Type of Functions based on Mapping Questions in English

(D) function $f: R \to R$ is an injection (one-one) if $f(x_1) = f(x_2) \implies x_1 = x_2$.
For $f(x) = x^2$,we have $f(1) = 1^2 = 1$ and $f(-1) = (-1)^2 = 1$.
Since $f(1) = f(-1)$ but $1 \neq -1$,the function is not an injection.
$A$ function $f: R \to R$ is a surjection (onto) if for every $y \in R$,there exists an $x \in R$ such that $f(x) = y$.
For $f(x) = x^2$,the range is $[0, \infty)$,which is a subset of the codomain $R$.
Since there are negative values in the codomain that have no pre-image in the domain (e.g.,$y = -1$),the function is not a surjection.
Therefore,the function is neither an injection nor a surjection.

(B) linear function is of the form $f(x) = ax + b$.
We want to map the interval $[-1, 1]$ to $[0, 2]$.
Case $1$: $f(-1) = 0$ and $f(1) = 2$.
$-a + b = 0 \implies a = b$.
$a + b = 2 \implies 2a = 2 \implies a = 1, b = 1$.
So,$f(x) = x + 1$.
Case $2$: $f(-1) = 2$ and $f(1) = 0$.
$-a + b = 2$.
$a + b = 0 \implies b = -a$.
$-a - a = 2 \implies -2a = 2 \implies a = -1, b = 1$.
So,$f(x) = -x + 1$.
Thus,there are two such linear functions.

(D) Given the function $f(x) = x^2 + x$.
Step $1$: Check for one-one or many-one.
We observe that $f(0) = 0^2 + 0 = 0$ and $f(-1) = (-1)^2 + (-1) = 1 - 1 = 0$.
Since $f(0) = f(-1)$ but $0 \neq -1$,the function is many-one.
Step $2$: Check for onto or into.
We can rewrite the function by completing the square: $f(x) = x^2 + x + \frac{1}{4} - \frac{1}{4} = (x + \frac{1}{2})^2 - \frac{1}{4}$.
Since $(x + \frac{1}{2})^2 \ge 0$ for all $x \in \mathbb{R}$,it follows that $f(x) \ge -\frac{1}{4}$.
This means the range of $f$ is $[-\frac{1}{4}, \infty)$.
Since the range $[-\frac{1}{4}, \infty)$ is not equal to the codomain $\mathbb{R}$,the function is into.
Conclusion: The function is many-one into.

(A) Let $x_1, x_2 \in R$. Then $f(x_1) = \cos x_1$ and $f(x_2) = \cos x_2$.
For the function to be one-one,$f(x_1) = f(x_2)$ must imply $x_1 = x_2$.
However,$\cos x_1 = \cos x_2$ implies $x_1 = 2n\pi \pm x_2$ for some integer $n$.
Since $x_1$ does not necessarily equal $x_2$ (e.g.,$\cos(0) = \cos(2\pi) = 1$),the function is not one-one.
For the function to be onto,the range must equal the codomain $R$.
The range of $f(x) = \cos x$ is $[-1, 1]$,which is a proper subset of $R$.
Since the range $\neq$ codomain,the function is not onto.
Therefore,the mapping is neither one-one nor onto.

(B) We have $f(x) = (x - 1)(x - 2)(x - 3)$.
First,check for one-one property:
$f(1) = (1 - 1)(1 - 2)(1 - 3) = 0$
$f(2) = (2 - 1)(2 - 2)(2 - 3) = 0$
$f(3) = (3 - 1)(3 - 2)(3 - 3) = 0$
Since $f(1) = f(2) = f(3) = 0$,the function is not one-one because distinct elements have the same image.
Next,check for onto property:
$f(x)$ is a polynomial of degree $3$. The range of any odd-degree polynomial function $f: R \to R$ is $(-\infty, \infty)$,which is the codomain $R$.
Therefore,for every $y \in R$,there exists at least one $x \in R$ such that $f(x) = y$.
Thus,the function is onto.
Conclusion: The function is onto but not one-one.

(D) Given the function $f: R \to R$ defined by $f(x) = |x|$.
$1$. Check for one-one: $A$ function is one-one if $f(x_1) = f(x_2) \implies x_1 = x_2$. Here,$f(-1) = |-1| = 1$ and $f(1) = |1| = 1$. Since $f(-1) = f(1)$ but $-1 \neq 1$,the function is not one-one (it is many-one).
$2$. Check for onto: $A$ function is onto if its range equals its codomain. The codomain is $R$. The range of $f(x) = |x|$ is $[0, \infty)$. Since the range is not equal to the codomain $([0, \infty) \neq R)$,the function is not onto.
Therefore,the function is neither one-one nor onto. The correct option is $(d)$.

(A) To check if $f$ is one-one,let $x, y \in N$ such that $f(x) = f(y)$.
$x^2 + x + 1 = y^2 + y + 1$
$x^2 - y^2 + x - y = 0$
$(x - y)(x + y) + (x - y) = 0$
$(x - y)(x + y + 1) = 0$
Since $x, y \in N$,$x + y + 1 > 0$,so we must have $x - y = 0$,which implies $x = y$. Thus,$f$ is one-one.
To check if $f$ is onto,we look for the range of $f$. For $x = 1$,$f(1) = 1^2 + 1 + 1 = 3$. For $x = 2$,$f(2) = 2^2 + 2 + 1 = 7$. Since $f(x) = x^2 + x + 1$,the minimum value for $x \in N$ is $3$. Values like $1, 2, 4, 5, 6$ are not in the range. Therefore,$f$ is not onto.
Hence,$f$ is one-one but not onto.

(C) function $f: X \to Y$ is one-one if $f(x_1) = f(x_2) \implies x_1 = x_2$ for all $x_1, x_2 \in X$.
For $f(x) = x^2$,if $X = R^+$,then $x_1^2 = x_2^2 \implies x_1 = x_2$ (since $x_1, x_2 > 0$),so $f$ is one-one.
$A$ function is onto if the range of $f$ equals the codomain $Y$.
Here,the range of $f$ is $R^+$. If $Y = R$,then the range $R^+ \subset R$,so $f$ is not onto.
Thus,for $X = R^+$ and $Y = R$,the function is one-one but not onto.

(C) An injective function (one-to-one function) from a set $A$ with $n$ elements to a set $B$ with $m$ elements exists if $n \le m$.
The number of such injective functions is given by the formula $^mP_n = \frac{m!}{(m-n)!}$.
Here,$n = 3$ and $m = 4$.
Therefore,the number of injections is $^4P_3 = \frac{4!}{(4-3)!} = \frac{4 \times 3 \times 2 \times 1}{1!} = 24$.
Thus,the correct option is $C$.

(B) For any $x, y \in R \setminus \{n\}$,we have $f(x) = f(y)$.
$\frac{x - m}{x - n} = \frac{y - m}{y - n}$
$(x - m)(y - n) = (y - m)(x - n)$
$xy - nx - my + mn = xy - ny - mx + mn$
$-nx - my = -ny - mx$
$mx - nx = my - ny$
$x(m - n) = y(m - n)$
Since $m \ne n$,we have $x = y$.
Therefore,$f$ is one-one.
Now,let $f(x) = y$ where $y \in R$.
$y = \frac{x - m}{x - n} \Rightarrow y(x - n) = x - m$
$yx - yn = x - m \Rightarrow x(y - 1) = yn - m$
$x = \frac{yn - m}{y - 1}$.
For $y = 1$,there is no $x \in R$ such that $f(x) = 1$. Thus,the range of $f$ is $R \setminus \{1\} \neq R$.
Therefore,$f$ is not onto. Hence,$f$ is one-one into.

(D) The function $f:R \to R$ is defined by $f(x) = e^x$.
To check for one-one: Let $x_1, x_2 \in R$ such that $f(x_1) = f(x_2)$.
Then $e^{x_1} = e^{x_2}$,which implies $x_1 = x_2$.
Thus,the function is one-one.
To check for onto: The range of $f(x) = e^x$ is $(0, \infty)$,which is a subset of the codomain $R$.
Since the range is not equal to the codomain (e.g.,negative values and zero have no pre-image),the function is not onto.
Therefore,the function is one-one but not onto.

(A) function $f: \mathbb{R} \to \mathbb{R}$ is bijective if it is both one-one (injective) and onto (surjective).
$1$. For $f(x) = |x|$,$f(1) = 1$ and $f(-1) = 1$. Since $f(1) = f(-1)$ but $1 \neq -1$,it is not one-one.
$2$. For $f(x) = x^2$,$f(1) = 1$ and $f(-1) = 1$. Since $f(1) = f(-1)$ but $1 \neq -1$,it is not one-one.
$3$. For $f(x) = x^2 + 1$,$f(1) = 2$ and $f(-1) = 2$. Since $f(1) = f(-1)$ but $1 \neq -1$,it is not one-one.
$4$. For $f(x) = 2x - 5$:
- One-one: Let $f(x) = f(y)$. Then $2x - 5 = 2y - 5$,which implies $2x = 2y$,so $x = y$. Thus,it is one-one.
- Onto: Let $y = 2x - 5$. Then $x = \frac{y + 5}{2}$. For every $y \in \mathbb{R}$,there exists $x = \frac{y + 5}{2} \in \mathbb{R}$ such that $f(x) = y$. Thus,it is onto.
Since $f(x) = 2x - 5$ is both one-one and onto,it is a bijective function.

(A) Given the function $f(x) = 2x + \sin x$ for $x \in R$.
To check for one-to-one,we find the derivative: $f'(x) = 2 + \cos x$.
Since $-1 \le \cos x \le 1$,we have $f'(x) = 2 + \cos x \ge 2 - 1 = 1 > 0$.
Since $f'(x) > 0$ for all $x \in R$,the function $f(x)$ is strictly increasing,which implies it is one-to-one.
To check for onto,we observe the range of $f(x)$. Since $f(x)$ is a continuous function and $\lim_{x \to \infty} f(x) = \infty$ and $\lim_{x \to -\infty} f(x) = -\infty$,the range of $f(x)$ is $(-\infty, \infty) = R$.
Since the range equals the codomain,the function is onto.
Therefore,$f$ is one-to-one and onto.

(C) Given the function $f: \mathbb{N} \to \mathbb{Z}$ defined as:
$f(n) = \begin{cases} \frac{n-1}{2}, & \text{if } n \text{ is odd} \\ -\frac{n}{2}, & \text{if } n \text{ is even} \end{cases}$
Let us calculate the images of the first few natural numbers:
$f(1) = \frac{1-1}{2} = 0$
$f(2) = -\frac{2}{2} = -1$
$f(3) = \frac{3-1}{2} = 1$
$f(4) = -\frac{4}{2} = -2$
$f(5) = \frac{5-1}{2} = 2$
$f(6) = -\frac{6}{2} = -3$
$1$. One-one check: For every distinct $n_1, n_2 \in \mathbb{N}$,we get distinct images in $\mathbb{Z}$. Since each input maps to a unique output,the function is one-one.
$2$. Onto check: The range of the function is $\{0, -1, 1, -2, 2, -3, 3, \dots\}$,which is the set of all integers $\mathbb{Z}$. Since the range equals the codomain,the function is onto.
Therefore,the function is both one-one and onto (bijective).

(B) Given $f(x) = \frac{x}{1+x}$ for $x \in [0, \infty)$.
To check for one-one: Let $f(x_1) = f(x_2)$,then $\frac{x_1}{1+x_1} = \frac{x_2}{1+x_2}$.
$x_1(1+x_2) = x_2(1+x_1) \implies x_1 + x_1x_2 = x_2 + x_1x_2 \implies x_1 = x_2$.
Thus,$f$ is one-one.
To check for onto: The range of $f(x) = \frac{x}{1+x} = \frac{1+x-1}{1+x} = 1 - \frac{1}{1+x}$.
As $x$ ranges from $0$ to $\infty$,$1+x$ ranges from $1$ to $\infty$,so $\frac{1}{1+x}$ ranges from $1$ to $0$.
Therefore,$f(x)$ ranges from $1-1=0$ to $1-0=1$,i.e.,$[0, 1)$.
Since the codomain is $[0, \infty)$ and the range is $[0, 1)$,the range $\neq$ codomain.
Thus,$f$ is not onto.
Therefore,$f$ is one-one but not onto.

(D) Let $f(x_1) = f(x_2)$.
This implies $[x_1] = [x_2]$,which does not necessarily imply $x_1 = x_2$.
For example,if $x_1 = 1.4$ and $x_2 = 1.5$,then $[1.4] = 1$ and $[1.5] = 1$.
Since $f(1.4) = f(1.5)$ but $1.4 \neq 1.5$,the function $f$ is not one-one.
Also,the range of the greatest integer function $f(x) = [x]$ is the set of integers $Z$,which is a proper subset of the co-domain $R$.
Since the range $\neq$ co-domain,the function $f$ is not onto.
Therefore,the function is neither one-one nor onto.

(D) Given the function $f(x) = x + \sqrt{x^2} = x + |x|$.
We can express this function as:
$f(x) = \begin{cases} x + x = 2x, & \text{if } x \ge 0 \\ x - x = 0, & \text{if } x < 0 \end{cases}$
$1$. Check for Injectivity (One-one):
For $x < 0$,$f(x) = 0$. For example,$f(-1) = 0$ and $f(-2) = 0$. Since $f(-1) = f(-2)$ but $-1 \neq -2$,the function is not injective.
$2$. Check for Surjectivity (Onto):
The range of $f(x)$ is $[0, \infty)$. Since the codomain is $R$ (all real numbers) and the range is not equal to the codomain,the function is not surjective.
Since the function is neither injective nor surjective,it is not bijective.
Therefore,the correct option is $(d)$.

(A) The function is defined as $f: (\mathbb{R} \setminus \{0\}) \times (\mathbb{R} \setminus \{0\}) \to \mathbb{R}$ given by $f(x, y) = \frac{x}{y}$.
For any real number $c \in \mathbb{R}$,we can choose $x = c$ and $y = 1$. Then $f(c, 1) = \frac{c}{1} = c$. Since for every $c$ in the codomain there exists at least one pair $(x, y)$ in the domain such that $f(x, y) = c$,the function is surjective (onto).
However,the function is not one-one because $f(2, 1) = 2$ and $f(4, 2) = 2$. Since $f(2, 1) = f(4, 2)$ but $(2, 1) \neq (4, 2)$,it is not injective.
Since it is not injective,it cannot be a bijection.
Therefore,the correct classification is a surjection.

(B) Given $f(x) = \sin (\log (x + \sqrt {x^2 + 1}))$.
To check if the function is even or odd,we find $f(-x)$:
$f(-x) = \sin (\log (-x + \sqrt {(-x)^2 + 1})) = \sin (\log (\sqrt {x^2 + 1} - x))$.
Multiply and divide the argument of the logarithm by $(\sqrt {x^2 + 1} + x)$:
$f(-x) = \sin \left( \log \left( (\sqrt {x^2 + 1} - x) \cdot \frac{\sqrt {x^2 + 1} + x}{\sqrt {x^2 + 1} + x} \right) \right)$.
Using $(a-b)(a+b) = a^2 - b^2$:
$f(-x) = \sin \left( \log \left( \frac{x^2 + 1 - x^2}{\sqrt {x^2 + 1} + x} \right) \right) = \sin \left( \log \left( \frac{1}{x + \sqrt {x^2 + 1}} \right) \right)$.
Using the property $\log(1/a) = -\log(a)$:
$f(-x) = \sin (-\log (x + \sqrt {x^2 + 1}))$.
Since $\sin(-\theta) = -\sin(\theta)$:
$f(-x) = -\sin (\log (x + \sqrt {x^2 + 1})) = -f(x)$.
Therefore,$f(x)$ is an odd function.

(B) To determine if the function $f(x) = \log (x + \sqrt {{x^2} + 1} )$ is even or odd,we evaluate $f(-x)$.
$f(-x) = \log (-x + \sqrt {(-x)^2 + 1} )$
$f(-x) = \log (-x + \sqrt {x^2 + 1} )$
We can rewrite the argument of the logarithm by multiplying and dividing by the conjugate $(\sqrt {x^2 + 1} + x)$:
$f(-x) = \log \left( \frac{(\sqrt {x^2 + 1} - x)(\sqrt {x^2 + 1} + x)}{\sqrt {x^2 + 1} + x} \right)$
$f(-x) = \log \left( \frac{(x^2 + 1) - x^2}{\sqrt {x^2 + 1} + x} \right)$
$f(-x) = \log \left( \frac{1}{\sqrt {x^2 + 1} + x} \right)$
Using the property $\log(1/a) = -\log(a)$:
$f(-x) = -\log (x + \sqrt {x^2 + 1} )$
$f(-x) = -f(x)$
Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.

(B) Given $g(x) = 1 + x - [x]$.
We know that $x - [x] = \{x\}$,where $\{x\}$ is the fractional part of $x$.
Thus,$g(x) = 1 + \{x\}$.
Since $0 \le \{x\} < 1$,it follows that $1 \le g(x) < 2$.
Therefore,$g(x) > 0$ for all $x \in \mathbb{R}$.
Now,$f(g(x))$ is defined as:
$f(g(x)) = \begin{cases} -1, & g(x) < 0 \\ 0, & g(x) = 0 \\ 1, & g(x) > 0 \end{cases}$.
Since $g(x) > 0$ for all $x$,we have $f(g(x)) = 1$ for all $x$.

(A) Let $h(x) = (f - g)(x)$.
For $x \in \mathbb{Q}$,$h(x) = f(x) - g(x) = x - 0 = x$.
For $x \notin \mathbb{Q}$,$h(x) = f(x) - g(x) = 0 - x = -x$.
Thus,$h(x) = \begin{cases} x, & x \in \mathbb{Q} \\ -x, & x \notin \mathbb{Q} \end{cases}$.
To check if $h(x)$ is one-one: Let $h(x_1) = h(x_2)$. If $x_1, x_2 \in \mathbb{Q}$,then $x_1 = x_2$. If $x_1, x_2 \notin \mathbb{Q}$,then $-x_1 = -x_2 \implies x_1 = x_2$. If $x_1 \in \mathbb{Q}$ and $x_2 \notin \mathbb{Q}$,then $x_1 = -x_2$,which is possible (e.g.,$x_1 = 1, x_2 = -1$,but $-1$ is rational,so this case is restricted). Since $h(x)$ maps every real number to a unique value,it is one-one.
To check if $h(x)$ is onto: For any $y \in \mathbb{R}$,if $y \in \mathbb{Q}$,$h(y) = y$. If $y \notin \mathbb{Q}$,$h(-y) = -(-y) = y$. Thus,for every $y \in \mathbb{R}$,there exists an $x$ such that $h(x) = y$. Hence,it is onto.

(A) The number of functions from a set $A$ to a set $B$ is given by the formula $(n(B))^{n(A)}$.
Given $n(A) = 5$ and $n(B) = 8$.
Therefore,the number of possible functions is $8^5$.
$8^5 = 8 \times 8 \times 8 \times 8 \times 8 = 32768$.

(C) Given the function $f(x) = |x| + |x - 1|$.
For the interval $0 \leq x \leq 1$,we have $x \geq 0$ and $x - 1 \leq 0$.
Therefore,$|x| = x$ and $|x - 1| = -(x - 1) = 1 - x$.
Substituting these into the function:
$f(x) = x + (1 - x) = 1$.
Since $f(x) = 1$ for all $x$ in the interval $[0, 1]$,the function is a constant function.

(B) For $x \in (1, 2)$,we have $x - 1 > 0$ and $x - 2 < 0$.
Thus,the function becomes $f(x) = 2(x - 1) - 3(x - 2)$.
Simplifying this,$f(x) = 2x - 2 - 3x + 6 = -x + 4$.
Now,finding the derivative,$f'(x) = \frac{d}{dx}(-x + 4) = -1$.
Since $f'(x) = -1 < 0$ for all $x \in (1, 2)$,the function $f(x)$ is monotonically decreasing in the interval $(1, 2)$.

(C) To check if $f(x) = x^3 + x^2 + 3x + \sin x$ is one-one,we find its derivative $f'(x)$.
$f'(x) = 3x^2 + 2x + 3 + \cos x$.
We know that $-1 \leq \cos x \leq 1$.
So,$f'(x) \geq 3x^2 + 2x + 3 - 1 = 3x^2 + 2x + 2$.
For the quadratic $3x^2 + 2x + 2$,the discriminant $D = b^2 - 4ac = (2)^2 - 4(3)(2) = 4 - 24 = -20 < 0$.
Since the coefficient of $x^2$ is positive $(3 > 0)$ and $D < 0$,$3x^2 + 2x + 2 > 0$ for all $x \in R$.
Thus,$f'(x) > 0$ for all $x \in R$.
Since $f'(x) > 0$,$f(x)$ is a strictly increasing function.
$A$ strictly increasing function is always a one-one function.
Therefore,Statement-$I$ is true.
Statement-$II$ claims $f(x)$ is a decreasing function,which is false because $f'(x) > 0$ implies it is increasing.

(C) Given $f(x) = x^3 + 5x + 1.$
First,we find the derivative: $f'(x) = 3x^2 + 5.$
Since $x^2 \ge 0$ for all $x \in R,$ it follows that $3x^2 + 5 \ge 5 > 0$ for all $x \in R.$
Because $f'(x) > 0$ for all $x,$ the function $f(x)$ is strictly increasing on $R.$
$A$ strictly increasing function is always one-one.
Next,we check if the function is onto. Since $f(x)$ is a polynomial of odd degree,its range is $(-\infty, \infty).$
Specifically,$\lim_{x \to -\infty} f(x) = -\infty$ and $\lim_{x \to \infty} f(x) = \infty.$
Since $f(x)$ is continuous and its range is the set of all real numbers $R,$ the function is onto.
Therefore,$f$ is both one-one and onto.

(D) Given $f:R \to \left[ { - \frac{1}{2},\frac{1}{2}} \right]$ defined by $f(x) = \frac{x}{1 + x^2}$.
First,check for injectivity:
$f'(x) = \frac{(1 + x^2)(1) - x(2x)}{(1 + x^2)^2} = \frac{1 - x^2}{(1 + x^2)^2} = \frac{(1 - x)(1 + x)}{(1 + x^2)^2}$.
Since $f'(x)$ changes sign at $x = 1$ and $x = -1$,the function is not monotonic,hence it is not injective.
Next,check for surjectivity:
Let $y = \frac{x}{1 + x^2}$. Then $yx^2 - x + y = 0$.
For $x$ to be a real number,the discriminant $D \ge 0$.
$D = (-1)^2 - 4(y)(y) = 1 - 4y^2 \ge 0$.
$4y^2 \le 1 \Rightarrow y^2 \le \frac{1}{4} \Rightarrow y \in \left[ -\frac{1}{2}, \frac{1}{2} \right]$.
Since the range is equal to the codomain,the function is surjective.
Therefore,the function is surjective but not injective.

(B) Let $A$ be a set with $n$ elements,where $n = 10$.
For a function $f: A \to A$,each of the $10$ elements in the domain $A$ can be mapped to any of the $10$ elements in the codomain $A$.
Since each element has $10$ choices,the total number of distinct functions is given by $n^n = 10^{10}$.

(A) The total number of functions from a set $E$ with $n$ elements to a set $F$ with $m$ elements is $m^n$.
Here,$n = |E| = 4$ and $m = |F| = 2$.
So,the total number of functions is $2^4 = 16$.
An onto function (surjective function) means every element in the codomain $F$ must have at least one pre-image in the domain $E$.
The only functions that are not onto are those where all elements of $E$ map to either ${1}$ or ${2}$.
There are exactly $2$ such constant functions: $f(x) = 1$ for all $x \in E$ and $f(x) = 2$ for all $x \in E$.
Therefore,the number of onto functions is $16 - 2 = 14$.

(B) Given $f(x) = x^3 - 8x^2 + 20x - 13$.
First,find the derivative $f'(x) = 3x^2 - 16x + 20$.
Setting $f'(x) = 0$,we get $3x^2 - 16x + 20 = 0$.
Using the quadratic formula,$x = \frac{16 \pm \sqrt{256 - 240}}{6} = \frac{16 \pm 4}{6}$.
So,$x_1 = 2$ and $x_2 = \frac{10}{3}$.
Since $f'(x) > 0$ for $x < 2$ and $x > 10/3$,and $f'(x) < 0$ for $2 < x < 10/3$,the function is not monotonic,hence it is many-one.
Since the range of a cubic polynomial is $(-\infty, \infty)$,the function is onto.
Now,calculate $f(2) = 8 - 32 + 40 - 13 = 3$.
Calculate $f(10/3) = (1000/27) - 8(100/9) + 20(10/3) - 13 = \frac{1000 - 2400 + 1800 - 351}{27} = \frac{49}{27}$.
Since $f(2) \cdot f(10/3) = 3 \cdot \frac{49}{27} = \frac{49}{9} > 0$,option $D$ is incorrect.
Since $f(2) > 0$ and $f(10/3) > 0$,and the local minimum is positive,the graph crosses the $x$-axis only once. Thus,it has only $1$ real root.
Therefore,the function is many-one and onto.

(D) function $f: R \to R$ is surjective if its range is $R$. $A$ polynomial of even degree with a positive leading coefficient has a range $[c, \infty)$,so it is not surjective.
For $f(x) = x^3 + x + 1$,$f'(x) = 3x^2 + 1 > 0$,so it is strictly increasing,making it both injective and surjective (bijective).
For $f(x) = \sqrt{1 + x^2}$,the range is $[1, \infty)$,so it is not surjective.
For $f(x) = x^3 + 2x^2 - x + 1$,the degree is odd,so the range is $R$ (surjective).
However,$f'(x) = 3x^2 + 4x - 1$. The discriminant $D = 16 - 4(3)(-1) = 28 > 0$,so $f'(x)$ changes sign,meaning the function is not monotonic and thus not injective.
Therefore,$f(x) = x^3 + 2x^2 - x + 1$ is surjective but not injective.

(D) function is called transcendental if it is not algebraic.
Algebraic functions are those that can be expressed as the root of a polynomial equation with polynomial coefficients.
Functions involving trigonometric,exponential,or logarithmic terms are generally transcendental.
In option $A$,$f(x) = 5 \sin \sqrt{x}$ involves a trigonometric function.
In option $B$,$f(x) = \frac{2 \sin 3x}{x^2 + 2x - 1}$ involves a trigonometric function.
In option $C$,$f(x) = (x^2 + 3) \cdot 2^x$ involves an exponential function.
Since all these functions are non-algebraic,they are all transcendental functions.
Therefore,the correct option is $D$.

(B) The domain of $f(x)$ is determined by $x-2 \ge 0$ and $4-x \ge 0$,which gives $x \in [2,4]$.
To find the range,let $y = \sqrt{x-2} + \sqrt{4-x}$. Squaring both sides,$y^2 = (x-2) + (4-x) + 2\sqrt{(x-2)(4-x)} = 2 + 2\sqrt{-x^2+6x-8}$.
The expression $-x^2+6x-8$ has a maximum value of $1$ at $x=3$. Thus,$y^2 \in [2, 4]$,so $y \in [\sqrt{2}, 2]$.
For $f(x)$ to be injective,the function must be monotonic. The function $f(x)$ increases on $[2,3]$ and decreases on $[3,4]$.
Therefore,$f(x)$ is injective on either $[2,3]$ or $[3,4]$.
On both intervals $[2,3]$ and $[3,4]$,the range of the function is $[\sqrt{2}, 2]$.
Thus,for $f(x)$ to be both injective and surjective,$X$ can be $[2,3]$ or $[3,4]$,and $Y$ must be $[\sqrt{2}, 2]$.
Comparing with the options,option $B$ is the correct choice.

(D) Given $f(x) = 3^{-|x|} - 3^x + \operatorname{sgn}(e^{-x}) + 2$.
Since $e^{-x} > 0$ for all $x \in R$,$\operatorname{sgn}(e^{-x}) = 1$.
Thus,$f(x) = 3^{-|x|} - 3^x + 1 + 2 = 3^{-|x|} - 3^x + 3$.
Case $1$: If $x < 0$,then $|x| = -x$,so $f(x) = 3^x - 3^x + 3 = 3$.
Since $f(x) = 3$ for all $x < 0$,the function is many-one,hence not injective.
Case $2$: If $x \ge 0$,then $|x| = x$,so $f(x) = 3^{-x} - 3^x + 3$.
As $x \to \infty$,$f(x) \to -\infty$,and at $x = 0$,$f(0) = 3^0 - 3^0 + 3 = 3$.
The range of $f(x)$ for $x \ge 0$ is $(-\infty, 3]$.
Combining both cases,the range of $f$ is $(-\infty, 3]$.
Since the range $(-\infty, 3] \neq R$,the function is not surjective.
Therefore,$f$ is neither injective nor surjective.

(D) Given $f(x) = |\sin x| + |\cos x|$.
We know that $f(x)$ is a periodic function with period $\frac{\pi}{2}$.
Let us evaluate the range of $f(x)$:
$f(x)^2 = (|\sin x| + |\cos x|)^2 = \sin^2 x + \cos^2 x + 2|\sin x \cos x| = 1 + |\sin 2x|$.
Since $0 \le |\sin 2x| \le 1$,we have $1 \le f(x)^2 \le 2$,which implies $1 \le f(x) \le \sqrt{2}$.
Now,$h(x) = g(f(x)) = [f(x)]$.
Since $1 \le f(x) \le \sqrt{2} \approx 1.414$,the value of $[f(x)]$ is always $1$ for all $x \in \mathbb{R}$.
Thus,$h(x) = 1$,which is a constant function.
$A$ constant function is periodic,but its fundamental period is not defined as it repeats for every real number $T > 0$.

(C) Given $f(x) = 5x - 3\cos x - 4\sin x$.
Find the derivative: $f'(x) = 5 + 3\sin x - 4\cos x$.
We know that the range of $a\sin x + b\cos x$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
Here,$3\sin x - 4\cos x$ has a range of $[-\sqrt{3^2 + (-4)^2}, \sqrt{3^2 + (-4)^2}] = [-5, 5]$.
Therefore,$f'(x) = 5 + (3\sin x - 4\cos x) \geq 5 - 5 = 0$.
Since $f'(x) \geq 0$ for all $x \in R$ and $f'(x)$ is not identically zero on any interval,$f(x)$ is a strictly increasing function.
As $x \rightarrow \infty$,$f(x) \rightarrow \infty$ and as $x \rightarrow -\infty$,$f(x) \rightarrow -\infty$.
Since the function is continuous and strictly increasing,its range is $(-\infty, \infty)$,which is equal to the codomain $R$.
Thus,the function is both one-one and onto (bijective).

(B) Given $f(x) = \frac{e^x - 1}{e^x + 1}$.
We can rewrite the function as $f(x) = \frac{e^x + 1 - 2}{e^x + 1} = 1 - \frac{2}{e^x + 1}$.
To check for one-one,we find the derivative: $f'(x) = \frac{d}{dx} (1 - 2(e^x + 1)^{-1}) = 0 - 2(-1)(e^x + 1)^{-2} \cdot e^x = \frac{2e^x}{(e^x + 1)^2}$.
Since $e^x > 0$ for all $x \in R$,$f'(x) > 0$ for all $x \in R$. Thus,$f(x)$ is strictly increasing and therefore one-one.
To check for onto,we look at the range. As $x \rightarrow \infty$,$f(x) \rightarrow 1$. As $x \rightarrow -\infty$,$f(x) \rightarrow -1$. Since the function is continuous and strictly increasing,the range is $(-1, 1)$,which is equal to the codomain.
Therefore,the function is one-one and onto.

(B) We are given the condition $|f(\alpha) - \alpha| \leqslant 1$ for each $\alpha \in \{1, 2, 3, 4\}$.
This implies $f(\alpha) \in \{\alpha - 1, \alpha, \alpha + 1\} \cap \{1, 2, 3, 4\}$.
For $\alpha = 1$: $f(1) \in \{1-1, 1, 1+1\} \cap \{1, 2, 3, 4\} = \{0, 1, 2\} \cap \{1, 2, 3, 4\} = \{1, 2\}$. ($2$ choices)
For $\alpha = 2$: $f(2) \in \{2-1, 2, 2+1\} \cap \{1, 2, 3, 4\} = \{1, 2, 3\}$. ($3$ choices)
For $\alpha = 3$: $f(3) \in \{3-1, 3, 3+1\} \cap \{1, 2, 3, 4\} = \{2, 3, 4\}$. ($3$ choices)
For $\alpha = 4$: $f(4) \in \{4-1, 4, 4+1\} \cap \{1, 2, 3, 4\} = \{3, 4\}$. ($2$ choices)
Since the choices for each $\alpha$ are independent,the total number of such functions is $2 \times 3 \times 3 \times 2 = 36$.

(B) Given the condition $|f(x) - f(y)| \geqslant |e^x - e^y|$ for all $x, y \in R$.
Suppose $f(x) = f(y)$ for some $x \neq y$.
Then $|f(x) - f(y)| = 0$.
Substituting this into the given inequality,we get $0 \geqslant |e^x - e^y|$.
Since the absolute value is always non-negative,this implies $|e^x - e^y| = 0$,which means $e^x = e^y$.
Since the exponential function $e^x$ is one-one,$e^x = e^y$ implies $x = y$.
This contradicts our assumption that $x \neq y$.
Therefore,$f(x)$ must be one-one.

(D) We are given that $f: A \to B$ is an onto function where $|A| = 7$ and $|B| = 3$.
It is given that exactly three elements in $A$ map to $y_2 \in B$.
The number of ways to choose these $3$ elements from $7$ is $^7C_3$.
Now,the remaining $4$ elements in $A$ must map to the remaining $2$ elements in $B$ (i.e.,${y_1, y_3}$) such that the function is onto.
For the function to be onto,both $y_1$ and $y_3$ must be mapped to by at least one element from the remaining $4$ elements.
The total number of functions from a set of $4$ elements to a set of $2$ elements is $2^4 = 16$.
The number of functions that are not onto (i.e.,mapping to only one element) is $^2C_1 = 2$.
Thus,the number of onto functions from the remaining $4$ elements to ${y_1, y_3}$ is $2^4 - 2 = 16 - 2 = 14$.
Therefore,the total number of such onto functions is $^7C_3 \times 14 = 14(^7C_3)$.

(A) Given $f(x) = \frac{ax^2 + ax + b}{ax + b}$.
For $f$ to be defined for all $x \in R$,the denominator $ax + b$ must not be zero for any $x \in R$. This implies $a = 0$ and $b \neq 0$.
Substituting $a = 0$ into the function,we get $f(x) = \frac{0 \cdot x^2 + 0 \cdot x + b}{0 \cdot x + b} = \frac{b}{b} = 1$.
Since $f(x) = 1$ for all $x \in R$,the function is a constant function.
$A$ constant function maps every element of the domain to the same element in the codomain,therefore it is many-one.
Thus,$f$ is many-one.

(A) Let us analyze each option:
$A$. If $A$ is a finite set with $n$ elements,a function $f: A \to A$ is one-one if and only if it is onto (by the Pigeonhole Principle). Thus,this statement is true.
$B$. By the Intermediate Value Theorem,if a continuous function changes sign between $x_1$ and $x_2$,there is at least one root. However,it does not guarantee an odd number of roots; there could be an even number of roots if the function crosses the axis multiple times.
$C$. If $A$ is an infinite set (e.g.,$f: \mathbb{N} \to \mathbb{N}$ defined by $f(x) = x + 1$),the function is one-one but not onto. Thus,this is false.
$D$. $A$ local maximum and a global minimum cannot occur at the same point unless the function is constant,which contradicts the definition of local maxima/minima in standard contexts. Thus,this is false.
Therefore,the correct statement is $A$.

(B) Given $f(x) = x|x| + x^3|x|$.
We can write this as $f(x) = |x|(x + x^3)$.
Case $1$: If $x \geq 0$,then $|x| = x$. So,$f(x) = x(x + x^3) = x^2 + x^4$.
Case $2$: If $x < 0$,then $|x| = -x$. So,$f(x) = -x(x + x^3) = -x^2 - x^4$.
Thus,$f(x) = \begin{cases} x^4 + x^2, & x \geq 0 \\ -(x^4 + x^2), & x < 0 \end{cases}$.
For $x \geq 0$,$f'(x) = 4x^3 + 2x > 0$ for $x > 0$. Since $f(0) = 0$,$f(x)$ is strictly increasing for $x \geq 0$.
For $x < 0$,$f'(x) = -(4x^3 + 2x) > 0$ for $x < 0$. Since $f(0) = 0$,$f(x)$ is strictly increasing for $x < 0$.
Since the function is strictly increasing on its entire domain $R$,it is a one-one function.
As $x \to \infty, f(x) \to \infty$ and as $x \to -\infty, f(x) \to -\infty$. Since $f(x)$ is continuous,the range is $(-\infty, \infty) = R$.
Since the range equals the codomain,the function is onto.
Therefore,the function is one-one onto.

(A) function $f: A \to B$ is injective (one-to-one) if $f(x_1) = f(x_2) \implies x_1 = x_2$. It is surjective (onto) if the range equals the codomain.
For option $A$: $f: N \to N$,$f(x) = 2x + 3$.
Injective: $2x_1 + 3 = 2x_2 + 3 \implies 2x_1 = 2x_2 \implies x_1 = x_2$. Thus,it is injective.
Surjective: For $y \in N$,$2x + 3 = y \implies x = \frac{y-3}{2}$. If $y=4$,$x = 0.5 \notin N$. Thus,it is not surjective.
For option $B$: $f: R \to R$,$f(x) = \frac{4x+3}{5}$. This is a linear function,which is both injective and surjective (bijective).
For option $C$: $f(x) = x^3 - x$. $f(0) = 0$,$f(1) = 0$,$f(-1) = 0$. Since $f(0) = f(1)$,it is not injective.
For option $D$: $f(x) = \ln(|x| + 1)$. $f(1) = \ln(2)$,$f(-1) = \ln(2)$. Since $f(1) = f(-1)$,it is not injective.
Therefore,the correct option is $A$.

(D) Given the function $f(x) = e^{x^2} + \cos x$.
Since $f(-x) = e^{(-x)^2} + \cos(-x) = e^{x^2} + \cos x = f(x)$,the function is an even function.
For any even function defined on $R$,$f(x) = f(-x)$ for all $x \neq 0$,which implies the function is many-one.
Next,we check the range of the function.
As $x \to \infty$,$f(x) \to \infty$.
At $x = 0$,$f(0) = e^0 + \cos(0) = 1 + 1 = 2$.
Since $e^{x^2} \ge 1$ and $\cos x \ge -1$,the minimum value of $f(x)$ is $2$ at $x=0$.
Because the range of $f(x)$ is $[2, \infty)$,which is a proper subset of the codomain $R$,the function is into.
Therefore,$f$ is many-one into.

(C) function $f(x)$ is even if $f(-x) = f(x)$.
Let us check option $C$:
$f(x) = \frac{x}{2^x - 1} + \frac{x}{2} + 1$
$f(-x) = \frac{-x}{2^{-x} - 1} + \frac{-x}{2} + 1$
$f(-x) = \frac{-x}{\frac{1}{2^x} - 1} - \frac{x}{2} + 1$
$f(-x) = \frac{-x \cdot 2^x}{1 - 2^x} - \frac{x}{2} + 1$
$f(-x) = \frac{x \cdot 2^x}{2^x - 1} - \frac{x}{2} + 1$
Since $\frac{x \cdot 2^x}{2^x - 1} = \frac{x(2^x - 1 + 1)}{2^x - 1} = x + \frac{x}{2^x - 1}$,
$f(-x) = x + \frac{x}{2^x - 1} - \frac{x}{2} + 1 = \frac{x}{2^x - 1} + \frac{x}{2} + 1 = f(x)$.
Thus,$f(x)$ is an even function.

(D) Given $f(x) = \begin{cases} x, & x \in \mathbb{Q} \\ 0, & x \notin \mathbb{Q} \end{cases}$ and $g(x) = \begin{cases} x, & x \in \mathbb{Q} \\ 0, & x \notin \mathbb{Q} \end{cases}$.
We need to find the function $h(x) = (f - g)(x) = f(x) - g(x)$.
For any $x \in \mathbb{Q}$,$h(x) = f(x) - g(x) = x - x = 0$.
For any $x \notin \mathbb{Q}$,$h(x) = f(x) - g(x) = 0 - 0 = 0$.
Thus,$h(x) = 0$ for all $x \in \mathbb{R}$.
This is a constant function.
$A$ constant function is neither one-one (as $h(1) = h(2) = 0$) nor onto (as the range is ${0}$,which is not equal to the codomain $\mathbb{R}$).
Therefore,the function $(f - g)$ is neither one-one nor onto.

(A) Given $f(x) = 2x + \cos x$.
To check for one-one,we find the derivative: $f'(x) = 2 - \sin x$.
Since $-1 \le \sin x \le 1$,it follows that $1 \le 2 - \sin x \le 3$.
Thus,$f'(x) > 0$ for all $x \in R$.
Since the derivative is strictly positive,$f(x)$ is a strictly increasing function,which implies $f$ is one-one.
To check for onto,we observe the limits: $\lim_{x \to \infty} (2x + \cos x) = \infty$ and $\lim_{x \to -\infty} (2x + \cos x) = -\infty$.
Since $f(x)$ is a continuous function and its range is $(-\infty, \infty) = R$,the function $f$ is onto.
Therefore,$f$ is one-one and onto.