus,$C-I$.
$D)$ $\sin ^{-1} x - \cos ^{-1} x + \sin ^{-1}(1-x) = 0$.
For $x=0$: $\sin ^{-1}(0) - \cos ^{-1}(0) + \sin ^{-1}(1) = 0 - \frac{\pi}{2} + \frac{\pi}{2} = 0$. (Correct)
For $x=\frac{1}{2}$: $\sin ^{-1}(\frac{1}{2}) - \cos ^{-1}(\frac{1}{2}) + \sin ^{-1}(\frac{1}{2}) = \frac{\pi}{6} - \frac{\pi}{3} + \frac{\pi}{6} = 0$. (Correct)
Thus,$D-II$.

(B) Identify the Domain and Use Logarithmic Forms:
First,note that the function $\cosh^{-1} x$ is only defined for $x \geq 1$. Therefore,we must have $x \geq 1$.
We use the logarithmic definitions of inverse hyperbolic functions:
$e^{\cosh^{-1} x} = x + \sqrt{x^2 - 1}$
$e^{\sinh^{-1} x} = x + \sqrt{x^2 + 1}$
Substitute and Simplify the Equation:
Substitute these expressions back into the original equation:
$\sqrt{2} + (x + \sqrt{x^2 - 1}) - (x + \sqrt{x^2 + 1}) = 0$
The $x$ terms cancel out,leaving:
$\sqrt{2} + \sqrt{x^2 - 1} - \sqrt{x^2 + 1} = 0$
$\sqrt{2} + \sqrt{x^2 - 1} = \sqrt{x^2 + 1}$
Solve for $x$:
Square both sides of the equation to eliminate the radicals:
$(\sqrt{2} + \sqrt{x^2 - 1})^2 = (\sqrt{x^2 + 1})^2$
$2 + (x^2 - 1) + 2\sqrt{2}\sqrt{x^2 - 1} = x^2 + 1$
$x^2 + 1 + 2\sqrt{2}\sqrt{x^2 - 1} = x^2 + 1$
Subtract $x^2 + 1$ from both sides:
$2\sqrt{2}\sqrt{x^2 - 1} = 0$
$\sqrt{x^2 - 1} = 0 \implies x^2 = 1 \implies x = \pm 1$
Since the domain requires $x \geq 1$,the only valid solution is $x = 1$. Thus,there is $1$ root.

(D) $\sinh x = \frac{e^x - e^{-x}}{2}$. Since $\sinh(-x) = \frac{e^{-x} - e^x}{2} = -\sinh x$,it is an odd function with range $\mathbb{R}$. Thus,$A-IV$.
$(B)$ $\text{sech } x = \frac{2}{e^x + e^{-x}}$. Since $\text{sech}(-x) = \frac{2}{e^{-x} + e^x} = \text{sech } x$,it is an even function. Thus,$B-III$.
$(C)$ $\tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}$. Since $\tanh(-x) = -\tanh x$,it is an odd function with range $(-1, 1)$. Thus,$C-V$.
$(D)$ $\text{cosech}^{-1} x = \ln\left(\frac{1}{x} + \frac{\sqrt{1+x^2}}{|x|}\right)$. The domain is $x \neq 0$ and it is neither even nor odd. Thus,$D-II$.
Therefore,the correct matching is $A-IV, B-III, C-V, D-II$.

(D) Given $f(x) = \begin{cases} x + 4, & x < -4 \\ 3x + 2, & -4 \leq x < 4 \\ x - 4, & x \geq 4 \end{cases}$
$(A) f(-5) + f(-4) = (-5 + 4) + (3(-4) + 2) = -1 + (-12 + 2) = -1 - 10 = -11$. Thus,$(A) \rightarrow (iii)$.
$(B) f(|f(-8)|) = f(|-8 + 4|) = f(|-4|) = f(4) = 4 - 4 = 0$. Thus,$(B) \rightarrow (vi)$.
$(C) f(f(-7) + f(3)) = f((-7 + 4) + (3(3) + 2)) = f(-3 + 11) = f(8) = 8 - 4 = 4$. Thus,$(C) \rightarrow (ii)$.
$(D) f(f(f(f(0)))) + 1$. First,$f(0) = 3(0) + 2 = 2$. Then $f(f(0)) = f(2) = 3(2) + 2 = 8$. Then $f(f(f(0))) = f(8) = 8 - 4 = 4$. Then $f(f(f(f(0)))) = f(4) = 4 - 4 = 0$. Finally,$0 + 1 = 1$. Thus,$(D) \rightarrow (v)$.
The correct matching is $(A)-(iii), (B)-(vi), (C)-(ii), (D)-(v)$.

(A) Given $e^{f(x)}=\frac{10+x}{10-x}$.
Taking the natural logarithm on both sides,we get $f(x)=\log \left(\frac{10+x}{10-x}\right)$.
We are given the relation $f(x)=k f\left(\frac{200 x}{100+x^2}\right)$.
Substituting the expression for $f(x)$ into the right side:
$f\left(\frac{200 x}{100+x^2}\right) = \log \left( \frac{10 + \frac{200x}{100+x^2}}{10 - \frac{200x}{100+x^2}} \right)$
$= \log \left( \frac{10(100+x^2) + 200x}{10(100+x^2) - 200x} \right)$
$= \log \left( \frac{1000 + 10x^2 + 200x}{1000 + 10x^2 - 200x} \right)$
$= \log \left( \frac{10(x^2 + 20x + 100)}{10(x^2 - 20x + 100)} \right)$
$= \log \left( \frac{(x+10)^2}{(10-x)^2} \right)$
$= 2 \log \left( \frac{10+x}{10-x} \right) = 2f(x)$.
Substituting this back into the given equation: $f(x) = k \cdot 2f(x)$.
Since $f(x) \neq 0$ for $x \in (-10, 10)$,we have $1 = 2k$,which implies $k = 0.5$.