Question 12 Marks
Examine whether the operation * defined on R by $\text{a}^*\text{b}=\text{ab}+1$ is (i) a binary or not. (ii) if a binary operation, is it associative or not?
AnswerThe given operation is $\text{a}^*\text{b}=\text{ab}+1$
If any operation is a binary operation then it must follow the closure property.
Let $\text{a}\in\text{R},\text{b}\in\text{R}$
then $\text{a}^*\text{b}\in\text{R}$
also $\text{ab}+1\in\text{R}$
i.e. $\text{a}^*\text{b}\in\text{R}$
So * on R satisfies the closure property
Now if this binary operation satisfies associative law then
$(\text{a}^*\text{b})^*\text{c}=\text{a}^*(\text{b}^*\text{c})$
$(\text{a}^*\text{b})^*\text{c}=(\text{ab}+1)^*\text{c}$
$=(\text{ab}+1)\text{c}+1$
$=\text{abc}+\text{c}+1$
$\text{a}^*(\text{b}^*\text{c})=\text{a}^*(\text{bc}+1)$
$=\text{a}(\text{bc}+1)+1$
$=\text{abc}+\text{a}+1$
$\therefore(\text{a}^*\text{b})^*\text{c}\neq\text{a}^*(\text{b}^*\text{c})$
i.e., * operation does not follow associative law.
View full question & answer→Question 22 Marks
Let $\text{f(x)=}\begin{cases}1+\text{x,}&0\leq\text{x}\leq2\\3-\text{x,}&2<\text{x}\leq3\end{cases}.$ Find fof.
Answer$\text{f(x)=}\begin{cases}1+\text{x,}&0\leq\text{x}\leq2\\3-\text{x,}&2<\text{x}\leq3\end{cases}$
$\therefore$ Range of $\text{f}=[0,3]\subseteq$ Domain of f.
$\therefore$ fof(x) = f(f(x))
$=\text{f}\begin{cases}1+\text{x,}&0\leq\text{x}\leq2\\3-\text{x,}&2<\text{x}\leq3\end{cases}$
$\text{fof(x)}=\begin{cases}2+\text{x},&0\leq\text{x}\leq1\\2-\text{x},&1<\text{x}\leq2\\4-\text{x},&2<\text{x}\leq3\end{cases}$
View full question & answer→Question 32 Marks
If $f : R → R$ is defined by $f(x) = x^2$, find $f^{-1}(-25)$.
Answer$f : R → R$ defined by $f(x) = x^2 \therefore f^{-1}(x^2)$ = x$\Rightarrow\ \text{f}^{-1}(-25)=\phi$ $[\because\ \sqrt{-25}\notin\text{R}]$
View full question & answer→Question 42 Marks
The binary operation *: R × R → R is defined as a * b = 2a + b. Find (2 * 3) * 4.
AnswerIt is given that, a * b = 2a + b
Now,
(2 * 3) = 2 × 2 + 3
= 4 + 3
= 7
(2 * 3) * 4 = 7 * 4 = 2 × 7 + 4
= 14 + 4
= 18
View full question & answer→Question 52 Marks
Relation R in the set N of natural numbers defined as R = {(x, y) : y = x + 5 and x < 4}
AnswerR = {( x, y) : y = x + 5 and x < 4} in set N of natural numbers.Clearly R = {(1, 6), (2, 7), (3, 8)}
| Now $(\text{x},\text{x})\notin\text{R},$ |
$\therefore$ |
R is not reflexive. |
| Again $(\text{x},\text{y})\in\text{R}\ \text{but}\ (\text{y},\text{x})\notin\text{R}$ |
$\therefore$ |
R is not symmetric. |
| Also $(1,6)\in\text{R}\ \text{and}\ (2, 7)\in\text{R}\ \text{but}(1,7)\notin\text{R},$ |
$\therefore$ |
R is not transitive. |
Therefore, R is neither reflexive, nor symmetric and nor transitive. View full question & answer→Question 62 Marks
Symmetric and transitive but not reflexive.
Answer“is brother of” R = {( x, y) : x is a brother of y}
| It is clear that $\text{x}\geq\text{x}$ |
$\therefore$ |
R is reflexive. |
| It is clear that x is not the brother of x. |
$\therefore$ |
R is not symmetric. |
| Also if x is brother of y and y is brother of z then |
|
|
| x can be brother of z |
$\therefore$ |
R is transitive. |
Therefore, R is symmetric and transitive but not reflexive. View full question & answer→Question 72 Marks
Define a symmetric relation.
AnswerA relation R on a set A is said to be symmetric if $\text{a, b}\in\text{R}$
Implies that, $\text{b, a}\in\text{R}$ for all $\text{a, b}\in\text{A}$
That is, aRb implies that bRa for all $\text{a, b}\in\text{A}$
View full question & answer→Question 82 Marks
$f: R → R$ defined by $f(x) = 1 + x^2$
Answer$f: R → R$ is defined as
$f(x) = 1 + x^2$
Let $\text{x}_1,\text{x}_2\in\text{R}$ such that $f(x_1) = f(x_2)$
$\Rightarrow1+\text{x}_{1}^{2}=1+\text{x}_{2}^{2}$
$\Rightarrow\text{x}_{1}^{2}=\text{x}_{2}^{2}$
$\Rightarrow\text{x}_{1}=\pm\text{x}_{2}$
$\therefore f(x_1) = f(x_2)$ does not imply that $x_1 = x_2.$
For instance,
$f(1) = f(-1) = 2$
$\therefore$ f is not one-one.
Consider an element $-2$ in co-domain $R.$
It is seen that $f(x) = 1 + x^2$ is positive for all $\text{x}\in\text{R}.$
Thus, there does not exist any x in domain $R $ such that $f(x) = -2.$
$\therefore$ $f$ is not onto.
Hence, $f$ is neither one-one nor onto.
View full question & answer→Question 92 Marks
Let $R$ = {$(x, y): |x^2 - y^2| < 1$} be a relation on set $A$ = {$1, 2, 3, 4, 5$}. Write $R$ as a set of ordered pairs.
AnswerGiven: $R$ = {$(x, y): |x^2 - y^2| < 1$} be a relation on $A$ = {$1, 2, 3, 4, 5$}
Then, $R$ = {$(1, 1), (2, 2), (3, 3), (4, 4), (5, 5)$}
View full question & answer→Question 102 Marks
Write the domain of the relation $R$ defined on the set $Z$ of integers as follows:
$(a, b) \in R ⇔ a^2 + b^2 = 25$
AnswerWe have,
$R =\{(a, b) ∈ R ⇔ a^2 + b^2 = 25\}$ be a relation on $Z.$
The domain of R is the value of $'a' ∈ Z$, that satisfies $a^2 + b^2 = 25$
$a^2 + b^2 = 25$
$\Rightarrow\ \text{a}=\pm\sqrt{25-\text{b}^2}$
$\therefore$ Domain of $\text{R}=\{0,\pm3,\pm4,\pm5\}$
View full question & answer→Question 112 Marks
If $A = \{1, 2, 3, 4\}$ and $B = \{a, b, c, d\}$ define any four bijections from $A$ to $B$. Also give their inverse functions.
Answer$f_1 = \{(1, a), (2, b), (3, c), (4, d)\} \Rightarrow\ \text{f}_1^{-1}=\{(\text{a},1), (\text{b},2), (\text{c},3),(\text{d},4)\}$
$f_2 = \{(1, b), (2, a), (3, c), (4, d)\} \Rightarrow\ \text{f}_2^{-1}=\{(\text{b},1), (\text{a},2), (\text{c},3),(\text{d},4)\}$
$f_3 = \{(1, a), (2, b), (4, c), (3, d)\} \Rightarrow\ \text{f}_3^{-1}=\{(\text{a},1), (\text{b},2), (\text{c},4),(\text{d},3)\}$
$f_4 = \{(1, b), (2, a), (4, c), (3, d)\} \Rightarrow\ \text{f}_3^{-1}=\{(\text{b},1), (\text{a},2), (\text{c},4),(\text{d},3)\}$
Clearly, all these are bijections because they are one-one and onto.
View full question & answer→Question 122 Marks
Is * defined on the set {1, 2, 3, 4, 5} by a * b = LCM of a and b a binary operation? Justify your answer.
Answer
|
LCM
|
1
|
2
|
3
|
4
|
5
|
|
1
|
1
|
2
|
3
|
4
|
5
|
|
2
|
2
|
2
|
6
|
4
|
10
|
|
3
|
3
|
5
|
3
|
12
|
15
|
|
4
|
4
|
4
|
12
|
4
|
20
|
|
5
|
5
|
10
|
15
|
20
|
5
|
In the given composition table, all the elements are not in the set {1, 2, 3, 4, 5}.
If we consider a = 2 and b = 3, a * b = LCM of a and b = 6 $\notin$ {1, 2, 3, 4, 5}.
Thus, * is not a binary operation on {1, 2, 3, 4, 5}. View full question & answer→Question 132 Marks
Let $f : R → R, g : R → R$ be two functions defined by $f(x) = x^2 + x + 1$ and $g(x) = 1 - x^2$. Write fog $(-2)$.
Answer$(fog)(-2) = f(g(-2))$
$= f(1 - (-2)^2)$
$= f(-3)$
$= (-3)^2 + (-3) + 1$
$= 9 - 3 + 1$
$= 7$
View full question & answer→Question 142 Marks
Relation R in the set Z of all integers defined as R = {(x, y) : x – y is an integer}
AnswerR = {( x, y) : x − y is an integer} in set Z of all integers.
| Now (x, x) i.e., (1, 1) = 1 - 1 = $0\in\text{Z},$ |
$\therefore$ |
R is reflexive. |
| Again $(\text{x},\ \text{y})\in\text{R},\ \text{and}(\text{y},\ \text{x})\in\text{R},$ i.e., x - y and y - x are an integer |
$\therefore$ |
R is symmetric. |
| $\text{Also}\ (\text{x}_1,\ \text{y}_1)=\text{x}_1-\text{y}_1\in\text{Z},\ \text{and}(\text{y}_1,\ \text{z}_1)=\text{y}_1-\text{z}_1\in\text{Z}\ \text{and}\\(\text{x}_1,\ \text{z}_1)\in\text{R}, $ |
$\therefore$ |
R is transitive. |
Therefore, R is reflexive, symmetric and transitive. View full question & answer→Question 152 Marks
Show that the relation R in R defined as R = {(a, b) : a ≤ b}, is reflexive and transitive but not symmetric.
Answer
| (i) |
$\text{a}\leq\text{a}$ which is true, so $(\text{a},\text{a})\in\text{R},$ |
$\therefore$ |
R is reflexive. |
| (ii) |
$\text{a}\leq\text{b}\ \text{but}\ \text{b}\leq\text{a}$ |
$\therefore$ |
R is not symmetric. |
| (iii) |
$\text{a}\leq\text{b}\ \text{and}\ \text{b}\leq\text{c}\Rightarrow\text{a}\leq\text{c}$ which is true. |
$\therefore$ |
R is transitive. |
Therefore, R is reflexive and transitive but not symmetric. View full question & answer→Question 162 Marks
If f : R → R is defined by f(x) = 3x + 2, find f(f(x)).
Answerf(f(x)) = f(3x + 2)
= 3(3x + 2) + 2
= 9x + 6 + 2
= 9x + 8
View full question & answer→Question 172 Marks
If $f : R → R, g : R → R$ are given by $f(x) = (x + 1)^2$ and $g(x) = x^2 + 1,$ then write the value of $fog(-3).$
Answer$(fog)(-3) = f(g(-3))$
$= f((-3)^2 + 1)$
$= f(10)$
$= (10 + 1)^2$
$= 121$
View full question & answer→Question 182 Marks
Let $*$ be a binary operation on the set $Q$ of rational numbers as follows:
$a * b = ab^2$
Answer$a * b = ab^2$ and $b * a = ba^2 \neq\text{a}*\text{b}$
$\therefore$ operation * is not commutative.
$(a * b) * c = (ab^2) * c = (ab^2)c^2 = ab^2c^2$
And $a * (b * c) = a * (bc^2) = a(bc^2)^2 = ab^2c^4$
Here, $(\text{a}*\text{b})*\text{c}=\text{a}*(\text{b}*\text{c})$
$\therefore$ operation $*$ not is associative.
View full question & answer→Question 192 Marks
Let A and B be two sets, each with a finite number of elements. Assume that there is an injective map from A to B and that there is an injective map from B to A. Prove that there is a bijection from A to B.
AnswerA and B are two non empty sets.
Let f be a function from A to B. It is given that there is injective map from A to B. That means f is one-one function. It is also given that there is injective map from B to A. That means every element of set B has its image in set A.
f is onto function or surjective.
$\therefore$ f is bijective.
If a function is both injective and surjective, then the function is bijective.
View full question & answer→Question 202 Marks
$f: Z → Z$ given by $f(x) = x^3$
Answer$f: Z → Z$ is given by,$f(x) = x^3$
It is seen that for $\text{x},\text{y}\in\text{Z},$
$f(x) = f(y) \Rightarrow x^3 = y^3 \Rightarrow x = y.$
$\therefore f$ is injective. Now, $2\in\text{N}.$
But, there does not exist any element $x$ in domain $Z$ such that $f(x) = x^3 = 2.$
$\therefore f $ is not surjective.
Hence, function f injective but not surjective.
View full question & answer→Question 212 Marks
Let $A = [-1, 1].$ Then, discuss whether the following functions defined on $A$ are one-one, onto or bijective:
$g(x) = |x|$
AnswerLet $g(x_1) = g(x_2)$
$\Rightarrow |x_1| = |x_2|$
$\Rightarrow\ \text{x}_1=\pm\text{x}_2$
So, $g(x)$ is not one-one.
Now, $\text{y}|\text{x}|\Rightarrow\ \text{x}=\pm\text{y}\notin\text{A},\ \forall\ \text{y}\in\text{A}$
So, $g(x)$ is not onto, also, $g(x)$ is not bijective.
View full question & answer→Question 222 Marks
If $f: R → R$ is defined by $f(x) = x^2 – 3x + 2,$ find $f(f(x)).$
AnswerIt is given that $f: R → R$ is defined by $f(x) = x^2 - 3x + 2.$
$f(f(x)) = f(x^2 - 3x + 2)$
$= (x^2 - 3x + 2)^2 - 3(x^2 - 3x + 2) + 2$
$= x^4 + 9x^2 + 4 - 6x^3 - 12x + 4x^2 - 3x^2 + 9x - 6 + 2$
$= x^4 - 6x^3 + 10x^2 - 3x$
View full question & answer→Question 232 Marks
Write the composition table for the binary operation $\times _5 ($multiplication modulo $5) $on the set $S = \{0, 1, 2, 3, 4\}.$
AnswerHere,
$1 \times _51 =$ Remainder obtained by dividing $1 \times 1$ by $5 = 1$
$3 \times _54 =$ Remainder obtained by dividing $3 \times 4$ by $5 = 2$
$4 \times _54 =$ Remainder obtained by dividing $4 \times 4$ by $5 = 1$
So, the composition table is as follows:
| $\times _5$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
| $0$ |
$0$ |
$0$ |
$0$ |
$0$ |
$0$ |
| $1$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
| $2$ |
$0$ |
$2$ |
$4$ |
$1$ |
$3$ |
| $3$ |
$0$ |
$3$ |
$1$ |
$4$ |
$2$ |
| $4$ |
$0$ |
$4$ |
$3$ |
$2$ |
$1$ |
View full question & answer→Question 242 Marks
Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective:
{(a, b): a is a person, b is an ancestor of a}
Answerg = {(a, b): a is a person, h is an ancestor of a}
Since,the ordered map (a, b) does not map 'a' - a person to a living person.
Therefore, g is not a function.
View full question & answer→Question 252 Marks
Let * be a binary operation on the set Q of rational numbers as follows:
a * b = a + ab
Answera * b = a + ab = a(1 + b) and b * a = b + ba = b(1 + a) $\neq\text{a}*\text{b}$Therefore, operation * is not commutative.
(a * b) * c = (a + ab) * c = (a + ab) + (a + ab)c
And a * (b * c) = a * (b + bc) = a + a(b + bc)
Here, $(\text{a}*\text{b})*\text{c}\neq\text{a}*(\text{b}*\text{c})$
$\therefore$ operation * is not associative.
View full question & answer→Question 262 Marks
Write the domain of the real function $\text{f(x)}=\sqrt{\text{x}-[\text{x}]}.$
AnswerWe have, $\text{f(x)}=\sqrt{\text{x}-[\text{x}]}$
The domain of will be real only if
$\text{x}-[\text{x}]\geq0$
⇒ Domain of f = x for all $\text{x}\in\text{R}$
$\therefore$ Domain of f = R
$[\because\ \text{f(x)}=\text{x}-[\text{x}]=\text{x}\ \forall\ \text{x}\in\text{R}]$
View full question & answer→Question 272 Marks
If A = {1, 2, 3, 4} define relations on A which have properties of being:
Symmetric but neither reflexive nor transitive.
AnswerThe relation on A having properties of being symmetric, but neither reflexive nor transitive is,
R = {(1, 2), (2, 1)}
The relation R on A is neither reflexive nor transitive, but symmetric.
View full question & answer→Question 282 Marks
Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.
Answer
| R = {(1, 2), (2, 1)}, so $(\text{a},\text{a}),(1,1)\notin\text{R}.$ |
$\therefore$ |
R is not reflexive. |
| Also if $(\text{a},\text{b})\in\ \text{then}\ (\text{b},\text{a})\in\text{R}$ |
$\therefore$ |
R is symmetric. |
| Now $(\text{a},\text{b})\in\text{R}\ \text{and}\ (\text{b},\text{c})\in$ then does not imply $(\text{a},\text{c})\notin\text{R}$ |
$\therefore$ |
R is not transitive. |
Therefore, R is symmetric but neither reflexive nor transitive. View full question & answer→Question 292 Marks
The following defines a relation on N:
xy is square of an integer, $\text{x, y}\in\text{N}$
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 $\text{a, b, c}\in\text{A.}$xy is square of an integer, $\text{x, y}\in\text{N}$
$(\text{x, y})\in\big\{(1, 1), (2, 2), (4, 1), (1, 4), (3, 3), (9, 1), (1, 9), (4, 4), (2, 8), \$8, 2), (16, 1), (1, 16), .....\big\}$ $$
This is reflexive as (1, 1), (2, 2), .... are present.
This is also symmetric because if aRb ⇒ bRa, for all $\text{a, b}\in\text{N.}$
This is transitive also because if aRb and bRc ⇒ aRc for all $\text{a, b, c}\in\text{N.}$
View full question & answer→Question 302 Marks
Is g = {(1, 1), (2, 3), (3, 5), (4, 7)} a function? If g is described by $\text{g}(\text{x})=\alpha\text{x}+\beta,$ then what value should be assigned to $\alpha$ and $\beta.$
AnswerYes, g is a function since every element in domain has a unique image.
Now, Let g(x) = ax + b then given,
g(1) = a + b = 1 and,
g(2) = 2a + b = 3
Subtracting g(1) from g(2) gives
(2a + b) - (a + b) = a = 2 and Substituting it into g(1)
We have b = -1.
View full question & answer→Question 312 Marks
$f: R → R$ given by $f(x) = x^2$
Answer$f: R → R$ is given by,
$f(x) = x^2$
It is seen that $f(-1) = f(1) = 1$, but $-1\neq1.$
$\therefore$ f is not injective.
Now, $-2\in\text{R}.$ But, there does not exist any element $\text{x}\in\text{R}$ such that $f(x) = x^2 = -2.$
$\therefore$ f is not surjective.
Hence, function f is neither injective nor surjective.
View full question & answer→Question 322 Marks
If $f : R → R$ is defined by $f(x) = x^2$, write $f^{-1}(25)$.
Answer$\text { Let } f^{-1}(25)=x \ldots .(1)$
$\Rightarrow f(x)=25$
$\Rightarrow x^2=25$
$\Rightarrow x^2-25=0$
$\Rightarrow(x-5)(x+5)=0$
$\Rightarrow x= \pm 5$
$\Rightarrow f^{-1}(25)=\{-5,5\}[\text { from (1)] }$
View full question & answer→Question 332 Marks
Let the relation R be defined on N by aRb if 2a + 3b = 30. Then write R as a set of ordered pairs.
AnswerIf $\text{a, b}\in\text{N}$ then b must be an even integer so that $\text{a}\in\text{N}$
Hence only possible values for b are 2, 4, 6, 8.
if b = 2, it gives a = 12
if b = 4, it gives a = 9
if b = 6, it gives a = 6
if b = 8, it gives a = 3
Hence $(\text{a, b})\in\big\{(3, 8), (6, 6), (9, 4), (12, 2)\big\}$ $$
View full question & answer→Question 342 Marks
Find fog (2) and gof (1) when : $f: R \rightarrow R ; f(x)=x^2+8$ and $g: R \rightarrow R ; g(x)=3 x^3+1$.
Answer$(f \circ g)(2)=f(g(2))=f\left(3 \times 2^3+1\right)=f(25)=25^2+8=633$
$(g \circ f)(1)=g(f(1))=g\left(1^2+8\right)=g(9)=3 \times 9^3+1=2188$
View full question & answer→Question 352 Marks
Let $R = \{(a, a^3):$ a is a prime number less than $5\}$ be a relation. Find the range of $R.$
AnswerWe have,
$R = \{(a, a^3): a$ is a prime number less than $5\}$
Or,
$R = \{(2, 8), (3, 27)\}$
So, the range of $R$ is $\{8, 27\}.$
View full question & answer→Question 362 Marks
Define an associative binary operation on a set.
AnswerAn operation * on a set A is called associative binary operation if and only if it is a binary operation as well as associative, i.e. it must satisfy the following two conditions:
- $\text{a}\times\text{b}\in\text{A},\forall\text{ a},\text{b}\in\text{A}$ (Binary operation)
- $\text{a}\times\text{b}\times\text{c}=\text{a}\times\text{b}\times\text{c},\forall\text{ a, b, c}\in\text{A}$ (Associative)
View full question & answer→Question 372 Marks
If $f : R → R$ is defined by $f(x) = 10x - 7$, then write $f^{-1}(x).$
AnswerLet $f^{-1}(x) = y ......(1)$
$\Rightarrow f(y) = x$
$\Rightarrow 10y - 7 = x$
$\Rightarrow 10y = x + 7$
$\Rightarrow\ \text{y}=\frac{\text{x}+7}{10}$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{\text{x}+7}{10}[ $from$ (1)]$
View full question & answer→Question 382 Marks
Let $\text{f}:\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\rightarrow\ \text{A}$ be defined by f(x)= sinx. If f is a bijection, write set A.
Answer$\because$ f is a bijection,Co-domain of f = range of f
As $-1\leq\sin\text{x}\leq1,$
$-1\leq\text{y}\leq1$
Therefore, A = [-1, 1]
View full question & answer→Question 392 Marks
Transitive but neither reflexive nor symmetric.
AnswerRelation R = {( x, y) : x > y}
We know that x > x is false. Also x > y but y > x is false and if x > y , y > z this implies x > z.
Therefore, R is transitive, but neither reflexive nor symmetric.
View full question & answer→Question 402 Marks
For each binary operation * defined below, determine whether * is commutative or associative.
On R - {-1}, define $\text{a}*\text{b}=\frac{\text{a}}{\text{b}+1}$
AnswerFor commutativity: $\text{a}*\text{b}=\frac{\text{a}}{\text{b}+1}\ \text{and}\ \text{b}*\text{a}=\frac{\text{b}}{\text{a}+1}\Rightarrow\ \ \text{a}*\text{b}\neq\text{b}*\text{a}$
For associativity: $\text{a}*(\text{b}*\text{c})=\text{a}*\Big(\frac{\text{b}}{\text{c}+1}\Big)=\frac{\text{a}}{\frac{\text{a}}{\text{c}+1}+1}=\frac{\text{a(c + a)}}{\text{b + c}+1}$
Also, $(\text{a}*\text{b})*\text{c}=\Big(\frac{\text{a}}{\text{b}+1}\Big)*\text{c}=\frac{\text{a}/\text{b}+1}{\text{c}+1/\text{c}}=\frac{\text{a}}{(\text{b}+1)(\text{c}+1)}$
$\therefore\ \ \text{a} * \text{(b} * \text{c)}\neq\text{(a} * \text{b)}* \text{c}$
Therefore, the operation * is neither commutative nor associative.
View full question & answer→Question 412 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 identity element in $Q_0.$
AnswerWe have,
$\text{a }^*\text{ b}=\frac{\text{ab}}{2}$ for all $\text{a},\text{b}\in\text{Q}_0$
Let $\text{e}\in\text{Q}_0$ be the identity element with respect to $*.$
By identity property, we have,
$a * e = e * a = a$ for all $\text{a}\in\text{Q}_0$
$\Rightarrow\frac{\text{ae}}{2}=\text{a}\Rightarrow\text{e}=2$
Thus the required identity element is $2.$
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If f(x) = x + 7 and g(x) = x - 7, x ∈ R, write fog (7).
Answer(fog)(7) = f(g(7))
= f(7 - 7)
= f(0)
= 0 + 7
= 7
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Define identity element for a binary operation defined on a set.
AnswerLet S be a non-empty set and * be a binary operation on S.
If there exist an element $\text{e}\in\text{S}$ such that
a * e = e * a = a for all $\text{e}\in\text{S}$
Then e is called the identity element for the binary operation * on S.
'0' is the identity element for '+' on Z
1 is the identity element for '×' on Z.
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Reflexive and transitive but not symmetric.
Answer“is greater or equal to” $\text{R}=\{(\text{x},\text{y}):\text{x}\geq\text{y}\}$
| It is clear that $\text{x}\geq\text{x}$ |
$\therefore$ |
R is reflexive. |
| And $\text{x}\geq\text{y}$ does not imply $\text{y}\geq\text{x}$ |
$\therefore$ |
R is not symmetric. |
| But $\text{x}\geq\text{y},\text{y}\geq\text{z}\Rightarrow\text{x}\geq\text{z}$ |
$\therefore$ |
R is transitive. |
Therefore, R is reflexive and transitive but not symmetric. View full question & answer→Question 452 Marks
Find the total number of binary operations on {a, b}.
AnswerWe have,
S = {a, b}
The total number of binary operation on S = {a, b} in $2^{2^{2}}= 2^4=16$
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Let * be a binary operation on Z defined by a * b = a + b - 4 for all a, b ∈ Z.
Find the identity element in Z.
AnswerLet e be the identity element in Z with respect to * such that
a * e = a = e * a, $\forall\ \text{a}\in\text{Z}$
a * e = a and e * a = a, $\forall\ \text{a}\in\text{Z}$
a + e - 4 = a and e + a - 4 = a, $\forall\ \text{a}\in\text{Z}$
e = 4, $\forall\ \text{a}\in\text{Z}$
Thus, 4 is the identity element in Z with respect to *.
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Let the relation R be defined on the set $A = \{1, 2, 3, 4, 5\}$ by $R = \{(a, b): |a^2 - b^2| < 8\}$. Write R as a set of ordered pairs.
AnswerGiven: $A = \{1, 2, 3, 4, 5\} R = \{(a, b): |a^2 - b^2| < 8\}R = \{(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (3, 3), (3, 4), (4, 3), (4, 4), (5, 5)\}$
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Write the smallest reflexive relation on set $A=\{1,2,3,4\}$.
AnswerThe smallest reflexive relation $R$ on any set $A$ is the identity relation $I_A$ on the set $A$. We are given, $A=\{1,2,3,4\}$
$\therefore R=\{(1,1),(2,2),(3,3),(4,4)\}$
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If $f : R → R$ is given by $f(x) = x^3,$ write $f^{-1}(1).$
AnswerLet $f^{-1}(1) = x .....(1)$
$\Rightarrow f(x) = 1$
$\Rightarrow x^3 = 1$
$\Rightarrow x^3 - 1 = 0$
$\Rightarrow (x - 1)(x^2 + x + 1) = 0 [$Using the identity$: a^3 - b^3 = (a - b)(a^2 + ab + b^2)]$
$\Rightarrow x = 1 (\text{as x}\in\text{R})$
$\Rightarrow f^{-1}(1) = {1} [$from $(1)]$
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Determine whether the following operations define a binary operation on the given set or not:
'*' on Q defined by $\text{a}\ ^*\ \text{b}=\frac{\text{a}-1}{\text{b}+1}$ for all $\text{a, b}\in\text{Q.}$
AnswerIf a = 2 and b = -1 in Q,
$\text{a}\ ^*\ \text{b}=\frac{\text{a}-1}{\text{b}+1}$
$=\frac{2-1}{-1+1}$
$=\frac{1}{0}$ [which is not defined]
For a = 2 and b = -1,
$\text{a}\ ^*\ \text{b}\notin\text{Q}$
Therefore,
* is a binary operation on Q.
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