Question 1513 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 1523 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 1533 Marks
Check the commutativity and associativity of the following binary operations:
'*' on N defined by a * b = gcd(a, b) for all a, b ∈ N.
AnswerCommutativity: Let $\text{a, b}\in\text{N}.$ Then,a * b = gcd(a, b)
= gcd(b, a)
= b * a
Therefore,
a * b = b * a, $\forall\ \text{a, b}\in\text{N}$
Thus '*' is commutative on N.
Associativity: Let $\text{a, b, c}\in\text{N}.$ Then,
a * (b * c) = a * [gcd(a, b)]
= gcd(a, b, c)
(a * b) * c = [gcd(a, b)] * c
= gcd(a, b, c)
Therefore,
a * (b * c) = (a * b) * c, $\forall\ \text{a, b, c}\in\text{N}$
Thus, '*' is not associative on N.
View full question & answer→Question 1543 Marks
Let $S$ be a relation on the set $R$ of all real numbers defined by $S=\left\{(a, b) \in R \times R: a^2+b^2=1\right\}$ 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 1553 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 → (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 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 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 1583 Marks
Give an example of a map:
- Which is one-one but not onto.
- Which is not one-one but onto.
- Which is neither one-one nor onto.
Answer
- Let f : N → N, be a mapping defined by f(x) = 2x
- Which is one-one
For $f(x_1) = f(x_2)$
$\Rightarrow 2x_1 = 2x_2$
$x_1 = x_2$
Further f is not onto, as for $1\in\text{N},$ there does not exist any x in N such that f(x) = 2x + 1.
- Let f : N → N, given by f(1) = f(2) = 1 and f(x) = x - 1 for every x > 2 is onto but not one-one. f is not one-one as f(1) = f(2) = 1. But f is onto.
- The mapping f : R → R defined as $f(x) = x^2$, is neither one-one not onto.
View full question & answer→Question 1593 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 1603 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\}$,
i. 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 R _1,(2,3) \in R _1$
$\Rightarrow(1,3) \in R _1$
Hence, $R_1$ is also transitive but $(1,2) \in R_1 \Rightarrow(2,1) \notin R_1$
So, it is not symmetric.
ii. Let $R_2=\{(1,2),(2,1)\}$
Now, $(1,2) \in R _2,(2,1) \in R _2$
So, it is symmetric.
iii. 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 1613 Marks
Find fog and gof if:$f(x) = \sin^{-1}x$, $g(x) = x^2$
Answer$f(x) = \sin^{-1}x, g(x) = x^2$$\text{f}:[-1,1]\rightarrow\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big];\ \text{g}:\text{R}\rightarrow[0,\infty)$
Computing fog: Clearly, the range of g is not a subset of the domain of f.
Domain(fog) = {x: $\text{x}\in$ domain of g and $\text{g(x)}\in$ domain of f}
Domain(fog) = {x: $\text{x}\in\text{R}$ and $\text{x}^2\in[-1,1]$}
Domain(fog) = {x: $\text{x}\in\text{R}$ and $\text{x}\in[-1,1]$}
Domain of (fog)$ = [-1, 1]$
fog :$ [-1, 1] \rightarrow R$
$(fog)(x) = f(g(x))$
$= f(x^2)$
$= \sin^{-1}(x^2)$
Computing gof: Clearly, the range of f is a subset of the domain of g.
fog : $[-1, 1] \rightarrow R$
$(gof)(x) = g(f(x))$
$= g(\sin^{-1}x)$
$= (\sin^{-1}x)^2$
View full question & answer→Question 1623 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 1633 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 1643 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 1653 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 1663 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 1673 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, $a \in R_3$
Implies that $a _2-4 a _2+3 a _2=0$
So, $R_3$ is reflexive.
Symmetry: Consider, $a, b \in R_3$
Implies that $a _2-4 a _2 b_2+3 b_2=0$
But $b _2-4 b_2 a _2+3 a _2 \neq 0$ for all $a , b \in R$
So, $R_3$ is not symmetric.
Transitivity: $1,2 \in R _3$ and $2,3 \in R _3$
Implies that $1-8+6=0$ and $4-24+27=0$
But $1-12+9 \neq 0$
So, $R_3$ is not transitive.
View full question & answer→Question 1683 Marks
Given A = {2, 3, 4}, B = {2, 5, 6, 7}. Construct an example of each of the following:
- An injective mapping from A to B.
- A mapping from A to B which is not injective.
- A mapping from B to A.
AnswerGiven that, A = {2, 3, 4}, B = {2, 5, 6, 7}
- Let f : A → B denote a mapping
$\text{f}\equiv\{(\text{x},\text{y}):\text{y}=\text{x}+3\}$
or $\text{f}\equiv\{(2,5),(3,6),(4,7)\},$ which is an injective mapping.
- Let g : A → B denote a mapping such that g = {(2, 2), (3, 2), (4, 5)}, which is not an injective mapping.
- Let h : B → A denote a mapping such that h = {(2, 2), (5, 3), (6, 4), (7, 4)}, which is one of the mappings from B to A.
View full question & answer→Question 1693 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 1703 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→MCQ 1713 Marks
Functions $f, g : R \rightarrow R$ are defined, respectively, by $f(x) = x^2 + 3x + 1, g(x) = 2x – 3$, find:
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 1723 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 1733 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 1743 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: $R \rightarrow R$ $f(x)=2 x+3$ and $g(x)=x^2+5$
Now, $(gof)(x) = g(f(x))$
$= g(2x + 3)$
$= (2x + 3)^2 + 5$
$= 4x^2 + 9 + 12x + 5$
$= 4x^2 + 12x + 14$
$(fog)(x) = f(g(x))$
$= f(x^2 + 5)$
$= 2(x^2 + 5) + 3$
$= 2x^2 + 10 + 3$
$= 2x^2 + 13$
View full question & answer→Question 1753 Marks
Suppose $f_1$ and $f_2$ are non-zero one-one functions from R to R. Is $\frac{\text{f}_1}{\text{f}_2}$ necessarily one-one? Justify your answer. Here, $\frac{\text{f}_1}{\text{f}_2}:\text{R}\rightarrow\ \text{R}$ is given by $\Big(\frac{\text{f}_1}{\text{f}_2}\Big)(\text{x})=\frac{\text{f}_1(\text{x})}{\text{f}_2(\text{x})}$ for all $\text{x}\in\text{R}.$
AnswerWe know that $f_1: R \rightarrow R$, given by $f_1(x)=x^3$ and $f_2(x)=x$ are one-one. Injectivity of $f_1$ : Let $x$ and $y$ be two elements in the domain $R$, such that $f_1(x)=f_2(y) \Rightarrow x^3=y$
$\Rightarrow\ \text{x}=\sqrt[3]{\text{y}}\in\text{R}$
So, $f_1$ is one-one.
Injectivity of $f_2$: Let x and y be two elements in the domain R, such that
$f_2(x) = f_2(y) \Rightarrow x = y$ $\text{x}\in\text{R.}$
So, $f_2$ is one-one.
Proving $\frac{\text{f}_1}{\text{f}_2}$ is not one-one.
Given that $\frac{\text{f}_1}{\text{f}_2}(\text{x})=\frac{\text{f}_1(\text{x})}{\text{f}_2(\text{x})}=\frac{\text{x}^3}{\text{x}}=\text{x}^2$
Let x and y be two elements in the domain R, such that
$\frac{\text{f}_1}{\text{f}_2}(\text{x})=\frac{\text{f}_1}{\text{f}_2}(\text{y})$
$\Rightarrow\ \text{x}^2=\text{y}^2$
$\Rightarrow\ \text{x}=\pm\text{y}$
So, $\frac{\text{f}_1}{\text{f}_2}$ is not one-one.
View full question & answer→Question 1763 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 1773 Marks
Test whether the following relations $R_1$ are:
- Reflexive.
- Symmetric.
- Transitive.
$R_1$ on $Q_0$ defined by $(\text{a, b})\in\text{R}_1\Leftrightarrow\ \text{a}=\frac{1}{\text{b}}.$ Answer$\text{R}_1=\Big\{(\text{x, y}),\ \text{x, y}\in\text{Q}_0, \text{x}=\frac{1}{\text{y}}\Big\}$
Reflexivity: Let, $\text{x}\in\text{Q}_0$
$\Rightarrow\ \text{x}\neq\frac{1}{\text{x}}$
$\Rightarrow\ (\text{x, x})\in\text{R}_1$
$\therefore$ $R_1$ is not reflexive.
Symmetric: Let, $(\text{x, y})\in\text{R}_1$
$\Rightarrow\ \text{x}=\frac{1}{\text{y}}$
$\Rightarrow\ \text{y}=\frac{1}{\text{x}}$
$\Rightarrow\ (\text{y, x})\in\text{R}_1$
$\therefore$ $R_1$ is Symmetric.
Transitive: Let, $(\text{x, y})\in\text{R}_1$ and $(\text{y, z})\in\text{R}_1$
$\Rightarrow\ \text{x}=\frac{1}{\text{y}}$ and $\text{y}=\frac{1}{\text{z}}$
$\Rightarrow\ \text{x}=\text{z}$
$\Rightarrow\ (\text{x, z})\notin\text{R}_1$
$\therefore$ $R_1$ is not Transitive.
View full question & answer→Question 1783 Marks
Classify the following functions as injection, surjection or bijection:f : R → R, defined by f(x) = 3 - 4x
Answerf : R → R, defined by f(x) = 3 - 4xInjection test: Let x and y be any two elements in the domain (R), such that f(x) = f(y).
f(x) = f(y)
3 - 4x = 3 - 4y
-4x = -4y
x = y
Therefore, 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
3 - 4x = y
4x = 3 - y
$\text{x}=\frac{3-\text{y}}{4}\in\text{R}$
Therefore, f is a surjection and f is a bijection.
View full question & answer→Question 1793 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 $f(x)=2 x+1$ and $g(x)=x^2-$ 2. Express $(g \circ f)^{-1}$ and $f^{-1} \circ g^{-1}$ as the sets of ordered pairs and verify that $(g \circ f)^{-1}=f^{-1} \circ g^{-1}$.
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,
$f^{-1}=3,1,5,2,7,3,9,4 \text { and } g^{-1}=7,3,23,5,47,7,79,9$
Now, $f ^{-1} og ^{-1}: C \rightarrow A , f ^{-1} og ^{-1}=7,1,23,2,47,3,79,4$
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
go$f(4) = g(f(4)) = 4 + 4 - 1 = 79$
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 1803 Marks
Consider the binary operation $*$ and o defined by the following tables on set $S = \{a, b, c, d\}.$
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$o$
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$a$
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$b$
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$c$
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$d$
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$a$
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$a$
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$a$
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$a$
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$a$
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$b$
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$a$
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$b$
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$c$
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$d$
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$c$
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$a$
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$c$
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$d$
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$b$
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$d$
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$a$
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$d$
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$b$
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$c$
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AnswerCommutativity:The table is symmetrical about the leading element. It means that o is commutative on S.
$a o (b o c) = a o c$
$= a$
$(a o b) o c = a o c$
$= a$
thus,
$a o (b o c) = (a o b) o$
$\text{c }\forall\text{ a, b, c}\in\text{S}$
Associativity:
Therefore, o is associative on $S.$
Finding identity element:
We observe that the second row of the composition table coincides with the top-most row and the first column coincides with the left-most column.
These two intersect at $b.$
Implies that $x o b = b o x$
$=\text{x, }\forall\text{ x}\in\text{S}$
Therefore,
$b$ is the identity element.
Finding inverse elements:
In the first row, we don't have b,
i.e. there does not exist an element x such that $a o x = x o a = b.$
Therefore,
$a^{-1}$ does not exists.
$b o b = b$
Implies that $b^{-1}= b$
$c o d = b$
Implies that $c^{-1} = d$
$d o c = b$
Implies that $d^{-1} = c$
View full question & answer→Question 1813 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 1823 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 1833 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 1843 Marks
Check the commutativity and associativity of the following binary operations:
'*' on Q defined by $\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{4}$ for all a, b ∈ Q.
AnswerCommutativity: Let $\text{a, b}\in\text{Q}.$ Then,$\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{4}$
$=\frac{\text{ba}}{4}$
$=\text{b}\ ^*\ \text{a}$
Therefore,
$\text{a}\ ^*\ \text{b}=\text{b}\ ^*\ \text{a},\ \forall\ \text{a, b}\in\text{Q}$
Thus '*' is commutative on Z.
Associativity: Let $\text{a, b, c}\in\text{Q}.$ Then,
$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=\text{a}\ ^*\ \Big(\frac{\text{bc}}{4}\Big)$
$=\frac{\text{a}\big(\frac{\text{bc}}{4}\big)}{4}$
$=\frac{\text{abc}}{16}$
$(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}=\frac{\text{ab}}{4}\ ^*\ \text{c}$
$=\frac{\big(\frac{\text{ab}}{4}\big)\text{c}}{4}$
$=\frac{\text{abc}}{16}$
Therefore,
$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=(\text{a}\ ^*\ \text{b})\ ^*\ \text{c},\ \forall\ \text{a, b, c}\in\text{Q}$
Thus, '*' is associative on Q.
View full question & answer→Question 1853 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 1863 Marks
Let $*$ be the binary operation defined on $Q$. Find which of the following binary operations are commutative:
- $a * b = a – b \forall a, b \in Q$
- $a * b = a^2 + b^2 \forall a, b \in Q$
- $a * b = a + ab \forall a, b \in Q$
- $a * b = (a – b)^2 \forall a, b \in Q$
AnswerGiven that $*$ be the binary operation defined on $Q.$
$i. a * b = a – b \forall a, b \in Q$
$= -b + a$
$= -(b - a)$
$= -b * a$
$\therefore\ \text{a}\ ^*\ \text{b}\neq\text{b}\ ^*\ \text{a}$
Hence, $*$ is not commutative.
$ii.a * b = a^2 + b^2$
$= b^2 + a^2$
$= b * a$
Hence $, *$ is commutative.
$iii$. We have $a^* b=a+a b$ and $b * a=b+a b$
Clearly, $a+a b \neq b+a b$
So $,*$ is not communicative.
$iv.$ We have $a ^* b=( a - b )^2 \forall a , b \in Q$
$= (-b + a)^2$
$= {-(b - a)}^2$
$= (b - a)^2$
$= b * a$
Hence $, *$ is communicative.
View full question & answer→Question 1873 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 1883 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 $\text{a},\text{b}\in\text{X}.\ \text{Then},\text{a}\neq-\text{b}.$
$\therefore$ 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 1893 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 1903 Marks
Prove that the function $f : N \rightarrow N$, defined by $f(x) = x^2 + x + 1$, is one-one but not onto.
Answerf : N → 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 1913 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 $f(x) = y$
$\Rightarrow x^2 + 4 = y$
$\Rightarrow x^2 = y - 4$
$\Rightarrow\ \text{x}=\text{y}-4\in\text{R}$
⇒ 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 1923 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 1933 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 1943 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 1953 Marks
If $\text{f}:\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\rightarrow\text{R}$ and $g : [-1, 1] → R$ be defined as f(x) = tanx 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] → R defined as f(x) = tanx and $\text{g(x)}=\sqrt{1-\text{x}^2}$Range of f: let y = f(x) ⇒ y = tanx
$\Rightarrow x = \tan^{-1}y$
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(tanx)
$=\sqrt{1-\tan^2\text{x}}$
View full question & answer→Question 1963 Marks
Let f be a real function given by $\text{f(x)}=\sqrt{\text{x}-2}.$ Find the following:
fofof
Also, show that fof ≠ $f^2$.
Answerfofof = (fof)of
We have, $\text{f}:[2,\infty)\rightarrow(0,\infty)$ and $\text{fof}:[6,\infty)\rightarrow\text{R}$
⇒ 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}
⇒ Domain ((fof)of) = {x : x $\in[2,\infty)$ and $\sqrt{\text{x}-2}\in[6,\infty)$}
⇒ Domain ((fof)of) = {x : x $\in[2,\infty)$ and $\sqrt{\text{x}-2}\geq6$}
⇒ Domain ((fof)of) = {x : x $\in[2,\infty)$ and $\text{x}-2\geq36$}
⇒ Domain ((fof)of) = {x : x $\in[2,\infty)$ and $\text{x}\geq38$}
⇒ Domain ((fof)of) = {x : x $\geq38$}
⇒ 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 1973 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 1983 Marks
Let $A = {-1, 0, 1}$ and $f = {(x, x^2): x \in A}$. Show that $f : A → A$ is neither one-one nor onto.
Answer$A=\{-1,0,1\}$ and $f=\left\{\left(x, x^2\right): x \in A\right\}$ 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 1993 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 2003 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 $f(x)=2 x, g(x)=\frac{1}{x}$ and $h(x)=e^x$.
AnswerWe have, $f : N \rightarrow Z _0, g: Z _0 \rightarrow Q$ and $h : Q \rightarrow R$
Also, $f(x)=2 x, g(x)=\frac{1}{x}$ and $h(x)=e^x$
Now, $f: N \rightarrow Z_0$ and hog: $Z_0 \rightarrow R$
$\therefore$ (hog)of: $N \rightarrow R$
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 $) \circ f(x)=($ hog $)(f(x))=($ hog $)(2 x)$
$=\text{h}\Big(\frac{1}{2\text{x}}\Big)$
$=\text{e}^\frac{1}{2\text{x}}$
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.
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