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
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 32 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 42 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 52 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 62 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 72 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 82 Marks
If $f : R \rightarrow R$ 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 92 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 102 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 112 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
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LCM
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1
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6
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10
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3
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3
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3
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12
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15
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12
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20
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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 122 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, $\text{x, y}\in\text{N}$
$(\text{x, y})\in\big\{(1, 1), (2, 2), (4, 1), (1, 4), (3, 3), (9, 1), (1, 9), (4, 4), (2, 8), \$8, 2), (16, 1), (1, 16), .....\big\}$ $$
This is reflexive as (1, 1), (2, 2), .... are present.
This is also symmetric because if aRb ⇒ bRa, for all $\text{a, b}\in\text{N.}$
This is transitive also because if aRb and bRc ⇒ aRc for all $\text{a, b, c}\in\text{N.}$
View full question & answer→Question 132 Marks
Determine whether or not the definition of $^*$ given below gives a binary operation. In the event that $^*$ is not a binary operation give justification of this. On $R,$ define by $a ^* b = ab^2.$
Here, $Z^+$ denotes the set of all non$-$negative integers.
Answer$\text{a, b}\in\text{R}$ Implies that $\text{a, b}^2\in\text{R}$
Implies that $\text{ab}^2\in\text{R}$
Implies that $\text{a}\ ^*\ \text{b}\in\text{R}$
Thus, $^*$ is a binary operation on $R.$
View full question & answer→Question 142 Marks
Write the identity element for the binary operation $^*$ on the set $R_0$ of all non$-$zero real numbers by the rule $\text{a}\times\text{b}=\frac{\text{ab}}{2}$ for all $a, b \in R_0.$
Answer$\because\ \text{a}\times\text{b}=\frac{\text{ab}}{2}$ for all $a, b \in R_0$
Let $e$ be the identity element, then $a ^* e = a$
$\Rightarrow\frac{\text{ae}}{2}=\text{a}$
$\Rightarrow\text{e}=2$
Thus, $e = 2$ is the identity element with respect to $^*.$
View full question & answer→Question 152 Marks
The following defines a relation on N:
x + 4y = 10, $\text{x, y}\in\text{N}$
Determine which of the above relations are reflexive, symmetric and transitive.
AnswerA relation R in A is said to be reflexive if aRa for all $\text{a}\in\text{A},$ R is symmetric if aRb ⇒ bRa, for all $\text{a, b}\in\text{A}$ and it is said to be transitive if aRb and bRc ⇒ aRc for all $\text{a, b, c}\in\text{A.}$$\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 162 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 172 Marks
Is * defined on the set {1, 2, 3, 4, 5} by a * b = L.C.M. of a and b a binary operation? Justify your answer.
AnswerLet A = {1, 2, 3, 4, 5} and a *' b= L.C.M. of a and b.
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*
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Here, $2*3=6\notin\text{A}$
Therefore, the operation * is not a binary operation. View full question & answer→Question 182 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 192 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 202 Marks
f: $Z \rightarrow Z$ given by $f(x) = x^3$
Answer$f: Z \rightarrow Z$ 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
If $f : R \rightarrow R$ is defined by $f(x) = x^2,$ write $f^{-1}(25).$
AnswerLet $f^{-1}(25) = x ....(1)$
$\Rightarrow f(x) = 25$
$\Rightarrow x^2 = 25$
$\Rightarrow x^2 - 25 = 0$
$\Rightarrow (x - 5)(x + 5) = 0$
$\Rightarrow\ \text{x}=\pm5$
$\Rightarrow f^{-1}(25) $$= {-5, 5} [$from $(1)]$
View full question & answer→Question 222 Marks
If $f : R \rightarrow 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 232 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 242 Marks
If $f : R \rightarrow 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 252 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)$
$\Rightarrow f(y) = x$
$\Rightarrow\ \frac{2\text{y}}{5\text{y}+3}=3\text{x}$
$\Rightarrow 2y = 5xy + 3x$
$\Rightarrow 2y - 5xy = 3x$
$\Rightarrow 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 262 Marks
If $f : R \rightarrow R$ is defined by $f(x) = x^2 - 3x + 2,$ write $f \{f(x)\} .$
AnswerWe have, $f(x) = x^2 - 3x + 2$
$\therefore 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$
$\therefore f \{f(x) \} = x^4 - 6x^3 + 10x^2 - 3x$
View full question & answer→Question 272 Marks
Let $f : R \rightarrow 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 282 Marks
$f: N \rightarrow N$ given by $f(x) = x^3$
Answer$f: R \rightarrow 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 292 Marks
Let $f : R - \{-1\} \rightarrow R - \{1\}$ be given by $\text{f(x)}=\frac{\text{x}}{\text{x}+1}.$ Write $f^{-1}(x).$
Answer$f : R - [-1] \rightarrow 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 302 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 312 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 322 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 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
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 352 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 362 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 372 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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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 392 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 402 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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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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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}.
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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 *.
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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.
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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.
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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 $(a, b), (c, d) \in A$
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 $A$.
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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 502 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
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