Type of Functions based on Mapping Questions in English

(A) Given function,$f(x) = \sqrt{3} \sin 2x - \cos 2x + 4$.
We can rewrite this as:
$f(x) = 2(\sin 2x \cdot \frac{\sqrt{3}}{2} - \cos 2x \cdot \frac{1}{2}) + 4$
$f(x) = 2(\sin 2x \cos \frac{\pi}{6} - \cos 2x \sin \frac{\pi}{6}) + 4$
$f(x) = 2 \sin(2x - \frac{\pi}{6}) + 4$.
$A$ function $f(x) = \sin(\theta)$ is one-one when the argument $\theta$ lies in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Therefore,for $f(x)$ to be one-one:
$-\frac{\pi}{2} \leq 2x - \frac{\pi}{6} \leq \frac{\pi}{2}$
Adding $\frac{\pi}{6}$ to all parts:
$-\frac{\pi}{2} + \frac{\pi}{6} \leq 2x \leq \frac{\pi}{2} + \frac{\pi}{6}$
$-\frac{2\pi}{6} \leq 2x \leq \frac{4\pi}{6}$
$-\frac{\pi}{3} \leq 2x \leq \frac{2\pi}{3}$
Dividing by $2$:
$-\frac{\pi}{6} \leq x \leq \frac{\pi}{3}$.
Thus,the function is one-one in the interval $[-\frac{\pi}{6}, \frac{\pi}{3}]$.

(D) Given sets are $A = \{1, 2, 3, 4, 5\}$ and $B = \{x \in Z \mid x^{2} - 5x + 6 = 0\}$.
Solving the quadratic equation $x^{2} - 5x + 6 = 0$,we get $(x - 2)(x - 3) = 0$,so $x = 2$ or $x = 3$.
Thus,$B = \{2, 3\}$.
The number of elements in $A$ is $n(A) = 5$ and the number of elements in $B$ is $n(B) = 2$.
The number of onto (surjective) functions from a set with $n$ elements to a set with $m$ elements is given by the formula $\sum_{k=0}^{m} (-1)^{m-k} \binom{m}{k} k^{n}$.
For $n = 5$ and $m = 2$,the number of onto functions is $2^{n} - 2 = 2^{5} - 2$.
$2^{5} - 2 = 32 - 2 = 30$.
Therefore,the number of onto functions is $30$.

(D) Given that,$n(A) = 4$ and $n(B) = 5$.
An injective mapping (one-to-one function) from set $A$ to set $B$ exists if each element of $A$ is mapped to a unique element of $B$.
The number of injective mappings from a set with $m$ elements to a set with $n$ elements (where $n \ge m$) is given by the formula $P(n, m) = \frac{n!}{(n-m)!}$.
Here,$n = 5$ and $m = 4$.
Therefore,the number of injective mappings is $P(5, 4) = \frac{5!}{(5-4)!} = \frac{5!}{1!} = 5 \times 4 \times 3 \times 2 \times 1 = 120$.

(C) Given the function $f: Z \rightarrow Z$ defined by $f(n) = \begin{cases} \frac{n}{2}, & n \text{ is even} \\ 0, & n \text{ is odd} \end{cases}$.
$1$. Check for Injectivity (One-to-One): $A$ function is injective if $f(n_1) = f(n_2) \implies n_1 = n_2$. Consider odd integers $n_1 = 1$ and $n_2 = 3$. Then $f(1) = 0$ and $f(3) = 0$. Since $f(1) = f(3)$ but $1 \neq 3$,the function is not injective (it is many-to-one).
$2$. Check for Surjectivity (Onto): $A$ function is surjective if for every $y \in Z$,there exists an $n \in Z$ such that $f(n) = y$. If we choose $y = 5$,there is no integer $n$ such that $f(n) = 5$ because even numbers map to $n/2$ (which would require $n=10$) and odd numbers map to $0$. However,the range of $f$ is the set of all integers $Z$ because for any $y \in Z$,we can choose $n = 2y$ (which is even),then $f(2y) = \frac{2y}{2} = y$. Thus,for every $y \in Z$,there exists $n = 2y$ such that $f(n) = y$. Therefore,the function is surjective.
Conclusion: The function is surjective but not injective.

(C) Given the function $f(x) = \begin{cases} 2x & : x > 3 \\ x^2 & : 1 < x \leq 3 \\ 3x & : x \leq 1 \end{cases}$.
To find $f(-1) + f(2) + f(4)$,we evaluate each term individually based on the given conditions:
$1$. For $f(-1)$,since $-1 \leq 1$,we use the third condition: $f(-1) = 3(-1) = -3$.
$2$. For $f(2)$,since $1 < 2 \leq 3$,we use the second condition: $f(2) = (2)^2 = 4$.
$3$. For $f(4)$,since $4 > 3$,we use the first condition: $f(4) = 2(4) = 8$.
Adding these values together: $f(-1) + f(2) + f(4) = -3 + 4 + 8 = 9$.

(A) bijection (one-to-one and onto function) exists between two sets if and only if they have the same number of elements (cardinality).
Since the number of elements in set $x$ is $7$ and the number of elements in set $y$ is $8$,the cardinalities are not equal.
Therefore,it is impossible to define a bijection from $x$ to $y$.
Thus,the number of bijections is $0$.

(D) Given the function $f(x) = \left\{\begin{array}{cc} 2x, & x > 3 \\ x^2, & 1 < x \leq 3 \\ 3x, & x \leq 1 \end{array}\right.$.
To find $f(-2) + f(3) + f(4)$,we calculate each term individually:
$1$. For $f(-2)$: Since $-2 \leq 1$,we use $f(x) = 3x$. Thus,$f(-2) = 3(-2) = -6$.
$2$. For $f(3)$: Since $1 < 3 \leq 3$,we use $f(x) = x^2$. Thus,$f(3) = (3)^2 = 9$.
$3$. For $f(4)$: Since $4 > 3$,we use $f(x) = 2x$. Thus,$f(4) = 2(4) = 8$.
Summing these values: $f(-2) + f(3) + f(4) = -6 + 9 + 8 = 11$.

(C) Given the function $f: N \rightarrow N$ defined by $f(n) = \begin{cases} \frac{n+1}{2} & n \text{ is odd} \\ \frac{n}{2} & n \text{ is even} \end{cases}$.
To check for one-one: Consider $n=1$ (odd) and $n=2$ (even).
$f(1) = \frac{1+1}{2} = 1$.
$f(2) = \frac{2}{2} = 1$.
Since $f(1) = f(2)$ but $1 \neq 2$,the function $f$ is not one-one.
To check for onto: For any $y \in N$,we need to find $n \in N$ such that $f(n) = y$.
If $y$ is any natural number,we can choose $n = 2y$ (which is even). Then $f(2y) = \frac{2y}{2} = y$.
Since for every $y \in N$,there exists an $n \in N$ such that $f(n) = y$,the function $f$ is onto.
Thus,$f$ is onto but not one-one.

(D) The total number of functions from set $A$ to set $B$ is given by $|B|^{|A|} = 2^4 = 16$.
An onto function (surjective function) means every element in $B$ must have at least one pre-image in $A$.
The only functions that are not onto are the constant functions: $f(x) = a$ for all $x \in A$ and $f(x) = b$ for all $x \in A$.
There are $2$ such constant functions.
Number of onto functions $= 16 - 2 = 14$.
Probability of onto function $= \frac{14}{16} = \frac{7}{8}$.

(D) Given that,$f: R \rightarrow R$ and $f(x)=x^{4}$.
For $f$ to be one-one,$f(x_{1}) = f(x_{2})$ must imply $x_{1} = x_{2}$.
Here,$x_{1}^{4} = x_{2}^{4} \Rightarrow x_{1} = \pm x_{2}$.
For example,$f(1) = 1^{4} = 1$ and $f(-1) = (-1)^{4} = 1$.
Since $f(1) = f(-1)$ but $1 \neq -1$,$f$ is not one-one.
For $f$ to be onto,the range must equal the codomain $R$.
Since $x^{4} \geq 0$ for all $x \in R$,the range is $[0, \infty)$.
Since the range $[0, \infty) \neq R$,$f$ is not onto.
Therefore,$f$ is neither one-one nor onto.

(D) Given,$|A| = m$ and $|B| = n$.
The total number of injective functions from $A$ to $B$ is given by the formula $^nP_m = \frac{n!}{(n-m)!} = 2520$.
We need to express $2520$ as a product of consecutive integers starting from $n$ down to $(n-m+1)$.
$2520 = 7 \times 6 \times 5 \times 4 \times 3$.
This is equivalent to $^7P_5 = \frac{7!}{(7-5)!} = \frac{7!}{2!} = 7 \times 6 \times 5 \times 4 \times 3 = 2520$.
Comparing $^nP_m$ with $^7P_5$,we get $n = 7$ and $m = 5$.
Thus,$m = 5$.

(C) Given $f(x) = \frac{ax + \sqrt{a^2 - x^2}}{bx}$. The domain is determined by $a^2 - x^2 \geq 0$,so $x \in [-|a|, |a|]$ (excluding $x=0$).
For $x \in (0, |a|]$,$f(x) = \frac{a}{b} + \frac{1}{b}\sqrt{\frac{a^2}{x^2} - 1}$.
Let $g(x) = \frac{a^2}{x^2} - 1$. As $x$ increases from $0$ to $|a|$,$g(x)$ decreases from $\infty$ to $0$.
Thus,$f(x)$ is a strictly monotonic function on its domain intervals.
Since the function is strictly monotonic on its domain,it is one-one.
For the range,as $x \to 0^+$,$f(x) \to \infty$,and as $x \to |a|$,$f(x) = \frac{a^2}{b|a|} = \frac{|a|}{b}$.
Since the function covers all real values in its codomain (assuming $R$ as codomain),it is onto.
Therefore,$f$ is both one-one and onto.

(A) Let $f(x) = \log \left( \frac{2x^2 - 3}{x} + \sqrt{\frac{4x^4 - 11x^2 + 9}{|x|}} \right)$.
To check if the function is odd,we evaluate $f(-x)$:
$f(-x) = \log \left( \frac{2(-x)^2 - 3}{-x} + \sqrt{\frac{4(-x)^4 - 11(-x)^2 + 9}{|-x|}} \right)$
$f(-x) = \log \left( -\left( \frac{2x^2 - 3}{x} \right) + \sqrt{\frac{4x^4 - 11x^2 + 9}{|x|}} \right)$
Now,consider $f(x) + f(-x) = \log \left( \sqrt{\frac{4x^4 - 11x^2 + 9}{|x|}} + \frac{2x^2 - 3}{x} \right) + \log \left( \sqrt{\frac{4x^4 - 11x^2 + 9}{|x|}} - \frac{2x^2 - 3}{x} \right)$
Using the property $\log(a) + \log(b) = \log(ab)$:
$f(x) + f(-x) = \log \left( \left( \sqrt{\frac{4x^4 - 11x^2 + 9}{|x|}} \right)^2 - \left( \frac{2x^2 - 3}{x} \right)^2 \right)$
$f(x) + f(-x) = \log \left( \frac{4x^4 - 11x^2 + 9}{|x|^2} - \frac{4x^4 - 12x^2 + 9}{x^2} \right)$
Since $|x|^2 = x^2$:
$f(x) + f(-x) = \log \left( \frac{4x^4 - 11x^2 + 9 - (4x^4 - 12x^2 + 9)}{x^2} \right)$
$f(x) + f(-x) = \log \left( \frac{x^2}{x^2} \right) = \log(1) = 0$
Since $f(x) + f(-x) = 0$,the function is an odd function.

(B) Given $f(x) = \frac{x}{e^x - 1} + \frac{x}{2} + 1$.
To check if the function is even or odd,we evaluate $f(-x)$.
$f(-x) = \frac{-x}{e^{-x} - 1} + \frac{-x}{2} + 1$.
Multiply the numerator and denominator of the first term by $e^x$:
$f(-x) = \frac{-x e^x}{1 - e^x} - \frac{x}{2} + 1 = \frac{x e^x}{e^x - 1} - \frac{x}{2} + 1$.
We can rewrite $\frac{x e^x}{e^x - 1}$ as $\frac{x(e^x - 1 + 1)}{e^x - 1} = x + \frac{x}{e^x - 1}$.
Substituting this back:
$f(-x) = x + \frac{x}{e^x - 1} - \frac{x}{2} + 1 = \frac{x}{e^x - 1} + \frac{x}{2} + 1$.
Since $f(-x) = f(x)$,the function is an even function.

(C) Given the function $f : R \rightarrow R$ defined by $f(x) = x - [x] + 3$.
We know that the fractional part function is defined as $\{x\} = x - [x]$.
Substituting this into the expression for $f(x)$,we get $f(x) = \{x\} + 3$.
The fractional part function $\{x\}$ is a periodic function with a fundamental period of $1$.
Since adding a constant to a periodic function does not change its period,$f(x) = \{x\} + 3$ is also a periodic function with period $1$.
Therefore,the correct option is $C$.

(A-III, B-VI, C-II, D-V) Given $f(x) = \begin{cases} x+4 & \text{for } x < -4 \\ 3x+2 & \text{for } -4 \leq x < 4 \\ x-4 & \text{for } x \geq 4 \end{cases}$
$(A)$ $f(-5) + f(-4) = (-5+4) + (3(-4)+2) = -1 + (-12+2) = -1 - 10 = -11$. Thus,$(A)$ matches $(iii)$.
$(B)$ $f(|f(-8)|) = f(|-8+4|) = f(|-4|) = f(4) = 4-4 = 0$. Thus,$(B)$ matches $(vi)$.
$(C)$ $f(f(-7) + f(3)) = f((-7+4) + (3(3)+2)) = f(-3 + 11) = f(8) = 8-4 = 4$. Thus,$(C)$ matches $(ii)$.
$(D)$ $f(f(f(f(0)))) + 1 = f(f(f(3(0)+2))) + 1 = f(f(f(2))) + 1 = f(f(3(2)+2)) + 1 = f(f(8)) + 1 = f(8-4) + 1 = f(4) + 1 = (4-4) + 1 = 1$. Thus,$(D)$ matches $(v)$.

(A) We analyze the function $f(x)$ in three intervals:
$1$. For $-3 < x \leq -1$,$f(x) = [x]$. Since $x \leq -1$,$[x] \leq -1$,so $f(x) < 0$.
$2$. For $-1 < x < 1$,$f(x) = |x|$. Since the absolute value is always non-negative,$f(x) \geq 0$ for all $x \in (-1, 1)$.
$3$. For $1 \leq x < 3$,$f(x) = |[x]|$. Since the absolute value of any integer is $\geq 0$,$f(x) \geq 0$ for all $x \in [1, 3)$.
Combining these intervals where $f(x) \geq 0$,we get $(-1, 1) \cup [1, 3) = (-1, 3)$.
Thus,the set is $(-1, 3)$.

(C) Given $f(x) = x - (-1)^x$ where $x \in Z$.
Case $1$: If $x$ is even,let $x = 2k$ for some $k \in Z$.
Then $f(2k) = 2k - (-1)^{2k} = 2k - 1$.
Case $2$: If $x$ is odd,let $x = 2k+1$ for some $k \in Z$.
Then $f(2k+1) = (2k+1) - (-1)^{2k+1} = 2k+1 - (-1) = 2k+2$.
For one-one: Let $f(x_1) = f(x_2)$. If $x_1$ is even and $x_2$ is odd,$2k_1 - 1 = 2k_2 + 2 \implies 2(k_1 - k_2) = 3$,which has no integer solution. If both are even,$2k_1 - 1 = 2k_2 - 1 \implies k_1 = k_2 \implies x_1 = x_2$. If both are odd,$2k_1 + 2 = 2k_2 + 2 \implies k_1 = k_2 \implies x_1 = x_2$. Thus,$f$ is one-one.
For onto: The range consists of all odd numbers (from even inputs) and all even numbers (from odd inputs). Since every integer $y \in Z$ is either even or odd,$f$ is onto.
Therefore,$f$ is both one-one and onto.

(D) Statement-$I$: $A$ function $f: A \rightarrow B$ is one-one (injective) if distinct elements in $A$ have distinct images in $B$. This is equivalent to the contrapositive statement: $f(x) = f(y) \Rightarrow x = y$. The given statement $f(x) \neq f(y) \Rightarrow x \neq y$ is the contrapositive of the definition $x = y \Rightarrow f(x) = f(y)$,which is the definition of a function. However,the definition of one-one is $x \neq y \Rightarrow f(x) \neq f(y)$. The statement $f(x) \neq f(y) \Rightarrow x \neq y$ is logically equivalent to $x = y \Rightarrow f(x) = f(y)$,which is the definition of a function,not one-one. Thus,Statement-$I$ is false.
Statement-$II$: $A$ relation $f: A \rightarrow B$ is a function if every element in $A$ has a unique image in $B$. The condition $x \neq y \Rightarrow f(x) \neq f(y)$ defines an injective (one-one) function,not the definition of a function itself. $A$ function can map different inputs to the same output (many-one). Thus,Statement-$II$ is false.
Therefore,neither statement is true.

(C) Given $f: R \rightarrow R$ such that $f(x)=x^3-x=x(x-1)(x+1)$.
For one-one check: $f(1) = 1^3 - 1 = 0$ and $f(0) = 0^3 - 0 = 0$. Since $f(1) = f(0)$ but $1 \neq 0$,the function is not one-one.
For onto check: $f(x)$ is a polynomial of odd degree. As $x \rightarrow \infty$,$f(x) \rightarrow \infty$ and as $x \rightarrow -\infty$,$f(x) \rightarrow -\infty$. Since the function is continuous,its range is $(-\infty, \infty)$,which is equal to the codomain $R$. Therefore,$f$ is onto.
Thus,$f$ is onto but not one-one.

(B) Given the function $f:[a, \infty) \rightarrow [b, \infty)$ where $f(x) = 2x^2 - 3x + 5$.
For a quadratic function to be a bijection (one-to-one and onto) on the interval $[a, \infty)$,it must be monotonic.
The derivative is $f'(x) = 4x - 3$.
Setting $f'(x) = 0$ gives $x = \frac{3}{4}$.
Thus,the function is increasing for $x \ge \frac{3}{4}$,so $a = \frac{3}{4}$.
Since the function is onto,the range $[b, \infty)$ must equal the image of $[a, \infty)$.
$b = f(a) = f\left(\frac{3}{4}\right) = 2\left(\frac{3}{4}\right)^2 - 3\left(\frac{3}{4}\right) + 5 = 2\left(\frac{9}{16}\right) - \frac{9}{4} + 5 = \frac{9}{8} - \frac{18}{8} + \frac{40}{8} = \frac{31}{8}$.
Finally,$3a + 2b = 3\left(\frac{3}{4}\right) + 2\left(\frac{31}{8}\right) = \frac{9}{4} + \frac{31}{4} = \frac{40}{4} = 10$.

(D) Given the function $f(x) = \begin{cases} a^x, & -\infty < x < 0 \\ b^x, & 0 \leq x < \infty \end{cases}$ with $a > 1$ and $0 < b < 1$.
For $x < 0$,$f(x) = a^x$. Since $a > 1$,as $x \to -\infty$,$f(x) \to 0$,and as $x \to 0^-$,$f(x) \to 1$. Thus,the range for this part is $(0, 1)$.
For $x \geq 0$,$f(x) = b^x$. Since $0 < b < 1$,as $x = 0$,$f(0) = b^0 = 1$,and as $x \to \infty$,$f(x) \to 0$. Thus,the range for this part is $(0, 1]$.
Combining these,the range of $f(x)$ is $(0, 1]$.
$1$. One-one check: For any $y \in (0, 1)$,there exist two values of $x$,one negative and one positive,such that $f(x) = y$. For example,if $y = 0.5$,there is an $x_1 < 0$ such that $a^{x_1} = 0.5$ and an $x_2 > 0$ such that $b^{x_2} = 0.5$. Thus,$f(x)$ is not one-one.
$2$. Onto check: The codomain is given as $[0, 1]$. Since the range is $(0, 1]$,the value $0$ is not in the range (as $a^x > 0$ and $b^x > 0$ for all $x$). Therefore,$f(x)$ is not onto.
Thus,$f(x)$ is neither one-one nor onto.

(B) Given $f(x) = | 2x + 1 | + | x - 2 |$.
We can write the function as:
$f(x) = \begin{cases} -(2x+1) - (x-2) = -3x + 1, & x < -\frac{1}{2} \\ (2x+1) - (x-2) = x + 3, & -\frac{1}{2} \leq x < 2 \\ (2x+1) + (x-2) = 3x - 1, & x \geq 2 \end{cases}$
At $x = -\frac{1}{2}$,$f(-\frac{1}{2}) = 0 + |-\frac{1}{2} - 2| = \frac{5}{2}$.
At $x = 2$,$f(2) = |4+1| + 0 = 5$.
The minimum value of the function is $\frac{5}{2}$ at $x = -\frac{1}{2}$.
Since the function is not strictly increasing or decreasing,it is not one-one (e.g.,$f(0) = 3$ and $f(1) = 4$,but there exist values where $f(x_1) = f(x_2)$ for $x_1 \neq x_2$).
Since the codomain is $[ \frac{5}{2}, \infty )$ and the range of the function is also $[ \frac{5}{2}, \infty )$,the function is onto.
Therefore,the function is onto but not one-one.

(D) Let $n(A) = 5$ and $n(B) = 7$.
Total number of functions from $A$ to $B$ is given by $|B|^{|A|} = 7^5$.
$A$ function is one-one if every element in $A$ maps to a distinct element in $B$. The number of one-one functions is given by $P(7, 5) = { }^7 P_5$.
$A$ function is many-one if it is not one-one.
Therefore,the number of many-one functions = (Total number of functions) - (Number of one-one functions) = $7^5 - { }^7 P_5$.

(C) The function $f: A \rightarrow B$ is defined by $f(x)=x^2-3x+2$.
To be a bijection,the function must be both one-one and onto.
The derivative is $f'(x) = 2x - 3$.
Setting $f'(x) = 0$ gives the vertex at $x = \frac{3}{2}$.
The minimum value of the function is $f\left(\frac{3}{2}\right) = \left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right) + 2 = \frac{9}{4} - \frac{9}{2} + 2 = \frac{9-18+8}{4} = -\frac{1}{4}$.
For the function to be one-one,we must restrict the domain to either $A = \left(-\infty, \frac{3}{2}\right]$ or $A = \left[\frac{3}{2}, \infty\right)$.
For the function to be onto,the codomain $B$ must be equal to the range,which is $\left[-\frac{1}{4}, \infty\right)$.
Comparing this with the given options,option $B$ and $C$ are candidates. However,standard convention for such problems usually selects the interval starting from the vertex. Thus,$A = \left[\frac{3}{2}, \infty\right)$ and $B = \left[-\frac{1}{4}, \infty\right)$ is a valid bijection.

(C) The total number of functions from a set $A$ with $n$ elements to itself is $n^n$.
$A$ function is one-one if every element in the domain has a unique image in the codomain. For a set with $n$ elements,the number of one-one functions is given by the number of permutations of $n$ elements,which is $n!$.
Therefore,the number of functions that are not one-one is the total number of functions minus the number of one-one functions.
Number of functions that are not one-one $= n^n - n!$.

(B) An injection (or one-to-one function) from set $A$ to set $B$ exists only if the number of elements in $A$ is less than or equal to the number of elements in $B$,i.e.,$m \le n$.
If $n < m$,it is impossible to map each element of $A$ to a unique element in $B$,so the number of injections is $0$.
If $n \ge m$,we need to choose $m$ distinct elements from $B$ and arrange them in a specific order to map to the $m$ elements of $A$.
The number of ways to do this is given by the permutation formula $^nP_m = \frac{n!}{(n-m)!}$.
Thus,the number of injections is $\begin{cases} ^nP_m, & n \ge m \\ 0, & n < m \end{cases}$.

(D) The function is given by $f(x) = \frac{x+3}{x-2}$.
Since the function is not defined when the denominator is zero,we have $x - 2 = 0$,which implies $x = 2$. Thus,the domain is $R - \{2\}$,so $l = 2$.
To find the range,let $y = \frac{x+3}{x-2}$.
Then $y(x - 2) = x + 3$,which gives $xy - 2y = x + 3$.
Rearranging for $x$,we get $x(y - 1) = 2y + 3$,or $x = \frac{2y+3}{y-1}$.
The function is not defined for $y = 1$,so the range is $R - \{1\}$. Thus,$m = 1$.
We need to calculate $3l - 2m$.
Substituting the values,$3(2) - 2(1) = 6 - 2 = 4$.

(D) Given the function $f: R \rightarrow R$ defined by $f(x) = 5x^4 + 2$.
To check for one-one:
Consider $f(x_1) = f(x_2)$.
$5x_1^4 + 2 = 5x_2^4 + 2$
$x_1^4 = x_2^4$
$x_1 = \pm x_2$.
Since $f(1) = 5(1)^4 + 2 = 7$ and $f(-1) = 5(-1)^4 + 2 = 7$,we have $f(1) = f(-1)$ but $1 \neq -1$.
Therefore,$f$ is not one-one.
To check for onto:
Since $x^4 \geq 0$ for all $x \in R$,it follows that $5x^4 \geq 0$.
Thus,$f(x) = 5x^4 + 2 \geq 2$.
The range of $f$ is $[2, \infty)$,which is not equal to the codomain $R$.
Therefore,$f$ is not onto.
Hence,$f$ is neither one-one nor onto.

(C) Given $f(x) = x^2 - 2x - 3$.
For one-one:
Let $f(x_1) = f(x_2)$.
$x_1^2 - 2x_1 - 3 = x_2^2 - 2x_2 - 3$
$x_1^2 - x_2^2 - 2(x_1 - x_2) = 0$
$(x_1 - x_2)(x_1 + x_2) - 2(x_1 - x_2) = 0$
$(x_1 - x_2)(x_1 + x_2 - 2) = 0$
This implies $x_1 = x_2$ or $x_1 + x_2 = 2$.
Since $x_1 + x_2 = 2$ allows distinct values (e.g.,$f(0) = -3$ and $f(2) = -3$),the function is not one-one.
For onto:
Let $y = x^2 - 2x - 3$.
$y = (x-1)^2 - 4$
$(x-1)^2 = y + 4$
Since $(x-1)^2 \geq 0$,we must have $y + 4 \geq 0$,which means $y \geq -4$.
The range of $f$ is $[-4, \infty)$,which is not equal to the codomain $R$.
Therefore,the function is not onto.
Thus,$f$ is neither one-one nor onto.

(C) Given,$f(x)=x|x|$.
We can write this as:
$f(x) = \begin{cases} x^2, & x \geq 0 \\ -x^2, & x < 0 \end{cases}$
$1$. One-one check:
Let $f(x_1) = f(x_2)$.
If $x_1, x_2 \geq 0$,then $x_1^2 = x_2^2 \Rightarrow x_1 = x_2$ (since $x \geq 0$).
If $x_1, x_2 < 0$,then $-x_1^2 = -x_2^2 \Rightarrow x_1^2 = x_2^2 \Rightarrow x_1 = x_2$ (since $x < 0$).
If one is positive and one is negative,$f(x)$ will have different signs,so $f(x_1) \neq f(x_2)$.
Thus,$f(x)$ is one-one.
$2$. Onto check:
For any $y \in R$,we can find $x$ such that $f(x) = y$.
If $y \geq 0$,$x = \sqrt{y}$. If $y < 0$,$x = -\sqrt{-y}$.
Since for every $y$ in the codomain $R$,there exists an $x$ in the domain $R$,the function is onto.
Therefore,$f$ is both one-one and onto.

(C) Given $f(x) = \max \{x+1, 1-x, 2\}$.
We can analyze the function by breaking it into intervals:
For $x < -1$,$x+1 < 0$ and $1-x > 2$,so $f(x) = 1-x$.
For $-1 \leq x \leq 1$,$x+1 \geq 0$,$1-x \geq 0$,and the maximum of these and $2$ is $2$ (since $x+1 \leq 2$ and $1-x \leq 2$ for $x \in [-1, 1]$).
For $x > 1$,$x+1 > 2$ and $1-x < 0$,so $f(x) = x+1$.
Thus,$f(x) = \begin{cases} 1-x, & x < -1 \\ 2, & -1 \leq x \leq 1 \\ x+1, & x > 1 \end{cases}$.
Since $f(x) = 2$ for all $x \in [-1, 1]$,the function is not one-one (many-to-one).
Since the range of $f(x)$ is $[2, \infty)$,which is a proper subset of the codomain $R$,the function is not onto.
Therefore,$f$ is neither one-one nor onto.

(C) scalar matrix $M$ of order $3 \times 3$ is of the form $M = \begin{bmatrix} m & 0 & 0 \\ 0 & m & 0 \\ 0 & 0 & m \end{bmatrix}$ for some $m \in R$.
The determinant of $M$ is given by $\operatorname{det}(M) = m^3$.
Thus,the function $f: A \rightarrow R$ is defined as $f(M) = m^3$.
Since $f(m) = m^3$ is a strictly increasing function,for any $m_1, m_2 \in R$,$f(m_1) = f(m_2) \implies m_1^3 = m_2^3 \implies m_1 = m_2$. Therefore,$f$ is one-one (injective).
For any real number $y \in R$,there exists a real number $m = \sqrt[3]{y}$ such that $f(m) = (\sqrt[3]{y})^3 = y$. Thus,the range of $f$ is $R$,which is equal to the codomain. Therefore,$f$ is onto (surjective).
Since $f$ is both one-one and onto,it is bijective.

(D) The function $f: N \times N \rightarrow N$ is defined by $f(m, n) = 2^{m-1}(2n-1)$.
To check for one-one: Let $f(m_1, n_1) = f(m_2, n_2)$.
Then $2^{m_1-1}(2n_1-1) = 2^{m_2-1}(2n_2-1)$.
Since any natural number $k$ can be uniquely written as $2^{m-1}(2n-1)$ where $2n-1$ is the odd part of $k$ and $2^{m-1}$ is the highest power of $2$ dividing $k$,we must have $m_1-1 = m_2-1$ and $2n_1-1 = 2n_2-1$.
This implies $m_1 = m_2$ and $n_1 = n_2$. Thus,$f$ is one-one.
To check for onto: For any $k \in N$,if $k$ is odd,$k = 2^{1-1}(2n-1)$ with $m=1$. If $k$ is even,$k = 2^p \cdot q$ where $q$ is odd,so $m-1 = p$ and $2n-1 = q$.
Since every $k \in N$ has a unique representation of this form,$f$ is onto.
Therefore,$f$ is both one-one and onto.

(D) Let $f(x) = \frac{e^{|x|} - e^{-x}}{e^x + e^{-x}} + \cos^3\left(\frac{x}{2}\right)$.
For $x \ge 0$,$|x| = x$,so $f(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} + \cos^3\left(\frac{x}{2}\right) = \tanh(x) + \cos^3\left(\frac{x}{2}\right)$.
For $x < 0$,$|x| = -x$,so $f(x) = \frac{e^{-x} - e^{-x}}{e^x + e^{-x}} + \cos^3\left(\frac{x}{2}\right) = 0 + \cos^3\left(\frac{x}{2}\right) = \cos^3\left(\frac{x}{2}\right)$.
Since $f(x)$ is not symmetric about the $y$-axis $(f(x) \neq f(-x))$,it is not an even function.
Since $f(x)$ is not symmetric about the origin $(f(-x) \neq -f(x))$,it is not an odd function.
As $x \to \infty$,$f(x) \to 1 + 0 = 1$. As $x \to -\infty$,$f(x) = \cos^3(x/2)$,which oscillates between $-1$ and $1$.
Since the function is not monotonic and its range is not $R$,it is not surjective.
Since it is not surjective,it cannot be bijective. Thus,the function is not bijective.

(C) The number of onto functions from a set $A$ with $n$ elements to a set $B$ with $m$ elements is given by $m^n - \binom{m}{1}(m-1)^n + \binom{m}{2}(m-2)^n - \ldots$.
Here,$m = 2$ and $n = n$. The number of onto functions is $2^n - \binom{2}{1}(2-1)^n = 2^n - 2$.
Given $2^n - 2 = 62$,we have $2^n = 64$,which implies $n = 6$.
The number of subsets of $A$ containing exactly three elements is given by $\binom{n}{3} = \binom{6}{3}$.
Calculating this,$\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.

(D) Given function $f(x)=\frac{x}{e^x-1}+\frac{x}{2}+2 \cos ^3 \frac{x}{2}$ on $R-\{0\}$.
To check if the function is even,we evaluate $f(-x)$:
$f(-x) = \frac{-x}{e^{-x}-1} + \frac{-x}{2} + 2 \cos ^3 \left(-\frac{x}{2}\right)$
Since $\cos(- \theta) = \cos(\theta)$,we have $\cos^3(-\frac{x}{2}) = \cos^3(\frac{x}{2})$.
$f(-x) = \frac{-x}{\frac{1}{e^x}-1} - \frac{x}{2} + 2 \cos ^3 \frac{x}{2}$
$f(-x) = \frac{-x e^x}{1-e^x} - \frac{x}{2} + 2 \cos ^3 \frac{x}{2}$
$f(-x) = \frac{x e^x}{e^x-1} - \frac{x}{2} + 2 \cos ^3 \frac{x}{2}$
$f(-x) = \frac{x(e^x-1+1)}{e^x-1} - \frac{x}{2} + 2 \cos ^3 \frac{x}{2}$
$f(-x) = x + \frac{x}{e^x-1} - \frac{x}{2} + 2 \cos ^3 \frac{x}{2}$
$f(-x) = \frac{x}{e^x-1} + \frac{x}{2} + 2 \cos ^3 \frac{x}{2} = f(x)$
Since $f(-x) = f(x)$ for all $x \in R-\{0\}$,the function is an even function.
Therefore,option $D$ is correct.

(A) For the function $f: R \rightarrow R$ defined by $f(x)=px+q$ $(p \neq 0)$,it is a linear function. Linear functions are bijections (both one-one and onto) on the set of real numbers $R$. Thus,$A \rightarrow II$.
$(B)$ For the function $f: R \rightarrow R^{+} \cup \{0\}$ defined by $f(x)=x^2$,we observe that $f(-1)=f(1)=1$,so it is not one-one. However,for every $y \in R^{+} \cup \{0\}$,there exists $x = \sqrt{y} \in R$ such that $f(x)=y$,so it is onto. Thus,$B \rightarrow IV$.
$(C)$ For $f: N \rightarrow N$ defined by $f(n)=n^2+2n+3$,it is one-one because $f(n_1)=f(n_2) \implies n_1^2+2n_1+3 = n_2^2+2n_2+3 \implies (n_1-n_2)(n_1+n_2+2)=0$,which implies $n_1=n_2$ for $n \in N$. It is not onto because for $f(n)=3$,$n^2+2n+3=3 \implies n(n+2)=0$,which gives $n=0$ or $n=-2$,neither of which is in $N$. Thus,$C \rightarrow III$.
$(D)$ For $f: R \rightarrow R$ defined by $f(x)=2(\cos^2 5x + \sin^2 5x) = 2(1) = 2$. This is a constant function. $A$ constant function is neither one-one nor onto (unless the domain and codomain are singletons). Thus,$D \rightarrow I$.
Therefore,the correct matching is $(A)-II, (B)-IV, (C)-III, (D)-I$.

(B) To check for one-one: Let $f(x_1) = f(x_2)$ for $x_1, x_2 \in [0, \infty)$.
$\frac{x_1}{1+x_1} = \frac{x_2}{1+x_2}$
$x_1(1+x_2) = x_2(1+x_1)$
$x_1 + x_1x_2 = x_2 + x_1x_2$
$x_1 = x_2$.
Thus,$f$ is one-one.
To check for onto: Let $y = \frac{x}{1+x}$.
$y(1+x) = x \implies y + xy = x \implies y = x(1-y) \implies x = \frac{y}{1-y}$.
For $x \in [0, \infty)$,we need $y \in [0, 1)$.
Since the codomain is $[0, \infty)$,values like $y = 2$ have no pre-image in the domain.
Thus,$f$ is not onto.
Therefore,the function is one-one but not onto.

(C) To check for one-one: Let $f(x_1) = f(x_2)$.
$\frac{4x_1-3}{x_1-1} = \frac{4x_2-3}{x_2-1}$
$(4x_1-3)(x_2-1) = (4x_2-3)(x_1-1)$
$4x_1x_2 - 4x_1 - 3x_2 + 3 = 4x_1x_2 - 4x_2 - 3x_1 + 3$
$-4x_1 - 3x_2 = -4x_2 - 3x_1$
$x_1 = x_2$.
Thus,the function is one-one.
To check for onto: Let $y = \frac{4x-3}{x-1}$.
$y(x-1) = 4x-3$
$yx - y = 4x - 3$
$yx - 4x = y - 3$
$x(y-4) = y-3$
$x = \frac{y-3}{y-4}$.
For every $y \in R-\{4\}$,there exists an $x = \frac{y-3}{y-4} \in R-\{1\}$.
Thus,the function is onto.
Therefore,the function is one-one and onto.

(D) Given the function $\sigma: N \rightarrow Z$ defined as:
$\sigma(n) = \begin{cases} \frac{n}{2} & \text{if } n \text{ is even} \\ -\frac{n-1}{2} & \text{if } n \text{ is odd} \end{cases}$
Case-$I$: If $n$ is even,let $n = 2k$ for $k \in N$. Then $\sigma(2k) = \frac{2k}{2} = k$. As $k$ takes values $1, 2, 3, \dots$,$\sigma(n)$ takes values $1, 2, 3, \dots$ (all positive integers).
Case-$II$: If $n$ is odd,let $n = 2k-1$ for $k \in N$. Then $\sigma(2k-1) = -\frac{(2k-1)-1}{2} = -\frac{2k-2}{2} = -(k-1) = 1-k$. As $k$ takes values $1, 2, 3, \dots$,$\sigma(n)$ takes values $0, -1, -2, \dots$ (all non-positive integers).
Combining both cases,the range of $\sigma$ is ${0, 1, -1, 2, -2, 3, -3, \dots} = Z$.
Since every distinct $n \in N$ maps to a distinct integer in $Z$,the function is one-one.
Since the range of $\sigma$ is equal to the codomain $Z$,the function is onto.
Therefore,$\sigma$ is both one-one and onto.

(C) We have $f(x) = |x-1| + |x-2| + |x-3|$.
For the interval $2 < x < 3$:
$|x-1| = x-1$ (since $x > 1$)
$|x-2| = x-2$ (since $x > 2$)
$|x-3| = -(x-3) = 3-x$ (since $x < 3$)
Therefore,$f(x) = (x-1) + (x-2) + (3-x) = x - 2 + 3 - 2 = x$.
In the interval $(2, 3)$,the function $f(x) = x$ is strictly increasing.
Since $f(x) = x$,for every $x \in (2, 3)$,the range is $(2, 3)$.
Thus,the function is one-one and onto,which means it is a bijection.

(A) Given,$f: Z \rightarrow Z$ defined as $f(x)=\begin{cases} \frac{x}{2}, & \text{if } x \text{ is even} \\ 0, & \text{if } x \text{ is odd} \end{cases}$.
For one-to-one check: 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 $1 \neq 3$,the function is not one-to-one.
For onto check: The range of $f$ consists of all integers. For any integer $y \in Z$,we can find an $x \in Z$ such that $f(x) = y$. If $y = 0$,$f(1) = 0$. If $y \neq 0$,let $x = 2y$,which is even,then $f(2y) = \frac{2y}{2} = y$. Since the range equals the codomain $(Z)$,the function is onto.
Therefore,$f$ is onto but not one-to-one.

(A) The function is $f(x) = x^2$.
For $f$ to be one-one,$f(x_1) = f(x_2) \implies x_1^2 = x_2^2$. This implies $x_1 = x_2$ only if $x_1, x_2$ have the same sign.
For $f$ to be onto,the range of $f$ must equal the codomain $B$. The range of $f(x) = x^2$ for $x \in A$ is ${x^2 : x \in A}$.
Analysis:
$(A)$ $A = B = R^{+}$: $f(x) = x^2$ maps positive reals to positive reals. It is one-one (since $x_1^2 = x_2^2$ and $x_1, x_2 > 0 \implies x_1 = x_2$) and onto (since for any $y \in R^{+}$,$x = \sqrt{y} \in R^{+}$ exists). Thus,$A \rightarrow 4$.
$(B)$ $A = R^{+}, B = R$: $f(x) = x^2$ is one-one (as $x > 0$),but not onto because negative values in $B$ are not covered by the range $(R^{+})$. Thus,$B \rightarrow 1$.
$(C)$ $A = R, B = R^{+}$: $f(x) = x^2$ is not one-one (as $f(1) = f(-1) = 1$) but is onto (since every $y \in R^{+}$ has a pre-image $x = \pm \sqrt{y} \in R$). Thus,$C \rightarrow 3$.
$(D)$ $A = B = R$: $f(x) = x^2$ is not one-one (as $f(1) = f(-1)$) and not onto (as negative values in $B$ are not covered). Thus,$D \rightarrow 2$.
The correct matching is $A-4, B-1, C-3, D-2$.

(D) Given the function $f(x) = e^{2ix} = \cos(2x) + i \sin(2x)$.
For $f$ to be one-one,$f(x_1) = f(x_2)$ must imply $x_1 = x_2$. However,$f(x + \pi) = e^{2i(x+\pi)} = e^{2ix} \cdot e^{2i\pi} = e^{2ix} \cdot 1 = f(x)$. Since $f(x) = f(x+\pi)$ for all $x \in R$,the function is many-one.
For $f$ to be onto,the range of $f$ must be equal to the codomain $C$. The range of $f(x) = \cos(2x) + i \sin(2x)$ is the set of all complex numbers with modulus $1$,i.e.,${z \in C : |z| = 1}$. Since this is a subset of $C$ and not equal to $C$,the function is not onto.
Therefore,$f$ is neither one-one nor onto.

(C) function is a bijection if it is both injective (one-to-one) and surjective (onto).
For the domain and codomain $A = [-1, 1]$:
$A) f(x) = x|x|$ is a bijection because it is strictly increasing and maps $[-1, 1]$ onto $[-1, 1]$.
$B) f(x) = x^3$ is a bijection because it is strictly increasing and maps $[-1, 1]$ onto $[-1, 1]$.
$C) f(x) = x^2$ is not a bijection. It is not injective because $f(1) = f(-1) = 1$. It is also not surjective because the range is $[0, 1]$,which is not equal to the codomain $[-1, 1]$.
$D) f(x) = \sin \left(\frac{\pi x}{2}\right)$ is a bijection because it is strictly increasing and maps $[-1, 1]$ onto $[-1, 1]$.
Thus,the correct option is $C$.

(C) The given function is $f(x) = \frac{ax + b}{cx + d}$.
For the function to be a constant function,its derivative with respect to $x$ must be zero,or the numerator must be a constant multiple of the denominator.
Let $f(x) = k$ (a constant).
Then $\frac{ax + b}{cx + d} = k$.
$ax + b = k(cx + d) = (kc)x + kd$.
Comparing the coefficients of $x$ and the constant terms,we get $a = kc$ and $b = kd$.
This implies $\frac{a}{c} = k$ and $\frac{b}{d} = k$ (assuming $c, d \neq 0$).
Therefore,$\frac{a}{c} = \frac{b}{d}$,which gives $ad = bc$.
Alternatively,if we take option $(c)$,i.e.,$ad = bc$,then $ad - bc = 0$.
If $ad = bc$,then $\frac{a}{c} = \frac{b}{d} = k$ (for $c, d \neq 0$).
Substituting $a = kc$ and $b = kd$ into the function:
$f(x) = \frac{(kc)x + kd}{cx + d} = \frac{k(cx + d)}{cx + d} = k$.
Since $f(x) = k$,the function is constant.