Type of Functions based on Mapping Questions in English

(B) Given $A = \{1, 2, 3, 4, 5, 6\}$.
We are given the condition $f(m) + f(n) = 7$ whenever $m + n = 7$.
The pairs $(m, n)$ such that $m + n = 7$ are $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$.
This implies the following constraints:
$f(1) + f(6) = 7$
$f(2) + f(5) = 7$
$f(3) + f(4) = 7$
For each pair,say $(f(1), f(6))$,the possible values for $f(1)$ are $\{1, 2, 3, 4, 5, 6\}$.
If $f(1) = 1$,then $f(6) = 6$.
If $f(1) = 2$,then $f(6) = 5$.
If $f(1) = 3$,then $f(6) = 4$.
If $f(1) = 4$,then $f(6) = 3$.
If $f(1) = 5$,then $f(6) = 2$.
If $f(1) = 6$,then $f(6) = 1$.
There are $6$ possible choices for each pair $(f(1), f(6)), (f(2), f(5)),$ and $(f(3), f(4))$.
Since there are $3$ such independent pairs,the total number of functions is $6 \times 6 \times 6 = 6^3 = 216$.

(D) Given $g(x)=1, \forall x \in R$ and the signum function $f(x)= \begin{cases}-1, & x < 0 \\ 0, & x=0 \\ 1, & x>0\end{cases}$.
For $x < 0$,$f(x)=-1$,so $g(x)=1 = 2+(-1) = 2+f(x)$.
For $x=0$,$f(x)=0$,so $g(x)=1 = 1+0 = 1+f(x)$.
For $x>0$,$f(x)=1$,so $g(x)=1 = 0+1 = 0+f(x) = f(x)$.
Thus,$g(x)= \begin{cases}f(x)+2, & x < 0 \\ 1+f(x), & x=0 \\ f(x), & x>0\end{cases}$.

(D) The function $f(x)$ is defined on the interval $[-1, 1]$.
For $x \in [-1, 0)$,$f(x) = 2^x + 1$. As $x \to 0^-$,$f(x) \to 2^0 + 1 = 2$. Since $2^x$ is strictly increasing,$f(x)$ increases from $f(-1) = 2^{-1} + 1 = 1.5$ to $2$.
At $x = 0$,$f(0) = 1$.
For $x \in (0, 1]$,$f(x) = 2^x - 1$. As $x \to 0^+$,$f(x) \to 2^0 - 1 = 0$. Since $2^x$ is strictly increasing,$f(x)$ increases from $0$ to $f(1) = 2^1 - 1 = 1$.
The range of the function is $[1.5, 2) \cup \{1\} \cup (0, 1]$.
Combining these,the range is $(0, 1] \cup [1.5, 2)$.
Since the range is not a closed interval and the function values do not attain a supremum or infimum within the domain (the values approach $2$ but never reach it,and approach $0$ but never reach it),the function has neither a maximum nor a minimum value on $[-1, 1]$.

(B) We are given $f(x) = 8$.
Case $1$: For $x < 2$,$f(x) = x^2 - 4x + 3$.
Setting $x^2 - 4x + 3 = 8$,we get $x^2 - 4x - 5 = 0$.
Factoring gives $(x - 5)(x + 1) = 0$,so $x = 5$ or $x = -1$.
Since the condition is $x < 2$,only $x = -1$ is a valid solution.
Case $2$: For $x \geq 2$,$f(x) = x - 3$.
Setting $x - 3 = 8$,we get $x = 11$.
Since $11 \geq 2$,this is a valid solution.
Thus,the solutions are $x = -1$ and $x = 11$.
The total number of real numbers $x$ for which $f(x) = 8$ is $2$.

(D) We are given $f(x) = [x]$ and $g(x) = 3[\frac{x}{3}]$.
Setting $f(x) = g(x)$,we have $[x] = 3[\frac{x}{3}]$.
Let $[\frac{x}{3}] = k$,where $k \in Z$.
Then,by the definition of the greatest integer function,$k \leq \frac{x}{3} < k + 1$,which implies $3k \leq x < 3k + 3$.
Also,we have $[x] = 3k$.
By the definition of the greatest integer function,$[x] = 3k$ implies $3k \leq x < 3k + 1$.
Since $3k \leq x < 3k + 1$ is a subset of $3k \leq x < 3k + 3$,the condition $[x] = 3[\frac{x}{3}]$ is satisfied for all $x$ in the interval $[3k, 3k + 1)$ for any integer $k$.
Thus,the set of all such real $x$ is $\{x \in R : 3k \leq x < 3k + 1, k \in Z\}$.

(D) The function is defined as $f(x) = 5^{-|x|} + \operatorname{sgn}(5^{-x})$.
Since $5^{-x} > 0$ for all $x \in R$,the signum function $\operatorname{sgn}(5^{-x}) = 1$ for all $x \in R$.
Thus,the function simplifies to $f(x) = 5^{-|x|} + 1$.
For $x \geq 0$,$f(x) = 5^{-x} + 1$,which is a strictly decreasing function with range $(1, 2]$.
For $x < 0$,$f(x) = 5^{x} + 1$,which is a strictly increasing function with range $(1, 2)$.
Since $f(x) = f(-x)$ for all $x$,the function is many-one.
Also,the range of the function is $(1, 2]$,which is a proper subset of the codomain $R$,so the function is not onto.
Therefore,$f$ is neither one-one nor onto.

(D) function is a bijection if it is both one-to-one (injective) and onto (surjective).
$(a)$ $f(x) = \sqrt{\{x\}}$. Since $\{x\}$ is periodic with period $1$,$f(0.1) = f(1.1)$,so it is many-to-one.
$(b)$ $f(x) = -(x^2-4x+4)+1 = 1-(x-2)^2$. This is a parabola opening downwards,which is many-to-one.
$(c)$ $f(x) = \frac{1}{\sqrt{x-5}}$. The range is $(0, \infty)$,which is not equal to the codomain $R \setminus \{0\}$,so it is into.
$(d)$ $f(x) = \sqrt{16-x^2}$. For $x \in [0,4]$,$f(x)$ is strictly decreasing from $4$ to $0$. Thus,it is one-to-one and onto. Hence,it is a bijection.

(A) We have $f(x) = x + 2|x + 1| + 2|x - 1|$.
Breaking the function into intervals:
For $x \leq -1$: $f(x) = x + 2(-x - 1) + 2(-x + 1) = x - 2x - 2 - 2x + 2 = -3x$.
For $-1 < x < 1$: $f(x) = x + 2(x + 1) + 2(-x + 1) = x + 2x + 2 - 2x + 2 = x + 4$.
For $x \geq 1$: $f(x) = x + 2(x + 1) + 2(x - 1) = x + 2x + 2 + 2x - 2 = 5x$.
The function is defined as:
$f(x) = \begin{cases} -3x, & x \leq -1 \\ x + 4, & -1 < x < 1 \\ 5x, & x \geq 1 \end{cases}$
At $x = -1$,$f(-1) = 3$. For $x < -1$,$f(x) > 3$. For $-1 < x < 1$,$f(x)$ ranges from $3$ to $5$. At $x = 1$,$f(1) = 5$. For $x > 1$,$f(x) > 5$.
The value $3$ is attained at $x = -1$ and as $x \to -1^+$,$f(x) \to 3$. However,looking at the graph,the function is continuous. The value $3$ is the minimum value of the function,occurring at $x = -1$. Since the function is strictly decreasing for $x < -1$ and strictly increasing for $x > -1$,the value $3$ is the unique minimum,thus it has a unique pre-image.

(B) Given the function $f(x)$ defined in pieces:
Step $1$: Calculate $f(-1.75)$. Since $-1.75 \leq -1$,we use $f(x) = x + 2$.
$f(-1.75) = -1.75 + 2 = 0.25$.
Step $2$: Calculate $f(0.5)$. Since $-1 < 0.5 < 1$,we use $f(x) = x^2$.
$f(0.5) = (0.5)^2 = 0.25$.
Step $3$: Calculate $f(1.5)$. Since $1.5 \geq 1$,we use $f(x) = 2 - x$.
$f(1.5) = 2 - 1.5 = 0.5$.
Step $4$: Sum the values:
$f(-1.75) + f(0.5) + f(1.5) = 0.25 + 0.25 + 0.5 = 1$.

(C) Given that $f(x) = ax^2 + bx + c$ is an even function,$f(x) = f(-x)$.
$ax^2 + bx + c = a(-x)^2 + b(-x) + c = ax^2 - bx + c$.
Comparing coefficients,we get $b = 0$.
Given that $g(x) = px^3 + qx^2 + rx$ is an odd function,$g(-x) = -g(x)$.
$p(-x)^3 + q(-x)^2 + r(-x) = -(px^3 + qx^2 + rx)$.
$-px^3 + qx^2 - rx = -px^3 - qx^2 - rx$.
Comparing coefficients,we get $q = 0$.
Now,$h(x) = f(x) + g(x) = ax^2 + bx + c + px^3 + qx^2 + rx$.
Since $b = 0$ and $q = 0$,$h(x) = px^3 + ax^2 + rx + c$.
We are given $h(-2) = 0$.
$p(-2)^3 + a(-2)^2 + r(-2) + c = 0$.
$-8p + 4a - 2r + c = 0$.
$4a + c = 8p + 2r$.
Since $b = 0$,$4a + 2b + c = 4a + 2(0) + c = 4a + c$.
Thus,$8p + 2r = 4a + 2b + c$.

(A) To check if the function is even or odd,we evaluate $f(-x)$.
$f(-x) = \log (-x + \sqrt{(-x)^2 + 1}) = \log (-x + \sqrt{x^2 + 1})$.
Now,multiply and divide by $(x + \sqrt{x^2 + 1})$ inside the logarithm:
$f(-x) = \log \left( \frac{(\sqrt{x^2 + 1} - x)(\sqrt{x^2 + 1} + x)}{\sqrt{x^2 + 1} + x} \right)$.
Using the identity $(a-b)(a+b) = a^2 - b^2$:
$f(-x) = \log \left( \frac{(x^2 + 1) - x^2}{\sqrt{x^2 + 1} + x} \right) = \log \left( \frac{1}{x + \sqrt{x^2 + 1}} \right)$.
Using the property $\log(1/a) = -\log(a)$:
$f(-x) = -\log(x + \sqrt{x^2 + 1}) = -f(x)$.
Since $f(-x) = -f(x)$,the function is an odd function.

(C) Given that $f(x)$ is an odd function,it must satisfy the condition $f(-x) = -f(x)$ for all $x$ in the domain.
For $x < 0$,we have $-x > 0$.
Since $f(x) = x^2 + 3x - 7$ for $x > 0$,we can find $f(-x)$ by substituting $-x$ into this expression:
$f(-x) = (-x)^2 + 3(-x) - 7 = x^2 - 3x - 7$.
Using the property of the odd function,$f(x) = -f(-x)$ for $x < 0$:
$h(x) = -(x^2 - 3x - 7) = -x^2 + 3x + 7$.
Thus,the correct option is $C$.

(D) To check for one-one: $A$ function is one-one if $f(x_1) = f(x_2) \implies x_1 = x_2$. Consider $f(1) = 0$ and $f(3) = 0$. Since $f(1) = f(3)$ but $1 \neq 3$,the function is not one-one.
To check for onto: $A$ function is onto if for every $y \in Z$,there exists $x \in Z$ such that $f(x) = y$. For any odd integer $y$ (where $y \neq 0$),there is no $x \in Z$ such that $f(x) = y$,because the range of $f$ only contains $0$ and even integers divided by $2$. Thus,the function is not onto.
Therefore,$f$ is neither one-one nor onto.

(C) Given the function $f(x) = x - [x] - \frac{1}{2}$.
We know that the fractional part of $x$ is defined as $\{x\} = x - [x]$.
Thus,$f(x) = \{x\} - \frac{1}{2}$.
We are given $f(x) = \frac{1}{2}$,so $\{x\} - \frac{1}{2} = \frac{1}{2}$.
This implies $\{x\} = 1$.
However,by definition,the fractional part $\{x\}$ always satisfies $0 \le \{x\} < 1$.
Since $\{x\}$ can never be equal to $1$,there is no value of $x$ that satisfies the equation.
Therefore,the set $\{x \in R: f(x) = \frac{1}{2}\}$ is the empty set,denoted by $\phi$.

(A) Given the function $f(x) = \begin{cases} [x], & -3 < x \leq -1 \\ |x|, & -1 < x < 1 \\ |[x]|, & 1 \leq x < 3 \end{cases}$.
Case $1$: For $-3 < x \leq -1$,$f(x) = [x]$. Since $x \leq -1$,$[x] \leq -1$,so $f(x) < 0$.
Case $2$: For $-1 < x < 1$,$f(x) = |x|$. Since the absolute value is always non-negative,$f(x) \geq 0$ for all $x \in (-1, 1)$.
Case $3$: For $1 \leq x < 3$,$f(x) = |[x]|$. Since the absolute value of the greatest integer function is always non-negative,$f(x) \geq 0$ for all $x \in [1, 3)$.
Combining the intervals from Case $2$ and Case $3$,we get $(-1, 1) \cup [1, 3) = (-1, 3)$.
Thus,the set $\{x : f(x) \geq 0\}$ is $(-1, 3)$.

(B) Given the function $f: N \rightarrow Z$ defined as:
$f(n) = \begin{cases} 2 & \text{if } n=3k, k \in Z \\ 10 & \text{if } n=3k+1, k \in Z \\ 0 & \text{if } n=3k+2, k \in Z \end{cases}$
We want to find the set $\{n \in N: f(n) > 2\}$.
Looking at the definition,$f(n) > 2$ only when $f(n) = 10$.
This occurs when $n = 3k + 1$ for $k \in Z$.
Since $n \in N$ (natural numbers),we consider $k \geq 0$ (assuming $N = \{1, 2, 3, \dots\}$):
For $k=0, n = 3(0) + 1 = 1$.
For $k=1, n = 3(1) + 1 = 4$.
For $k=2, n = 3(2) + 1 = 7$.
Thus,the set is $\{1, 4, 7, \dots\}$.

(B) To determine the nature of the function $f(x)$,we analyze its behavior in two intervals.
For $x \leq \frac{4}{3}$,$f(x) = 2x+3$. This is a strictly increasing linear function. The range for this part is $(-\infty, 2(\frac{4}{3})+3] = (-\infty, \frac{17}{3}]$.
For $x > \frac{4}{3}$,$f(x) = -3x^2+8x$. This is a downward-opening parabola with vertex at $x = -\frac{b}{2a} = -\frac{8}{2(-3)} = \frac{4}{3}$. Since the interval is $x > \frac{4}{3}$,the function is strictly decreasing in this domain. The value at $x = \frac{4}{3}$ is $-3(\frac{16}{9}) + 8(\frac{4}{3}) = -\frac{16}{3} + \frac{32}{3} = \frac{16}{3}$. As $x \rightarrow \infty$,$f(x) \rightarrow -\infty$. Thus,the range for this part is $(-\infty, \frac{16}{3})$.
Combining these,the function is not one-one because the value $\frac{16}{3}$ is attained at $x = \frac{4}{3}$ and also at some $x > \frac{4}{3}$ (since the parabola decreases from $\frac{16}{3}$).
Since the range is $(-\infty, \frac{17}{3}]$,which is not equal to the codomain $R$,the function is not onto. Therefore,it is not bijective. The correct description is that it is not onto.

(A) Given the function $f(x) = 2x + \sin x$.
To check for one-one,we find the derivative: $f'(x) = 2 + \cos x$.
Since $-1 \leq \cos x \leq 1$,it follows that $f'(x) = 2 + \cos x \geq 2 - 1 = 1 > 0$.
Since $f'(x) > 0$ for all $x \in R$,the function $f(x)$ is strictly increasing,which implies $f(x)$ is one-one.
To check for onto,we observe the range of $f(x)$. As $x \to \infty$,$f(x) \to \infty$,and as $x \to -\infty$,$f(x) \to -\infty$.
Since $f(x)$ is a continuous function,by the Intermediate Value Theorem,its range is $(-\infty, \infty) = R$.
Therefore,$f(x)$ is onto.
Thus,$f$ is one-one and onto.

(D) The function is defined as $f(x) = \begin{cases} 2x-3, & x < -2 \\ x^2-1, & -2 \leq x \leq 2 \\ 3x+2, & x > 2 \end{cases}$.
$1$. Checking for Injectivity (One-to-One):
$A$ function is injective if $f(x_1) = f(x_2) \implies x_1 = x_2$.
Consider the interval $[-2, 2]$,where $f(x) = x^2 - 1$.
We observe that $f(-1) = (-1)^2 - 1 = 0$ and $f(1) = (1)^2 - 1 = 0$.
Since $f(-1) = f(1)$ but $-1 \neq 1$,the function is not injective.
$2$. Checking for Surjectivity (Onto):
$A$ function is surjective if its range equals its codomain $(R)$.
- For $x < -2$,$f(x) < 2(-2) - 3 = -7$. So,$f(x) \in (-\infty, -7)$.
- For $-2 \leq x \leq 2$,$f(x) = x^2 - 1$. The minimum value is $-1$ (at $x=0$) and the maximum is $f(-2) = f(2) = 3$. So,$f(x) \in [-1, 3]$.
- For $x > 2$,$f(x) > 3(2) + 2 = 8$. So,$f(x) \in (8, \infty)$.
The range of $f$ is $(-\infty, -7) \cup [-1, 3] \cup (8, \infty)$.
Since the range is not equal to the codomain $R$,the function is not surjective.
Conclusion: The function is neither injective nor surjective.

(C) For Statement $I$: Given $f(x) = \sec x + \tan x$ on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
$f'(x) = \sec x \tan x + \sec^2 x = \sec x(\tan x + \sec x)$.
Since $\sec x + \tan x = \frac{1+\sin x}{\cos x} = \frac{(\cos(x/2) + \sin(x/2))^2}{\cos^2(x/2) - \sin^2(x/2)} = \frac{\cos(x/2) + \sin(x/2)}{\cos(x/2) - \sin(x/2)} = \tan(\frac{\pi}{4} + \frac{x}{2})$.
For $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,$\frac{x}{2} \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$,so $\frac{\pi}{4} + \frac{x}{2} \in (0, \frac{\pi}{2})$.
In this interval,$\sec x > 0$ and $\tan(\frac{\pi}{4} + \frac{x}{2}) > 0$,so $f'(x) > 0$. Thus,$f(x)$ is strictly increasing and one-one.
For Statement $II$: Given $f(x) = x^2$ on $[0, \infty)$.
If $f(x_1) = f(x_2)$,then $x_1^2 = x_2^2$. Since $x_1, x_2 \geq 0$,this implies $x_1 = x_2$.
Thus,$f(x)$ is one-one.
Both statements are true.

(D) Given $f(x) = \begin{cases} -2, & -2 \leq x \leq 0 \\ x^2-2, & 0 \leq x \leq 2 \end{cases}$.
For $f(x)$,the range is $[-2, 2]$. Since $f(x) = -2$ for all $x \in [-2, 0]$,$f$ is not injective. Thus,$f$ is not bijective.
Now,$g(x) = |f(x)| + f(|x|)$.
If $-2 \leq x \leq 0$,then $|x| \in [0, 2]$,so $f(|x|) = |x|^2 - 2 = x^2 - 2$. Also $f(x) = -2$,so $|f(x)| = 2$. Thus $g(x) = 2 + x^2 - 2 = x^2$.
If $0 \leq x \leq 2$,then $|x| = x$,so $f(|x|) = f(x) = x^2 - 2$. Thus $g(x) = |x^2 - 2| + x^2 - 2$.
For $0 \leq x \leq \sqrt{2}$,$x^2 - 2 \leq 0$,so $g(x) = -(x^2 - 2) + x^2 - 2 = 0$.
For $\sqrt{2} < x \leq 2$,$x^2 - 2 > 0$,so $g(x) = (x^2 - 2) + x^2 - 2 = 2(x^2 - 2)$.
Thus,$g(x) = \begin{cases} x^2, & -2 \leq x \leq 0 \\ 0, & 0 \leq x \leq \sqrt{2} \\ 2(x^2-2), & \sqrt{2} < x \leq 2 \end{cases}$.
The range of $g(x)$ is $[0, 4]$,which is equal to the codomain,so $g$ is surjective. Since $g(x) = 0$ for $x \in [0, \sqrt{2}]$,$g$ is not injective.

(C) To check for injectivity,let $f(x_1) = f(x_2)$ for $x_1, x_2 \in R$.
$\frac{x_1}{\sqrt{1+x_1^2}} = \frac{x_2}{\sqrt{1+x_2^2}}$
Squaring both sides:
$\frac{x_1^2}{1+x_1^2} = \frac{x_2^2}{1+x_2^2}$
$x_1^2(1+x_2^2) = x_2^2(1+x_1^2)$
$x_1^2 + x_1^2x_2^2 = x_2^2 + x_1^2x_2^2$
$x_1^2 = x_2^2$
Since $f'(x) = \frac{1}{(1+x^2)^{3/2}} > 0$ for all $x \in R$,the function is strictly increasing.
Therefore,$f(x_1) = f(x_2) \implies x_1 = x_2$. Thus,$f$ is injective.
To check for surjectivity,let $y = \frac{x}{\sqrt{1+x^2}}$.
Since $x^2 < 1+x^2$,we have $\frac{|x|}{\sqrt{1+x^2}} < 1$.
Thus,the range of $f$ is $(-1, 1)$,which is not equal to the codomain $R$.
Therefore,$f$ is not surjective.
Hence,the function is injective but not surjective.

(C) Given the function $f(x) = \begin{cases} \frac{60-5x}{3}, & 0 \leq x \leq 6 \\ 10, & 6 \leq x \leq 7 \\ 31-3x, & 7 \leq x \leq 10 \end{cases}$.
For $x \in [6, 7]$,$f(x) = 10$. Since the function takes the same value for all $x$ in the interval $[6, 7]$,it is not one-one.
Now,let us find the range of $f(x)$:
For $0 \leq x \leq 6$,$f(x) = \frac{60-5x}{3}$. As $x$ goes from $0$ to $6$,$f(x)$ goes from $\frac{60}{3} = 20$ to $\frac{60-30}{3} = 10$. So,the range is $[10, 20]$.
For $6 \leq x \leq 7$,$f(x) = 10$. So,the range is ${10}$.
For $7 \leq x \leq 10$,$f(x) = 31-3x$. As $x$ goes from $7$ to $10$,$f(x)$ goes from $31-21 = 10$ to $31-30 = 1$. So,the range is $[1, 10]$.
The union of these ranges is $[1, 10] \cup {10} \cup [10, 20] = [1, 20]$.
Since the range $[1, 20]$ is equal to the co-domain $[1, 20]$,the function is onto.
Thus,$f(x)$ is onto but not one-one.

(B) Given that $A_n = \{(n+1)k \mid k \in N\}$.
For $n=1$,$A_1 = \{2k \mid k \in N\} = \{2, 4, 6, 8, \dots\}$.
For $n=2$,$A_2 = \{3k \mid k \in N\} = \{3, 6, 9, 12, \dots\}$.
For $n=3$,$A_3 = \{4k \mid k \in N\} = \{4, 8, 12, 16, \dots\}$.
Now,$X = \bigcup_{n \in N} A_n = A_1 \cup A_2 \cup A_3 \cup \dots = \{2, 3, 4, 5, 6, \dots\}$.
Here,$f: X \rightarrow N$ is defined by $f(x) = x$ for all $x \in X$.
Since $f(x) = x$ is an identity function on the domain $X$,it is clearly one-one.
However,the codomain is $N = \{1, 2, 3, 4, \dots\}$.
For $f$ to be onto,the range must equal the codomain. Here,the range is $X = \{2, 3, 4, 5, \dots\}$.
Since $1 \in N$ but $1 \notin X$,there is no $x \in X$ such that $f(x) = 1$.
Therefore,$f$ is not onto.

(A) Given the functional equation $f(x+y) = f(x) + f(y)$ for all $x, y \in Z$.
Setting $x=0, y=0$,we get $f(0) = f(0) + f(0)$,which implies $f(0) = 0$.
For any $n \in Z^+$,by induction,$f(nx) = nf(x)$.
Let $f(1) = k$,where $k \in Z$. Then $f(n) = nk$ for all $n \in Z$.
Since $f(x+y) = f(x) + f(y)$,it follows that $f(x) = kx$ for all $x \in Z$.
For $f$ to be a bijection from $Z$ to $Z$,it must be both injective and surjective.
If $f(x) = kx$,then $f$ is injective if $k \neq 0$.
For $f$ to be surjective,the range of $f$ must be $Z$.
The range of $f(x) = kx$ is the set of all multiples of $k$,i.e.,$\{..., -2k, -k, 0, k, 2k, ...\}$.
For this set to be equal to $Z$,we must have $k = 1$ or $k = -1$.
If $k = 1$,$f(x) = x$,which is the identity function (bijective).
If $k = -1$,$f(x) = -x$,which is also bijective.
Thus,there are exactly two such functions.

(C) We are given $f(n) = \left(r+1, \frac{q+1}{2}\right)$ where $n = q \times 2^r$,$q$ is an odd integer,and $r \geq 0$.
For one-one mapping:
Suppose $f(n_1) = f(n_2)$.
Then $\left(r_1+1, \frac{q_1+1}{2}\right) = \left(r_2+1, \frac{q_2+1}{2}\right)$.
This implies $r_1+1 = r_2+1 \Rightarrow r_1 = r_2$ and $\frac{q_1+1}{2} = \frac{q_2+1}{2} \Rightarrow q_1 = q_2$.
Since $n_1 = q_1 \times 2^{r_1}$ and $n_2 = q_2 \times 2^{r_2}$,it follows that $n_1 = n_2$. Thus,$f$ is one-one.
For onto mapping:
Let $(a, b) \in N \times N$. We need to find $n \in N$ such that $f(n) = (a, b)$.
$r+1 = a \Rightarrow r = a-1$. Since $a \in N$,$a \geq 1$,so $r \geq 0$.
$\frac{q+1}{2} = b \Rightarrow q = 2b-1$. Since $b \in N$,$b \geq 1$,so $q \geq 1$ and $q$ is odd.
Thus,for any $(a, b) \in N \times N$,there exists $n = (2b-1) \times 2^{a-1} \in N$ such that $f(n) = (a, b)$.
Therefore,$f$ is onto.
Since $f$ is both one-one and onto,$f$ is a bijection.

(B) Given $f: Z \rightarrow N$ defined by $f(n) = \begin{cases} 2n, & \text{if } n > 0 \\ 1, & \text{if } n = 0 \\ -2n-1, & \text{if } n < 0 \end{cases}$.
For $n > 0$,$f(n) \in \{2, 4, 6, 8, \dots\}$.
For $n = 0$,$f(0) = 1$.
For $n < 0$,let $n = -k$ where $k > 0$. Then $f(n) = -2(-k) - 1 = 2k - 1$. As $k$ takes values $1, 2, 3, \dots$,$f(n)$ takes values $1, 3, 5, 7, \dots$.
Combining these,the range of $f$ is $\{1, 2, 3, 4, \dots\} = N$. Since Range = Codomain,$f$ is onto.
Now,check for one-one: $f(0) = 1$ and $f(-1) = -2(-1) - 1 = 2 - 1 = 1$.
Since $f(0) = f(-1)$ but $0 \neq -1$,the function is not one-one.
Therefore,$f$ is onto but not one-one.

(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 $f(x) = \begin{cases} -x^2 & -2 \leq x < 0 \\ x^2 & 0 \leq x \leq 2 \end{cases}$. This function is strictly increasing on $[-2, 2]$,so it is injective. The range is $[-4, 4]$,which equals the codomain,so it is surjective. Thus,$B \rightarrow III$.
$(C)$ $f: R \rightarrow R$,$f(x) = (x-2)(x-3)(x-5)$. 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,$D \rightarrow I$.

(C) Given $f(x) = \frac{x}{1+x^2}$ and $g(x) = \frac{x^2}{1+x^2}$ for $x \in R$.
For $f(x)$,$f'(x) = \frac{(1+x^2)(1) - x(2x)}{(1+x^2)^2} = \frac{1-x^2}{(1+x^2)^2}$.
Since $f'(x)$ changes sign at $x = \pm 1$,$f(x)$ is not monotonic,hence not one-one.
The range of $f(x)$ is $[-\frac{1}{2}, \frac{1}{2}]$,which is not equal to the codomain $R$,so $f(x)$ is not onto.
For $g(x)$,$g(-x) = \frac{(-x)^2}{1+(-x)^2} = \frac{x^2}{1+x^2} = g(x)$,so $g(x)$ is an even function,hence not one-one.
The range of $g(x)$ is $[0, 1)$,which is not equal to the codomain $R$,so $g(x)$ is not onto.
Therefore,both $f$ and $g$ are neither one-one nor onto.

(B) We have $f(x) = \frac{x}{1+x}$ where $f: [0, \infty) \rightarrow [0, \infty)$.
For one-one:
Let $f(x_1) = f(x_2)$.
$\frac{x_1}{1+x_1} = \frac{x_2}{1+x_2}$
$x_1(1+x_2) = x_2(1+x_1)$
$x_1 + x_1x_2 = x_2 + x_1x_2$
$x_1 = x_2$.
Since $f(x_1) = f(x_2) \implies x_1 = x_2$,the function is one-one.
For onto:
Let $y = f(x) = \frac{x}{1+x}$.
$y(1+x) = x \implies y + xy = x \implies y = x(1-y) \implies x = \frac{y}{1-y}$.
Since $x \in [0, \infty)$,we must have $\frac{y}{1-y} \geq 0$.
This implies $y \in [0, 1)$.
The codomain is $[0, \infty)$,but the range is $[0, 1)$.
Since Range $\neq$ Codomain,the function is not onto.
Thus,$f$ is one-one but not onto.

(C) Given,$f: N \rightarrow R$ with $f(1)=-1$ and the recurrence relation $f(n+1)=3f(n)+2$ for $n \geq 1$.
For $n=1$,$f(2) = 3f(1)+2 = 3(-1)+2 = -3+2 = -1$.
For $n=2$,$f(3) = 3f(2)+2 = 3(-1)+2 = -3+2 = -1$.
By mathematical induction,if $f(k)=-1$,then $f(k+1) = 3f(k)+2 = 3(-1)+2 = -1$.
Since $f(n)=-1$ for all $n \in N$,the function $f$ maps every input to the same output $-1$.
Therefore,$f$ is a constant function.

(A) Given,$f: Z \rightarrow Z$,$f(x) = \begin{cases} \frac{x}{2}, & \text{if } x \text{ is even} \\ 0, & \text{if } x \text{ is odd} \end{cases}$.
For $f$ to be one-to-one,$f(x_1) = f(x_2)$ must imply $x_1 = x_2$.
Consider $x_1 = 1$ and $x_2 = 3$. Both are odd,so $f(1) = 0$ and $f(3) = 0$.
Since $f(1) = f(3)$ but $1 \neq 3$,$f$ is not one-to-one.
For $f$ to be onto,the range must equal the codomain $Z$.
If $x$ is even,let $x = 2k$ for some $k \in Z$. Then $f(2k) = \frac{2k}{2} = k$.
Since $k$ can be any integer in $Z$,the range of $f$ is $Z$.
Thus,$f$ is onto.
Therefore,$f$ is onto but not one-to-one.

(C) Given the function $f: R \rightarrow R$ defined by $f(x)=3^{-x}$.
$I$. For one-one check: Let $f(x_1) = f(x_2)$.
$3^{-x_1} = 3^{-x_2} \Rightarrow -x_1 = -x_2 \Rightarrow x_1 = x_2$.
Since $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$,the function is one-one.
$II$. For onto check: The range of $f(x) = 3^{-x}$ is $(0, \infty)$,which is a subset of the codomain $R$. Since the range $\neq$ codomain,the function is not onto.
$III$. For decreasing check: Differentiating $f(x)$ with respect to $x$,we get $f'(x) = -3^{-x} \ln 3$. Since $3^{-x} > 0$ and $\ln 3 > 0$,$f'(x) < 0$ for all $x \in R$. Thus,the function is strictly decreasing.
Therefore,statements $I$ and $III$ are true.

(C) Given the function $f(x) = \frac{ax + b}{cx + d} \quad \dots(i)$
For the function to be a constant function,the derivative $f'(x)$ must be zero for all $x$ in its domain.
Using the quotient rule,$f'(x) = \frac{a(cx + d) - c(ax + b)}{(cx + d)^2} = \frac{acx + ad - acx - bc}{(cx + d)^2} = \frac{ad - bc}{(cx + d)^2}$.
For $f(x)$ to be constant,$f'(x) = 0$,which implies $ad - bc = 0$,or $ad = bc$.
Alternatively,if $ad = bc$,let $\frac{a}{c} = \frac{b}{d} = k$. Then $a = ck$ and $b = dk$.
Substituting these into the function: $f(x) = \frac{ckx + dk}{cx + d} = \frac{k(cx + d)}{cx + d} = k$,which is a constant.

(A) Given $f(x) = \begin{cases} 2x-5 & x < -3 \\ x+2 & -3 \leq x < 5 \\ 3x+1 & x \geq 5 \end{cases}$
$(A) f(-5)+f(0)+f(-1) = (2(-5)-5) + (0+2) + (-1+2) = -15 + 2 + 1 = -12$. Thus $(A) \rightarrow (IV)$.
$(B) f(f(5)+10f(-3)) = f((3(5)+1) + 10(-3+2)) = f(16 - 10) = f(6) = 3(6)+1 = 19$. Thus $(B) \rightarrow (V)$.
$(C) f(f(-4)) = f(2(-4)-5) = f(-13) = 2(-13)-5 = -31$. Thus $(C) \rightarrow (III)$.
$(D) f(f(f(1))) = f(f(1+2)) = f(f(3)) = f(3+2) = f(5) = 3(5)+1 = 16$. Thus $(D) \rightarrow (I)$.
Therefore,the correct match is $A-IV, B-V, C-III, D-I$.

(B) To find the fixed point,we set $f(c) = c$.
$1 + \sqrt{c} = c$
$\sqrt{c} = c - 1$
Squaring both sides (where $c \geq 1$):
$c = (c - 1)^2$
$c = c^2 - 2c + 1$
$c^2 - 3c + 1 = 0$
Using the quadratic formula $c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$c = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2}$
Since $c \geq 1$,we check the values:
$\frac{3 + \sqrt{5}}{2} \approx \frac{3 + 2.236}{2} = 2.618 \geq 1$ (Valid)
$\frac{3 - \sqrt{5}}{2} \approx \frac{3 - 2.236}{2} = 0.382 < 1$ (Invalid)
Thus,there is only one fixed point in the domain $[1, \infty)$.

(D) We have,$f(x) = x^{2} - \frac{x^{2}}{1+x^{2}}$.
First,check for one-one property:
$f(-x) = (-x)^{2} - \frac{(-x)^{2}}{1+(-x)^{2}} = x^{2} - \frac{x^{2}}{1+x^{2}} = f(x)$.
Since $f(-x) = f(x)$ for all $x \in R$,the function is not one-one (it is many-one).
Next,simplify the expression for range:
$f(x) = \frac{x^{2}(1+x^{2}) - x^{2}}{1+x^{2}} = \frac{x^{2} + x^{4} - x^{2}}{1+x^{2}} = \frac{x^{4}}{1+x^{2}}$.
Since $x^{4} \ge 0$ and $1+x^{2} > 0$ for all $x \in R$,$f(x) \ge 0$.
The range of $f(x)$ is $[0, \infty)$.
Since the co-domain is $R$ and the range $[0, \infty) \neq R$,the function is not onto.
Therefore,$f$ is neither one-one nor onto.

(B) To determine if the function $f(x)$ is even or odd,we check $f(-x)$.
Let $g(x) = \log \left( x + \sqrt{1 + x^2} \right)$.
Then $g(-x) = \log \left( -x + \sqrt{1 + (-x)^2} \right) = \log \left( \sqrt{1 + x^2} - x \right)$.
Multiplying and dividing by $\sqrt{1 + x^2} + x$,we get $g(-x) = \log \left( \frac{(\sqrt{1 + x^2} - x)(\sqrt{1 + x^2} + x)}{\sqrt{1 + x^2} + x} \right) = \log \left( \frac{1 + x^2 - x^2}{\sqrt{1 + x^2} + x} \right) = \log \left( \frac{1}{\sqrt{1 + x^2} + x} \right) = -\log \left( x + \sqrt{1 + x^2} \right) = -g(x)$.
Thus,$g(x)$ is an odd function.
Now,$f(x) = \sec(g(x))$.
Since $\sec(- \theta) = \sec(\theta)$,we have $f(-x) = \sec(g(-x)) = \sec(-g(x)) = \sec(g(x)) = f(x)$.
Therefore,$f(x)$ is an even function.

(D) Given the function $f(x) = \frac{e^{|x|} - e^{-x}}{e^x + e^{-x}}$.
Case $I$: If $x \geq 0$,then $|x| = x$. Thus,$f(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} = \tanh(x)$. As $x$ increases from $0$ to $\infty$,$f(x)$ increases from $0$ to $1$.
Case $II$: If $x < 0$,then $|x| = -x$. Thus,$f(x) = \frac{e^{-x} - e^{-x}}{e^x + e^{-x}} = \frac{0}{e^x + e^{-x}} = 0$.
Since $f(x) = 0$ for all $x < 0$ and $f(0) = 0$,the function is not one-one because $f(-1) = f(0) = 0$.
Since the range of $f$ is $[0, 1)$,which is a proper subset of the codomain $R$,the function is not onto.
Therefore,$f$ is neither one-one nor onto.

(D) Step $1$: Check for one-one property.
$f(x) = e^x + e^{-x}$.
Since $f(-x) = e^{-x} + e^{-(-x)} = e^{-x} + e^x = f(x)$, the function is an even function.
For an even function, $f(x_1) = f(x_2)$ does not imply $x_1 = x_2$ (e.g., $f(1) = f(-1)$).
Therefore, the function is not one-one.
Step $2$: Check for onto property.
We know that $e^x > 0$ for all $x \in R$.
By the Arithmetic Mean-Geometric Mean inequality, $\frac{e^x + e^{-x}}{2} \geq \sqrt{e^x \cdot e^{-x}} = 1$, which implies $e^x + e^{-x} \geq 2$.
Thus, the range of $f(x)$ is $[2, \infty)$.
Since the codomain is $R$ and the range is $[2, \infty)$, the range $\neq$ codomain.
Therefore, the function is not onto.
Conclusion: Since the function is neither one-one nor onto, it is not bijective.

(D) To check for one-one: Let $f(x_1) = f(x_2)$.
$\frac{2x_1}{x_1-1} = \frac{2x_2}{x_2-1}$
$x_1(x_2-1) = x_2(x_1-1)$
$x_1x_2 - x_1 = x_1x_2 - x_2$
$-x_1 = -x_2 \Rightarrow x_1 = x_2$.
Thus,$f$ is one-one.
To check for onto: Let $y = \frac{2x}{x-1}$.
$y(x-1) = 2x \Rightarrow yx - y = 2x \Rightarrow x(y-2) = y \Rightarrow x = \frac{y}{y-2}$.
Since $y \in R-\{2\}$,$x$ is always defined and $x \neq 1$. Thus,for every $y$ in the codomain,there exists an $x$ in the domain.
Therefore,$f$ is onto.
Hence,$f$ is both one-one and onto.

(C) Let $t = 2x-1$. Since $0 < x < 1$,we have $-1 < 2x-1 < 1$,so $-1 < t < 1$.
The function becomes $f(t) = \frac{t}{1-|t|}$ for $t \in (-1, 1)$.
We can write this as:
$f(t) = \begin{cases} \frac{t}{1+t}, & -1 < t \leq 0 \\ \frac{t}{1-t}, & 0 < t < 1 \end{cases}$
Since $f$ is continuous and $\lim_{t \to -1^+} f(t) = -\infty$ and $\lim_{t \to 1^-} f(t) = +\infty$,the range of $f$ is $(-\infty, \infty) = R$. Thus,$f$ is surjective.
Now,check for injectivity by finding the derivative:
$f'(t) = \begin{cases} \frac{1}{(1+t)^2}, & -1 < t < 0 \\ \frac{1}{(1-t)^2}, & 0 < t < 1 \end{cases}$
Since $f'(t) > 0$ for all $t \in (-1, 1)$,the function is strictly increasing.
Therefore,$f$ is injective.
Since $f$ is both injective and surjective,it is bijective.

(D) The function $f: S \rightarrow R$ is defined by $f(A) = \det(A)$,where $A \in S$ and $S$ is the set of all non-singular matrices of order $2 \times 2$. $A$ matrix is non-singular if its determinant is non-zero. Thus,the range of $f$ is $R \setminus \{0\}$.
$1$. Check for one-one: Consider two matrices $A_1 = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$ and $A_2 = \begin{bmatrix} 4 & 0 \\ 0 & 1 \end{bmatrix}$. Both $A_1, A_2 \in S$ because $\det(A_1) = 4 \neq 0$ and $\det(A_2) = 4 \neq 0$. Here,$f(A_1) = 2(2) - 0(0) = 4$ and $f(A_2) = 4(1) - 0(0) = 4$. Since $f(A_1) = f(A_2)$ but $A_1 \neq A_2$,the function $f$ is not one-one.
$2$. Check for onto: The codomain is $R$. For any $y = 0 \in R$,there exists no matrix $A \in S$ such that $f(A) = 0$,because $S$ only contains non-singular matrices (where $\det(A) \neq 0$). Thus,$f$ is not onto.
Therefore,$f$ is neither one-one nor onto.