Defination of Function Questions in English

(D) relation is a function if every input $x$ has exactly one output $y$. This is tested using the Vertical Line Test. If any vertical line intersects the graph at more than one point,it is $NOT$ a function.
$A$: $A$ parabola is a function.
$B$: $A$ step function is a function.
$C$: $A$ cubic function is a function.
$D$: $A$ vertical line passing through the origin represents the relation $x = 0$. For $x = 0$,there are infinitely many values of $y$. Thus,it fails the Vertical Line Test and is $NOT$ a function.

(B) relation $f: A \to B$ is a function if every element in the domain $A$ has a unique image in the codomain $B$.
$1$. For option $A$: $y = \sqrt{x} - |x|$. If $x < 0$,$\sqrt{x}$ is not defined in the set of real numbers $R$. Thus,it is not a function on $R$.
$2$. For option $B$: $y = \sqrt{x} - |x|$ with $x \ge 1$. For every $x$ in the domain $[1, \infty)$,there exists a unique real value $y$. Therefore,this represents a function.
$3$. For option $C$: $x = y^2$. This implies $y = \pm \sqrt{x}$. For a single value of $x > 0$,there are two values of $y$. Thus,it is not a function.
Conclusion: Option $B$ is the correct representation of a function.

(B) Statement-$1$: The number of functions from a set $A$ with $n(A) = p$ elements to a set $B$ with $n(B) = q$ elements is given by $q^p$. This is a standard result in set theory. Thus,Statement-$1$ is true.
Statement-$2$: The number of ways to select $p$ objects from $q$ distinct objects is given by the combination formula ${}^qC_p = \frac{q!}{p!(q-p)!}$. This is a true statement.
However,Statement-$2$ describes the number of combinations,which is a fundamental counting principle,but it does not explain why the number of functions from $A$ to $B$ is $q^p$. The number of functions is derived from the fact that each of the $p$ elements in $A$ has $q$ choices in $B$. Therefore,Statement-$2$ is not the correct explanation for Statement-$1$.

(N/A) The domain of $R$ is the set of natural numbers $N$.
The codomain of $R$ is the set of natural numbers $N$.
The range of $R$ is the set of even natural numbers,i.e.,$\{2, 4, 6, \dots \}$.
Since every element $x \in N$ has a unique image $y = 2x$ in $N$,this relation is a function.

(A) relation $R$ from a set $A$ to a set $B$ is a function if every element of $A$ has one and only one image in $B$.
In the given relation $R = \{(2, 1), (3, 1), (4, 2)\}$,the domain is $\{2, 3, 4\}$.
For each element in the domain:
- The image of $2$ is $1$.
- The image of $3$ is $1$.
- The image of $4$ is $2$.
Since each element in the domain has a unique image in the codomain,the relation $R$ is a function.

(B) relation $R$ from a set $A$ to a set $B$ is a function if every element of $A$ has one and only one image in $B$.
In the given relation $R = \{(2, 2), (2, 4), (3, 3), (4, 4)\}$,the element $2$ is associated with two different images,$2$ and $4$.
Since the same first element $2$ corresponds to two different images,this relation is not a function.

(A) relation $R$ from a set $A$ to a set $B$ is a function if every element of $A$ has one and only one image in $B$.
In the given relation $R = \{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7)\}$,each element of the domain $\{1, 2, 3, 4, 5, 6\}$ is associated with a unique element in the range $\{2, 3, 4, 5, 6, 7\}$.
Since no two ordered pairs have the same first element,every input has exactly one output.
Therefore,the relation is a function.

(N/A) To complete the table,we substitute each value of $x$ into the function $f(x) = 2x + 1$:

$x$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$y$	$f(1) = 3$	$f(2) = 5$	$f(3) = 7$	$f(4) = 9$	$f(5) = 11$	$f(6) = 13$	$f(7) = 15$

(N/A) The given relation is $R = \{(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)\}$.
Since each element in the domain $\{2, 5, 8, 11, 14, 17\}$ has a unique image in the codomain,this relation is a function.
Domain $= \{2, 5, 8, 11, 14, 17\}$
Range $= \{1\}$

(N/A) The given relation is $R = \{(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)\}$.
$A$ relation is a function if every element of the domain has a unique image in the codomain.
In this relation,each first element $(2, 4, 6, 8, 10, 12, 14)$ is associated with exactly one unique second element ($1, 2, 3, 4, 5, 6, 7$ respectively).
Therefore,the given relation is a function.
The domain is the set of all first elements: $\text{Domain} = \{2, 4, 6, 8, 10, 12, 14\}$.
The range is the set of all second elements: $\text{Range} = \{1, 2, 3, 4, 5, 6, 7\}$.

(A) The given function is $f(x) = 2x - 5$.
To find $f(0)$,substitute $x = 0$:
$f(0) = 2(0) - 5 = 0 - 5 = -5$.
To find $f(7)$,substitute $x = 7$:
$f(7) = 2(7) - 5 = 14 - 5 = 9$.
To find $f(-3)$,substitute $x = -3$:
$f(-3) = 2(-3) - 5 = -6 - 5 = -11$.

(A) The given function is $f(x) = 2x - 5$.
To find $f(7)$,substitute $x = 7$ into the function:
$f(7) = 2(7) - 5$
$f(7) = 14 - 5$
$f(7) = 9$

(A) The given function is $f(x) = 2x - 5$.
To find $f(-3)$,substitute $x = -3$ into the function:
$f(-3) = 2(-3) - 5$
$f(-3) = -6 - 5$
$f(-3) = -11$

(A) The given function is $t(C) = \frac{9C}{5} + 32$.
To find $t(0)$,substitute $C = 0$ into the function:
$t(0) = \frac{9(0)}{5} + 32$
$t(0) = 0 + 32$
$t(0) = 32$

(A) The given function is $t(C) = \frac{9C}{5} + 32$.
To find $t(-10)$,substitute $C = -10$ into the function:
$t(-10) = \frac{9 \times (-10)}{5} + 32$
$t(-10) = 9 \times (-2) + 32$
$t(-10) = -18 + 32$
$t(-10) = 14$

(A) The given function is $t(C) = \frac{9C}{5} + 32$.
It is given that $t(C) = 212$.
Substituting the value in the equation:
$212 = \frac{9C}{5} + 32$
Subtracting $32$ from both sides:
$212 - 32 = \frac{9C}{5}$
$180 = \frac{9C}{5}$
Multiplying both sides by $5$:
$180 \times 5 = 9C$
$900 = 9C$
Dividing by $9$:
$C = \frac{900}{9} = 100$
Thus,the value of $C$ when $t(C) = 212$ is $100$.

(N/A) For a relation to be a function,every element in the domain must have a unique image.
For $f(x)$:
At $x = 3$,the first part gives $f(3) = 3^2 = 9$.
The second part gives $f(3) = 3 \times 3 = 9$.
Since both parts yield the same value $9$ at $x = 3$,$f(x)$ is a function.
For $g(x)$:
At $x = 2$,the first part gives $g(2) = 2^2 = 4$.
The second part gives $g(2) = 3 \times 2 = 6$.
Since $x = 2$ has two different images ($4$ and $6$),$g(x)$ is not a function.

(B) relation $f$ from a set $A$ to a set $B$ is a function if every element of $A$ has one and only one image in $B$.
Given $f = \{(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)\}$.
Here,the element $2 \in A$ is associated with two different images in $B$,which are $9$ and $11$ (i.e.,$f(2) = 9$ and $f(2) = 11$).
Since an element of the domain cannot have more than one image in the codomain,$f$ is not a function.

(N/A) The relation $f$ is defined as $f = \{(ab, a+b) : a, b \in Z\}$.
$A$ relation $f$ from a set $A$ to a set $B$ is a function if every element of $A$ has a unique image in $B$.
Consider the elements $a=2, b=6$ and $a=-2, b=-6$ in $Z$.
For $a=2, b=6$,we have $(ab, a+b) = (2 \times 6, 2+6) = (12, 8) \in f$.
For $a=-2, b=-6$,we have $(ab, a+b) = (-2 \times -6, -2-6) = (12, -8) \in f$.
Since the same first element $12$ corresponds to two different images $8$ and $-8$,the relation $f$ is not a function.

(C) For any polynomial $f(x)$ with integer coefficients,the property $(a-b)$ divides $(f(a)-f(b))$ holds for any distinct integers $a$ and $b$.
Given $f(1)=5$ and $f(2)=7$,we have $(2-1)$ divides $(f(2)-f(1))$,which is $1$ divides $(7-5)=2$. This is always true.
For $f(12)$,we have $(12-1)$ divides $(f(12)-f(1)) \implies 11$ divides $(f(12)-5)$.
Also,$(12-2)$ divides $(f(12)-f(2)) \implies 10$ divides $(f(12)-7)$.
Let $f(12) = k$. Then $k \equiv 5 \pmod{11}$ and $k \equiv 7 \pmod{10}$.
From $k \equiv 7 \pmod{10}$,$k$ can be $7, 17, 27, 37, \dots$.
Checking these values for $k \equiv 5 \pmod{11}$:
$7 \equiv 7 \pmod{11}$
$17 \equiv 6 \pmod{11}$
$27 \equiv 5 \pmod{11}$.
Thus,the smallest positive value is $27$.

(A) The given condition $(a, b_1) \in R$ and $(a, b_2) \in R \Rightarrow b_1 = b_2$ is the definition of a function from set $A$ to set $B$.
In a function,every element $a \in A$ must be mapped to exactly one element $b \in B$.
Since there are $m$ elements in set $A$ and each element has $n$ choices in set $B$,the total number of such relations (which are functions) is $n \times n \times \dots \times n$ ($m$ times).
Therefore,the total number of such relations is $n^m$.

(D) relation $f$ from set $A$ to set $B$ is a function if every element of $A$ has a unique image in $B$.
For $R_1$: Each element of $A$ has exactly one image in $B$. Thus,$R_1$ is a function.
For $R_2$: Each element of $A$ has exactly one image in $B$. Thus,$R_2$ is a function.
For $R_3$: Each element of $A$ has exactly one image in $B$. Thus,$R_3$ is a function.
For $R_4$: The element $a \in A$ is mapped to two different values,$1$ and $2$ (i.e.,$(a, 1) \in R_4$ and $(a, 2) \in R_4$).
Since an element cannot have two distinct images,$R_4$ is not a function.
Therefore,only $R_4$ is not a function.

(C) To find the pre-image of $17$,we set $f(x) = 17$:
$x^2 + 1 = 17$
$x^2 = 16$
$x = \pm 4$
So,the pre-image of $17$ is $\{4, -4\}$.
To find the pre-image of $-3$,we set $f(x) = -3$:
$x^2 + 1 = -3$
$x^2 = -4$
Since the square of a real number cannot be negative,there is no real value of $x$ such that $f(x) = -3$.
Thus,the pre-image of $-3$ is $\phi$ (the empty set).
Therefore,the pre-images are $\{4, -4\}$ and $\phi$.

(C) Given $f(x) = x^4 - 2x^3 + 3x^2 - ax + b$.
By the Remainder Theorem,$f(1) = 5$ and $f(-1) = 19$.
For $f(1) = 5$:
$1 - 2 + 3 - a + b = 5 \implies 2 - a + b = 5 \implies b - a = 3$ (Eq. $i$).
For $f(-1) = 19$:
$1 + 2 + 3 + a + b = 19 \implies 6 + a + b = 19 \implies a + b = 13$ (Eq. $ii$).
Adding Eq. $i$ and Eq. $ii$:
$(b - a) + (a + b) = 3 + 13 \implies 2b = 16 \implies b = 8$.
Substituting $b = 8$ into Eq. $ii$:
$a + 8 = 13 \implies a = 5$.
Thus,$f(x) = x^4 - 2x^3 + 3x^2 - 5x + 8$.
To find the remainder when $f(x)$ is divided by $(x - 2)$,we calculate $f(2)$:
$f(2) = (2)^4 - 2(2)^3 + 3(2)^2 - 5(2) + 8$
$f(2) = 16 - 16 + 12 - 10 + 8 = 10$.
Therefore,the remainder is $10$.

(D) Given the relation $f: R_{+} \rightarrow R_{+}$ defined by $f(x) = 3x^2 - 2$.
For $f$ to be a function,every element $x$ in the domain $R_{+}$ must have a unique image $f(x)$ in the codomain $R_{+}$.
Given $x \in (0, \infty)$,we have $x^2 > 0$,so $3x^2 > 0$.
Thus,$f(x) = 3x^2 - 2 > -2$.
This means the range of $f$ is $(-2, \infty)$.
Since the codomain is given as $R_{+} = (0, \infty)$,and the range $(-2, \infty)$ is not a subset of the codomain $(0, \infty)$ (for example,if $x=0.5$,$f(0.5) = 3(0.25) - 2 = 0.75 - 2 = -1.25$,which is not in $R_{+}$),the relation $f$ does not map every element of the domain to an element in the codomain.
Therefore,$f$ is not a function.

(C) Given $f: X \rightarrow Y$ is a function.
By definition,$A_y = f^{-1}(\{y\}) = \{x \in X : f(x) = y\}$.
This set $A_y$ represents the preimage of the element $y$ under the function $f$.
For any two distinct elements $i, j \in Y$ where $i \neq j$,the sets $A_i$ and $A_j$ are disjoint because a function maps each element of the domain to exactly one element in the codomain. Thus,$A_i \cap A_j = \phi$.
Furthermore,the union of all preimages $A_y$ for all $y \in Y$ covers the entire domain $X$,i.e.,$\bigcup_{y \in Y} A_y = X$.
These properties hold true for any function $f: X \rightarrow Y$.
Therefore,option $(C)$ is correct.

(C) Given,$f\left(\frac{p}{q}\right)=\sqrt{p^2-q^2}$ for $\frac{p}{q} \in Q$.
If $p < q$,then $p^2 - q^2 < 0$.
Since the square root of a negative number is not a real number,statement $I$ is false.
However,the square root of a negative number is a complex number (specifically,an imaginary number),so statement $II$ is true.
Therefore,$I$ is false and $II$ is true.