Question 1513 Marks
Each of the following defines a relation on $N: x, y$ is square of an integer $\text{x},\text{y}\in\text{N}$
Determine which of the above relations are reflexive, symmetric and transitive.
AnswerGiven $x, y$ is square of an integer $\text{x},\text{y}\in\text{N}$
$\therefore R = \{(x, y): xy$ is a square of an integer $\text{x},\text{y}\in\text{N}\}$
Clearly $(\text{x},\text{x})\in\text{R},\ \forall\ \text{x}\in\text{N}$
As $x^2$ is square of an integer for any $\text{x}\in\text{N}$
Hence, $R$ is reflexive.
If $(\text{x},\text{y})\in\text{R}$
$\Rightarrow\ (\text{y},\text{x})\in\text{R}$
So, $R$ is symmetric.
Now if $xy$ is square of an integer and $yz$ is square of an integer.
Then let $xy = m^2$ and $yz = n^2$ for some $\text{m, n}\in\text{Z}$
$\Rightarrow\ \text{x}=\frac{\text{m}^2}{\text{y}}$ and $\text{z}=\frac{\text{x}^2}{\text{y}}$
$\Rightarrow\ \text{xz}=\frac{\text{m}^2\text{n}^2}{\text{y}^2},$
which is square of an integer.
So, $R$ is transitive.
View full question & answer→Question 1523 Marks
Let $\text{f}:\text{R}-\Big\{-\frac{4}{3}\Big\}\rightarrow\text{R}$ be a function defined as $\text{f(x)}=\frac{4\text{x}}{3\text{x}+4}.$ Show that $\text{f}:\text{R}-\Big\{-\frac{4}{3}\Big\}\rightarrow\text{Range (f)}$ is one$-$one and onto. Hence find $f^{-1}$.
AnswerWe have given that
$f : R \rightarrow (0, 2)$ defined by
$\text{f(x)}=\frac{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+1$ is invertible.
let $f(x) = y$
$\Rightarrow\ \frac{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+1=\text{y}$
$\Rightarrow\ \frac{2\text{e}^{\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}=\text{y}$
$\Rightarrow\ \frac{2\text{e}^{2\text{x}}}{\text{e}^{2\text{x}}+1}=\text{y}$
$\Rightarrow\ 2\text{e}^{2\text{x}}=\text{y}(\text{e}^{2\text{x}}+1)$
$\Rightarrow\ \text{e}^{2\text{x}}(2-\text{y})=\text{y}$
$\Rightarrow\ \text{e}^{2\text{x}}=\frac{\text{y}}{2-\text{y}}$
$\Rightarrow\ \text{x}=\frac{1}{2}\log_\text{e}\Big(\frac{\text{y}}{2-\text{y}}\Big)$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{1}{2}\log_\text{e}\Big(\frac{\text{x}}{2-\text{x}}\Big)$
View full question & answer→Question 1533 Marks
Let O be the origin. We define a relation between two points P and Q in a plane if OP = OQ. Show that the relation, so defined is an equivalence relation.
AnswerLet A be set of points on plane. Let R = {(P, Q): OP = OQ} be a relation on A where O is the origin. To prove R is an equivalence relation, we need to show that R is reflexive, symmetric and transitive on A. Now, Reflexivity: Let $\text{P}\in\text{A}$ Since OP = OP $\Rightarrow\ (\text{P, P})\in\text{R}$ ⇒ R is reflexive. Symmetric: Let $(\text{P, Q})\in\text{R}$ for $\text{P, Q}\in\text{A}$ Then OP = OQ ⇒ OQ = OP$\Rightarrow\ (\text{Q, P})\in\text{R}$
⇒ R is symmetric. Transitive: Let $(\text{P, Q})\in\text{R}$ and $(\text{Q, S})\in\text{R}$ ⇒ OP = OQ and OQ = OS ⇒ OP = OS $\Rightarrow\ (\text{P, S})\in\text{R}$ ⇒ R is transitive. Thus, R is an equivalence relation on A.
View full question & answer→Question 1543 Marks
If $f : R \rightarrow R$ be defined by $f(x) = x^3 - 3,$ then prove that $f^{-1}$ exists and find a formula for $f^{-1}$. Hence, find $f^{-1}(24)$ and $f^{-1}(5)$.
AnswerInjectivity of $f$ : Let $x$ and $y$ be two elements in domain $(R),$
Such that$, x^3 - 3 = y^3 - 3$
$\Rightarrow x^3 = y^3 $
$\Rightarrow x = y$
So$, f$ is one$-$one.
Surjectivity of $f$ : Let $y$ be in the co$-$domain $(R)$ such that $f(x) = y$
$\Rightarrow x^3 - 3 = y$
$\Rightarrow x^3 = y + 3$
$\Rightarrow\ \text{x}=(\text{y}+3)^3\in\text{R}$
$\Rightarrow f$ is onto.
So$, f$ is a bijection and hence, it is invertible.
Finding $f^{-1}$ : Let $f^{-1}(x) = y ....(1)$
$\Rightarrow x = y^3 - 3$
$\Rightarrow x + 3 = y^3$
$\Rightarrow y = (x + 3)^3 = f^{-1}(x)\ [$from $1]$
So$, f^{-1}(x) = (x + 3)^3$
Now$, f^{-1}(24) = (24 + 3)^3 = 27^3 = 19683$ and$, f^{-1}(5) = (5 + 3)^3 = 8^3 = 512$
View full question & answer→Question 1553 Marks
If $A = \{1, 2, 3, 4\},$ define relations on $A$ which have properties of being:
- reflexive, transitive but not symmetric.
- symmetric but neither reflexive nor transitive.
- reflexive, symmetric and transitive.
AnswerGiven that, $A = \{1, 2, 3, 4\},$
- Let $R_1 = \{(1, 1), (1, 2), (2, 3), (2, 2), (1, 3), (3, 3)\}$
$R_1$ is reflexive, since, $(1,1) (2,2) (3,3)$ lie is $R_1$
Now, $(1, 2)\in\text{R}_1,\ (2,3)\in\text{R}_1$
$\Rightarrow\ (1,3)\in\text{R}_1$
Hence, $R_{1 }$ is also transitive but $(1, 2)\in\text{R}_1\Rightarrow\ (2,1)\notin\text{R}_1$
So, it is not symmetric.
- Let $R_2 = \{(1, 2), (2, 1)\}$
Now, $(1, 2)\in\text{R}_2,\ (2,1)\in\text{R}_2$
So, it is symmetric.
- Let $R_3 = \{(1, 2), (2, 1), (1, 1), (2, 2), (3, 3), (1, 3), (3, 1), (2, 3)\}$
Hence, $R_3$ is reflexive, symmetric and transitive. View full question & answer→Question 1563 Marks
Let R be a relation on the set A of ordered pair of integers defined by (x, y)R(u, v) if xv = yu. Show that R is an equivalence relation.
AnswerWe observe the following properties of R.
Reflexivity: Consider a, b be an arbitrary element of the set A.
Then $\text{a, b}\in\text{A}$
Implies that ab = ba
Implies that a, bRa, b
Thus, R is reflexive on A.
Symmetry: Consider x, y and u, v $\in\text{A}$ such that x, yRu, v.
Then xv = yu
Implies that vx = uy
Implies that uy = vx
Implies that u, vRx, y.
So, R is symmetric on A.
Transitivity: Let x, y, u, v and p, q $\in\text{R}$ such that x, yRu, v and u, vRp, q.
Implies that xv = yu and uq = vp.
Multiplying the corresponding sides,
We get xv × uq = yu × vp
Implies that xq = yp
Implies that x, yRp, q.
So, R is transitive on A.
Hence, R is an equivalence relation on A.
View full question & answer→Question 1573 Marks
Let R be a relation defined on the set of natural numbers N as,
R = {(x, y): x, y ∈ N, 2x + y = 41}
Find the domain and range of R. Also, verify whether R is:
- Reflexive.
- Symmetric.
- Transitive.
Answer{(x, y): x, y ∈ N, 2x + y = 41}
Domain of R = {1, 2, 3, 4, ....., 20}
Then Domain of R is x ∈ N such that
2x + y = 41
$\text{x}=\frac{41-\text{y}}{2}$
Therefore, Domain of R is:
R = {1, 2, 3, 4, ..., 20}
Range of R is y ∈ N
Such that range of R = {1, 3, 5, ...., 37, 39}
Let x be an arbitrary element of R.
Since, $(2,2)\notin\text{R,}$ R is not reflexive.
Hence, R is not symmetric.
Finally, since, $(15,11)\notin\text{R}$ and $(11,19)\notin\text{R}$ but $(15,19)\notin\text{R}$
Thus, R is not transitive.
View full question & answer→Question 1583 Marks
Let $S$ be a relation on the set $R$ of all real numbers defined by $S = \{(a, b) \in R \times R: a^2 + b^2 = 1\}$ prove that $S$ is not an equivalence relation on $R.$
AnswerWe observe the following properties of $S.$
Reflexivity: Consider a be an arbitrary element of $R.$
Then, $\text{a}\in\text{R}$
Implies that $\text{a}^2+\text{a}^2\neq1\ \forall\ \text{a}\in\text{R}$
Implies that $\text{a, a}\notin\text{S.}$
So, $S$ is not reflexive on $R.$
Symmetry: Consider $\text{a, b}\in\text{R}$
Implies that $a^2 + b^2 = 1$
Implies that $b^2 + a^2= 1$
Implies that $\text{b, a}\in\text{S}$ for all $\text{a, b}\in\text{R}$
So, $S$ is symmetric on $R.$
Transitivity: Let $a, b$ and $\text{b, c}\in\text{R}$
Implies that $a^2 + b^2 = 1$ and $b^2 + c^2 = 1$
Adding the above two,
we get $\text{a}^2+\text{c}^2=2-2\text{b}^2\neq1$ for all $\text{a, b, c}\in\text{R}.$
So, $S$ is not transitive on $R.$
Hence, $S$ is not an equivalence relation on $R.$
View full question & answer→Question 1593 Marks
On Z, the set of all integers, a binary operation * is defined by a * b = a + 3b - 4. Prove that * is neither commutative nor associative on Z.
AnswerCommutativity: Let $\text{a, b}\in\text{Z.}$ Then,
a * b = a + 3b - 4
b * a = b + 3a - 4
$\text{a}\ ^*\ \text{b}\neq\text{b}\ ^*\ \text{a}$
Let a = 1, b = 2
1 * 2 = 1 + 6 - 4
= 3
2 * 1 = 2 + 3 - 4
= 1
Therefore, $\exists\ \text{a}=1,\text{b}=2\in\text{Z}$ such that $\text{a}\ ^*\ \text{b}\neq\text{b}\ ^*\ \text{a}$
Thus, * is not commutative on Z.
Associativity: Let $\text{a, b, c}\in\text{Z.}$ Then,
a * (b * c) = a * (b + 3c - 4)
= a + 3(b + 3c - 4) - 4
= a + 3b + 9c - 12 - 4
= a + 3b + 9c - 16
(a * b) * c = (a + 3b - 4) * c
= a + 3b - 4 + 3c - 4
= a + 3b + 3c - 8
Thus, $\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})\neq(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}$
If a = 1, b = 2, c = 3
1 * (2 * 3) = 1 * (2 + 9 - 4)
= 1 * 7
= 1 + 21 - 4
= 18
(1 * 2) * 3 = (1 + 6 - 4) * 3
= 3 * 3
= 3 + 9 - 4
= 8
Therefore, $\exists\ \text{a}=1,\text{b}=2,\text{c}=3\in\text{Z}$ such that $\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})\neq(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}$
Thus, * is not associative on Z.
View full question & answer→Question 1603 Marks
Let n be a fixed positive integer. Define a relation R in Z as follows $\forall\ \text{a},\ \text{b}\in\text{Z},$ aRb if and only if a - b is divisible by n. Show that R is an equivalance relation.
AnswerGiven that, $\forall\ \text{a},\ \text{b}\in\text{Z},$ aRb if and only if a - b is divisible by n. Now, I. Reflexive aRa ⇒ (a - a) is divisible by n, which is true for any integer 'a' as ‘0’ is divisible by n.Hence, R is reflexive.
II. Symmetric.
aRb ⇒ a - b is divisible by n. ⇒ -(b - a) is divisible by n. ⇒ (b - a) is divisible by n. ⇒ bRaHence, R is symmetric.
III. Transitive.
Let aRb and bRc
⇒ (a - b) is divisible by n and (b - c) is divisible by n. ⇒ (a - b) + (b - c) is divisible by n. ⇒ (a - c) is divisible by n. ⇒ aRcHence, R is transitive.
So, R is an equivalence relation.
View full question & answer→Question 1613 Marks
Show that if $f_1$ and $f_2$ are one$-$one maps from $R$ to $R,$ then the product $f_1 \times f_2 : R \rightarrow R$ defined by $(f_1 \times f_2)(x) = f_1(x)f_2(x)$ need not be one$-$one.
AnswerWe know that $f_1 : R \rightarrow R,$ given by $f_1(x) = x,$ and $f_2(x) = x$ are one$-$one.
Proving $f_{1 }$ is one$-$one: Let $x$ and $y$ be two elements in the domain $R,$
such that $f_1(x) = f_1(y) \Rightarrow x = y$
So, $f_{1 }$ is one$-$one.
Proving $f_{2 }$ is one$-$one: Let $x$ and $y$ be two elements in the domain $R,$
such that $f_2(x) = f_2(y) \Rightarrow x = y$
So, $f_2$ is one$-$one.
Proving $f_1 \times f_2$ is not one$-$one:
Given: $(f_1 \times f_2)(x) = f_1(x) \times f_2(x) = x \times x = x^2$
Let $x$ and $y$ be two elements in the domain $R,$
such that $(f_1 \times f_2)(x) = (f_1 \times f_2)(y) $
$\Rightarrow x^2 = y^2 $
$\Rightarrow\ \text{x}=\pm\text{y}$
So, $(f_1 \times f_2)$ is not one$-$one.
View full question & answer→Question 1623 Marks
On Q, the set of all rational numbers a binary operation * is defined by $\text{a}\ ^*\ \text{b}=\frac{\text{a}+\text{b}}{2}.$ Show that * is not associative on Q.
AnswerLet $\text{a, b, c}\in\text{Q}.$ Then,
$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=\text{a}\ ^*\ \Big(\frac{\text{b}+\text{c}}{2}\Big)$
$=\frac{\text{a}+\big(\frac{\text{b}+\text{c}}{2}\big)}{2}$
$=\frac{2\text{a}+\text{b}+\text{c}}{4}$
$(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}=\Big(\frac{\text{a}+\text{b}}{2}\Big)\ ^*\ \text{c}$
$=\frac{\big(\frac{\text{a}+\text{b}}{2}\big)+\text{c}}{2}$
$=\frac{\text{a}+\text{b}+2\text{c}}{4}$
Thus, $\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})\neq(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}$
If a = 1, b = 2, c = 3
$1\ ^*\ (2\ ^*\ 3)=1\ ^*\ \Big(\frac{2+3}{2}\Big)$
$=1\ ^*\ \frac{5}{2}$
$=\frac{1+\frac{5}{2}}{2}$
$=\frac{7}{4}$
$(1\ ^*\ 2)\ ^*3=\Big(\frac{1+2}{2}\Big)\ ^*\ 3$
$=\frac{3}{2}\ ^*\ 3$
$=\frac{\frac{3}{2}+3}{2}$
$=\frac{9}{4}$
Therefore, $\exists\text{ a}=1,\text{b}=2,\text{c}=3\in\text{Q}$ such that $\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})\neq(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}$
Thus, * is not associative on Q.
View full question & answer→Question 1633 Marks
Let A = [-1, 1]. Then, discuss whether the following functions from A to itself are one-one, onto or bijective:
$\text{f(x)}=\frac{\text{x}}{2}$
Answerf : A → A, given by $\text{f(x)}=\frac{\text{x}}{2}$
Injection test: Let x and y be any two elements in the domain (A), such that f(x) = f(y).
f(x) = f(y)
$\frac{\text{x}}{2}=\frac{\text{y}}{2}$
x = y
So, f is one-one.
Surjection test: Let y be any element in the co-domain (A), such that f(x) = y for some element x in A (domain).
f(x) = y
$\frac{\text{x}}{2}=\text{y}$
x = 2y, which may not be in A.
For example, if y = 1, then
x = 2, which is not in A.
So, f is not onto.
So, f is not bijective.
View full question & answer→Question 1643 Marks
Given A = {2, 3, 4}, B = {2, 5, 6, 7}. Construct an example of each of the following:
- An injective map from A to B.
- A mapping from A to B which is not injective.
- A mapping from A to B.
AnswerGiven A = {2, 3, 4}, B = {2, 5, 6, 7}
Let f : A → B, f : A → B be a mapping from A to B f = {(2, 5), (3, 6), (4, 7)}
f is an injective mapping.
Since for every element $\text{a}\in\text{A}$ there is an unique element $\text{b}\in\text{B}$
Let us define a mapping: A → B given by g = {(2, 2), (2, 5), (3, 6), (4, 7)}
g is not an injective mapping.
Since the element $2\in\text{A}$ is not uniquely mapped
Since (2, 2) and (2, 5) both belong to the mapping g, g is not injective.
Let us define a mapping h : A → B
h : A → B given by h = {(2, 2), (5, 3), (7, 4)}
h is a mapping from A to B.
B to A since the every ordered puts {2, 5, 7} $\in\text{B}$ to elements in {2, 3, 4} $\in\text{A}$
View full question & answer→Question 1653 Marks
Using the definition, prove that the function $f : A \rightarrow B$ is invertible if and only if $f$ is both one$-$one and onto.
AnswerLet $f : A \rightarrow B$ be many$-$one function.
Let $f(a) = p$ and $f(b) = p$
So for inverse function we will have $f^{-1}(p) = a$ and $f^{-1}(p) = b$
Hence in this case inverse function is not defined as we have two images $'a\ '$ and $'b\ '$ for one pre$-$image $'p\ '.$
So for $f$ to be invertible it must be one$-$one.
Now let $f : A \rightarrow B$ is not onto function.
Let $B = \{p, q, r\}$ and range of $f$ be $\{p, q\}$
Here image $'r\ '$ has not any pre$-$image from set $A$ associated.
But when $f^{-1}$ is defined, $'r\ '$ becomes pre$-$image, which will have no image in set $A.$
So for $f $ to be invertible it must be onto.
Therefore $'f\ '$ is invertible if and only if $'f\ '$ is both one$-$one and onto.
A function $f = X \rightarrow Y$ is invertible iff $f$ is a bijective function.
View full question & answer→Question 1663 Marks
Functions $f, g : R \rightarrow R$ are defined, respectively, by $f(x) = x^2 + 3x + 1, g(x) = 2x – 3,$ find:
- $fog$
- $gof$
- $fof$
- $gog$
AnswerGiven that $f(x) = x^2 + 3x + 1, g(x) = 2x - 3,$
- $fog = f{g(x)} = f(2x - 3)$
$= (2x - 3)^2 + 3(2x - 3) + 1$
$= 4x^2 + 9 -12x + 6x - 9 + 1$
$= 4x^2 - 6x + 1$
- $gof = g{f(x)} = g(x^2 + 3x +1)$
$= 2(x^2 + 3x + 1) - 3$
$= 2x^2 + 6x + 2 - 3$
$= 2x^2 + 6x - 1$
- $fof = f{f(x)} = f(x^2 + 3x + 1)$
$= (x^2 + 3x + 1)^2 + 3(x^2 + 3x + 1) + 1$
$= x^4 + 9x^2 + 1 + 6x^3 + 6x + 2x^2 + 3x^2 + 9x + 3 + 1$
$= x^4 + 6x^3 + 14x^2 + 15x + 5$
- $gog = g{g(x)} = g(2x - 3)$
$= 2(2x - 3) - 3$
$= 4x - 6 - 3 = 4x - 9$ View full question & answer→Question 1673 Marks
Show that the relation R on the set A = {x ∈ Z; 0 ≤ x ≤ 12}, given by R = {(a, b): a = b}, is an equivalence relation. Find the set of all elements related to 1.
AnswerWe have, A = {x ∈ Z; 0 ≤ x ≤ 12} be a set and R = {(a, b): a = b} be a relation on A Now, Reflexivity: Let $\text{a}\in\text{A}$ ⇒ a = a$\Rightarrow\ (\text{a, a})\in\text{R}$
⇒ R is reflexive. Symmetric: Let $\text{a, b}\in\text{A}$ and $(\text{a, b})\in\text{R}$ ⇒ a = b ⇒ b = a $\Rightarrow\ (\text{b, a})\in\text{R}$ ⇒ R is symmetric. Transitive:$\text{Let a, b & c}\in\text{A}$ $$ and Let $(\text{a, b})\in\text{R}\ \text{ and }\ (\text{b, c})\in\text{R}$ $$ ⇒ a = b and b = c ⇒ a = c $\Rightarrow\ (\text{a, c})\in\text{R}$ ⇒ R is transitive. Since R is being reflexive, symmetric and transitive, so R is an equivalance relation. Also, we need to find the set of all elements related to 1. Since the relation is given by, R = {(a, b): a = b}, and 1 is an element of A, R = {(1, 1): 1 = 1} Thus, the set of all elements related to 1 is 1.
View full question & answer→Question 1683 Marks
Construct the composition table for $\times _4$ on set $S = \{0, 1, 2, 3\}.$
AnswerHere,
$1\times _41 =$ Remainder obtained by dividing $1 \times 1$ by $4 = 1$
$0\times _41 =$ Remainder obtained by dividing $0 \times 1$ by $4 = 0$
$2\times _43 =$ Remainder obtained by dividing $2 \times 3$ by $4 = 2$
$3\times _43 =$ Remainder obtained by dividing $3 \times 3$ by $4 = 1$
Therefore,
The composition table is as follows:
| $x_4$ |
$0$ |
$1$ |
$2$ |
$3$ |
| $0$ |
$0$ |
$0$ |
$0$ |
$0$ |
| $1$ |
$0$ |
$1$ |
$2$ |
$3$ |
| $2$ |
$0$ |
$2$ |
$0$ |
$2$ |
| $3$ |
$0$ |
$3$ |
$2$ |
$1$ |
View full question & answer→Question 1693 Marks
Give an example of a relation which is,Symmetric but neither reflexive nor transitive.
AnswerLet A = {5, 6, 7}.
Define a relation R on A as R = {(5, 6), (6, 5)}.
Relation R is not reflexive as $(5, 5), (6, 6), (7, 7)\notin\text{R.}$ $$
Now, as $(5, 6)\in\text{R}$ and also $(6,5)\in\text{R,}$ R is symmetric.
$\Rightarrow(5, 6), (6, 5)\in\text{R,}\text{but}(5, 5)\notin\text{R}$ $$ $$ $$
Therefore, R is not transitive.
Hence, relation R is symmetric but not reflexive or transitive.
View full question & answer→Question 1703 Marks
Find gof and fog when $f : R \rightarrow R$ and $g : R \rightarrow R$ are defined by: $f(x) = 2x + 3$ and $g(x) = x^2 + 5$
AnswerGiven: $f : R \rightarrow R$ and $g : R \rightarrow R$
Therefore, gof : $R \rightarrow R$ and fog :$ R \rightarrow R$
$f(x) = 2x + 3$ and $g(x) = x^2 + 5$
Now, $(g$ of$)(x) = g(f(x))$
$= g(2x + 3)$
$= (2x + 3)^2 + 5$
$= 4x^2 + 9 + 12x + 5$
$= 4x^2 + 12x + 14$
$($fo $g)(x) = f(g(x))$
$= f(x^2 + 5)$
$= 2(x^2 + 5) + 3$
$= 2x^2 + 10 + 3$
$= 2x^2 + 13$
View full question & answer→Question 1713 Marks
Classify the following functions as injection, surjection or bijection: $f : Z \rightarrow Z,$ defined by $f(x) = x^2 + x$
Answer$f : Z \rightarrow Z,$ given by $f(x) = x^2 + x$
Injection test: Let $x$ and $y$ be any two elements in the domain $(Z),$ such that $f(x) = f(y).$
$f(x) = f(y)$
$x^2 + x = y^2 + y$
Here, we cannot say that $x = y.$
For example, $x = 2$ and $y = -3$
Then, $x^2 + x = 2^2 + 2 = 6$
$y^2 + y = (-3)^2 - 3 = 6$
So, we have two numbers $2$ and $-3$ in the domain $Z$ whose image is same as $6.$
Surjection test: Let $y$ be any element in the co$-$domain $(Z),$ such that $f(x) = y$ for some element $x$ in $Z$ $($domain$).$
$f(x) = y$
$x^2 + x = y$
Here, we cannot say $\text{x}\in\text{Z}$
For example, $y = -4$
$x^2 + x = -4$
$x^2 + x + 4 = 0$
$\text{x}=\frac{-1\pm\sqrt{-15}}{2}$
$=\frac{-1\pm\text{i}\sqrt{15}}{2}$ which is not in $Z.$
So, $f$ is not a surjection and $f$ is not a bijection.
View full question & answer→Question 1723 Marks
Let $A = \{1, 2, 3, 4\}; B = \{3, 5, 7, 9\}; C = \{7, 23, 47, 79\}$ and$ f : A \rightarrow B,\ g : B \rightarrow C$ be defined as $f(x) = 2x + 1$ and $g(x) = x^2 - 2$. Express $(gof)^{-1}$ and $f^{-1}og^{-1}$ as the sets of ordered pairs and verify that $(gof)^{-1} = f^{-1} og^{-1}.$
Answer$= 2x + 1$
$\Rightarrow f = 1, 21 + 1, 2, 22 + 1, 3, 23 + 1, 4, 24 + 1 = 1, 3, 2, 5, 3, 7, 4, 9$
$g(x) = x^2 - 2$
$\Rightarrow g = 3, 32 - 2, 5, 52 - 2, 7, 72 - 2, 9, 92 - 2$
$= 3, 7, 5, 23, 7, 47, 9, 79$
Clearly $f$ and $g$ are bijections and,
hence,$ f^{-1} : B $
$\rightarrow A$ and $g^{-1} : C \rightarrow B$ exist.
So, $f^{-1} = 3, 1, 5, 2, 7, 3, 9, 4$ and $g^{-1} = 7, 3, 23, 5, 47, 7, 79, 9$
Now, $f^{-1}og^{-1} : C$
$\rightarrow A, f^{-1}og^{-1} = 7, 1, 23, 2, 47, 3, 79, 4 .....(1)$
Also, $f\ : A \rightarrow B$ and $g\ : B \rightarrow C,$
$\Rightarrow gof : A \rightarrow C, gof^{-1} : C \rightarrow A$
So, $f^{-1}og^{-1}$ and $gof^{-1}$ have same domains.
$= gfx = g(2x + 1) = 2x + 12 - 2$
$= 4 \times 2 + 4x + 1 - 2$
$= 4 \times 2 + 4x - 1$
Then,
$gof(1) = g(f(1)) = 4 + 4 - 1 = 7,$
$gof(2) = g(f(2)) = 4 + 4 - 1 = 23,$
$gof(3) = g(f(3)) = 4 + 4 - 1 = 47$ and
$gof(4) = g(f(4)) = 4 + 4 - 1 = 79$
So, $= 1, 7, 2, 23, 3, 47, 4, 79$
$\Rightarrow gof^{-1} = 7, 1, 23, 2, 47, 3, 79, 4 ....(2)$
We get: $gof^{-1} = f^{-1}og^{-1}$
View full question & answer→Question 1733 Marks
Let R be relation defined on the set of natural number N as follows: $\text{R}=\{(\text{x},\text{y}):\text{x}\in\text{N},\ \text{y}\in\text{N},\ 2\text{x}+\text{y}=41\}.$ Find the domain and range of the relation R. Also verify whether R is reflexive, symmetric and transitive.
AnswerWe are given that, $\text{R}=\{(\text{x},\text{y}):\text{x}\in\text{N},\ \text{y}\in\text{N},\ 2\text{x}+\text{y}=41\}$
Now, 2x + y = 41
⇒ y = 41 - 2x ....(1)
The domain of the relation R is {1, 2, 3, ...., 20}
Using (1), the range of the relation is{1, 3, 5, 7, ...., 39}
Hence, the relation R = {(1, 39), (2, 37), (3, 35),...., (19, 3), (20, 1)}
Since, $(2,2)\notin\text{R}$
Hence, R is not reflexive.
Since $(1, 39)\in\text{R}$ but $(39,1)\notin\text{R}$
Hence, R is not symmetric.
Since $(11,9)(19, 3)\in\text{R}$ but $(11,3)\notin\text{R}$
Hence, R is not transitive.
Therefore, R is neither reflexive, nor symmetric and nor transitive.
View full question & answer→Question 1743 Marks
Let $A = {\text{x}\in\text{R} | −1 \leq x \leq 1}$ and let $f : A \rightarrow A, g : A \rightarrow A$ be two functions defined by $f(x) = x^2$ and $g(x) = \sin \Big(\frac{\pi\text{x}}{2}\Big).$ Show that $g^{-1}$ exists but $f^{-1}$ does not exist. Also, find $g^{-1}.$
Answer$f$ is not one$-$one because $f^{-1} = (-1)^2 = 1$ and $f^1 = 1^2 = 1$
$\Rightarrow -1$ and $1$ have the same image under $f.$
$\Rightarrow f$ is not a bijection.
So, $f^{-1}$ does not exist.
Injectivity of $g:$ Let $x$ and $y$ be any two elements in the domain $(A),$
such that $\Rightarrow\ \sin\pi\text{x}^2=\sin\pi\text{y}^2$
$\Rightarrow\ \pi\text{x}^2=\pi\text{y}^2$
$\Rightarrow x = y$
So, $g$ is one$-$one.
Surjectivity of $g:$ Range of $\text{g}=\sin\pi^{-1}2,\ \sin\pi^{1}2=\sin-\pi^2,\sin\pi^2=-1,1=\text{A} ($co$-$domain of $g)$
$\Rightarrow g$ is onto.
$\Rightarrow g$ is a bijection.
So, $g^{-1}$ exists.
Also, let $g^{-1}x = y ......(1)$
$\Rightarrow gy = x$
$\Rightarrow\ \sin\pi\text{y}^2=\text{x}$
$\Rightarrow\ \pi\text{y}^2=\sin^{-1}\text{x}$
$\Rightarrow\ \text{y}=2\pi\sin^{-1}\text{x}$
$\Rightarrow\ \text{g}^{-1}\text{x}=2\pi\sin^{-1}\text{x} [$ from $(1)]$
View full question & answer→Question 1753 Marks
Let A be the set of all human beings in a town at a particular time. Determine whether the following relations are reflexive, symmetric and transitive:
R = {(x, y): x and y live in the same locality}
AnswerA be the set of human beings.
R = {(x, y): x and y live in the same locality}
Reflexive: Since x and x lives in the same locality.
$\Rightarrow\ (\text{x, x})\in\text{R}$
⇒ R is Reflexive.
Symmetric: Let $(\text{x, y})\in\text{R}$
⇒ x and y lives in the same locality.
⇒ y and x lives in the same locality.
$\Rightarrow\ (\text{y, x})\in\text{R}$
Transitive: Let $(\text{x, y})\in\text{R}$ and $(\text{y, z})\in\text{R}$
$(\text{x, y})\in\text{R}$
⇒ x and y lives in the same locality and $(\text{y, z})\in\text{R}$
⇒ y and z lives in the same locality.
⇒ x and z lives in the same locality.
$\Rightarrow\ (\text{x, z})\in\text{R}$
⇒ R is transitive.
View full question & answer→Question 1763 Marks
Let A be the set of all human beings in a town at a particular time. Determine whether the following relations are reflexive, symmetric and transitive:
R = {(x, y): x is wife of y}
AnswerReflexivity: Let x be an element of R. Then,
x is wife of x cannot be true.
$\Rightarrow\ (\text{x, x})\notin\text{R}$ so, R is not a reflexive relation.
View full question & answer→Question 1773 Marks
Define a binary operation$ *$ on the set $\{0, 1, 2, 3, 4, 5\}$ as $\text{a}*\text{b}=\begin{cases}\text{a + b},&\text{if a + b}<6\\\text{a + b}-6&\text{if a + b}\geq6\end{cases}$
Show that zero is the identity for this operation and each element a of the set is invertible with $6 – a$ being the inverse of a.
AnswerLet $X = \{0, 1, 2, 3, 4, 5\}.$
The operation $*$ on $X$ is defined as:
$\text{a}*\text{b}=\begin{cases}\text{a + b},&\text{if a + b}<6\\\text{a + b}-6\ \&\ \text{if a + b}\geq6\end{cases}$
An element $\text{e}\in\text{X}$ is the identity element for the operation *, if $\text{a}*\text{e}=\text{a}=\text{e}*\text{a}\ \forall\text{a}\in\text{X}.$
For $\text{a}\in\text{X},$ we observed that:
$\text{a}*0=\text{a}+0=\text{a}\ \ \ [\text{a}\in\text{X}\Rightarrow\text{a}+0<6]$
$0*\text{a}=0+\text{a}=\text{a}\ \ \ [\text{a}\in\text{X}\Rightarrow0+\text{a}<6]$
$\therefore\text{A}*0=\text{a}=0*\text{a}\ \forall\text{a}\in\text{X}$
Thus, $0$ is the identity element for the given operation*.
An element $\text{a}\in\text{X}$ is invertible if there exists $\text{b}\in\text{X}$ such that $a * b = 0 = b * a.$
$\text{i.e.},\begin{cases}\text{a + b}=0=\text{b + a},&\text{if a + b}<6\\\text{a + b}-6=0=\text{b + a}-6,&\text{if a + b}\geq6\end{cases}$
i.e.,
$a = -b$ or $b = 6 - a$
But, $X = \{0, 1, 2, 3, 4, 5\}$ and $\text{a},\text{b}\in\text{X}$ Then $\text{a}\neq-\text{b}.$
$\therefore b = 6 - a$ is the inverse of $\text{a }\Box\text{ a}\in\text{X}.$
Hence, the inverse of an element $\text{a}\in\text{X},\text{a}\neq0$ is $6 - a$ i.e., $a^{-1} = 6 - a.$
View full question & answer→Question 1783 Marks
If $\text{f}:\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\rightarrow\text{R}$ and $g : [-1, 1] \rightarrow R$ be defined as $f(x) = \tan x$ and $\text{g(x)}=\sqrt{1-\text{x}^2}$ respectively, describe fog and gof.
Answer$\text{f}:\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\rightarrow\text{R}$ and $g : [-1, 1] \rightarrow R$ defined as $f(x) = \tan x$ and $\text{g(x)}=\sqrt{1-\text{x}^2}$
Range of $f:$ let $y = f(x)$
$\Rightarrow y = \tan x$
$\Rightarrow x = \tan^{-1}y$
Since, $\text{x}\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big),\ \text{y}\in(-\infty,\infty)$
$\therefore$ Range of f $\subset$ domain of $g = [-1, 1]$
$\therefore$ gof exists.
By similar argument fog exists.
Now,
fog$(x) = f(g(x))$
$=\text{f}(\sqrt{1-\text{x}^2})$
$=\tan(\sqrt{1-\text{x}^2})$
Again,
gof$(x) = g(f(x))$
$= g(\tan x)$
$=\sqrt{1-\tan^2\text{x}}$
View full question & answer→Question 1793 Marks
If $f : R \rightarrow R$ be the function defined by $f(x) = 4x^3 + 7,$ show that $f$ is a bijection.
AnswerInjectivity:
Let $x$ and $y$ be any two elements in the domain $(R),$ such that $f(x) = f(y)$
$\Rightarrow 4x^3 + 7 = 4y^3 + 7$
$\Rightarrow 4x^3 = 4y^3$
$\Rightarrow x^3 = y^3$
$\Rightarrow x = y$
So$, f$ is one$-$one.
Surjectivity:
Let $y$ be any element in the co$-$domain $(R),$ such that $f(x) = y$ for some element $x$ in $R\ ($domain$)$.
$f(x) = y$
$\Rightarrow 4x^3 + 7 = y$
$\Rightarrow 4x^3 = y - 7$
$\Rightarrow\ \text{x}^3=\frac{\text{y}-7}{4}$
$\Rightarrow\ \text{x}=\sqrt[3]{\frac{\text{y}-7}{4}}\in\text{R}$
So, for every element in the co$-$domain, there exists some pre$-$image in the domain.
$\Rightarrow f$ is onto.
Since$, f$ is both one$-$to$-$one and onto, it is a bijection.
View full question & answer→Question 1803 Marks
Prove that the function $f : N \rightarrow N,$ defined by $f(x) = x^2 + x + 1,$ is one$-$one but not onto.
Answer$f : N \rightarrow N,$ defined by $f(x) = x^2 + x + 1$
Injectivity:
Let $x$ and $y$ be any two elements in the domain $(N),$ such that $f(x) = f(y).$
$\Rightarrow x^2 + x + 1 = y^2 + y + 1$
$\Rightarrow (x^2 - y^2) + (x - y) = 0$
$\Rightarrow (x + y)(x - y) + (x - y) = 0$
$\Rightarrow (x - y)(x + y + 1) = 0$
$\Rightarrow x - y = 0 [(x + y + 1)$ cannot be zero because $x$ and $y$ are natural numbers$]$
$\Rightarrow x = y$
So$, f$ is one$-$one.
Surjectivity:
The minimum number in $N$ is $1$.
When $x = 1, x^2 + x + 1 = 1 + 1 + 1 = 3$
$\Rightarrow\ \text{x}^2 + \text{x} + 1\geq3,$ for every $x$ in $N$.
$\Rightarrow f(x)$ will not assume the values $1$ and $2$.
So$, f$ is not onto.
View full question & answer→Question 1813 Marks
Consider $\text{f}:\text{R}\rightarrow\text{R}_+\rightarrow[4,\infty]$ given by $f(x) = x^2 + 4.$ Show that $f$ is invertible with inverse of $f$ given by $\text{f}^{-1}(\text{x})=\sqrt{\text{x}-4,}$ where $R^+$ is the set of all non-negative real numbers.
AnswerInjectivity of $f:$ Let $x$ and $y$ be two elements of the domain $(Q),$
such that $f(x) = f(y)$
$\Rightarrow x^2 + 4 = y^2 + 4$
$\Rightarrow x^2 = y^2$
$\Rightarrow x = y$ as co$-$domain as $R^+$
So, $f$ is one$-$one.
Surjectivity of $f:$ Let $y$ be in the co$-$domain $(Q),$ such that$ f(x) = y$
$\Rightarrow x^2 + 4 = y$
$\Rightarrow x^2 = y - 4$
$\Rightarrow\ \text{x}=\text{y}-4\in\text{R}$
$\Rightarrow f$ is onto.
So, $f$ is a bijection and,
hence, it is invertible.
Finding $f^{-1}:$
Let $f^{-1}(x) = y .......(1)$
$\Rightarrow x = y^2 + 4$
$\Rightarrow x - 4 = y^2$
$\Rightarrow y = x - 4$
So, $f^{-1}(x) = x - 4 [$from $(1)]$
View full question & answer→Question 1823 Marks
Give examples of two functions f : N → Z and g : Z → Z, such that gof is injective but g is not injective.
AnswerDefine f : N → Z as f(x) = x and g : Z → Z as g(x) = |x|. We first show that g is not injective. It can be observed that:g(-1) = |-1| = 1
g(1) = |1| = 1
Therefore, g(-1) = g(1), but $-1\neq1.$
Therefore, g is not injective.
Now, gof : N → Z is defined as gof(x) = g(f(x)) = g(x) = |x|.
Let $\text{x, y}\in\text{N}$ such that gof(x) = gof(y). ⇒ |x| = |y| Since x and y $\in\text{N,}$ both are positive. $\therefore$ |x| = |y| ⇒ x = y Hence, gof is injective.
View full question & answer→Question 1833 Marks
Let A be the set of all human beings in a town at a particular time. Determine whether the following relations are reflexive, symmetric and transitive:
R = {(x, y): x and y work at the same place}
AnswerReflexivity: Consider x be an arbitrary element of R .
Then, $\text{x}\in\text{R}$
Implies that, x and x work at the same place is true since they are the same.
Implies that x, $\text{x}\in\text{R},$ therefore
R is a reflexive relation.
Symmetry: Consider $\text{x},\text{y}\in\text{R}$
Implies that x and y work at the same place
Implies that $\text{x},\text{y}\in\text{R}$
R is a symmetric relation.
Transitivity: Consider $\text{x},\text{y}\in\text{R}$ and $\text{y},\text{z}\in\text{R}.$
Then x and y works at the same place.
y and z also work at the same place
Implies that, x, y and z all work at the same place
Implies that x and z work at the same place.
Implies that $\text{x},\text{z}\in\text{R}$
Therefore, R is a transitive relation.
View full question & answer→Question 1843 Marks
Let $f(x) = x^2 + x + 1$ and $g(x) = \sin x.$ Show that $fog \neq gof.$
Answer$(fog)(x) = f(g(x))$
$f(\sin x) = \sin^2x + \sin x + 1$
and$, (gof)(x) = g(f(x))$
$= g(x^2 + x + 1)$
$= \sin x^2 + x + 1$
Therefore, $fog \neq gof.$
View full question & answer→Question 1853 Marks
Let f : R → R be the function defined by $\text{f}(\text{x})=\frac{1}{2-\cos\text{x}},\ \forall\ \text{x}\in\text{R}.$ Then, find the range of f.
AnswerWe are given that, $\text{f}(\text{x})=\frac{1}{2-\cos\text{x}},\ \forall\ \text{x}\in\text{R}$ Let us suppose, $\text{y}=\frac{1}{2-\cos\text{x}}$ $\Rightarrow\ 2\text{y}-\text{y}\cos\text{x}=1$ $\Rightarrow\ \text{y}\cos\text{x}=2\text{y}-1$ $\Rightarrow\ \cos\text{x}=\frac{2\text{y}-1}{\text{y}}=2-\frac{1}{\text{y}}$ $\Rightarrow\ \cos\text{x}=2-\frac{1}{\text{y}}$ We know that, the range of cosine function is [-1, 1] $\therefore\ -1\leq\cos\text{x}\leq1$$\Rightarrow\ -1\leq2-\frac{1}{\text{y}}\leq1$ $\Big[\because\ \cos\text{x}=2-\frac{1}{\text{y}}\Big]$
$\Rightarrow\ -3\leq\frac{1}{\text{y}}\leq-1$
$\Rightarrow\ 1\leq\frac{1}{\text{y}}\leq3$
$\Rightarrow\ \frac{1}{3}\leq\text{y}\leq1$
Therefore, the range of y is $\Big[\frac{1}{3},1\Big].$
View full question & answer→Question 1863 Marks
Let $f$ be a real function given by $\text{f(x)}=\sqrt{\text{x}-2}.$ Find the following: $fofof$ Also, show that $fof \neq f^2.$
Answer$fofof = (fof)of$
We have, $\text{f}:[2,\infty)\rightarrow(0,\infty)$ and $\text{fof}:[6,\infty)\rightarrow\text{R}$
$\Rightarrow$ Range of $f$ is not a subset of the domain of $fof.$
Then, domain $((fof)of) = \{x : x \in\}$ doamin of fand $f(x) \in$ domain of $fof$
$\Rightarrow$ Domain $((fof)of) = \{x : x \in[2,\infty)$ and $\sqrt{\text{x}-2}\in[6,\infty)\}$
$\Rightarrow$ Domain $((fof)of) = \{x : x \in[2,\infty)$ and $\sqrt{\text{x}-2}\geq6\}$
$\Rightarrow$ Domain $((fof)of) = \{x : x \in[2,\infty)$ and $\text{x}-2\geq36\}$
$\Rightarrow$ Domain $((fof)of) = \{x : x \in[2,\infty)$ and $\text{x}\geq38\}$
$\Rightarrow$ Domain $((fof)of) = \{x : x \geq38\}$
$\Rightarrow$ Domain $((fof)of) = [38,\infty)$
$\text{fof}:[38,\infty)\rightarrow\text{R}$
So, $((fof)of)(x) = (fof)(f(x))$
$=(\text{fof})(\sqrt{\text{x}-2})$
$=\sqrt{\sqrt{\sqrt{\text{x}-2}-2}-2}$
View full question & answer→Question 1873 Marks
Let X = {1, 2, 3} and Y = {4, 5}. Find whether the following subsets of X × Y are functions from X to Y or not.
- f = {(1, 4), (1, 5), (2, 4), (3, 5)}
- g = {(1, 4), (2, 4), (3, 4)}
- h = {(1,4), (2, 5), (3, 5)}
- k = {(1,4), (2, 5)}.
AnswerWe have X = {1, 2, 3} and Y = {4, 5} $\therefore$ X × Y = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}
- f = {(1, 4), (1, 5), (2, 4), (3, 5)}
f is not a function as f(1) = 4 and f(1) = 5
Thus pre-image '1' has not unique image.
- g = {(1, 4), (2, 4), (3, 4)}
Clearly g is a function in which each element of the domain has unique image.
- h = {(1,4), (2, 5), (3, 5)}
Clearly h is a function as each pre-image is mapped with a unique image.
Function is many-one as h(2) = h(3) = 5
- k = {(1,4), (2, 5)}
k is not a function as '3' has not any image under the mapping. View full question & answer→Question 1883 Marks
Classify the following functions as injection, surjection or bijection : $f : N \rightarrow N$ given by $f(x) = x^3$
Answer$f : N \rightarrow N$ given by $f(x) = x^3$
Injection test:
Let $x$ and $y$ be any two elements in the domain $(N),$ such that $f(x) = f(y)$.
$f(x) = f(y)$
$x^3 = y^3$
$x = y$
Therefore$, f$ is an injection.
Surjection test:
Let $y$ be any element in the co$-$domain $(N),$ such that $f(x) = y$ for some element $x$ in $N\ ($domain$)$.
$f(x) = y$
$x^3 = y$
$\text{x}=\sqrt[3]{\text{y}}$ which may not be in $N$.
For example, if $y = 3,$
$\text{x}=\sqrt[3]{3}$ is not in $N$.
Therefore$, f$ is not a surjection and $f$ is not a bijection.
View full question & answer→Question 1893 Marks
Let $A = \{-1, 0, 1\}$ and $f = \{(x, x^2): x \in A\}.$ Show that $f : A \rightarrow A$ is neither one$-$one nor onto.
Answer$A = \{-1, 0, 1\}$ and $f = \{(x, x^2): x \in A\}$
Given, $f(x) = x^2$ Injectivity:
$f(1) = 1^2 = 1$ and $f(-1) = (-1)^2 = 1$
Implies that $1$ and $-1$ have the same images.
Therefore, $f$ is not one$-$one.
Surjectivity: Co$-$domain of $f = {-1, 0, 1}$
$f(1) = 1^2 = 1,$
$f(-1) = (-1)^2 = 1$ and
$f(0) = 0$
Implies that Range of $f = {0, 1}$
Therefore, both are not same.
Hence$, f$ is not onto.
View full question & answer→Question 1903 Marks
Let * be the binary operation on N defined by,
a * b = H.C.F. of a and b.
Does there exist identity for this binary operation one N?
AnswerThe binary operation * on N is defined as:
a * b = H.C.F. of a and b.
it is known that:
H.C.F. of a and b = H.C.F. of b and a. $\text{a, b}\in\text{N}$.
Therefore, a * b = b * a
Thus, the operation * is commutative.
For $\text{a, b, c}\in\text{N}$, we have:
(a * b) * c = (H.C.F of a and b) * c = H.C.F. of a, b and c
a * (b * c) = a * (H.C.F. of b and c) = H.C.F. of a, b and c
Therefore, (a * b) * c = a * (b * c)
Thus, the operation * is associative.
Now, an element $\text{e}\in\text{N}$ will be the identity for the operation
if a * e = a = e * a, $\forall\text{ a}\in\text{N}$.
But this relation is not true for any $\text{a}\in\text{N}$.
Thus, the operation * does not have any identity in N.
View full question & answer→Question 1913 Marks
Verify associativity for the following three mappings $: f : N \rightarrow Z_0 ($the set of non$-$zero integers$), g : Z_0 \rightarrow Q$ and $h : Q \rightarrow R$ given by $f(x) = 2x, \text{g(x)}=\frac{1}{\text{x}}$ and $h(x) = e^x.$
AnswerWe have, $f : N \rightarrow Z_0, g : Z_0 \rightarrow Q$ and $h : Q \rightarrow R$
Also, $f(x) = 2x, \text{g(x)}=\frac{1}{\text{x}}$ and $h(x) = e^x$
Now, $f : N \rightarrow Z_0$ and $hog : Z_0 \rightarrow R$
$\therefore (hog) f : N \rightarrow R$
also, $gof : N \rightarrow Q$ and $h : Q \rightarrow R$
$\therefore ho(gof) : N \rightarrow R$
Thus, $(hog) of$ and $ho(gof)$ exist and are function from $N$ to set $R.$
Finally, $(hog)of(x) = (hog)(f(x)) = (hog)(2x)$
$=\text{h}\Big(\frac{1}{2\text{x}}\Big)$
$=\text{e}^\frac{1}{2\text{x}}$
Now, $ho(gof)(x) = ho(g(2x))$
$=\text{h}\Big(\frac{1}{2\text{x}}\Big)$
$=\text{e}^\frac{1}{2\text{x}}$
Hence, associativity verified.
View full question & answer→Question 1923 Marks
If $\text{f(x)}=\sqrt{\text{x}+3}$ and $g(x) = x^2 + 1$ be two real functions, then find fog and gof.
Answer$\text{f(x)}=\sqrt{\text{x}+3}$ For domain,
$\text{x}+3\geq0$
$\Rightarrow\ \text{x}\geq-3$
Domain of $\text{f}=[-3,\infty)$
Since $f$ is a square root fuction, range of $\text{f}=(0,\infty)$
$\text{f}:(-3,\infty)\rightarrow(0,\infty)$
$g(x) = x^2 + 1$ is a polynomial.
$\Rightarrow g : R \rightarrow R$
Computation of fog: Range of $g$ is not a subset of the doamin of $f.$
and domain $(fog) = x : x \in$ domain of $g$ and $g(x) \in$ domain of $f(x)$
$\Rightarrow$ Domain $(fog) = x : x \in\text{R}$ and $x^2 + 1 \in(-3,\infty)$
$\Rightarrow$ Domain $(fog) = x : x \in\text{R}$ and $x^2 + 1 \geq-3$
$\Rightarrow$ Domain $(fog) = x : x \in\text{R}$ and $x^2 + 4 \geq0$
$\Rightarrow$ Domain $(fog) = x : x \in\text{R}$ and $\text{x}\in\text{R}$
$\Rightarrow$ Domain $(fog) = R$
fog : $R \rightarrow R$
$(fog)(x) = f(g(x))$
$= f(x^2 + 1)$
$=\sqrt{\text{x}^2+1+3}$
$=\sqrt{\text{x}^2+4}$
Computation of $gof:$ Range of $f$ is a subset of the doamin of $g.$
$\text{gof}:(-3,\infty)\rightarrow\text{R}$
$\Rightarrow\ \text{(gof)(x)}=\text{g(f(x)})$
$=\text{g}(\sqrt{\text{x}+3})$
$=(\sqrt{\text{x}+3})^2+1$
$=\text{x}+3+1$
$=\text{x}+4$
View full question & answer→Question 1933 Marks
Test whether the following relations $R_{2 }$ are:
- Reflexive.
- Symmetric.
- Transitive.
$R_2$ on $Z$ defined by $(\text{a, b})\in\text{R}_2\Leftrightarrow\ |\text{a}-\text{b}|\leq5$ AnswerReflexivity: Let a be an arbitrary element of $R_2$.
Then,$\text{a}\in\text{R}_2$
$\Rightarrow\ |\text{a}-\text{a}|=0\leq5$
So, $R_2$ is reflexive.
Symmetry: Let $(\text{a, b})\in\text{R}_2$
$\Rightarrow\ |\text{a}-\text{b}|\leq5$
$\Rightarrow\ |\text{b}-\text{a}|\leq5 [$Since, $|a - b| = |b - a|]$
$\Rightarrow\ (\text{b, a})\in\text{R}_2$
So, $R_2$ is symmetric.
Transitivity:
Let $(1, 3)\in\text{R}_2$ and $(3,7)\in\text{R}_2$
$\Rightarrow\ |1-3|\leq5$ and $|3-7|\leq5$
But $|1-7|\nleq5$
$\Rightarrow\ (1,7)\notin\text{R}_2$
So, $R_2$ is transitive.
View full question & answer→Question 1943 Marks
Let the function f : R → R be defined by f(x) = cosx, ∀ x ∈ R. Show that f is neither one-one nor onto.
AnswerWe are given, f(x) = cosx, ∀ x ∈ RFor $\frac{\pi}{2},$ we have
$\text{f}\Big(\frac{\pi}{2}\Big)=\cos\frac{\pi}{2}=0$
For $-\frac{\pi}{2},$ we have
$\text{f}\Big(\frac{-\pi}{2}\Big)=\cos\frac{\pi}{2}=0$
$\therefore\ \text{f}\Big(\frac{\pi}{2}\Big)=\text{f}\Big(\frac{-\pi}{2}\Big)$
But, $\frac{\pi}{2}\neq\frac{-\pi}{2}$
Hence, f(x) is not one-one.
We also know that, there is not any pre-image for any real number which does not belong to the range of cosine function i.e., [-1, 1].
View full question & answer→Question 1953 Marks
Let $A = R_0 \times R,$ where $R_0$ denote the set of all non$-$zero real numbers. A binary operation $'⊙\ '$ is defined on $A$ as follows: $(a, b) ⊙ (c, d) = (ac, bc + d)$ for all $(a, b), (c, d) \in R_0 \times R.$ Find the identity element in $A.$
AnswerLet $E = (x, y)$ be the identity element in $A$ with respect to $⊙,\forall\ \text{x}\in\text{R}_0\ \ \text{ y}\in\text{R}$ such that
$x ⊙ E = X = E ⊙ X, \forall\text{ x}\in\text{A}$
$\Rightarrow X ⊙ E = X$ and $E ⊙ X = X$
$\Rightarrow (ax, bx + y) = (a, b)$ and $(xa, ya + b) = (a, b)$
Considering $(ax, bx + y) = (a, b)$
$\Rightarrow ax = a$
$\Rightarrow x = 1$
$bx + y = b$
$\Rightarrow y = 0 [\because\text{ x}=1]$
Considering $(xa, ya + b) = (a, b)$
$\Rightarrow xa = a$
$\Rightarrow x = 1$
$ya + b = b$
$\Rightarrow y = 0 [\because\text{ x}=1]$
$[\because\ 1,0] $ is the identity element in $A$ with respect to $⊙.$
View full question & answer→Question 1963 Marks
Test whether the following relations $R_{3 }$ are:
- Reflexive.
- Symmetric.
- Transitive.
$R_3$ on $R$ defined by $(\text{a, b})\in\text{R}_3\Leftrightarrow\ \text{a}^2-4\text{ab}+3\text{b}^2=0$ AnswerReflexivity: Consider a be an arbitrary element of $R_3$
Then, $\text{a}\in\text{R}_3$
Implies that $a_2 - 4a_2 + 3a_2 = 0$
So, $R_3$ is reflexive.
Symmetry: Consider, $\text{a, b}\in\text{R}_3$
Implies that $a_2 - 4a_2b_2 + 3b_2 = 0$
But $\text{b}_2-4\text{b}_2\text{a}_2+3\text{a}_2\neq0$ for all $\text{a, b}\in\text{R}$
So, $R_3$ is not symmetric.
Transitivity: $1,2\in\text{R}_3$ and $2,3\in\text{R}_3$
Implies that $1 - 8 + 6 = 0$ and $4 - 24 + 27 = 0$
But $1 - 12 + 9 \neq0$
So, $R_3$ is not transitive.
View full question & answer→Question 1973 Marks
Classify the following functions as injection, surjection or bijection $:f : R \rightarrow R,$ defined by $f(x) = 5x^3 + 4$
Answer$f : R \rightarrow R,$ defined by $f(x) = 5x^3 + 4$
Injection test: Let $x$ and $y$ be any two elements in the domain $(R),$ such that $f(x) = f(y).$
$f(x) = f(y)$
$5x^3 + 4 = 5y^3 + 4$
$5x^3 = 5y^3$
$x^3 = y^3$
$x = y$
So, $f$ is an injection.
Surjection test: Let $y$ be any element in the co$-$domain $(R),$ such that $f(x) = y$ for some element $x$ in $R ($domain$).$
$f(x) = y$
$5x^3 + 4 = y$
$5x^3 = y - 4$
$\text{x}^3=\frac{\text{y}-4}{5}$
$\text{x}=\sqrt[3]{\frac{\text{y}-4}{5}}\in\text{R}$
So, $f$ is a surjection and $f $ is a bijection.
View full question & answer→Question 1983 Marks
If f : {5, 6} → {2, 3} and g : {2, 3} → {5, 6} are given by f = {(5, 2), (6, 3)} and g = {(2, 5), (3, 6)}, then find fog.
AnswerWe have, f : {5, 6} → {2, 3} and g : {2, 3} → {5, 6} are given by f = {(5, 2), (6, 3)} and g = {(2, 5), (3, 6)}
As,
fog(2) = f(g(2)) = f(5) = 2,
fog(3) = f(g(3)) = f(6) = 3,
Therefore, fog : {2, 3} → {2, 3} is defined as fog = {(2, 2), (3, 3)}
View full question & answer→Question 1993 Marks
Classify the following functions as injection, surjection or bijection: $f : Z \rightarrow Z$ given by $f(x) = x^2$
Answer$f : Z \rightarrow Z$ given by $f(x) = x^2$
Injection test: Let $x$ and $y$ be any two elements in the domain $(Z),$
such that $f(x) = f(y). f(x) = f(y) x^2 = y^2 \text{x}=\pm\text{y}$
So, $f$ is not an injection.
Surjection test: Let $y$ be any element in the co$-$domain $(Z),$
such that $f(x) = y$ for some element $x$ in $Z ($domain$).$
$f(x) = y x^2 = y\text{x}=\pm\sqrt{\text{y}}$ which may not be in $Z.$
For example, if $y = 3,$
$\text{x}=\pm\sqrt{3}$ is not in $Z.$
So, $f$ is not a bijection.
View full question & answer→Question 2003 Marks
Check whether the relation $R$ on $R$ defined by $R = \{(a, b): a \leq b^3\}$ is reflexive, symmetric or transitive.
AnswerThe relation $R$ on $R$ is defined by $R.$
We observe that $(-2)\leq(-2)^3$ is not true.
Therefore$, R$ is not reflexive.
Since $1\leq\Big(3^{\frac{1}{3}}\Big)^3$ but $3^\frac{1}{3}\leq1$ i.e. $\Big(1,3^\frac{1}{3}\Big)\in\text{R.}$
Therefore, $R$ is not symmetric.
Hence, $R$ is not transitive because $(5,2)\in\text{R}$ and $\Big(2,2^\frac{1}{3}\Big)\in\text{R}$ but $\Big(5,2^\frac{1}{3}\Big)\notin\text{R.}$
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