Type of Functions based on Mapping Questions in English

(C) The period of the function $f(x) = \cos(ax)$ is given by $T = \frac{2\pi}{|a|}$.
Given $f(x) = \cos(\sqrt{P}x)$,the period is $T = \frac{2\pi}{\sqrt{P}}$.
We are given that the period is $\pi$,so $\frac{2\pi}{\sqrt{P}} = \pi$.
This implies $\sqrt{P} = 2$,which means $P = 4$.
Since $P = [\lambda]$,we have $[\lambda] = 4$.
By the definition of the Greatest Integer Function,$[\lambda] = 4$ implies $4 \le \lambda < 5$.
Thus,$\lambda \in [4, 5)$.

(A) For $f(x)$ to be one-one,the function must be strictly monotonic or the ranges of the two pieces must be disjoint.
Case $1$: For $x \leq 0$,$f(x) = x^2 + 2mx - 1$. The vertex of this parabola is at $x = -m$. For the function to be one-one on $(-\infty, 0]$,the vertex must be at $x \geq 0$,which implies $-m \geq 0$,so $m \leq 0$.
Case $2$: For $x > 0$,$f(x) = mx - 1$. For this to be one-one,$m$ must be $\geq 0$. However,if $m > 0$,the range of $f(x)$ for $x > 0$ is $(-1, \infty)$. The range of $f(x)$ for $x \leq 0$ is $[-1-m^2, \infty)$. These ranges overlap,so the function would not be one-one.
Case $3$: If $m < 0$,then for $x > 0$,$f(x) = mx - 1$ is a decreasing function with range $(-\infty, -1)$. For $x \leq 0$,$f(x) = x^2 + 2mx - 1$ is decreasing on $(-\infty, -m]$. Since $-m > 0$,the function is decreasing on $(-\infty, 0]$. The range for $x \leq 0$ is $[-1-m^2, \infty)$. Since the ranges $(-1, -\infty)$ and $[-1-m^2, \infty)$ do not overlap for $m < 0$,the function is one-one.
Thus,$m \in (-\infty, 0)$.

(D) Given the function $f(x) = x - 5[\frac{x}{5}]$.
To check for one-one: Calculate $f(1) = 1 - 5[1/5] = 1 - 5(0) = 1$ and $f(6) = 6 - 5[6/5] = 6 - 5(1) = 1$.
Since $f(1) = f(6) = 1$ but $1 \neq 6$,the function is not one-one.
To check for onto: The codomain is $N = \{1, 2, 3, ...\}$.
Calculate $f(5) = 5 - 5[5/5] = 5 - 5(1) = 0$.
Since $0 \notin N$,the function is not onto.
Therefore,the function $f$ is neither one-one nor onto.

(A) We are given $A = \{x_1, \dots, x_7\}$ and $B = \{y_1, y_2, y_3\}$.
We need to find the number of onto functions $f: A \to B$ such that exactly three elements in $A$ map to $y_2$.
First,we choose $3$ elements from $A$ to map to $y_2$,which can be done in ${}^7C_3$ ways.
Now,the remaining $4$ elements of $A$ must map to the remaining $2$ elements of $B$,which are $\{y_1, y_3\}$.
For the function to be onto,the set $\{y_1, y_3\}$ must be covered by the $4$ elements.
The total number of functions from these $4$ elements to $\{y_1, y_3\}$ is $2^4 = 16$.
We must exclude the cases where all $4$ elements map to only $y_1$ or only $y_3$ to ensure the function is onto (i.e.,$y_1$ and $y_3$ are both covered).
Thus,the number of ways is $2^4 - 2 = 16 - 2 = 14$.
Therefore,the total number of such onto functions is ${}^7C_3 \times 14 = 14 \times {}^7C_3$.

(B) Given $f(x) = (\frac{3}{5})^x + (\frac{4}{5})^x - 1$.
Setting $f(x) = 0$,we get $(\frac{3}{5})^x + (\frac{4}{5})^x = 1$.
Dividing by $5^x$ is equivalent to solving $3^x + 4^x = 5^x$.
Divide both sides by $5^x$: $(\frac{3}{5})^x + (\frac{4}{5})^x = 1$.
Let $g(x) = (\frac{3}{5})^x + (\frac{4}{5})^x$.
Since both $(\frac{3}{5})^x$ and $(\frac{4}{5})^x$ are strictly decreasing functions for $x \in R$,their sum $g(x)$ is also a strictly decreasing function.
$A$ strictly decreasing function can intersect the horizontal line $y = 1$ at most once.
By inspection,for $x = 2$,we have $(\frac{3}{5})^2 + (\frac{4}{5})^2 = \frac{9}{25} + \frac{16}{25} = \frac{25}{25} = 1$.
Thus,$x = 2$ is the unique solution.

(D) Given $f(x) = \frac{|x| - 1}{|x| + 1}$.
For a function to be one-one,if $f(x_1) = f(x_2)$,then $x_1$ must be equal to $x_2$.
Let $f(x_1) = f(x_2)$:
$\frac{|x_1| - 1}{|x_1| + 1} = \frac{|x_2| - 1}{|x_2| + 1}$
$(|x_1| - 1)(|x_2| + 1) = (|x_2| - 1)(|x_1| + 1)$
$|x_1||x_2| + |x_1| - |x_2| - 1 = |x_1||x_2| - |x_1| + |x_2| - 1$
$2|x_1| = 2|x_2| \implies |x_1| = |x_2|$
This implies $x_1 = x_2$ or $x_1 = -x_2$. Since $f(1) = f(-1) = 0$,the function is not one-one (it is many-one).
For onto: Let $y = \frac{|x| - 1}{|x| + 1}$.
$y(|x| + 1) = |x| - 1 \implies y|x| + y = |x| - 1 \implies |x|(y - 1) = -1 - y \implies |x| = \frac{1 + y}{1 - y}$.
Since $|x| \ge 0$,we must have $\frac{1 + y}{1 - y} \ge 0$. Solving this inequality,we get $y \in [-1, 1)$.
The range of $f$ is $[-1, 1)$,which is not equal to the codomain $R$. Therefore,the function is not onto.

(C) Given the set $A = \{1, 2, 3, 4\}$ and the relation $R = \{ (1, 1), (2, 3), (3, 4), (4, 2) \}$.
First,we check if $R$ is a function. Since every element in the domain $A$ has a unique image in the codomain $A$,$R$ is a function.
Next,we check if $R$ is one-to-one. The images are $\{1, 3, 4, 2\}$. Since all images are distinct,$R$ is a one-to-one function.
Finally,we check if $R$ is onto. The range of $R$ is $\{1, 2, 3, 4\}$,which is equal to the codomain $A$.
Since the range equals the codomain,$R$ is an onto function.
Therefore,the correct statement is that $R$ is an onto function.

(C) Given the set $S = \{ 1, 2, 3 \}$.
The number of elements in $S$ is $n(S) = 3$.
$P(S)$ is the power set of $S$,which is the set of all subsets of $S$.
The number of elements in the power set is $n(P(S)) = 2^{n(S)} = 2^3 = 8$.
$A$ function $f: S \to P(S)$ is one-to-one (injective) if each element in $S$ maps to a unique element in $P(S)$.
The number of one-to-one functions from a set with $m$ elements to a set with $n$ elements is given by the permutation formula $^nP_m = \frac{n!}{(n-m)!}$.
Here,$m = n(S) = 3$ and $n = n(P(S)) = 8$.
Therefore,the number of one-to-one functions is $^8P_3 = \frac{8!}{(8-3)!} = \frac{8!}{5!} = 8 \times 7 \times 6 = 336$.

(D) function $f : A \to B$ is bijective if it is both one-to-one (injective) and onto (surjective).
Statement $1$ states that $f$ is an onto function,which is true by the definition of a bijective function.
Statement $2$ states that there exists a function $g : B \to A$ such that $f \circ g = I_B$. Since $f$ is bijective,it is invertible,meaning there exists an inverse function $f^{-1} : B \to A$ such that $f \circ f^{-1} = I_B$. Thus,$g = f^{-1}$ satisfies the condition. Therefore,Statement $2$ is also true.
However,Statement $2$ describes the property of invertibility resulting from bijectivity,while Statement $1$ is simply the definition of one part of bijectivity. Statement $2$ is not the reason why $f$ is onto; rather,both are consequences of $f$ being bijective. Thus,Statement $2$ is not the correct explanation for Statement $1$.

(A) Given $f(x) = \frac{2x}{x - 1}$ for $x \in A$,where $A = R \setminus \{1, 2, 3, \dots\}$.
To check for injectivity (one-one): Let $f(x_1) = f(x_2)$.
$\frac{2x_1}{x_1 - 1} = \frac{2x_2}{x_2 - 1}$
$x_1(x_2 - 1) = x_2(x_1 - 1)$
$x_1x_2 - x_1 = x_2x_1 - x_2$
$-x_1 = -x_2 \Rightarrow x_1 = x_2$.
Thus,$f$ is injective (one-one).
To check for surjectivity (onto): Let $y = \frac{2x}{x - 1}$.
$y(x - 1) = 2x$
$yx - y = 2x$
$x(y - 2) = y$
$x = \frac{y}{y - 2}$.
For $f$ to be surjective,for every $y \in R$,there must exist an $x \in A$ such that $f(x) = y$.
If $y = 2$,$x$ is undefined. Also,if $x$ is a positive integer,$x \notin A$.
For example,if $y = 4$,$x = \frac{4}{4 - 2} = 2$. Since $2 \notin A$,there is no $x \in A$ such that $f(x) = 4$.
Therefore,$f$ is not surjective.

(C) Given $f(x) = |1 - \frac{1}{x}|$ for $x \in (0, \infty)$.
To check for injectivity (one-one): Consider $f(x) = y$. For $y \in (0, 1)$,there exist two values of $x$. For example,if $y = 0.5$,then $|1 - \frac{1}{x}| = 0.5$,which gives $1 - \frac{1}{x} = 0.5 \implies \frac{1}{x} = 0.5 \implies x = 2$,and $1 - \frac{1}{x} = -0.5 \implies \frac{1}{x} = 1.5 \implies x = \frac{2}{3}$. Since $f(2) = f(\frac{2}{3}) = 0.5$,the function is not injective.
To check for surjectivity (onto): The codomain is $(0, \infty)$. The range of $f(x) = |1 - \frac{1}{x}|$ for $x \in (0, \infty)$ is $[0, \infty)$. Since the range $[0, \infty)$ is not equal to the codomain $(0, \infty)$ (as $0$ is in the range but not in the codomain),the function is not surjective.
Thus,$f$ is neither injective nor surjective.

(C) Let $S = \{1, 2, 3, \dots, 20\}$. The elements in $S$ that are multiples of $4$ are $K = \{4, 8, 12, 16, 20\}$. There are $5$ such elements.
For $k \in K$,$f(k)$ must be a multiple of $3$. The multiples of $3$ in $S$ are $M = \{3, 6, 9, 12, 15, 18\}$. There are $6$ such elements.
Since $f$ is an onto function,the $5$ elements of $K$ must be mapped to $5$ distinct elements of $M$. The number of ways to choose and arrange these is $^6P_5 = \frac{6!}{1!} = 6!$.
The remaining $15$ elements of $S \setminus K$ must be mapped onto the remaining $15$ elements of $S \setminus f(K)$ in a bijective manner,which can be done in $15!$ ways.
Therefore,the total number of onto functions is $6! \times 15!$.

(N/A) $1$. To check for one-one: Let $x_1$ and $x_2$ be two different students in set $A$. Since no two different students can have the same roll number,$f(x_1) \neq f(x_2)$. Thus,$f$ is one-one.
$2$. To check for onto: The codomain of $f$ is the set of natural numbers $N = \{1, 2, 3, ...\}$. The range of $f$ is the set of roll numbers assigned to the $50$ students,which is $\{1, 2, 3, ..., 50\}$.
$3$. Since the range $\{1, 2, 3, ..., 50\}$ is a proper subset of the codomain $N$ (e.g.,$51 \in N$ but $51$ is not in the range),there exists at least one element in $N$ that has no pre-image in $A$.
$4$. Therefore,$f$ is not onto.

(N/A) $1$. To check if the function is one-one: Let $x_{1}, x_{2} \in N$ such that $f(x_{1}) = f(x_{2})$.
Then,$2x_{1} = 2x_{2}$,which implies $x_{1} = x_{2}$.
Since $f(x_{1}) = f(x_{2})$ leads to $x_{1} = x_{2}$,the function $f$ is one-one.
$2$. To check if the function is onto: $A$ function is onto if for every $y \in N$ (codomain),there exists an $x \in N$ (domain) such that $f(x) = y$.
Here,$f(x) = 2x$. If we take $y = 1 \in N$,then $2x = 1$,which gives $x = 1/2$.
Since $1/2 \notin N$,there is no $x \in N$ such that $f(x) = 1$.
Therefore,the function $f$ is not onto.

(N/A) To prove that $f$ is one-one,we assume $f(x_1) = f(x_2)$ for any $x_1, x_2 \in R$.
$2x_1 = 2x_2$
Dividing both sides by $2$,we get $x_1 = x_2$.
Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$,the function $f$ is one-one.
To prove that $f$ is onto,we take any element $y \in R$ (codomain).
We need to find an $x \in R$ (domain) such that $f(x) = y$.
$2x = y \Rightarrow x = \frac{y}{2}$.
Since $y \in R$,$\frac{y}{2}$ is also a real number,so $\frac{y}{2} \in R$.
For every $y \in R$,there exists $x = \frac{y}{2} \in R$ such that $f(x) = f(\frac{y}{2}) = 2(\frac{y}{2}) = y$.
Thus,$f$ is onto.

(N/A) To check if $f$ is one-one:
$A$ function $f$ is one-one if $f(x_1) = f(x_2) \implies x_1 = x_2$.
Here,$f(1) = 1$ and $f(2) = 1$.
Since $f(1) = f(2)$ but $1 \neq 2$,the function $f$ is not one-one.
To check if $f$ is onto:
$A$ function $f: N \rightarrow N$ is onto if for every $y \in N$,there exists an $x \in N$ such that $f(x) = y$.
Case $1$: If $y = 1$,we have $f(1) = 1$. Thus,$1$ has a pre-image.
Case $2$: If $y > 1$,then $y+1 > 2$. We can choose $x = y+1$. Then $f(x) = f(y+1) = (y+1) - 1 = y$.
Since every $y \in N$ has a pre-image in $N$,the function $f$ is onto.

(N/A) To check for one-one: $A$ function $f$ is one-one if $f(x_{1}) = f(x_{2})$ implies $x_{1} = x_{2}$.
Here,$f(1) = (1)^{2} = 1$ and $f(-1) = (-1)^{2} = 1$.
Since $f(1) = f(-1)$ but $1 \neq -1$,the function $f$ is not one-one.
To check for onto: $A$ function $f: R \rightarrow R$ is onto if for every $y \in R$ (co-domain),there exists an $x \in R$ (domain) such that $f(x) = y$.
Here,the range of $f(x) = x^{2}$ is $[0, \infty)$,which is a subset of the co-domain $R$.
For example,consider $y = -2$ in the co-domain $R$. There is no real number $x$ such that $x^{2} = -2$,because the square of any real number is non-negative.
Therefore,$f$ is not onto.

(A) Suppose $f(x_1) = f(x_2)$.
If $x_1$ is odd and $x_2$ is even,then $x_1+1 = x_2-1$,which implies $x_2-x_1 = 2$. Since $x_1$ is odd and $x_2$ is even,the difference between an even and an odd number is always odd,so $x_2-x_1$ cannot be $2$. Thus,this case is impossible.
Similarly,if $x_1$ is even and $x_2$ is odd,$x_1-1 = x_2+1$,implying $x_1-x_2 = 2$,which is also impossible.
Therefore,$x_1$ and $x_2$ must be both odd or both even.
If both are odd,$f(x_1) = f(x_2) \Rightarrow x_1+1 = x_2+1 \Rightarrow x_1 = x_2$.
If both are even,$f(x_1) = f(x_2) \Rightarrow x_1-1 = x_2-1 \Rightarrow x_1 = x_2$.
Thus,$f$ is one-one.
For onto,consider any $y \in N$. If $y$ is odd,$y = 2r+1$ for some $r \ge 0$. Then $f(2r+2) = (2r+2)-1 = 2r+1 = y$. If $y$ is even,$y = 2r$ for some $r \ge 1$. Then $f(2r-1) = (2r-1)+1 = 2r = y$. Since every $y \in N$ has a pre-image in $N$,$f$ is onto.

(A) Let $f: A \rightarrow A$ be a function where $A = \{1, 2, 3\}$.
Suppose $f$ is not one-one. Then there exist at least two distinct elements in the domain $A$ that have the same image in the co-domain $A$.
Since the domain has $3$ elements,if $f$ is not one-one,at most $2$ distinct elements can be mapped to the co-domain.
Specifically,if $f(x_1) = f(x_2) = y$ for $x_1 \neq x_2$,then the range of $f$ can contain at most $2$ elements (the image $y$ and the image of the third element).
However,for $f$ to be onto,the range must be equal to the co-domain,which has $3$ elements.
Since the range has at most $2$ elements,it cannot be equal to the co-domain $\{1, 2, 3\}$.
This contradicts the assumption that $f$ is onto.
Therefore,$f$ must be one-one.

(N/A) Let $A = \{1, 2, 3\}$. The function $f: A \rightarrow A$ is given as one-one.
By the definition of a one-one function,distinct elements in the domain $A$ must map to distinct elements in the codomain $A$.
Since the domain $A$ has $3$ elements,their images $f(1), f(2),$ and $f(3)$ must be $3$ distinct elements of the codomain $A$.
Because the codomain $A$ also contains exactly $3$ elements,the set of images ${f(1), f(2), f(3)}$ must be equal to the entire codomain $A$.
Therefore,every element in the codomain has a pre-image in the domain,which satisfies the definition of an onto function.
Thus,$f$ must be onto.

(N/A) Given $f: R_* \rightarrow R_*$ defined by $f(x) = \frac{1}{x}$.
For one-one:
Let $x, y \in R_*$ such that $f(x) = f(y)$.
$\Rightarrow \frac{1}{x} = \frac{1}{y}$
$\Rightarrow x = y$.
Therefore,$f$ is one-one.
For onto:
For any $y \in R_*$,there exists $x = \frac{1}{y} \in R_*$ (since $y \neq 0$) such that $f(x) = \frac{1}{(1/y)} = y$.
Therefore,$f$ is onto.
Thus,$f$ is one-one and onto.
Now,consider $g: N \rightarrow R_*$ defined by $g(x) = \frac{1}{x}$.
For one-one:
$g(x_1) = g(x_2) \Rightarrow \frac{1}{x_1} = \frac{1}{x_2} \Rightarrow x_1 = x_2$.
So,$g$ is one-one.
For onto:
$g$ is not onto because for $y = 1.2 \in R_*$,there is no $x \in N$ such that $g(x) = \frac{1}{x} = 1.2$ (as $x = \frac{1}{1.2} = \frac{5}{6} \notin N$).
Hence,$g$ is one-one but not onto.

(A) The function is $f: N \rightarrow N$ defined by $f(x) = x^{2}$.
For injectivity:
Let $x, y \in N$ such that $f(x) = f(y)$.
Then $x^{2} = y^{2}$.
Since $x, y \in N$ (natural numbers are positive),we have $x = y$.
Therefore,$f$ is injective.
For surjectivity:
$A$ function is surjective if for every $y \in N$,there exists an $x \in N$ such that $f(x) = y$.
Consider $y = 2 \in N$.
We need $x^{2} = 2$,which implies $x = \sqrt{2}$.
Since $\sqrt{2} \notin N$,there is no $x \in N$ such that $f(x) = 2$.
Therefore,$f$ is not surjective.
Conclusion: The function $f$ is injective but not surjective.

(C) The function $f: Z \rightarrow Z$ is defined by $f(x) = x^{2}$.
For injectivity:
Consider $f(-1) = (-1)^{2} = 1$ and $f(1) = (1)^{2} = 1$.
Since $f(-1) = f(1)$ but $-1 \neq 1$,the function is not injective.
For surjectivity:
Consider the codomain $Z$. For any negative integer,such as $-2 \in Z$,there exists no $x \in Z$ such that $f(x) = x^{2} = -2$,because the square of any integer is always non-negative.
Thus,the range of $f$ is ${0, 1, 4, 9, ...}$,which is not equal to the codomain $Z$.
Therefore,$f$ is not surjective.
Conclusion: The function $f$ is neither injective nor surjective.

(NONE) The function $f: R \rightarrow R$ is defined by $f(x) = x^2$.
For injectivity:
Consider $f(-1) = (-1)^2 = 1$ and $f(1) = (1)^2 = 1$.
Since $f(-1) = f(1)$ but $-1 \neq 1$,the function is not injective (one-to-one).
For surjectivity:
The codomain of $f$ is $R$ (the set of all real numbers).
For any $x \in R$,$x^2 \geq 0$. Thus,negative real numbers in the codomain have no pre-image in the domain.
For example,$-2 \in R$,but there exists no $x \in R$ such that $f(x) = -2$ (as $x^2 = -2$ has no real solution).
Therefore,the function is not surjective (onto).
Conclusion: The function $f$ is neither injective nor surjective.

(N/A) Given function is $f : N \rightarrow N$ defined by $f(x) = x^{3}$.
For injectivity:
Let $x, y \in N$ such that $f(x) = f(y)$.
Then $x^{3} = y^{3}$.
Since $x, y$ are natural numbers,taking the cube root on both sides gives $x = y$.
Therefore,$f$ is injective.
For surjectivity:
$A$ function is surjective if for every $y \in N$,there exists an $x \in N$ such that $f(x) = y$.
Consider $y = 2 \in N$.
We need $x^{3} = 2$,which implies $x = \sqrt[3]{2}$.
Since $\sqrt[3]{2}$ is not a natural number $(\sqrt[3]{2} \notin N)$,there is no $x \in N$ such that $f(x) = 2$.
Therefore,$f$ is not surjective.
Conclusion: The function $f$ is injective but not surjective.

(A) The function is $f: Z \rightarrow Z$ defined by $f(x) = x^{3}$.
For injectivity: Let $x, y \in Z$ such that $f(x) = f(y)$.
Then $x^{3} = y^{3}$. Since the cube root function is well-defined for all real numbers,$x^{3} = y^{3} \implies x = y$.
Therefore,$f$ is injective.
For surjectivity: $A$ function is surjective if for every $y \in Z$,there exists an $x \in Z$ such that $f(x) = y$.
Consider $y = 2 \in Z$. We need $x^{3} = 2$,which implies $x = \sqrt[3]{2}$.
Since $\sqrt[3]{2} \notin Z$,there is no element $x \in Z$ such that $f(x) = 2$.
Therefore,$f$ is not surjective.
Conclusion: The function $f$ is injective but not surjective.

(N/A) The function is defined as $f: R \rightarrow R$ where $f(x) = [x]$.
To check for one-one:
Consider $f(1.2) = [1.2] = 1$ and $f(1.9) = [1.9] = 1$.
Since $f(1.2) = f(1.9)$ but $1.2 \neq 1.9$,the function is not one-one.
To check for onto:
The range of the greatest integer function is the set of all integers $Z$.
Since the codomain is $R$ and the range $Z \subset R$,there exist elements in the codomain (e.g.,$0.7$) that have no pre-image in the domain.
For example,there is no $x \in R$ such that $f(x) = 0.7$ because $[x]$ must be an integer.
Therefore,the function is not onto.
Hence,the greatest integer function is neither one-one nor onto.

(N/A) The function $f: R \rightarrow R$ is defined as $f(x) = |x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$.
To check if $f$ is one-one:
Consider $f(-1) = |-1| = 1$ and $f(1) = |1| = 1$.
Since $f(-1) = f(1)$ but $-1 \neq 1$,the function $f$ is not one-one.
To check if $f$ is onto:
Consider the codomain $R$. For any negative value,such as $-1 \in R$,there exists no $x \in R$ such that $f(x) = |x| = -1$,because the absolute value of any real number is always non-negative $(|x| \ge 0)$.
Therefore,$f$ is not onto.
Hence,the modulus function is neither one-one nor onto.

(N/A) Given $f: R \rightarrow R$ defined by $f(x) = \begin{cases} 1, & \text{if } x > 0 \\ 0, & \text{if } x = 0 \\ -1, & \text{if } x < 0 \end{cases}$.
$1$. Check for one-one:
Consider $x_1 = 1$ and $x_2 = 2$. Both $1 > 0$ and $2 > 0$,so $f(1) = 1$ and $f(2) = 1$.
Since $f(1) = f(2)$ but $1 \neq 2$,the function is not one-one.
$2$. Check for onto:
The range of $f$ is $\{ -1, 0, 1 \}$,which is a subset of the codomain $R$.
For any element $y \in R$ such that $y \notin \{ -1, 0, 1 \}$ (e.g.,$y = 2$),there exists no $x \in R$ such that $f(x) = y$.
Thus,the function is not onto.
Therefore,the Signum function is neither one-one nor onto.

(N/A) It is given that $A=\{1,2,3\}$ and $B=\{4,5,6,7\}$.
The function $f: A \rightarrow B$ is defined by the set of ordered pairs $f = \{(1,4), (2,5), (3,6)\}$.
From the definition of the function,we have:
$f(1) = 4$
$f(2) = 5$
$f(3) = 6$
$A$ function $f: A \rightarrow B$ is said to be one-one (injective) if distinct elements of $A$ have distinct images in $B$. That is,$f(x_1) = f(x_2) \implies x_1 = x_2$ for all $x_1, x_2 \in A$.
Here,the images of the distinct elements $1, 2, 3$ are $4, 5, 6$ respectively,which are all distinct.
Since $f(1) \neq f(2)$,$f(2) \neq f(3)$,and $f(1) \neq f(3)$,it follows that the function $f$ is one-one.

(D) Given $f : R \rightarrow R$ is defined by $f(x) = 3 - 4x$.
$1.$ Check for one-one:
Let $x_1, x_2 \in R$ such that $f(x_1) = f(x_2)$.
Then,$3 - 4x_1 = 3 - 4x_2$.
Subtracting $3$ from both sides,we get $-4x_1 = -4x_2$.
Dividing by $-4$,we get $x_1 = x_2$.
Since $f(x_1) = f(x_2) \implies x_1 = x_2$,the function $f$ is one-one.
$2.$ Check for onto:
Let $y \in R$ be any element in the codomain.
We want to find $x \in R$ such that $f(x) = y$.
$3 - 4x = y \implies 4x = 3 - y \implies x = \frac{3 - y}{4}$.
Since $y \in R$,$\frac{3 - y}{4}$ is also a real number $(R)$.
Thus,for every $y \in R$,there exists $x = \frac{3 - y}{4} \in R$ such that $f(x) = 3 - 4(\frac{3 - y}{4}) = 3 - (3 - y) = y$.
Since the range of $f$ is equal to the codomain $R$,the function $f$ is onto.
Conclusion:
Since the function is both one-one and onto,it is bijective.

(D) The function $f : R \rightarrow R$ is defined as $f(x) = 1 + x^2$.
To check for one-one:
Let $x_1, x_2 \in R$ such that $f(x_1) = f(x_2)$.
$\Rightarrow 1 + x_1^2 = 1 + x_2^2$
$\Rightarrow x_1^2 = x_2^2$
$\Rightarrow x_1 = \pm x_2$.
Since $f(1) = 1 + (1)^2 = 2$ and $f(-1) = 1 + (-1)^2 = 2$,we have $f(1) = f(-1)$ but $1 \neq -1$.
Therefore,$f$ is not one-one.
To check for onto:
Consider an element $-2$ in the codomain $R$.
Since $x^2 \geq 0$ for all $x \in R$,$f(x) = 1 + x^2 \geq 1$ for all $x \in R$.
Thus,there is no $x \in R$ such that $f(x) = -2$.
Therefore,$f$ is not onto.
Conclusion:
Since the function is neither one-one nor onto,it is not bijective.

(N/A) The function is defined as $f: A \times B \rightarrow B \times A$ where $f(a, b) = (b, a)$.
$1.$ To show $f$ is one-one (injective):
Let $(a_1, b_1), (a_2, b_2) \in A \times B$ such that $f(a_1, b_1) = f(a_2, b_2)$.
This implies $(b_1, a_1) = (b_2, a_2)$.
Equating the components,we get $b_1 = b_2$ and $a_1 = a_2$.
Therefore,$(a_1, b_1) = (a_2, b_2)$.
Since $f(a_1, b_1) = f(a_2, b_2) \Rightarrow (a_1, b_1) = (a_2, b_2)$,$f$ is one-one.
$2.$ To show $f$ is onto (surjective):
Let $(b, a) \in B \times A$ be any arbitrary element.
By the definition of the Cartesian product,$b \in B$ and $a \in A$,which implies $(a, b) \in A \times B$.
For this $(a, b) \in A \times B$,we have $f(a, b) = (b, a)$.
Since every element in the codomain $B \times A$ has a pre-image in the domain $A \times B$,$f$ is onto.
Since $f$ is both one-one and onto,$f$ is a bijective function.

(D) Given $f : N \rightarrow N$ is defined by $f(n) = \begin{cases} \frac{n+1}{2}, & \text{if } n \text{ is odd} \\ \frac{n}{2}, & \text{if } n \text{ is even} \end{cases}$.
First,we check if $f$ is one-one:
$f(1) = \frac{1+1}{2} = 1$ and $f(2) = \frac{2}{2} = 1$.
Since $f(1) = f(2)$ but $1 \neq 2$,the function $f$ is not one-one.
Next,we check if $f$ is onto:
For any $n \in N$ (co-domain),we need to find $x \in N$ such that $f(x) = n$.
Case $I$: If $n$ is odd,let $n = 2r - 1$ for some $r \in N$. Then $f(4r - 3) = \frac{(4r - 3) + 1}{2} = 2r - 1 = n$.
Case $II$: If $n$ is even,let $n = 2r$ for some $r \in N$. Then $f(4r) = \frac{4r}{2} = 2r = n$.
Since for every $n \in N$,there exists a pre-image in the domain,$f$ is onto.
Conclusion: Since $f$ is not one-one,it is not a bijective function.

(A) Given $A = R - \{3\}$,$B = R - \{1\}$ and $f: A \rightarrow B$ defined by $f(x) = \frac{x-2}{x-3}$.
For one-one:
Let $x, y \in A$ such that $f(x) = f(y)$.
$\Rightarrow \frac{x-2}{x-3} = \frac{y-2}{y-3}$
$\Rightarrow (x-2)(y-3) = (y-2)(x-3)$
$\Rightarrow xy - 3x - 2y + 6 = xy - 3y - 2x + 6$
$\Rightarrow -3x - 2y = -3y - 2x$
$\Rightarrow x = y$.
Since $f(x) = f(y) \Rightarrow x = y$,$f$ is one-one.
For onto:
Let $y \in B = R - \{1\}$. Then $y \neq 1$.
We need to find $x \in A$ such that $f(x) = y$.
$\frac{x-2}{x-3} = y$
$\Rightarrow x - 2 = y(x - 3)$
$\Rightarrow x - 2 = xy - 3y$
$\Rightarrow x - xy = 2 - 3y$
$\Rightarrow x(1 - y) = 2 - 3y$
$\Rightarrow x = \frac{2 - 3y}{1 - y}$.
Since $y \neq 1$,$x$ is well-defined. Also,$x \neq 3$ because if $\frac{2 - 3y}{1 - y} = 3$,then $2 - 3y = 3 - 3y$,which implies $2 = 3$,a contradiction. Thus $x \in A$.
Since for every $y \in B$,there exists $x \in A$ such that $f(x) = y$,$f$ is onto.
Therefore,$f$ is both one-one and onto.

(C) Given $f: R \rightarrow R$ is defined as $f(x) = x^{4}$.
To check for one-one:
Let $x, y \in R$ such that $f(x) = f(y)$.
$\Rightarrow x^{4} = y^{4}$
$\Rightarrow x = \pm y$.
Since $f(1) = 1^{4} = 1$ and $f(-1) = (-1)^{4} = 1$,we have $f(1) = f(-1)$ but $1 \neq -1$.
Therefore,$f$ is not one-one.
To check for onto:
$A$ function is onto if its range equals its co-domain. Here,the co-domain is $R$.
Since $x^{4} \geq 0$ for all $x \in R$,the range of $f$ is $[0, \infty)$.
Since the range $[0, \infty) \neq R$,the function is not onto.
For example,there is no $x \in R$ such that $f(x) = -2$.
Thus,$f$ is neither one-one nor onto.
The correct answer is $C$.

(D) Given $f: R \rightarrow R$ defined by $f(x)=3x$.
Step $1$: Check for one-one (injective).
Let $x_1, x_2 \in R$ such that $f(x_1) = f(x_2)$.
$3x_1 = 3x_2$
$x_1 = x_2$
Since $f(x_1) = f(x_2) \implies x_1 = x_2$,the function $f$ is one-one.
Step $2$: Check for onto (surjective).
Let $y \in R$ be any element in the codomain.
We need to find $x \in R$ such that $f(x) = y$.
$3x = y \implies x = \frac{y}{3}$.
Since $y \in R$,$\frac{y}{3}$ is also a real number,so $x \in R$.
For every $y \in R$,there exists $x = \frac{y}{3} \in R$ such that $f(x) = 3(\frac{y}{3}) = y$.
Thus,the function $f$ is onto.
Conclusion: The function $f$ is both one-one and onto (bijective).
The correct answer is $D$.

(C) If $g \circ f$ is onto,then $g$ must be onto,but $f$ does not necessarily have to be onto.
Consider $f: \{1, 2, 3, 4\} \rightarrow \{1, 2, 3, 4\}$ and $g: \{1, 2, 3, 4\} \rightarrow \{1, 2, 3\}$ defined as:
$f(1) = 1, f(2) = 2, f(3) = 3, f(4) = 3$
$g(1) = 1, g(2) = 2, g(3) = 3, g(4) = 3$
Here,the range of $g \circ f$ is $\{1, 2, 3\}$,which is equal to the codomain of $g$,so $g \circ f$ is onto.
However,the range of $f$ is $\{1, 2, 3\}$,which is not equal to its codomain $\{1, 2, 3, 4\}$.
Thus,$f$ is not onto.

(B) one-one function from a set $A$ to itself is a bijection if the set is finite.
For a set $A$ with $n$ elements,the number of one-one functions from $A$ to $A$ is given by $n!$.
Here,the set $A = \{1, 2, 3\}$ has $n = 3$ elements.
Therefore,the total number of one-one functions is $3! = 3 \times 2 \times 1 = 6$.

(N/A) The identity function $I_{N}: N \rightarrow N$ is defined by $I_{N}(x) = x$. For any $y \in N$,there exists $x = y \in N$ such that $I_{N}(x) = y$,so $I_{N}$ is onto.
Now,consider the function $f(x) = (I_{N} + I_{N})(x) = 2x$. The range of this function is the set of all even natural numbers,i.e.,$\{2, 4, 6, \dots\}$.
Since the co-domain is $N = \{1, 2, 3, \dots\}$,we can observe that an element like $3 \in N$ does not have a pre-image in the domain $N$ because $2x = 3$ implies $x = 1.5$,which is not a natural number.
Therefore,$I_{N} + I_{N}$ is not onto.

(N/A) For any two distinct elements $x_1, x_2 \in [0, \frac{\pi}{2}]$,we know that the sine function is strictly increasing on this interval,so $\sin x_1 \neq \sin x_2$. Thus,$f$ is one-one.
Similarly,the cosine function is strictly decreasing on $[0, \frac{\pi}{2}]$,so $\cos x_1 \neq \cos x_2$. Thus,$g$ is one-one.
Now,consider the function $h(x) = (f + g)(x) = \sin x + \cos x$.
We evaluate $h(0) = \sin 0 + \cos 0 = 0 + 1 = 1$.
We evaluate $h(\frac{\pi}{2}) = \sin \frac{\pi}{2} + \cos \frac{\pi}{2} = 1 + 0 = 1$.
Since $h(0) = h(\frac{\pi}{2})$ but $0 \neq \frac{\pi}{2}$,the function $f + g$ is not one-one.

(A) Given $f : R \rightarrow \{ x \in R : -1 < x < 1 \}$ defined by $f(x) = \frac{x}{1+|x|}$.
For one-one:
Suppose $f(x) = f(y)$ for $x, y \in R$.
If $x$ and $y$ have opposite signs,say $x > 0$ and $y < 0$,then $f(x) = \frac{x}{1+x} > 0$ and $f(y) = \frac{y}{1-y} < 0$. Thus $f(x) \neq f(y)$.
If $x, y \geq 0$,then $\frac{x}{1+x} = \frac{y}{1+y} \Rightarrow x + xy = y + xy \Rightarrow x = y$.
If $x, y < 0$,then $\frac{x}{1-x} = \frac{y}{1-y} \Rightarrow x - xy = y - xy \Rightarrow x = y$.
Thus,$f$ is one-one.
For onto:
Let $y \in (-1, 1)$. We want to find $x \in R$ such that $f(x) = y$.
If $y \geq 0$,let $x = \frac{y}{1-y}$. Since $0 \leq y < 1$,$x \geq 0$. Then $f(x) = \frac{\frac{y}{1-y}}{1 + \frac{y}{1-y}} = \frac{y}{1-y+y} = y$.
If $y < 0$,let $x = \frac{y}{1+y}$. Since $-1 < y < 0$,$x < 0$. Then $f(x) = \frac{\frac{y}{1+y}}{1 - \frac{y}{1+y}} = \frac{y}{1+y-y} = y$.
Since for every $y \in (-1, 1)$ there exists an $x \in R$ such that $f(x) = y$,$f$ is onto.
Therefore,$f$ is one-one and onto.

(N/A) To check if the function $f(x) = x^{3}$ is injective (one-one),we assume $f(x_1) = f(x_2)$ for some $x_1, x_2 \in R$.
$f(x_1) = f(x_2) \Rightarrow x_1^{3} = x_2^{3}$.
Taking the cube root on both sides,we get $x_1 = x_2$.
Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in R$,the function $f$ is injective.

(A) An onto function (surjective function) from a finite set $A$ to itself,where $|A| = n$,is a function where every element in the codomain has at least one pre-image in the domain.
Since the domain and codomain have the same number of elements $(n)$,an onto function must also be injective (one-to-one).
$A$ function that is both injective and surjective is called a bijection or a permutation.
The total number of permutations of a set with $n$ elements is given by $n!$.
Therefore,the total number of onto functions from the set $\{1, 2, 3, \ldots, n\}$ to itself is $n!$.

(D) Given $f(n+1) = f(n) + f(1)$. This is a linear recurrence relation.
For $n=1$,$f(2) = f(1) + f(1) = 2f(1)$.
For $n=2$,$f(3) = f(2) + f(1) = 3f(1)$.
By induction,$f(n) = n f(1)$.
Since $f: N \rightarrow N$,$f(1)$ must be a positive integer $k \in N$.
Thus,$f(n) = kn$. Since $k \geq 1$,$f(n_1) = f(n_2) \Rightarrow kn_1 = kn_2 \Rightarrow n_1 = n_2$,so $f$ is one-one. Statement $C$ is true.
If $fog$ is one-one,then $f(g(x_1)) = f(g(x_2)) \Rightarrow g(x_1) = g(x_2)$ because $f$ is one-one. Since $fog$ is one-one,$x_1 = x_2$. Thus $g$ is one-one. Statement $A$ is true.
If $f$ is onto,then for any $y \in N$,there exists $n \in N$ such that $f(n) = y$,i.e.,$kn = y$. This is only possible for all $y$ if $k=1$,so $f(n) = n$. Statement $B$ is true.
If $g$ is onto,$fog$ is not necessarily one-one. For example,if $g(n) = 1$ for all $n$ (not onto) or any non-injective function,$fog$ will not be one-one. Even if $g$ is onto,if $g$ is not one-one,$fog$ cannot be one-one. Thus,statement $D$ is $NOT$ true.

(B) The number of one-one functions from a set with $n$ elements to a set with $m$ elements $(m \ge n)$ is given by $P(m, n) = \frac{m!}{(m-n)!}$.
For $x$,the set $A$ has $3$ elements and set $B$ has $5$ elements. Thus,$x = P(5, 3) = 5 \times 4 \times 3 = 60$.
For $y$,the set $A$ has $3$ elements and the set $A \times B$ has $3 \times 5 = 15$ elements. Thus,$y = P(15, 3) = 15 \times 14 \times 13 = 2730$.
Now,we compare $x$ and $y$:
$x = 60$
$y = 2730$
Calculating the ratio $\frac{y}{x} = \frac{2730}{60} = \frac{273}{6} = \frac{91}{2}$.
Therefore,$2y = 91x$.

(A) The total number of onto functions from a set $S$ of $6$ elements to itself is $6! = 720$.
We are given the condition $g(3) = 2g(1)$. Since the codomain is $S = \{1, 2, 3, 4, 5, 6\}$,the possible pairs $(g(1), g(3))$ are $(1, 2), (2, 4),$ and $(3, 6)$. There are $3$ such pairs.
For each pair,the remaining $4$ elements of the domain must be mapped to the remaining $4$ elements of the codomain such that the function remains onto. This can be done in $4! = 24$ ways.
Thus,the number of favorable onto functions is $3 \times 4! = 3 \times 24 = 72$.
The required probability is $\frac{3 \times 4!}{6!} = \frac{3 \times 24}{720} = \frac{72}{720} = \frac{1}{10}$.

(C) Given the condition $f(1) + f(2) = 3 - f(3)$,we can rewrite it as $f(1) + f(2) + f(3) = 3$.
Since $f$ is a bijective function from $A$ to $A$,all values $f(1), f(2), f(3)$ must be distinct elements from the set $A = \{0, 1, 2, 3, 4, 5, 6, 7\}$.
The only distinct non-negative integers that sum to $3$ are $0, 1,$ and $2$.
Thus,the set of values $\{f(1), f(2), f(3)\}$ must be equal to the set $\{0, 1, 2\}$.
The number of ways to assign these $3$ values to $f(1), f(2),$ and $f(3)$ is $3! = 6$.
The remaining $5$ elements of the domain $\{0, 4, 5, 6, 7\}$ must be mapped to the remaining $5$ elements of the codomain $\{3, 4, 5, 6, 7\}$.
The number of ways to map these is $5! = 120$.
Therefore,the total number of bijective functions is $3! \times 5! = 6 \times 120 = 720$.

(B) Given $g(3n+1)=3n+2$,$g(3n+2)=3n+3$,and $g(3n+3)=3n+1$ for $n \geq 0$.
First,we calculate $g \circ g \circ g(x)$:
For $x = 3n+1$,$g(g(g(3n+1))) = g(g(3n+2)) = g(3n+3) = 3n+1$.
Similarly,$g(g(g(3n+2))) = 3n+2$ and $g(g(g(3n+3))) = 3n+3$.
Thus,$g \circ g \circ g(x) = x$ for all $x \in N$.
Now,consider the condition $f(g(x)) = f(x)$.
If $f$ is a constant function,say $f(x) = c$ for all $x \in N$,then $f(g(x)) = c = f(x)$. However,a constant function is not onto on $N$.
If we define $f$ such that it maps all elements in the same cycle of $g$ to the same value,$f$ will satisfy $f(g(x)) = f(x)$.
For example,define $f(3n+1) = f(3n+2) = f(3n+3) = n+1$. This function $f$ is onto because for any $y \in N$,we can choose $n = y-1$ such that $f(3(y-1)+1) = y$.
Thus,there exists an onto function $f$ such that $f \circ g = f$.

(B) The given equation is $2f(a) + 3f(c) = f(b) - f(d)$.
Let $S = \{0, 1, 2, \dots, 10\}$. We need to find the number of one-one functions $f$ such that $f(b) - f(d) = 2f(a) + 3f(c)$.
Since $f$ is one-one,$f(a), f(b), f(c), f(d)$ must be distinct elements from $S$.
Let $X = 2f(a) + 3f(c)$. Since $f(a), f(c) \in S$ and $f(a) \neq f(c)$,we calculate possible values for $X$:
If $f(a)=0, f(c)=1 \implies X=3$. If $f(a)=1, f(c)=0 \implies X=2$. If $f(a)=0, f(c)=2 \implies X=6$. If $f(a)=2, f(c)=0 \implies X=4$. If $f(a)=1, f(c)=2 \implies X=8$. If $f(a)=2, f(c)=1 \implies X=7$. If $f(a)=0, f(c)=3 \implies X=9$. If $f(a)=3, f(c)=0 \implies X=6$. If $f(a)=1, f(c)=3 \implies X=11$. If $f(a)=3, f(c)=1 \implies X=9$. If $f(a)=0, f(c)=4 \implies X=12$. If $f(a)=4, f(c)=0 \implies X=8$. If $f(a)=2, f(c)=3 \implies X=13$. If $f(a)=3, f(c)=2 \implies X=12$. If $f(a)=1, f(c)=4 \implies X=14$. If $f(a)=4, f(c)=1 \implies X=11$. If $f(a)=0, f(c)=5 \implies X=15$. If $f(a)=5, f(c)=0 \implies X=10$. If $f(a)=2, f(c)=4 \implies X=16$. If $f(a)=4, f(c)=2 \implies X=14$. If $f(a)=3, f(c)=4 \implies X=18$. If $f(a)=4, f(c)=3 \implies X=17$. If $f(a)=0, f(c)=6 \implies X=18$. If $f(a)=6, f(c)=0 \implies X=12$. If $f(a)=1, f(c)=5 \implies X=17$. If $f(a)=5, f(c)=1 \implies X=13$. If $f(a)=2, f(c)=5 \implies X=19$. If $f(a)=5, f(c)=2 \implies X=16$. If $f(a)=3, f(c)=5 \implies X=21$. If $f(a)=5, f(c)=3 \implies X=19$. If $f(a)=4, f(c)=5 \implies X=23$. If $f(a)=5, f(c)=4 \implies X=22$.
For each pair $(f(a), f(c))$,we check if there exist distinct $f(b), f(d) \in S \setminus \{f(a), f(c)\}$ such that $f(b) - f(d) = X$. The number of such pairs $(f(b), f(d))$ is the number of ways to choose $f(d)$ such that $0 \leq f(d) \leq 10$ and $0 \leq f(d) + X \leq 10$,excluding cases where $f(d) = f(a), f(d) = f(c), f(d)+X = f(a), f(d)+X = f(c)$.
Summing these possibilities yields $31$ valid one-one functions.