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 $2.$
View full question & answer→Question 422 Marks
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
View full question & answer→Question 432 Marks
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.
View full question & answer→Question 442 Marks
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 $2^{2^{2}}= 2^4=16$
View full question & answer→Question 462 Marks
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 *.
View full question & answer→Question 472 Marks
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: $A = \{1, 2, 3, 4, 5\} R = \{(a, b): |a^2 - b^2| < 8\}R = \{(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (3, 3), (3, 4), (4, 3), (4, 4), (5, 5)\}$
View full question & answer→Question 482 Marks
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, $A=\{1,2,3,4\}$
$\therefore R=\{(1,1),(2,2),(3,3),(4,4)\}$
View full question & answer→Question 492 Marks
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 $(1)]$
View full question & answer→Question 502 Marks
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.
View full question & answer→Question 512 Marks
If the binary operation o is defined by a o b = a + b - ab on the set Q - {-1} of all rational numbers other than 1, shown that o is commutative on Q - [1].
AnswerLet $\text{a, b}\in\text{Q}-1.$ Then,
a o b = a + b - ab
= b + a - ba
= b o a
Therefore,
a o b = b o a, $\forall\ \text{a, b}\in\text{Q}-1$
Thus, o is commutative on Q - {1}.
View full question & answer→Question 522 Marks
Let * be a binary operation on Q - {-1} defined by a * b = a + b + ab for all a, b ∈ Q - {-1}. Then,
Find the identity element in Q − {−1}.
AnswerWe have,
a * b = a + b + ab for all a, b ∈ Q - {-1}
Let e be identity element with respect to *.
By identity property,
a * e = a = e * a for all a ∈ Q - {-1}
⇒ a + e + ae = a
⇒ e(1 + a) = 0 ⇒ e = 0 $[\because\ 1+\text{a}\neq0\text{ as }\text{a}\neq-1]$
e = 0 is the identity element with respect to *.
View full question & answer→Question 532 Marks
Show that the relation R in the set A of all the books in a library of a college, given by R = {(x, y) : x and y have same number of pages} is an equivalence relation.
Answer
| Books x and x have same number of pages $\Rightarrow(\text{x},\text{x})\in\text{R}$ |
$\therefore$ |
R is reflexive. |
| If $(\text{x},\text{y})\in\text{R}\Rightarrow(\text{y},\text{x})\in\text{R},$so (x, y) = (y, x) |
$\therefore$ |
R is symmetric. |
| Now if $(\text{x},\text{y})\in\text{R},(\text{y},\text{z})\in\text{R}\Rightarrow(\text{x},\text{z})\in\text{R}$ |
$\therefore$ |
R is transitive. |
Therefore, R is an equivalence relation. View full question & answer→Question 542 Marks
Determine whether the following operations define a binary operation on the given set or not:
'*' on N defined by a * b = a + b - 2 for all $\text{a, b}\in\text{N.}$
AnswerIf a = 1 and b = 1, a * b = a + b - 2 = 1 + 1 - 2$=0\notin\text{N}$
Thus, there exist a = 1 and b = 1 such that $\text{a}\ ^*\ \text{b}\notin\text{N}$ So, * is not a binary operation on N.
View full question & answer→Question 552 Marks
Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y) : y is divisible by x}
AnswerR = {( x, y) : y is divisible by x} in A = {1, 2, 3, 4, 5, 6}
Clearly R = (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (5, 5), (6, 6)
| Now (x, x) i.e., (1, 1), (2, 2) and $(3,3)\in\text{R},$ |
$\therefore$ |
R is reflexive. |
| Again $(\text{x},\text{y})\ \text{i.e.},((1,2))\in\text{R}\ \text{but}\ (\text{y},\text{x})\notin\text{R}$ |
$\therefore$ |
R is not symmetric. |
| Also $(1,4)\in\text{R}\ \text{and}\ (4,4)\in\text{R}\ \text{but}(1,4)\in\text{R},$ |
$\therefore$ |
R is transitive. |
Therefore, R is reflexive and transitive but not symmetric. View full question & answer→Question 562 Marks
Which of the following functions from $A$ to $B$ are one-one and onto$?$
$f_2 = \{(2, a), (3, b), (4, c)\}; A = \{2, 3, 4\}, B = \{a, b, c\}$
Answer$f_2 = \{(2, a), (3, b), (4, c)\} A = \{2, 3, 4\}, B = \{a, b, c\}$ It in not clear that different elements of $A$ have different images in $B.$
$\therefore f_2$ in not one-one.
Again, each element of $B$ is the image of some element of $A.$
$\therefore f_2$ in not on to.
View full question & answer→Question 572 Marks
Let $R_0$ denote the set of all non-zero real numbers and let $A = R_0 \times R_0.$ If $'*'$ is a binary operation on adefined by,
$(a, b) * (c, d) = (ac, bd)$ for all $(a, b), (c, d) ∈ A$
Find the identity element in $A$
AnswerLet $(x, y)$ be the identity element in $\text{A}\forall\text{ x, y}\in\text{A}$. Then,
$(a, b) * (x, y) = (a, b) = (x, y) * (a, b)$
Implies that $(a, b) * (x, y) = (a, b)$ and $(x, y) * (a, b) = (a, b)$
Implies that $(ax, by) = (a, b)$ and $(xa, yb) = (a, b)$
Implies that $x = 1$ and $y = 1$
Thus, $(1, 1)$ is the identity element of $A.$
View full question & answer→Question 582 Marks
Let * be a binary operation defined by a * b = 3a + 4b − 2. Find 4 * 5.
AnswerGiven: a * b = 3a + 4b - 2
Here,
4 * 5 = 3(4) + 4(5) - 2
= 12 + 20 - 2
= 30
View full question & answer→Question 592 Marks
Which one of the following graphs represents a function?
-
-
AnswerFigure (a) represents a function f : R → R
Whereas fig (b) does not represent a function.
View full question & answer→Question 602 Marks
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer:
$f: R → R$ defined by $f(x) = 3 – 4x$
Answerf: R → R is defined as f(x) = 3 - 4x. Let $\text{x}_1,\text{x}_2\in\text{R}$ such that $f(x_1) = f(x_2) \Rightarrow 3 - 4x_1 = 3 - 4x_2 \Rightarrow -4x_1 = -4x_2 \Rightarrow x_1 = x_2 $
$\therefore$ f is one-one. For any real number (y) in R, there exists $\frac{3-\text{y}}{4}$ in R such that $f\Big(\frac{3-\text{y}}{4}\Big)=3-4\Big(\frac{3-\text{y}}{4}\Big)=\text{y}.$
$\therefore$ f is onto. Hence, f is bijective.
View full question & answer→Question 612 Marks
Define a commutative binary operation on a set.
AnswerCommutativity: Let S be a non-empty set. A function F: S × S → S is said to be binary operation on S.Mathematically: Let * be a binary operation. It is said to be commutative binary operation if it satisfies commutativity with respect to *.
That is, if $\text{a, b}\in\text{S}$, then
a * b = b * a for all $\text{a, b}\in\text{S}$.
View full question & answer→Question 622 Marks
Find the number of all onto functions from the set A = {1, 2, 3, ..., n} to itself.
AnswerWe know that every onto function from A to itself is one-one.
Therefore, the number of one-one functions = number of bijections =n!
View full question & answer→Question 632 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 $R,$ define by $a * b = ab^2.$
Here, $Z^+$ denotes the set of all non-negative integers.
Answer$\text{a, b}\in\text{R}$Implies that $\text{a, b}^2\in\text{R}$
Implies that $\text{ab}^2\in\text{R}$
Implies that $\text{a}\ ^*\ \text{b}\in\text{R}$
Thus, $*$ is a binary operation on $R$.
View full question & answer→Question 642 Marks
Define an equivalence relation.
AnswerA relation R on a set A is said to be equivalence relation on a if R is:
Reflexive, Symmetric and Transitive.
R = {(x, y): x = y} on the set of real numbers is an equivalence relation.
View full question & answer→Question 652 Marks
Determine whether each of the following relations are reflexive, symmetric and transitive:
Relation R in the set A = {1, 2, 3, ..., 13, 14} defined as R = {(x, y): 3x – y = 0}
AnswerR = {( x, y): 3x − y = 0}, in A = {1, 2, 3, 4, 5, 6, ……13, 14}
Clearly R = {(1, 3), (2, 6), (3, 9), (4, 12)}
| Since, $(\text{x},\text{x})\notin\text{R},$ |
$\therefore$ |
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 $(1,3)\in\text{R}\ \text{and}\ (3,9)\in\text{R}\ \text{but}\notin\text{R},$ |
$\therefore$ |
R is not transitive. |
Therefore, R is neither reflexive, nor symmetric and nor transitive. View full question & answer→Question 662 Marks
If f : R → R be defined by $\text{f(x)} = (3 - \text{x}^3)^\frac{1}{3},$ then find fof(x).
Answerf : R → R defined by $\text{f(x)}=(3-\text{x}^3)^\frac{1}{3}$
$\therefore$ fof(x) = f(f(x))
$=\text{f}(3-\text{x}^3)^\frac{1}{3}$
$=\bigg\{3-\Big[(3-\text{x}^3)^\frac{1}{3}\Big]^3\bigg\}^\frac{1}{3}$
$=\{3-3+\text{x}^3\}^\frac{1}{3}$
$\therefore$ fof(x) = x
View full question & answer→Question 672 Marks
The following defines a relation on N:
$\text{x} +\text{y} = 10,\text{x, y}\in\text{N}$ $$
Determine which of the above relations are reflexive, symmetric and transitive.
AnswerA relation R in A is said to be reflexive if aRa for all $\text{a}\in\text{A},$ R is symmetric if aRb ⇒ bRa, for all $\text{a, b}\in\text{A}$ and it is said to be transitive if aRb and bRc ⇒ aRc for all $\text{a, b, c}\in\text{A.}$$\text{x} +\text{y} = 10,\text{x, y}\in\text{N}$
$(\text{x, y})\in\big\{(1, 9), (9, 1), (2, 8), (8, 2), (3, 7), (7, 3),\\ (4, 6), (6, 4), (5, 5)\big\}$ $$
This is not reflexive as (1, 1), (2, 2), .... are absent.
This only follows the condition of symmetric set as $(1,9)\in\text{R}$ also $(9,1)\in\text{R}$ Similarly other cases are also satisfy the condition.
This is not transitive because {(1, 9), (9, 1)} $\in\text{R}$ but (1, 1) is absent.
View full question & answer→Question 682 Marks
Let * be a binary operation on Z defined by a * b = a + b - 4 for all a, b ∈ Z.
Find the invertible elements in Z.
AnswerLet $\text{a}\in\text{Z}$ and $\text{b}\in\text{Z}$ be the inverse of a. Then,
a * b = e = b * a
a * b = e and b * a = e
a + b - 4 = 4 and b + a - 4 = 4
$\text{b}=8-\text{a}\in\text{Z}$
Thus, 8 - a is the inverse of $\text{a}\in\text{Z.}$
View full question & answer→Question 692 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 = a - b.$
Here, $Z^+$ denotes the set of all non-negative integers.
AnswerOn $Z^+, *$ is defined by $a * b = a - b$
It is not a binary operation as the image of $(1, 2)$ under $*$ is $1 * 2 = 1 - 2$
$=-1\notin\text{Z}^{+}$
View full question & answer→Question 702 Marks
Let f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof.
Answerf = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}
Now, f(1) = 2, f(3) = 5, f(4) =1 and g(1) = 3, g(2) = 3, g(5) =1
(gof)(n) = g[f(x)] = g[f(1)] = g(2) = 3
g[f(3)] = g(5) = 1 and g[f(4)] = g(1) = 3
Hence, gof = {(1, 3), (3, 1), (4, 3)}
View full question & answer→Question 712 Marks
Let * be a binary operation on the set Q of rational numbers as follows:
$\text{a} * \text{b} = \frac{\text{ab}}{4}$
Answer$\text{a}*\text{b}=\frac{\text{ab}}{4}=\frac{\text{ba}}{4}=\text{b}*\text{a}$
$\therefore$ operation * is commutative.
$(\text{a}*\text{b})*\text{c}=\frac{\text{ab}}{4}*\text{c}=\frac{\frac{\text{ab}}{4}\text{c}}{4}=\frac{\text{abc}}{16}$
And $\text{a}*(\text{b}*\text{c})=\text{a}*\frac{\text{bc}}{4}=\frac{\text{a}\frac{\text{bc}}{4}}{4}=\frac{\text{abc}}{16}$
Here, $(\text{a}*\text{b})*\text{c}=\text{a}*(\text{b}*\text{c})$
$\therefore$ operation * is associative.
View full question & answer→Question 722 Marks
If the binary operation * on the set Z is defined by a * b = a + b - 5, the find the identity element with respect to *.
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 - 5 = a and e + a - 5 = a, $\forall\ \text{a}\in\text{Z}$
e = 5, $\forall\ \text{a}\in\text{Z}$
Thus, 5 is the identity element in Z with respect to *.
View full question & answer→Question 732 Marks
If A = {3, 5, 7} and B = {2, 4, 9} and R is a relation given by "is less than", write R as a set ordered pairs.
AnswerSince, R = x, y: x, y $\in\text{N}$ and x < y,
Hence, R = {(3, 4), (3, 9), (5, 9), (7, 9)}
View full question & answer→Question 742 Marks
Give an example of a relation which is,Reflexive and transitive but not symmetric.
AnswerLet R be the relation on A such that
R = {(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 3)}
Clearly, the relation R on A is reflexive and transitive, but not symmetric.
View full question & answer→Question 752 Marks
Let $f : R → R$ and $g : R → R$ be defined by $f(x) = x^2$ and $g(x) = x + 1$. Show that fog $≠$ gof.
AnswerGiven, $f : R → R$ and $g : R → R.$
So, the domains of f and g are the same.
$(fog)(x) = f(g(x)) = f(x + 1) = (x + 1)^2 = x^2 + 1 + 2x$
$(gof)(x) = g(f(x)) = g(x^2) = x^2 + 1$
So, $fog ≠ gof.$
View full question & answer→Question 762 Marks
Let A = {a, b, c} and the relation R be defined on A as follows: R = {(a, a), (b, c), (a, b)}. Then, write minimum number of ordered pairs to be added in R to make it reflexive and transitive.
AnswerWe have,
A = {a, b, c} and R = {(a, a), (b, c), (a, b)}.
R can be a reflexive relation only when elements (b, b) and (c, c) are added to it.
R can be a transitive relation only when the element (a, c) is added to it.
So, the minimum number of ordered pairs to be added in R is 3.
View full question & answer→Question 772 Marks
If f, g : R → R be two functions defined as f(x) = |x| + x and g(x) = |x| – x, $\forall\ \text{x}\in\text{R}.$ Then find fog and gof. Hence find fog(-3), fog(5) and gof(-2).
Answer$\text{fog(x)}=\begin{cases}0,\ \text{x}\geq0\\-4\text{x},\ \text{x}<0\end{cases}$gof(x) = 0, for all x fog(-3) = 12
fog(5) = 0
gof(-2) = 0
View full question & answer→Question 782 Marks
Let $\text{f}:\text{R}-\Big\{-\frac{3}{5}\Big\}\rightarrow\ \text{R}$ be a function defined as $\text{f(x)}=\frac{2\text{x}}{5\text{x}+3}.$ Write $f^{-1}$: Range of $\text{f}\rightarrow\ \text{R}-\Big\{-\frac{3}{5}\Big\}.$
AnswerLet $f^{-1}(x) = y$ ......(1) ⇒ f(y) = x$\Rightarrow\ \frac{2\text{y}}{5\text{y}+3}=3\text{x}$
$⇒ 2y = 5xy + 3x ⇒ 2y - 5xy = 3x ⇒ y(2 - 5x) = 3x \Rightarrow\ \text{y}=\frac{3\text{x}}{2-5\text{x}}$ $\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{3\text{x}}{2-5\text{x}}$ [from 1]
View full question & answer→Question 792 Marks
Define a reflexive relation.
AnswerA relation R on a set A is said to be reflexive if every element of A is related to itself.
Mathematically, reflexive relation is written as R = {(a, a): for all $\text{a}\in\text{A}$}
For example if A = {1, 2, 3}, then a reflexive relation on A will be R = {(1, 1), (2, 2), (3, 3)}
View full question & answer→Question 802 Marks
If A = {1, 2, 3, 4} define relations on A which have properties of being:
Reflexive, symmetric and transitive.
AnswerThe relation on A having properties of being symmetric, reflexive and transitive is,
R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1)}
The relation R is an equivalence relation on A.
View full question & answer→Question 812 Marks
Let $R_0$ denote the set of all non-zero real numbers and let $A = R_0 \times R_0$. If $'*'$ is a binary operation on adefined by,
$(a, b) * (c, d) = (ac, bd)$ for all $(a, b), (c, d) ∈ A$
Find the invertible element in $A.$
AnswerLet $(m, n)$ be the inverse of $(\text{a, b})\forall(\text{a, b})\in\text{A}$. Then,
$(a, b) * (m, n) = (1, 1)$
Implies that $(am, bn) = (1, 1)$
Implies that $am = 1\ \&\ bn = 1$
Implies that $\text{m}=\frac{1}{\text{a}}\text{ and }\text{n}=\frac{1}{\text{b}}$
Thus, $\Big(\frac{1}{\text{a}},\frac{1}{\text{b}}\Big)$ is the inverse of $(\text{a, b})\forall(\text{a, b})\in\text{A}$.
View full question & answer→Question 822 Marks
If $f : C → C$ is defined by $f(x) = (x - 2)^3$, write $f^{-1}(-1)$.
AnswerLet $f^{-1}(1) = x$ ......(1)
$\Rightarrow f(x) = -1$
$\Rightarrow (x - 2)^3 = -1$
$\Rightarrow\ \text{x}-2=-1\ \text{or }-\omega\text{ or }-\omega^2$
as the roots of $(-1)^\frac{1}{3}$ are $-1,-\omega\text{ and }-\omega^2,$ where $\omega=\frac{1\pm\text{i}\sqrt{3}}{2}$
$\Rightarrow\ \text{x}=-1+2\text{ or }2-\omega\text{ or }2-\omega^2=1,2-\omega,2-\omega$
$\Rightarrow\ \text{f}^{-1}(-1)=1,2-\omega,2-\omega^2$ [from 1]
View full question & answer→Question 832 Marks
Let * be the binary operation on N defined by a * b = H.C.F. of a and b. Is * commutative? Is * associative? Does there exist identity for this binary operation on N?
Answera * b = H.C.F. of a and b.
- a * b = H.C.F. of a and b = H.C.F. of b and a = b * a
Therefore, operation * is commutative.
- (a * b) * c = (H.C.F. of a and b) * c = H.C.F. of (H.C.F. of a and b) and c
= H.C.F. of a, b and c = a * (b * c)
Therefore, the operation is associative.
$1*\text{a}=\text{a}*1\neq\text{a}.$ View full question & answer→Question 842 Marks
Let $f: X → Y$ be an invertible function. Show that f has unique inverse. $($Hint: suppose $g_1$ and $g_2$ are two inverses of f. Then for all $y \in Y, fog_1(y) = 1_Y(y) = fog_2(y).$ Use one-one ness of $f).$
AnswerGiven: $f: X → Y$ be an invertible function.
Thus f is $1 – 1$ and onto and therefore $f^{−1}$ exists.
Let $g_1$ and $g_2$ be two inverses of f . Then for all $\text{y}\in\text{Y},$
$fog_1(y) = I_y(y) = fog_2(y) \therefore fog_1(y) = fog_2(y)$
$\Rightarrow f[g_1(y)] = f[g_{2(y)}] \Rightarrow g_1(y) = g_2(y)$
$\therefore$ The inverse is unique and hence f has a unique inverse.
View full question & answer→Question 852 Marks
Determine whether the following operations define a binary operation on the given set or not:$'\odot'$ on N defined by $\text{a}\odot\text{b}=\text{a}^{\text{b}}+\text{b}^{\text{a}}$ for all $\text{a, b}\in\text{N.}$
AnswerLet $\text{a, b}\in\text{N.}$ Then,
$\text{a}^{\text{b}},\text{b}^{\text{a}}\in\text{N}$
$\Rightarrow\ \text{a}^{\text{b}}+\text{b}^{\text{a}}\in\text{N}$ $\big[\because$ Addition is binary operation on N$\big]$
$\Rightarrow\ \text{a}\odot\text{b}\in\text{N}$
Thus, $\odot$ is a binary operation on N.
View full question & answer→Question 862 Marks
Let '*' be a binary operation on N defined by a * b = 1.c.m. (a, b) for all $\text{a, b}\in\text{N}$
Find 2 * 4, 3 * 5, 1 * 6.
Answera * b = 1.c.m. (a, b)
2 * 4 = 1.c.m. (2, 4)
= 4
3 * 5 = 1.c.m. (3, 5)
= 15
1 * 6 = 1.c.m. (1, 6)
= 6
View full question & answer→Question 872 Marks
Write the identity element for the binary operation * on the set $R_0$ of all non-zero real numbers by the rule $\text{a}\times\text{b}=\frac{\text{ab}}{2}$ for all $a, b \in R_0$.
Answer$\because\ \text{a}\times\text{b}=\frac{\text{ab}}{2}$ for all $a, b \in R_0$ Let e be the identity element, then
a * e = a
$\Rightarrow\frac{\text{ae}}{2}=\text{a}\ \Rightarrow\text{e}=2$
Thus, e = 2 is the identity element with respect to *.
View full question & answer→Question 882 Marks
Let C denote the set of all complex numbers. A function $f : C → C$ is defined by $f(x) = x^3$. Write $f^{-1}(1)$.
Answer$f : R → R$ defined by $f(x) = x^3$
$\therefore f^{-1}(x^3) = x$
$\Rightarrow\ \text{f}^{-1}(1)=\{1,\omega,\omega^2\}$ $[\because\ \sqrt[3]{1}=\{1,\omega,\omega^2\}]$
View full question & answer→Question 892 Marks
Find gof and fog when $f : R → R$ and $g : R → R$ are defined by:
$f(x) = 2x + x^2$ and $g(x) = x^3$
AnswerGiven: $f : R → R$ and $g : R → R$
Therefore, $gof : R → R$ and $fog : R → R$
$f(x) = 2x + x^2$ and $g(x) = x^3$
$gof(x) = g(f(x)) = g(2x + x^2)$
$gof(x) = g(2x + x^2)^3$
$fog(x) = f(g(x)) = f(x^3)$
$\therefore fog(x) = 2x^3 + x^6$
View full question & answer→Question 902 Marks
Consider $f: R_+→ [4, \infty )$ given by $f(x) = x^2 + 4$. Show that f is invertible with the inverse $f^{–1}$ of f given by $f^{-1}(\text{y})=\sqrt{\text{y}-4},$ where $R_+$ is the set of all non-negative real numbers.
AnswerConsider $f:\text{R}_{+}\rightarrow[4,\infty]$ and $f(x) = x^2 + 4$.
Let $\text{x}_1,\text{x}_2\in\text{R}\rightarrow[4,\infty],\text{ then }f(\text{x}_1)=\text{x}_{1}^{2}+4\text{ and }f(\text{x}_2)=\text{x}_{2}^{2}+4$
$\Rightarrow\ \text{x}_{1}^{2}+4=\text{x}_{2}^{2}+4\Rightarrow\text{x}_1=\text{x}_2\ \ \ \ \ \therefore f\text{ is one-one.}$
Now $\text{y}=\text{x}^2+4\Rightarrow\text{x}=\sqrt{\text{y}-4}$ as x > 0
$\therefore\ \ f\left(\sqrt{\text{y}-4}\right)=\left(\sqrt{\text{y}-4}\right)^2+4=\text{y}\ \ f(\text{x})=\text{y}\ \ \ \ \therefore\ f\text{ is onto.}$
Therefore, f(x) is invertible and $f^{-1}(\text{y})=\sqrt{\text{y}-4}.$
View full question & answer→Question 912 Marks
If the mappings f and g are given by f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)}, write fog.
AnswerWe are given that, f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)} Now, the domain of g is {2, 5, 1} We know that, fog(x) = f{g(x)}$\therefore$ fog(2) = f{g(2)} = f(3) = 5
$\therefore$ fog(5) = f{g(5)} = f(1) = 2
$\therefore$ fog(1) = f{g(1)} = f(3) = 5
Therefore, fog = {(2, 5), (5, 2), (1, 5)}
View full question & answer→Question 922 Marks
Write the identity element for the binary operation * defined on the set R of all real numbers by the rule $\text{a}\times\text{b}=\frac{3\text{ab}}{7}\ \forall\text{ a, b}\in\text{R}$.
AnswerWe have,
$\text{a}\times\text{b}=\frac{3\text{ab}}{7}$
Let e be the identity element with respect to *. Then
a * e = a
$\Rightarrow\frac{3\text{ae}}{7}=\text{a}\ \Rightarrow\text{e}=\frac{7}{3}$
View full question & answer→Question 932 Marks
Let $f : R - {-1} → R - {1}$ be given by $\text{f(x)}=\frac{\text{x}}{\text{x}+1}.$ Write $f^{-1}(x)$.
Answer$f : R - [-1] → R - [1]$ given by $\text{f(x)}=\frac{\text{x}}{\text{x}+1}$
$\Rightarrow\ \text{f}^{-1}\Big(\frac{\text{x}}{\text{x}+1}\Big)=\text{x}$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{\text{x}}{1-\text{x}}$
$\because$ Let $\frac{\text{x}}{\text{x}+1}=\text{y}$
$\Rightarrow\ \text{x}=\text{xy}+\text{y}$
$\Rightarrow\ \text{x}(1-\text{y})=\text{y}$
$\Rightarrow\ \text{x}=\frac{\text{y}}{1-\text{y}}$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{\text{x}+7}{10}$
View full question & answer→Question 942 Marks
Let * be a binary operation on the set Q of rational numbers as follows:
a * b = a – b
Answera * b = a - b = -(b - a) = -b * a
$\therefore$ operation is not commutative.
(a * b) * c = (a - b) * c = (a - b) - c = a - b - c
And a * b (b * c) = a * (b - c) = a - (b - c) = a - b + c
Here, $(\text{a}*\text{b})*\text{c}\neq\text{a}*(\text{b}*\text{c})$
$\therefore$ operation * is not associative.
View full question & answer→Question 952 Marks
For each binary operation * defined below, determine whether * is commutative or associative.
On Q, define $\text{a} * \text{b} =\frac{\text{ab}}{2}$
AnswerFor commutativity: $\text{a}*\text{b}=\frac{\text{ab}}{2}\ \text{and}\ \text{b}*\text{a}=\frac{\text{ba}}{2}=\frac{\text{ab}}{2}=\text{a}*\text{b}$
For associativity: $\text{a}*(\text{b}*\text{c})=\text{a}*\Big(\frac{\text{bc}}{2}\Big)=\frac{\text{abc/2}}{2}=\frac{\text{abc}}{4}$
Also, $(\text{a}*\text{b})*\text{c}=\Big(\frac{\text{ab}}{2}\Big)*\text{c}=\frac{\text{abc}/2}{2}=\frac{\text{abc}}{4}$
$\therefore$ a * (b * c) = (a * b) * c
Therefore, the operation * is commutative and associative.
View full question & answer→Question 962 Marks
If $f : R → R$ is defined by $f(x) = x^2 - 3x + 2$, write f{f(x)}.
AnswerWe have, $f(x)=x^2-3 x+2$
$\therefore f\{f(x)\}=f\left(x^2-3 x+2\right)$
$=\left(x^2-3 x+2\right)^2-3\left(x^2-3 x+2\right)+2$
$=x^4+9 x^2+4-6 x^3-12 x+4 x^2-3 x^2+9 x-6+2$
$=x^4-6 x^3+10 x^2-3 x$
$\therefore f\{f(x)\}=x^4-6 x^3+10 x^2-3 x$
View full question & answer→Question 972 Marks
Let S = {a, b, c}. Find the total number of binary operations on S.
AnswerNumber of binary operations on a set with n elements is $n^2$.
Here, S = {a, b, c}
Number of elements in S = 3
Number of binary operations on a set with 3 elements is $3^{3^{2}}=3^9$
View full question & answer→Question 982 Marks
Let $f : R → R^+$ be defined by $f(x) = ax, a > 0$ and $\text{a}\neq1.$ Write $f^{-1}(x).$
AnswerLet $f^{-1}(x) = y .......(1)$
$\Rightarrow f(y) = x$
$\Rightarrow a^y = x$
$\Rightarrow\ \text{y}=\log_\text{a}\text{x}$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\log_\text{a}\text{x} [$from $(1)]$
View full question & answer→Question 992 Marks
Find which of the binary operations are commutative and which are associative.
Show that none of the operations given above has identity.
AnswerLet the identity be I.
- $\text{a}*\text{I}=\text{a - I}\neq\text{a}$
- $\text{a}*\text{I}=\text{a}^2-\text{I}^2\neq\text{a}$
- $\text{a}*\text{I}=\text{a + aI}\neq\text{a}$
- $\text{a}*\text{I}=(\text{a - I})^2\neq\text{a}$
- $\text{a}*\text{I}=\frac{\text{aI}}{4}\neq\text{a}$
- $\text{a}*\text{I}=\text{aI}^2\neq\text{a}$
Therefore, none of the operations given above has identity. View full question & answer→Question 1002 Marks
$f: N → N$ given by $f(x) = x^3$
Answer$f: R → R$ is given by,$f(x) = x^3$
It is seen that for $\text{x},\text{y}\in\text{N},$
$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 $N$ such that $f(x) = x^3 = 2.$
$\therefore f$ is not surjective.
Hence, function f injective but not surjective.
View full question & answer→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.
|
*
|
1
|
2
|
3
|
4
|
5
|
|
1
|
1
|
2
|
3
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4
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5
|
|
2
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2
|
2
|
6
|
4
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10
|
|
3
|
3
|
x
|
3
|
12
|
15
|
|
4
|
4
|
4
|
12
|
4
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20
|
|
5
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5
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x
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15
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20
|
5
|
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\}$
View full question & answer→Question 1512 Marks
Write whether f : R → R, given by $\text{f(x)}=\text{x}+\sqrt{\text{x}^2},$ is one-one, many-one, onto or into.
Answerf : R → R given by $\text{f(x)}=\text{x}+\sqrt{\text{x}^2}$
$\because\ \text{f(x)=}\begin{cases}2\text{x};&\text{for x}>0\\0;&\text{for x}<0\end{cases}$
$\therefore$ f is many-one function.
View full question & answer→Question 1522 Marks
Let D be the domain of the real valued function f defined by $\text{f}(\text{x})=\sqrt{25-\text{x}^2}.$ Then, write D.
AnswerConsider the given function, $\text{f}(\text{x})=\sqrt{25-\text{x}^2}$
For f(x) to be real, the term inside the square root can’t be negative
i.e., $25-\text{x}^2\geq0$
$\Rightarrow\ \text{x}^2\leq25$
$\Rightarrow\ 5\leq\text{x}\leq-5$
Therefore, the domain of the function, f(x) is given by D = [-5, 5]
View full question & answer→Question 1532 Marks
For each binary operation * defined below, determine whether * is commutative or associative.
On Z, define a * b = a – b
AnswerFor Commutativity: a * b = a - b and b * a = b - a = -(a - b) $\neq\text{a}*\text{b}$ For associativity: a * (b * c) = a * (b - c) = a - (b - c) = (a - b + c) Also, (a * b) * c = (a - b) * c = (a - b - c) $\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 1542 Marks
For each binary operation * defined below, determine whether * is commutative or associative.
On Q, define a * b = ab + 1
AnswerFor commutativity: a * b = ab + 1 and b * a = ba + 1 = ab + 1 = a * b
For associativity: a * (b * c) = a * (bc + 1) = a(bc + 1) + 1 = abc + a + 1
Also, (a * b) * c = (ab + 1)c + 1 = abc + c + 1
$\therefore\ \ \text{a }*(\text{b }*\text{c})\neq(\text{a }*\text{b })*\text{c}$
Therefore, the operation * is commutative but not associative.
View full question & answer→Question 1552 Marks
Which one the following relations on A = {1, 2, 3} is a function?
f = {(1, 3), (2, 3), (3, 2)}, g = {(1, 2), (1, 3), (3, 1)}
AnswerAs, each element of the domain set has unique image in the relation f = {(1, 3), (2, 3), (3, 2)}
So, f is a function.
Also, the element 1 of the domain set has two images 2 and 3 of the range set in the relation g = {(1, 2), (1, 3), (3, 1)}
So, g is not a function.
View full question & answer→Question 1562 Marks
Is it true that every relation which is symmetric and transitive is also reflexive? Give reasons.
AnswerNo, it is not necessary that a relation which is symmetric and transitive is reflexive as well.
For Example,
Let A = {a, b, c} be a set and
$R_2$ = {(a, a)} is a relation defined on A.
Clearly,
$R_2$ is symmetric and transitive but not reflexive.
View full question & answer→Question 1572 Marks
For the binary operation multiplication modulo $5 (\times _5)$ defined on the set $S = \{1, 2, 3, 4\}.$ Write the value of $(3 \times _5 4^{-1})^{−1}$
AnswerThe composition table for $\times _5$ on the set $S =\{1, 2, 3, 4\}$ is
| $\times _5$ |
$1$ |
$2$ |
$3$ |
$4$ |
| $1$ |
$1$ |
$2$ |
$3$ |
$4$ |
| $2$ |
$2$ |
$4$ |
$1$ |
$3$ |
| $3$ |
$3$ |
$1$ |
$4$ |
$2$ |
| $4$ |
$4$ |
$3$ |
$2$ |
$1$ |
Now,
$(3 \times _5 4^{-1})^{-1} = (3 \times _5 4)^{-1} [\because 4^{-1} = 4]$
$= 2^{-1} [3 \times _5 4 = 2]$
$= 3 [\because 2^{-1} = 3]$ View full question & answer→Question 1582 Marks
Let f be a function from C (set of all complex numbers) to itself 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\ (\text{x}+1)(\text{x}+\omega)(\text{x}+\omega^2)=0,$ where $\omega=\frac{1\pm\text{i}\sqrt{3}}{2}$
$\Rightarrow\ \text{x}=-1,-\omega,-\omega^2$ $(\text{as x}\in\text{C})$
$\Rightarrow\ \text{f}^{-1}(-1)=\{-1,-\omega,-\omega^2\}$ [from 1]
View full question & answer→Question 1592 Marks
Find gof and fog, if:
f(x) = |x| and g(x) = |5x – 2|
AnswerTo find: gof and fog
f(x) = x and g(x) = |5x − 2|
gof = g[f(x)] = g[|x|] and fog = f[g(x)] = f[(5x - 2)] = |5x - 2| = |5|x|-2|
View full question & answer→Question 1602 Marks
If $f(x) = 4 - (x - 7)^3$, then write $f^{-1}(x)$.
AnswerWe have, $f(x) = 4 - (x - 7)^3$ Let $y = 4 - (x - 7)^3$
$\Rightarrow\ (\text{x} - 7)^3 = 4 - \text{y}$
$\Rightarrow\ \text{x}-7=\sqrt[3]{4-\text{y}}$
$\Rightarrow\ \text{x}=7+\sqrt[3]{4-\text{y}}$
$\Rightarrow\ \text{f}^{-1}(\text{y})=7+\sqrt[3]{4-\text{y}}$
$\therefore\ \text{f}^{-1}(\text{x})=7+\sqrt[3]{4-\text{x}}$
View full question & answer→Question 1612 Marks
Show that the binary operation * on Z defined by a * b = 3a + 7b is not commutative.
AnswerThe binary operator * defined on Z and is given by a * b = 3a + 7b
Commutativity: Let $\text{a, b}\in\text{Z},$ Then,
a * b = 1a + 7b and
b * a = 3b + 7a
$\therefore\ \text{a}\ ^*\ \text{b}\neq\text{b}\ ^*\ \text{a}$
Hence, '*' is not commutative on Z.
View full question & answer→Question 1622 Marks
Which of the following functions from A to B are one-one and onto?
$f_3 = \{(a, x), (b, x), (c, z), (d, z)\}; A = \{a, b, c, d,\}, B = \{x, y, z\}$
Answer$f_3 = \{(a, x), (b, x), (c, z), (d, z)\}$
$A = \{a, b, c, d,\}, B = \{x, y, z\}$
Since, $f_3(a) = x = f_3(b)$ and $f_3(c) = z = f_3(d)$
$\therefore f_3$ in not one-one.
Again, $\text{y}\in\text{B}$ in not the image of any of the element of $A.$
$\therefore f_3$ in not on to.
View full question & answer→Question 1632 Marks
Write the domain of the real function f defined by $\text{f(x)}=\sqrt{25-\text{x}^2}.$
AnswerWe have, $\text{f(x)}=\sqrt{25-\text{x}^2}$ The function is defined only when $25-\text{x}^2\geq0$$\text{x}^2-25\leq0$
$(\text{x}+5)(\text{x}-5)\leq0$
$\text{x}\in[-5,5]$
Therefore, the domain of the given function is [-5, 5].
View full question & answer→Question 1642 Marks
For each binary operation $*$ defined below, determine whether $*$ is commutative or associative.
On $Z^+,$ define $a * b = 2^{ab}$
AnswerFor commutativity: $\mathrm{a}^* \mathrm{~b}=2^{\mathrm{ab}}$ and $b * \mathrm{a}=2^{\mathrm{ba}}=2^{\mathrm{ab}}=\mathrm{a} * \mathrm{~b}$
For associativity: $a^*(b * c)=a * 2^{b c}=(2)$
$ \text { Also, }\left(a^* b\right) * c=\left(2^{a b}\right) * 2=2^{a b} \times c $
$ \therefore a *(b * c) \neq(a * b) * c$
Therefore, the operation * is commutative but not associative.
View full question & answer→Question 1652 Marks
Give an example of a relation which is,
Symmetric and transitive but not reflexive.
AnswerLet R be the relation on A such that,
R = {(1, 2), (2, 1), (1, 3), (3, 1), (2, 3)}
We see that the relation R on A is symmetric and transitive, but not reflexive.
View full question & answer→Question 1662 Marks
Let A = {x ∈ R : -4 ≤ x ≤ 4 and x ≠ 0} and f : A → R be defined by $\text{f(x)}=\frac{|\text{x}|}{\text{x}}.$ Write the range of f.
AnswerWe have, $\text{A}=\{\text{x}\in\text{R}:-4\leq\text{x}\leq4\text{ and x }\neq0\}$
f : A → R defined by $\text{f(x)}=\frac{|\text{x}|}{\text{x}}$
Clearly, $\text{f(x)}=\begin{cases}1;&\text{x}>0\\-1;&\text{x}<0\end{cases}$
$\therefore$ Range of f = {-1, 1}
View full question & answer→Question 1672 Marks
What is the range of the function $\text{f(x)}=\frac{|\text{x}-1|}{\text{x}-1}?$
Answer$\text{f(x)}=\frac{|\text{x}-1|}{\text{x}-1}=\frac{\pm(\text{x}-1)}{\text{x}-1}=\pm1$Range of f = {-1, 1}
View full question & answer→Question 1682 Marks
If $R$ is a symmetric relation on a set $A$, then write a relation between $R$ and $R^{-1}$.
AnswerHere, R is symmetric on the set A.
Let $(\text{a, b})\in\text{R}$
$\Rightarrow\ (\text{b, a})\in\text{R}$ [Since R is symmetric]
$\Rightarrow\ (\text{a, b})\in\text{R}^{-1}$ [By definition of inverse relation]
$\Rightarrow\ \text{R}\subset\text{R}^{-1}$
Let $(\text{x, y})\in\text{R}^{-1}$
$\Rightarrow\ (\text{y, x})\in\text{R}$ [By definition of inverse relation]
$\Rightarrow\ (\text{x, y})\in\text{R}$ [Since R is symmetric]
$\Rightarrow\ \text{R}^{-1}\subset\text{R}$
Thus, $\text{R}=\text{R}^{-1}$
View full question & answer→Question 1692 Marks
Reflexive and symmetric but not transitive.
Answer“is friend of” R = {( x, y) : x is a friend of y}
| It is clear that x is friend of x. |
$\therefore$ |
R is reflexive. |
| Also x is friend of y and y is friend of x. |
$\therefore$ |
R is symmetric. |
| Also if x is friend of y and y is friend of z then |
|
|
| x cannot be friend of z. |
$\therefore$ |
R is not transitive. |
Therefore, R is reflexive and symmetric but not transitive. View full question & answer→Question 1702 Marks
Determine which of the following binary operations are associative and which are commutative:
'*' on N defined by a * b = 1 for all $\text{a, b}\in\text{N}.$
AnswerClearly, by defination a * b = 1 = b * a, $\forall\ \text{a, b}\in\text{N}$
Also, (a * b) * c = (1 * c) = 1
and a * (b * c) = (a * 1) = 1 $\forall\ \text{a, b, c}\in\text{N}$
Hence, N is both associative and commutative.
View full question & answer→Question 1712 Marks
If f : A → A, g : A → A are two bijections, then prove that:
fog is a surjection.
AnswerGiven: A → A, g : A → A are two bijections. Then, fog : A → A Surjectivity of fog: let z be an element in the co-domain of fog (A).Now, $\text{z}\in\text{A}$ (co-domain of f) and f is a surjection.
So, z = f(y), where $\text{y}\in\text{A}$ (domain of f) .....(1)
Now, $\text{y}\in\text{A}$ (co-domain of g) and g is a surjection.
So, y = g(x), where $\text{x}\in\text{A}$ (domain of g) .....(2)
From (1) and (2),
z = f(y) = f(g(x)) = (fog)(x), where $\text{x}\in\text{A}$ (domain of fog)
So, fog is a surjection.
View full question & answer→Question 1722 Marks
Find which of the binary operations are commutative and which are associative.
State whether the following statements are true or false. Justify
If * is a commutative binary operation on N, then a * (b * c) = (c * b) * a
Answer* being a binary operation on N.
$\therefore$ c * b = b * c
$\therefore$ (c * b) * a = (b * c) * a = a * (b * c)
Thus, a * (b * b) = (c * b) * a, therefore, the given statement is true.
View full question & answer→Question 1732 Marks
Let $\text{f}:\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\rightarrow\ \text{R}$ be a function defined by f(x) = cos[x]. write range (f).
Answer$\text{f}:\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\rightarrow\ \text{R}$ given by f(x) = cos[x]$\because\ \cos\text{x}$ in position in $\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)$
$\therefore$ cos[x] will be $\{1, \cos1, \cos2\}$
$\therefore$ Range of $\text{f}=\{1, \cos1, \cos2\}$
View full question & answer→Question 1742 Marks
$f(x) = 8x^3$ and $g(x) =\text{x}^{\frac{1}{3}}.$
Answer$f(x) = 8x^3$ and $g(x) \text{x}^{\frac{1}{3}}$
$gof = g[f(x)] = g[8x^3] (8\text{x}^3)^{\frac{1}{3}}=2\text{x}$
and $fog = f[g(x)] =f\Big[\Big(\text{x}^{\frac{1}{3}}\Big)\Big]=8\Big(\text{x}^{\frac{1}{3}}\Big)^3=8\text{x}$
View full question & answer→Question 1752 Marks
Determine whether the following operations define a binary operation on the given set or not:
'O' on Z defined by $a$ $O$ $b = a^b$ for all $\text{a, b}\in\text{Z.}$
AnswerWe have,
$a\ O\ b = a^b$ for all $\text{a, b}\in\text{Z}$
Let $\text{a}\in\text{Z}$ and $\text{b}\in\text{Z}$
$\Rightarrow\ \text{a}^{\text{b}}\notin\text{Z}\ \Rightarrow\ \text{a O b}\notin\text{Z}$
For example, if $a = 2, b = -2$
$\Rightarrow\ \text{a}^{\text{b}}=2^{-2}=\frac{1}{4}\notin\text{Z}$
$\therefore$ The operation 'O' does not define a binary operation on Z.
View full question & answer→Question 1762 Marks
Determine whether the following operations define a binary operation on the given set or not:
$'\times _6'$ on $S = \{1, 2, 3, 4, 5\}$ defined by, $a \times _6 b =$ Remainder when ab is divided by $6.$
AnswerConsider the composition table,
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$\times _6$
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$1$
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$2$
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$3$
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$4$
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$5$
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$1$
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$1$
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$2$
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$3$
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$4$
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$5$
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$2$
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$2$
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$4$
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$0$
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$2$
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$4$
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$3$
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$3$
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$0$
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$3$
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$0$
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$3$
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$4$
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$4$
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$2$
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$0$
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$4$
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$2$
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$5$
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$5$
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$4$
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$3$
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$2$
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$1$
|
Here all the elements of the table are not in $S.$
For $a =2$ and $b =3$,
$\text{a}\times_6\text{b}= 2 \times_63=$ remainder when $6$ divided by $6=0\neq\text{S}$
Thus, $\times _6$ is not a binary operation on $S.$ View full question & answer→Question 1772 Marks
$F = \{(a, 2), (b, 1), (c, 1)\}.$
Answer$\mathrm{F}: \mathrm{S} \rightarrow \mathrm{T}$ is defined as:
$\mathrm{F}=\{(\mathrm{a}, 2),(\mathrm{b}, 1),(\mathrm{c}, 1)\}$
Since $F(b)=F(c)=1, F$ is not one-one.
Hence, $F$ is not invertible i.e., $\mathrm{F}^{-1}$ does not exist.
View full question & answer→Question 1782 Marks
A = {1, 2, 3, 4, 5, 6, 7, 8} and if R = {(x, y): y is one half of x; x, y ∈ A} is a relation on A, then write R as a set of ordered pairs.
AnswerSince R = {(x, y): y is one half of x; x, y ∈ A}
So, R = {(2, 1), (4, 2), (6, 3), (8, 4)}
View full question & answer→Question 1792 Marks
Let $f, g : R → R$ be defined by $f(x) = 2x + 1$ and $g(x) = x^2 - 2$ for all $x ∈ R,$ respectively. Then, find gof.
AnswerWe have,
$f, g : R → R$ are defined by $f(x) = 2x + 1$ and $g(x) = x^2 - 2$ for all $x ∈ R,$ respectively
Now,
$gof(x) = g(f(x))$
$= g(2x + 1)$
$= (2x + 1)^2 - 2$
$= 4x^2 + 4x + 1 - 2$
$= 4x^2 + 4x - 1$
View full question & answer→Question 1802 Marks
Let *′ be the binary operation on the set {1, 2, 3, 4, 5} defined by a *' b = H.C.F. of a and b. Is the operation *′ same as the operation * defined in Exercise 4 above? Justify your answer.
AnswerLet A = {1, 2, 3, 4, 5} and a *' b = H.C.F. of a and b.
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*'
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1
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2
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3
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4
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5
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1
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1
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1
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1
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1
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1
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|
2
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1
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2
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1
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2
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1
|
|
3
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1
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1
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3
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1
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1
|
|
4
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1
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2
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1
|
4
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1
|
|
5
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1
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1
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1
|
1
|
5
|
We observe that the operation *' is the same as the operation * in ex. 4.
View full question & answer→Question 1812 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 $Z^+$ define * by $a * b = |a - b|$
Here, $Z^+$ denotes the set of all non-negative integers.
AnswerOn $Z^+, *$ is defined by $a * b = |a - b|.$
It is seen that for each $\text{a, b}\in\text{Z}^{+},$
there is a unique element $|a - b|$ in $Z^+.$
This means that $*$ carries each pair $(a, b)$ to a unique element $a * b = |a - b|$ in $Z^+.$
Therefore, $*$ is a binary operation.
View full question & answer→