Question 1012 Marks
Let R be the equivalence relation on the set Z of the integers given by R = {(a, b): 2 divides a - b}. Write the equivalence class [0].
Answer$\text{a, b}\in\text{Z}$ and R is given by R = {(a, b): 2 divides a - b}.The equivalence classes can be taken as [0], [1].
Note that, $\text{for}\ 0\leq\text{i}\leq1,$ [i] = {2n + i: $\text{n}\in\text{Z}$}
So equivalence class [0] = {2n: $\text{n}\in\text{Z}$}
It is clear that all the elements of equivalence class [0] are even.
Hence, equivalence class $[0]=\{0,\pm2,\pm4,\pm6\ ...\}$
View full question & answer→Question 1022 Marks
Let $*$ be a binary operation on the set $Q$ of rational numbers as follows:
$a * b = a^2+ b^2$
Answer$a * b = a^2 + b^2 = b^2 + a^2 = b * a$
$\therefore$ operation is commutative.
$(a * b) * c = (a^2 + b^2) * c = (a^2 + b^2) + c^2 = a^2 + b^2 + c^2$
And $a * b (b * c) = a * (b^2 + c^2) = a^2 + (b^2 + c^2)^2$
Here, $(\text{a}*\text{b})*\text{c}\neq\text{a}*(\text{b}*\text{c})$
$\therefore$ operation $*$ is not associative.
View full question & answer→Question 1032 Marks
Give an example of a relation which is,
Transitive but neither reflexive nor symmetric.
AnswerLet R be the relation on A such that
R = {(1, 2), (2, 3), (1, 3)}
The relation R on A is transitive, but neither symmetric nor reflexive.
View full question & answer→Question 1042 Marks
Let 'o' be a binary operation on the set $Q_0$ of all non-zero rational numbers defined by $\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{2}$ for all $\text{a},\text{b}\in\text{Q}_0.$
Find the invertible elements of $Q_0$.
AnswerWe have,
$\text{a }^*\text{ b}=\frac{\text{ab}}{2}$ for all $\text{a},\text{b}\in\text{Q}_0$
Let $\text{b}\in\text{Q}_0$ be the inverse of $\text{a}\in\text{Q}_0$ with respect to *, then,
a * b = b * a = e for all $\text{a}\in\text{Q}_0$
$\Rightarrow\frac{\text{ab}}{2}=\text{e}\Rightarrow\frac{\text{ab}}{2}=2$
$\Rightarrow\text{b}=\frac{4}{\text{a}}$
Thus, $\text{b}=\frac{4}{\text{a}}$ is the inverse of a with respect to *.
View full question & answer→Question 1052 Marks
Find the identity element in the set $I^+$ of all positive integers defined by $a * b = a + b$ for all $a, b \in I^+.$
AnswerLet $e$ be the identity element in $I^+$ with respect to $*$ such that
$a * e = a = e * a, \forall\ \text{a}\in\text{I}^{+}$
$a * e = a$ and $e * a = a, \forall\ \text{a}\in\text{I}^{+}$
$a + e = a$ and $e + a = a, \forall\ \text{a}\in\text{I}^{+}$
$e = 0, \forall\ \text{a}\in\text{I}^{+}$
Thus, 0 is the identity element in $I^+$ with respect to $*$.
View full question & answer→Question 1062 Marks
Let C be the set of complex numbers. Prove that the mapping f : C → R given by f(z) = |z|, ∀ z ∈ C, is neither one-one nor onto.
AnswerThe mapping f : C → R Given, f(z) = |z|, ∀ z ∈ C f(1) = |1| = 1 f(-1) = |-1| = 1 f(1) = f(-1)$\text{But}\ 1\neq-1$
So, f(z) is not one-one. Also, f(z) is not onto as there is no pre-image for any negative element of R under the mapping f(z).
View full question & answer→Question 1072 Marks
Let $S = \{a, b, c\}$ and $T = \{1, 2, 3\}.$ Find $F^{–1}$ of the following functions $F$ from $S$ to $T,$ if it exists:
$F = \{(a, 3), (b, 2), (c, 1)\}$
Answer$S = \{a, b, c\}, T = \{1, 2, 3\}$
$F: S → T$ is defined as:
$F =\{(a, 3), (b, 2), (c, 1)\}$
$\Rightarrow F(a) = 3, F(b) = 2, F(c) = 1$
Therefore, $F^{-1}: T → S$ is given by
$F^{-1} = \{(3, a), (2, b), (1, c)\}.$
View full question & answer→Question 1082 Marks
The following defines a relation on N:
$\text{x}>\text{y, x, y}\in\text{N}$
Determine which of the above relations are reflexive, symmetric and transitive.
AnswerA relation R in A is said to be reflexive if aRa for all $\text{a}\in\text{A},$ R is symmetric if aRb ⇒ bRa, for all $\text{a, b}\in\text{A}$ and it is said to be transitive if aRb and bRc ⇒ aRc for all $\text{a, b, c}\in\text{A.}$$\text{x}>\text{y, x, y}\in\text{N}$
$(\text{x, y})\in\big\{(2, 1), (3, 1),..., (3, 2), (4, 2),....\big\}$ $$
This is not reflexive as (1, 1), (2, 2), .... are absent.
This is not symmetric as (2, 1) is present but (1, 2) is absent.
This is transitive as $(3,2)\in\text{R}$ and $(2,1)\in\text{R}$ also $(3,1)\in\text{R},$ similarly this property satisfies all cases.
View full question & answer→Question 1092 Marks
Let $'o'$ be a binary operation on the set $Q_0$ of all non-zero rational numbers defined by $\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{2}$ for all $\text{a},\text{b}\in\text{Q}_0.$
Show that $'o'$ is both commutative and associate.
AnswerWe have, $\text{a }^*\text{ b}=\frac{\text{ab}}{2}$ for all $\text{a},\text{b}\in\text{Q}_0$
Commutativity:
Let $\text{a},\text{b}\in\text{Q}_0,$ then
$\Rightarrow\text{a }^*\text{ b}=\frac{\text{ab}}{2}=\frac{\text{ba}}{2}=\text{a }^*\text{ b}$
$\Rightarrow\text{a }^*\text{ b}=\text{b }^*\text{ a}$
Thus, $*$ is commutative on $Q_0.$
Associativity:
Let $\text{a},\text{b},\text{c}\in\text{Q}_0,$ then
$\Rightarrow(\text{a }^*\text{ b})\ ^*\ \text{c}=\frac{\text{ab}}{2}\ ....(1)$ and,
$\text{a }^*\ (\text{b }^*\text{ c})=\text{a }^*\ \frac{\text{bc}}{2}=\frac{\text{abc}}{4}\ ....(2)$ From $(1)\ \&\ (2)$
$(\text{a }^*\text{ b})\ ^*\ \text{c}=\text{a }^*\ (\text{b }^*\text{ c})$
$ \Rightarrow * $ is accosiative on $Q_0.$
View full question & answer→Question 1102 Marks
If f : A → A, g : A → A are two bijections, then prove that:
fog is an injection.
AnswerGiven: A → A, g : A → A are two bijections.
Then, fog : A → A
Injectivity of fog: Let x and y be two elements of the domain (A), such that
(fog)(x) = (fog)(y)
⇒ f(g(x)) = f(g(y))
⇒ g(x) = g(y) (As, f is one-one)
⇒ x = y (As, g is one-one)
So, fog is an injection.
View full question & answer→Question 1112 Marks
Write the total number of binary operations on a set consisting of two elements.
AnswerNumber of binary operations on a set with n elements $=\text{n}^{\text{n}^2}$
Here, Number of binary operations on a set with 2 elements $=2^{2^2}$
$= 2^4$
$=16$
View full question & answer→Question 1122 Marks
Check the injectivity and surjectivity of the following functions:
$f: N → N$ given by $f(x) = x^2$
Answerf: $\mathbf{N} \rightarrow \mathbf{N}$ is given by,
$f(x)=x^2$
It is seen that for $x, y \in N, f(x)=f(y) \Rightarrow x^2=y^2 \Rightarrow x=y$.
$\therefore \mathrm{f}$ is injective.
Now, $2 \in \mathbf{N}$. But, there does not exist any x in $\mathbf{N}$ such that $\mathrm{f}(\mathrm{x})=\mathrm{x}^2=2$.
$\therefore \mathrm{f}$ is not surjective.
Hence, function $f$ is injective but not surjective.
View full question & answer→Question 1132 Marks
Which of the following functions from A to B are one-one and onto?
$f_1 = \{(1, 3), (2, 5), (3, 7)\}; A = \{1, 2, 3\}, B = \{3, 5, 7\}$
Answer$ f_1=\{(1,3),(2,5),(3,7)\}$
$ A=\{1,2,3\}, B=\{3,5,7\}$
We can earily observe that in $f_1$ every element of $A$ has different image from $B$.
$\therefore \mathrm{f}_1$ in not one-one.
Also, each element of $B$ is the image of some element of $A$.
$\therefore \mathrm{f}_1$ in not on to.
View full question & answer→Question 1142 Marks
If $f : R → R$ be defined by $f(x) = x^4,$ write $f^{-1}(1).$
AnswerLet $f^{-1}(1) = x ......(1)$
$\Rightarrow f(x) = 1$
$\Rightarrow x^4 = 1$
$\Rightarrow x^4 - 1 = 0$
$\Rightarrow (x^2 - 1)(x^2 + 1) = 0 [$Using identity: $a^2 - b^2 = (a - b)(a + b)]$
$\Rightarrow (x - 1)(x + 1)(x^2 + 1) = 0 [$Using identity: $a^2 - b^2 = (a - b)(a + b)]$
$\Rightarrow\ \text{x}=\pm1\ [\text{as x}\in\text{R}]$
$\Rightarrow f^{-1}(1) = {-1, 1} [$from $(1)]$
View full question & answer→Question 1152 Marks
Determine whether or not the definition of $*$ given below gives a binary operation. In the event that $*$ is not a binary operation give justification of this.
On $Z^+,$ defined * by $a * b = ab.$
Here, $Z^+$ denotes the set of all non-negative integers.
Answer$\text{a, b}\in\text{Z}^{+}$$\Rightarrow\ \text{ab}\in\text{Z}^+$
$\Rightarrow\ \text{a}\ ^*\ \text{b}\in\text{Z}^+$
Therefore,
$\text{a}\ ^*\ \text{b}\in\text{Z}^+,\ \forall\ \text{a, b}\in\text{Z}^+$
Thus, $*$ is a binary operation on $Z^+.$
View full question & answer→Question 1162 Marks
f: Z → Z given by $f(x) = x^2$
Answerf: Z → Z is given by,
$f(x) = x^2$
It is seen that for f(= 1) = f(1) = 1, but $-1\neq1.$
$\therefore$ f is not injective.
Now, $-2\in\text{Z}.$ But, there does not exist any element $\text{x}\in\text{Z}$ such that $f(x) = x^2 = -2$.
$\therefore$ f is not surjective.
Hence, function f is neither injective but not surjective.
View full question & answer→Question 1172 Marks
Show that the relation R in the set R of real numbers, defined as $R$ = {$(a, b) : a \leq b^2$} is neither reflexive nor symmetric nor transitive.
Answer$\text{R}=\big\{(\text{a},\text{b}):\text{a}\leq\text{b}^2\big\},$ Relation R is defined as the set of real numbers.
| (i) |
Whether $(\text{a},\text{a})\in\text{R},\ \text{then}\ \text{a}\leq\text{a}^2$ which is false. |
$\therefore$ |
R is not reflexive. |
| (ii) |
Whether (a,b) = (b, a) , then $\text{a}\leq\text{b}^2\ \text{and}\ \text{b}\leq\text{a}^2,$ it is false |
$\therefore$ |
R is not symmetric. |
| (iii) |
$\text{Now }\text{a}\leq\text{b}^2\ \text{and}\ \text{b}\leq\text{c}^2\Rightarrow\text{a}\leq\text{c}^4,$ which is false |
$\therefore$ |
R is not transitive. |
Therefore, R is neither reflexive, nor symmetric and nor transitive. View full question & answer→Question 1182 Marks
If $A=\{2,3,4\}, B=\{1,3,7\}$ and $R=\{(x, y): x \in A, y \in B$ and $x<y\}$ is a relation from $A$ to $B$, then write $R^{-1}$.
Answer$ \text { Since } R=\{(x, y): x \in A, y \in B \text { and } x<y\} $
$ R=\{(2,3),(2,7),(3,7),(4,7)\} $
$ \text { Hence, } R^{-1}=\{(3,2),(7,2),(7,3),(7,4)\}$
View full question & answer→Question 1192 Marks
If f : A → B is an injection, such that range of f = {a}, determine the number of elements in A.
AnswerRange of f = {a}
Therefore, the number of images of f = 1
Since, is an injection, there will be exactly one image for each element of f.
Therefore, number of element in A = 1.
View full question & answer→Question 1202 Marks
Let A = {2, 3, 4, 5} and B = {1, 3, 4}. If R is the relation from A to B given by a R b if "a is a divisor of b". Write R as a set of ordered pairs.
AnswerWe have, A = {2, 3, 4, 5}, B = {1, 3, 4} and relation from A to B is given by aRb if ''is divisor of'' B$\therefore$ R can be written as ordered pair as R = {(2, 4), (3, 3), (4, 4)}
View full question & answer→Question 1212 Marks
Write the inverse of 5 under multiplication modulo 11 on the set $\{1,2, \ldots, 10\}$.
AnswerAs, e $=1: 5 \times 9 \equiv 1(\bmod 11)$
So, the inverse of 5 i.e. $5^{-1}=9$
View full question & answer→Question 1222 Marks
If the binary operation $*$ on the set $Z$ of integers is defined by $a * b = a + 3b^2,$ find the value of $2 * 4.$
AnswerGiven: $a * b = a + 3b^2$
Here,
$2 * 4 = 2 + 3(4)^2$
$= 2 + 3(16)$
$= 2 + 48$
$= 50$
View full question & answer→Question 1232 Marks
Prove that the operation $*$ on the set $\text{M}=\Bigg\{\begin{bmatrix}\text{a} & 0 \\0 & \text{b} \end{bmatrix};\text{ a, b}\in\text{R}-\{0\}\Bigg\}$ defined by $A * B = AB$ is a binary operation.
AnswerGiven that $*$ is an operation that is valid on the set $\text{M}=\Bigg\{\left(\begin{array}{c}\text{a}&0\\ 0&\text{b}\end{array}\right):\text{b}\in \text{R}-\big\{0\big\}\Bigg\}$ and it is defined as given: $A * B = AB.$
According to the problem it is given that on applying the operation $*$ fore two given numbers in the set $'M\ '$ it gives a number in the set $'M\ '$ as a result of the operation.
$\Rightarrow \text{A}*\text{B}\in \text{M}...(1)$
Let us take $\text{A}=\left(\begin{array}{c}\text{a}&0\\ 0&\text{b}\end{array}\right)\text{ and }\text{B}=\left(\begin{array}{c}\text{c}&0\\ 0&\text{d}\end{array}\right)$ here $\text{a}\in \text{R},\ \text{c}\in \text{R},\ \text{d}\in \text{R}$ then,
$\Rightarrow \text{AB}=\left(\begin{array}{c}\text{a}&0\\ 0&\text{b}\end{array}\right)\times\left(\begin{array}{c}\text{c}&0\\ 0&\text{d}\end{array}\right)$
$\Rightarrow \text{AB}=\begin{pmatrix}((\text{a}\times\text{c})+(0\times 0))&((\text{a}\times0)+(0\times \text{d}))(0\times\text{c})+(\text{b}\times 0))&((0\times0)+(\text{b}\times\text{d})) \end{pmatrix}$
$\Rightarrow \text{Ab}=\begin{pmatrix}(\text{ac}+0)(0+0)(0+0)&(0+\text{bd}) \end{pmatrix}$
$\Rightarrow \text{AB}=\begin{pmatrix} \text{ac}&0\\0&\text{bd}\end{pmatrix}$
Since $\text{b}\in \text{R}$ and $\text{c}\in \text{R}$ then $\text{ac}\in \text{R}$
And also $\text{b}\in \text{R}$ and $\text{d}\in \text{R}$ then $\text{bd}\in \text{R}$
$\Rightarrow \text{AB}\in \text{R}$
View full question & answer→Question 1242 Marks
Let $A = \{3, 5, 7\}, B = \{2, 6, 10\}$ and R be a relation from $A$ to $B$ defined by $R = \{(x, y): x$ and $y$ are relatively prime$\}.$ Then, write $R$ and $R^{-1}.$
Answer$R=\{(x, y): x$ and $y$ are relatively prime $\}$
Then,
$ R=\{(3,2),(5,2),(7,2),(3,10),(7,10),(5,6),(7,6)\}$
$ \text { So, } R^{-1}=\{(2,3),(2,5),(2,7),(10,3),(10,7),(6,5),(6,7)\}$
View full question & answer→Question 1252 Marks
Write the domain of the real function $\text{f(x)}=\frac{1}{\sqrt{|\text{x}|-\text{x}}}.$
AnswerCase-1: When x > 0
|x| = x
$\Rightarrow\ \frac{1}{\sqrt{|\text{x}|-\text{x}}}=\frac{1}{\sqrt{\text{x}-\text{x}}}=\frac{1}{0}=\infty$
Case-2: When x < 0
|x| = -x
$\Rightarrow\ \frac{1}{\sqrt{|\text{x}|-\text{x}}}=\frac{1}{\sqrt{-\text{x}-\text{x}}}=\frac{1}{\sqrt{-2\text{x}}}$ (exists because when x < 0, -2x > 0)
⇒ f(x) is defined when x < 0
So, domain $=(-\infty,0)$
View full question & answer→Question 1262 Marks
If $\mathrm{A}=\{a, b, c, d\}$ and the function $f=\{(a, b),(b, d),(c, a),(d, c)\}$, write $f^{-1}$.
AnswerWe are given that, $\mathrm{f}=\{(\mathrm{a}, \mathrm{b}),(\mathrm{b}, \mathrm{d}),(\mathrm{c}, \mathrm{a}),(\mathrm{d}, \mathrm{c})\}$
An inverse relation is the set of ordered pairs obtained by interchanging the first and second elements of each pair in the original relation.
$\therefore f^{-1}=\{(b, a),(d, b),(a, c),(c, d)\}$
View full question & answer→Question 1272 Marks
Let A = {0, 1, 2, 3} and R be a relation on A defined as R = {(0, 0), (0, 1), (0, 3), (1, 0), (1, 1), (2, 2), (3, 0), (3, 3)}. Is R reflexive? symmetric? transitive?
AnswerWe have, R = {(0, 0), (0, 1), (0, 3), (1, 0), (1, 1), (2, 2), (3, 0), (3, 3)}As, $(\text{a, a})\in\text{R}\ \forall\ \text{a}\in\text{A}$
So, R is a reflexive relation. Also, $(\text{a, b})\in\text{R}$ and $(\text{b, a})\in\text{R}$ So, R is a symmetric as well And, $(0,1)\in\text{R}$ but $(1,2)\notin\text{R}$ and $(2,3)\notin\text{R}$ So, R is not a transitive relation.
View full question & answer→Question 1282 Marks
Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.
AnswerIt is given that a = {1, 2, 3}, B = {4, 5, 6, 7}
f: A → B is defined as f = {(1, 4), (2, 5), (3, 6)}.
$\therefore$ f(1) = 4, f(2) = 5, f(3) = 6
It is seen that the images of distinct elements of A under f are distinct.
Hence, function f is one-one.
View full question & answer→Question 1292 Marks
Let * be a binary operation, on the set of all non-zero real numbers, given by
$\text{a}\times\text{b}=\frac{\text{ab}}{5}\ \forall\text{ a, b}\in\text{R}-\{0\}$
Write the value of x given by 2 * (x * 5) = 10.
AnswerGiven: 2 * (x * 5) = 10
Here,
$2\times\Big(\frac{5\text{x}}{5}\Big)=10$
Implies that 2 * x = 10
Implies that $\frac{2\text{x}}{5}=10$
Implies that $\text{x}=\frac{10\times5}{2}$
Implies that x = 25
View full question & answer→Question 1302 Marks
Write the total number of one-one functions from set A = {1, 2, 3, 4} to set B = {a, b, c}.
AnswerA has 4 elements and B has 3 elements.
Also, one-one function is only possible from A to B if nA ≤ nB.
But, here nA > nB
So, the number of one-one functions from A to B is 0.
View full question & answer→Question 1312 Marks
Consider $f:\{1,2,3\} \rightarrow\{a, b, c\}$ given by $f(1)=a, f(2)=b$ and $f(3)=c$. Find $f^{-1}$ and show that $\left(f^{-1}\right)^{-1}=f$.
Answer$f=\{(1, a),(2, b),(3, c)\}$, then it is clear that $f$ is $1-1$ and onto and therefore $f^{-1}$ exists.
Also, $f^{-1}=\{(1, a),(b, 2),(c, 3)\}$ and $\left(f^{-1}\right)^{-1}=\{(1, a),(2, b),(3, c)\}=f$
Hence, $\left(f^{-1}\right)^{-1}=f$.
View full question & answer→Question 1322 Marks
The following defines a relation on N:
x + 4y = 10, $\text{x, y}\in\text{N}$
Determine which of the above relations are reflexive, symmetric and transitive.
AnswerA relation R in A is said to be reflexive if aRa for all $\text{a}\in\text{A},$ R is symmetric if aRb ⇒ bRa, for all $\text{a, b}\in\text{A}$ and it is said to be transitive if aRb and bRc ⇒ aRc for all $\text{a, b, c}\in\text{A.}$$\text{x} + 4\text{y} = 10, \ \text{x, y}\in\text{N}$, $$
$(\text{x, y})\in\big\{(6, 1), (2, 2)\big\}$ $$
This is not reflexive as (1, 1), (2, 2), .... are absent.
This is also symmetric because $(6,1)\in\text{R}$ but (1, 6) is absent.
This is not transitive as there are only two elements in the set having no element common.
View full question & answer→Question 1332 Marks
Let A = {1, 2, 3, 4} and B = {a, b} be two sets. Write the total number of onto functions from A to B.
AnswerWhen two sets A and B have m and n elements respectively, then the number of onto functions from A to B is,$\begin{cases}\sum_{\text{r}=1}^\text{n}(-1)^\text{r}\text{ nC}_\text{r}\text{r}^\text{m},&\text{if m}\geq\text{n}\\0,&\text{if m}<\text{n}\end{cases}$
Here, number of elements in A = 4 = m Number of elements in B = 2 = n So, m > n Number of onto functions$=\sum_{\text{r}=1}^2(-1)^\text{r}2\text{C}_\text{r}\text{r}^4$
$= (-1)^12\text{C}_11^4 + (-1)^22\text{C}_22^4$
$= -2 + 16$$= 14$
View full question & answer→Question 1342 Marks
If functions $f: A \rightarrow B$ and $g: B \rightarrow A$ satisfy gof $=I_A$, then show that $f$ is oneone and $g$ is onto.
AnswerGiven that, $f: A \rightarrow B$ and $g: B \rightarrow A$ satisfy gof $=I_A, $
$\because$ gof $=I_A \Rightarrow \operatorname{gof}\left\{f\left(x_1\right)\right\}=\operatorname{gof}\left\{f\left(x_2\right)\right\}$
$\Rightarrow g\left(x_1\right)=g\left(x_2\right)\left[\because\right.$ gof $\left.=I_A\right]$
$\therefore \mathrm{x}_1=\mathrm{x}_2$
Hence, f is one-one and g is onto.
View full question & answer→Question 1352 Marks
Let $A = \{a, b, c, d\}$ and $f : A → A$ be given by $f = \{(a, b), (b, d), (c, a), (d, c)\}.$ Write $f ^{-1}$.
AnswerWe have,
$A = \{a, b, c, d\}$ and $f : A → A$ be given by
$f = \{(a, b), (b, d), (c, a), (d, c)\}$
$($Since, the elements of a function when interchanged gives inverse function. Therefore, $f^{-1} = \{(b, a), (d, b), (a, c), (c, d)\})$
View full question & answer→Question 1362 Marks
Let * be a binary operation on N given by a * b = LCM (a, b) for all $\text{a, b}\in\text{N.}$ Find 5 * 7.
AnswerAs, a * b = LCM (a, b)
So, 5 * 7 = LCM (5, 7) = 35
View full question & answer→Question 1372 Marks
Define a transitive relation.
AnswerA relation R on a set A is said to be transitive if
$(\text{a, b})\in\text{R}$ and $(\text{b, c})\in\text{R}$
$\Rightarrow\ (\text{a, c})\in\text{R}$ for all $\text{a, b, c}\in\text{R}$
i.e., aRb and bRc
⇒ aRc for all $\text{a, b, c}\in\text{R}$
View full question & answer→Question 1382 Marks
Determine whether the following operations define a binary operation on the given set or not:
$'+6'$ on $S = \{0, 1, 2, 3, 4, 5\}$ defined by, $\text{a}+_6\text{b}=\begin{cases}\text{a}+\text{b},&\text{if a}+\text{b}<6\\\text{a}+\text{b}-6,&\text{if a}+\text{b}\geq6\end{cases}$
AnswerWe have, $S = \{0, 1, 2, 3, 4, 5\}$ and, $\text{a}+_6\text{b}=\begin{cases}\text{a}+\text{b},&\text{if a}+\text{b}<6\\\text{a}+\text{b}-6,&\text{if a}+\text{b}\geq6\end{cases}$
Let $\text{a}\in\text{S}$ and $\text{b}\in\text{S}$ such that $a + b < 6$
Then $\text{a}+_6\text{b}=\text{a}+\text{b}\in\text{S}$ $\big[\because a + b < 6 = 0, 1, 2, 3, 4, 5 \big]$
Let $\text{a}\in\text{S}$ and $\text{b}\in\text{S}$ such that $a + b > 6$
Then $\text{a}+_6\text{b}=\text{a}+\text{b}-6\in\text{S}$ $\big[\because\ \text{if a}+\text{b}\geq6$ then $\text{a}+\text{b}-6\geq6 = 0, 1, 2, 3, 4, 5 \big]$
$\therefore\ \text{a}+_6\text{b}\in\text{S}$ for $\text{a, b}\in\text{S}$
$\therefore +_6$ defined a binary operation on $S.$
View full question & answer→Question 1392 Marks
Let f, g and h be functions from R to R. Show that:
(f + g)oh = foh + goh
(f.g)oh = (foh).(goh)
Answer
- To prove: (f + g)oh = foh + goh
L.H.S. = (f + g)oh = (f + g)[h(x)] = f[h(x)] + g[h(x)] = foh + goh = R.H.S.
- (b) To prove: (f.g)oh = (foh).(goh)
L.H.S. = (f.g)oh = (f.g)[h( x)] = f[h(x)].g[h(x)] = foh.goh = R.H.S.
View full question & answer→Question 1402 Marks
Is * defined on the set {1, 2, 3, 4, 5} by a * b = L.C.M. of a and b a binary operation? Justify your answer.
AnswerLet A = {1, 2, 3, 4, 5} and a *' b= L.C.M. of a and b.
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*
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1
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2
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1
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2
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6
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10
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Here, $2*3=6\notin\text{A}$
Therefore, the operation * is not a binary operation. View full question & answer→Question 1412 Marks
If $f : R → R$ defined by $f(x) = 3x - 4$ is invertible, then write $f^{-1}(x)$.
AnswerLet $f^{-1}(x) = y .....(1)\Rightarrow f(y) = x$
$\Rightarrow 3y - 4 = x$
$\Rightarrow 3y = x + 4$
$\Rightarrow\ \text{y}=\frac{\text{x}+4}{3}$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{\text{x}+4}{3}$ [from (1)]
View full question & answer→Question 1422 Marks
A fair die is rolled. Consider events E = $\{1,\ 3,\ 5\},\ \text{F}=\{2,\ 3\}\ \text{and}\ \text{G}=\{2,\ 3,\ 4,\ 5\}.\ \text{Find}:$
$\text{P}(\text{E}|\text{G})\ \text{and}\ \text{P}(\text{G}|\text{E})$
Answer$\text{P}\left(\text{E}\right)=\frac{\text{n}\left(\text{E}\right)}{\text{n}\left(\text{S}\right)}=\frac{3}{6}\ \ \ \ \ \ \ \ \text{P}\left(\text{G}\right)=\frac{\text{n}\left(\text{G}\right)}{\text{n}\left(\text{S}\right)}=\frac{4}{6}$
$\text{E}\ \cap\ \text{G}=(3,\ 5)\ \Rightarrow\ \ \ \ \ \text{n}\left(\text{E}\cap\text{G}\right)=2$
$\text{P}\left(\text{E}\cap\text{G}\right)=\frac{\text{n}\left(\text{E}\ \cap\ \text{G}\right)}{\text{n}\left(\text{S}\right)}=\frac{2}{6}$
$\text{P}\left(\text{E}|\text{G}\right)=\frac{\text{P}\left(\text{E}\ \cap\ \text{G}\right)}{\text{P}\left(\text{G}\right)}=\frac{\frac{2}{6}}{\frac{4}{6}}=\frac{2}{4}=\frac{1}{2}\ \ \ \\ \text{and}\ \ \ \text{P}\left(\text{G}|\text{E}\right)=\frac{\text{P}\left(\text{E}\ \cap\ \text{G}\right)}{\text{P}\left(\text{E}\right)}=\frac{\frac{2}{6}}{\frac{3}{6}}=\frac{2}{3}$
View full question & answer→Question 1432 Marks
If $R=\left\{(x, y): x^2+y^2 \leq 4 ; x, y \in Z\right\}$ is a relation on $Z$, write the domain of $R$.
AnswerDomain of R is the set of values of x that satisfies the relation R.
Because x must be an integer, the provided values of x are:
$0,\pm1,\pm2$
Thus, Domain of R is $0,\pm1,\pm2$
View full question & answer→Question 1442 Marks
If $A = \{1, 2, 3\}$ and $B = \{a, b\}$, write the total number of functions from $A$ to $B.$
AnswerIf set $A$ has $m$ elements and set $B$ has $n$ elements, then the number of functions from $A$ to $B$ is nm.
Given: $A = \{1, 2, 3\}$ and $B = \{a, b\}$
$⇒ n(A) = 3$ and $n(B) = 2$
$\therefore$ Number of functions from $A$ to $B = 2^3 = 8$
View full question & answer→Question 1452 Marks
If $f(x) = 2x + 5$ and $g(x) = x^2 + 1$ be two real functions, then describe the following functions:
gof
Also, show that fof ≠ $f^2$
Answerf(x) and g(x) are polynomials.
⇒ f : R → R and g : R → R.
So, fog : R → R and gof : R → R.
(gof)(x) = g(f(x))
$= g(2x + 5)$
$= (2x + 5)^2 + 1$
$= 4x^2 + 20x + 26$
View full question & answer→Question 1462 Marks
If $f : C → C$ is defined by $f(x) = x^2$, write $f^{-1}(-4)$. Here, C denotes the set of all complex numbers.
Answer$f: C \rightarrow C \text { defined by } f(x)=x^2 \Rightarrow f^{-1}\left(x^2\right)=x$
$\Rightarrow f^{-1}(-4)=f^{-1}\left[(2 i,-2 i)^2\right]=(2 i,-2 i)$
$\therefore f^{-1}(-4)=(2 i,-2 i)$
View full question & answer→Question 1472 Marks
Write the identity relation on set A = {a, b, c}.
AnswerIdentity set of A is:
I = {(a, a), (b, b), (c, c)}
Every element of this relation is related to itself.
View full question & answer→Question 1482 Marks
Let * be a binary operation on the set I of integers, defined by a * b = 2a + b - 3. Find the value of 3 * 4.
AnswerIt is given that, a * b = 2a + b - 3 Now, 3 * 4 = 2 × 3 + 4 - 3 = 10 - 3= 7
View full question & answer→Question 1492 Marks
For the set A = {1, 2, 3}, define a relation R on the set A as follows:
R = {(1, 1), (2, 2), (3, 3), (1, 3)}
Write the ordered pairs to be added to R to make the smallest equivalence relation.
Answer(3, 1) is the single ordered pair which needs to be added to R to make it the smallest equivalence relation.
View full question & answer→Question 1502 Marks
Let A = {1, 2, 3} and R = {(1, 2), (1, 1), (2, 3)} be a relation on A. What minimum number of ordered pairs may be added to R so that it may become a transitive relation on A.
AnswerWe have, A = {1, 2, 3} and R = {(1, 2), (1, 1), (2, 3)}
To make R transitive we shall add (1, 3) only.
$$ $\therefore \text{R}' = \big\{(1, 2), (1, 1), (2, 3), (1, 3)\big\}$
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