M.C.Q (1 Marks)

Question 11 Mark

If a binary operation * is defined on the set Z of integers as a * b = 3a − b, then the value of (2 * 3) * 4 is:

Answer

Solution:
Given: a * b = 3a - b
2 * 3 = 3 (2) - 3
= 6 - 3
= 3
(2 * 3) * 4 = 3 * 4
= 3(3) - 4
= 9 - 4
= 5

Question 21 Mark

Mark the correct alternative in the following question for the binary operation * on Z defined by a * b = a + b + 1, the identity element is:

0
-1
1
2

Answer

-1

Solution:
We have,
a * b = a + b + 1
Let e be the identity element of *. Then,
a * e = a = e * a
a + e + 1 = a
e = a - a - 1
e = -1

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Question 31 Mark

On the power set P of a non-empty set A, we define an operation $\triangle \text{ by }\text{X}\triangle\text{Y}=(\text{X}\cap\text{Y})∪(\text{X}∩\text{Y})\text{X}\triangle\text{Y}=\text{X}∩\text{Y}∪\text{X}∩\text{Y}$
Then which are of the following statements is true about $\triangle$

Commutative and associative without an identity.
Commutative but not associative with an identity.
Associative but not commutative without an identity.
Associative and commutative with an identity.

Answer

Associative and commutative with an identity.

Solution:
Commutativity:
$\text{X}\triangle\text{Y}=(\overline{\text{X}}\cap\text{Y})\cup(\text{X}\cap\overline{\text{Y}})$
$=(\overline{\text{Y}}\cap\text{X})\cup(\text{Y}\cap\overline{\text{X}})$
$=\text{Y}\triangle\text{X}$
Thus,
$\text{X}\triangle\text{Y}=\text{Y}\triangle\text{X}$
Hence, $\triangle$ is commutative on A.
Let $\phi$ be the identity element for $\triangle$ on P.
$\text{A}\triangle\phi=\big(\overline{\text{A}}\cap\phi\big)\cup\big(\text{A}\cap\overline{\phi}\big)$
$=\phi\cup\text{A}$
$=\text{A}$
and,
$\phi\triangle\text{A}=\big(\overline{\phi}\cap\text{A}\big)\cup\big(\phi\cap\overline{\text{A}}\big)$
$=\text{A}\cup\phi$
$=\text{A}$

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MCQ 41 Mark

Let * be a binary operation defined on $Q^+$ by the rule $\text{a}*\text{b}=\frac{\text{ab}}3\forall\text{ a, b}\in \text{Q}^+$. The inverse of $4 * 6$ is:

✓
$\frac{9}{8}$
B
$\frac{2}3$
C
$\frac{3}2$
D
None of these.

Answer

Correct option: A.

$\frac{9}{8}$

Let e be the identity element in $Q^+$ with respect to * such thata * $e = a = e * a, \forall\text{ a}\in\text{Q}^+$
$a * e = a$ and $e * a = a, \forall\text{ a}\in\text{Q}^+$
Then,
$\frac{\text{ae}}{3}=\text{a}\text{ and }\frac{\text{ea}}{3}=\text{a},\forall\text{ a}\in\text{Q}^+$
$e = 3, \forall\text{ a}\in\text{Q}^+$
Thus$, 3$ is the identity element in $Q^+$ with respect to *.
Let $\text{a}\in\text{Q}^+$ and $\text{b}\in\text{Q}^+$ be the inverse of a. Then,
$a * b = e = b * a$
$a * b = e$ and $b * a = e$
$\therefore\ \frac{\text{ab}}3=3\text{ and }\frac{\text{ba}}3=3$
$\text{b}=\frac{9}{\text{a}}\in\text{Q}^+$
Thus,$ \frac{9}{\text{a}}$ is the inverse of $\text{a}\in\text{Q}^+.$
Given: $\text{a}*\text{b}=\frac{\text{ab}}3$
$4*6=\frac{4\times6}3=8$
Now,
$\text{a}^{-1}=\frac{9}{\text{a}}$
$(4*6)^{-1}=8^{-1}$
$=\frac{9}8$

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Question 51 Mark

Let * be a binary operation defined on set Q − {1} by the rule a * b = a + b − ab. Then, the identify element for * is:

$1$
$\frac{\text{a}-1}{\text{a}}$
$\frac{\text{a}}{\text{a}-1}$
$0$

Answer

Solution:
Let e be the identity element in Q - {1} with respect to * such that
a * e = a = e * a, $\forall\text{ a}\in\text{Q}-\{-1\}$
a * e = a and e * a = a, $\forall\text{ a}\in\text{Q}-\{-1\}$
a + e - ae = a and e + a - ea = a, $\forall\text{ a}\in\text{Q}-\{-1\}$
e(1 - a) = 0, $\forall\text{ a}\in\text{Q}-\{-1\}$ $[\because \text{a}\neq1]$
Thus, 0 is the identity element in Q - {1} with respect to *.

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Question 61 Mark

The number of commutative binary operation that can be defined on a set of 2 elements is:

Answer

Solution:
The number of commutative binary operations on a set of n elements is $\text{n}\frac{\text{n}(\text{n}-1)}{2}$.
Therefore,
Number of commutative binary operations an a set of 2 elements $=2\frac{2(2-1)}{2}=2^1$
$=2$

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MCQ 71 Mark

On the set $Q^+$ of all positive rational numbers a binary operation $*$ is defined by $\text{a}*\text{b}=\frac{\text{ab}}2\forall\text{ a, b}\in \text{Q}^+$. The inverse of $8$ is:

A
$\frac{1}{8}$
✓
$\frac{1}2$
C
$2$
D
$4$

Answer

Correct option: B.

$\frac{1}2$

Let e be the identity element in $Q^+$ with respect to $*$ such that
$a * e = a = e * a, \forall\text{ a}\in\text{Q}^+$
$a * e = a$ and $e * a = a, \forall\text{ a}\in\text{Q}^+$
Then,
$\frac{\text{ae}}{2}=\text{a}$ and $\frac{\text{ea}}{2}=\text{a},\forall\text{ a}\in\text{Q}^+$
$e = 2, \forall\text{ a}\in\text{Q}^+$
Thus, $2$ is the identity element in $Q^+$ with respect to $*.$
Let $\text{b}\in\text{Q}^+$ be the inverse of $8.$ Then,
$8 * b = e = b * 8$
$8 * b = e$ and $b * 8 = e$
$\frac{(8)\text{b}}2=2$ and $\frac{\text{b}(8)}2=2 [\because\ \text{e}=2]$
$b = 12$
Thus, $\frac{1}2$ is the inverse of $8.$

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Question 81 Mark

For the multiplication of matrices as a binary operation on the set of all matrices of the form $\begin{bmatrix}\text{a}&\text{b}\\-\text{b}&\text{a}\end{bmatrix},\text{a, b}\in\text{R}$ the inverse of $\begin{bmatrix}2&3\\-3&2\end{bmatrix}$ is:

$\begin{bmatrix}-2&3\\-3&-2\end{bmatrix}$
$\begin{bmatrix}2&3\\-3&2\end{bmatrix}$
$\begin{bmatrix}\frac{2}{13}&\frac{-3}{13}\\\frac{3}{13}&\frac{2}{13}\end{bmatrix}$
$\begin{bmatrix}1&0\\0&1\end{bmatrix}$

Answer

$\begin{bmatrix}\frac{2}{13}&\frac{-3}{13}\\\frac{3}{13}&\frac{2}{13}\end{bmatrix}$

Solution:
Let the identity of $\begin{bmatrix}2&3\\-3&2\end{bmatrix}$ be $\begin{bmatrix}\text{e}&\text{f}\\-\text{f}&\text{e}\end{bmatrix}$, then
$\begin{bmatrix}2&3\\-3&2\end{bmatrix}*\begin{bmatrix}\text{e}&\text{f}\\-\text{f}&\text{e}\end{bmatrix}=\begin{bmatrix}2&3\\-3&3\end{bmatrix}$
⇒ 2e - 3f = 2 →(1)
2f + 3e = 3 →(2)
Solving (1) and (2) we get e = 1 and f = 0
So, the identity is $\begin{bmatrix}1&0\\0&1\end{bmatrix}$.
Let the inverse be $\begin{bmatrix}\text{a}&\text{b}\\-\text{b}&\text{a}\end{bmatrix}$, then
$\begin{bmatrix}2&3\\-3&2\end{bmatrix}*\begin{bmatrix}\text{a}&\text{b}\\-\text{b}&\text{a}\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
⇒ 2a - 3b = 1 →(1)
2b + 3a = 0 →(2)
Solving (1) and (2), we get $\text{a}=\frac{2}{13}$ and $\text{b}=\frac{-3}{13}$
So, the inverse of $\begin{bmatrix}2&3\\-3&2\end{bmatrix}$ is $\begin{bmatrix}\frac{2}{13}&\frac{-3}{13}\\\frac{3}{13}&\frac{2}{13}\end{bmatrix}$

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Question 91 Mark

If G is the set of all matrices of the form $\begin{bmatrix}\text{x}&\text{x}\\\text{x}&\text{x}\end{bmatrix}$, where $\text{x}\in\text{R}-\{0\}$, then the identity element with respect to the multiplication of matrices as binary operation, is:

$\begin{bmatrix}1&1\\1&1\end{bmatrix}$
$\begin{bmatrix}-\frac{1}2&-\frac{1}2\\-\frac{1}2&-\frac{1}2\end{bmatrix}$
$\begin{bmatrix}\frac{1}2&\frac{1}2\\\frac{1}2&\frac{1}2\end{bmatrix}$
$\begin{bmatrix}-1&-1\\-1&-1\end{bmatrix}$

Answer

$\begin{bmatrix}\frac{1}2&\frac{1}2\\\frac{1}2&\frac{1}2\end{bmatrix}$

Solution:
Let $\begin{bmatrix}\text{x}&\text{x}\\\text{x}&\text{x}\end{bmatrix}\in\text{G}$ and $\begin{bmatrix}\text{e}&\text{e}\\\text{e}&\text{e}\end{bmatrix}\in\text{G}$ such that
$\begin{bmatrix}\text{x}&\text{x}\\\text{x}&\text{x}\end{bmatrix}\begin{bmatrix}\text{e}&\text{e}\\\text{e}&\text{e}\end{bmatrix}=\begin{bmatrix}\text{x}&\text{x}\\\text{x}&\text{x}\end{bmatrix}$
$\begin{bmatrix}\text{x}&\text{x}\\\text{x}&\text{x}\end{bmatrix}\begin{bmatrix}\text{e}&\text{e}\\\text{e}&\text{e}\end{bmatrix}=\begin{bmatrix}\text{x}&\text{x}\\\text{x}&\text{x}\end{bmatrix}$
$\begin{bmatrix}2\text{ex}&2\text{ex}\\2\text{ex}&2\text{ex}\end{bmatrix}=\begin{bmatrix}\text{x}&\text{x}\\\text{x}&\text{x}\end{bmatrix}$
$2\text{ex}=\text{x}$
$\text{e}=\frac{1}2\in\text{R}-\{0\}$
Thus, $\begin{bmatrix}\frac{1}2&\frac{1}2\\\frac{1}2&\frac{1}2\end{bmatrix}\in\text{G}$, is the identity element in G.

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Question 101 Mark

For the binary operation * defined on R − {1} by the rule a * b = a + b + ab for all a, b ∈ R − {1}, the inverse of a is:

$-\text{a}$
$-\frac{\text{a}}{\text{a}-1}$
$\frac{1}{\text{a}}$
$\text{a}^2$

Answer

$-\frac{\text{a}}{\text{a}-1}$

Solution:
Let e be the identity element in R - {1} with respect to * such that
a * e = a = e * a, $\forall\text{ a}\in\text{R}-\{1\}$
a * e = a and e * a = a, $\forall\text{ a}\in\text{R}-\{1\}$
Then,
a + e + ae = a and e + a + ea = a, $\forall\text{ a}\in\text{R}-\{1\}$
e(1 + a) = 0, $\forall\text{ a}\in\text{R}-\{1\}$
$\text{e}=0\in\text{R}-\{1\}$
Thus, 0 is the identity element in R - {1} with respect to *.
Let $\text{a}\in\text{R}-\{1\}$ and $\text{b}\in\text{R}-\{1\}$ be the inverse of a. Then,
a * b = e = b * a
a * b = e and b * a = e
⇒ a + b + ab = 0 and b + a + ba = 0
$\Rightarrow\text{b}(1+\text{a})=-\text{a}\in\text{R}-\{1\}$
$\Rightarrow\text{b}=\frac{-\text{a}}{\text{a}-1}\in\text{R}-\{1\}$
Thus, $\frac{-\text{a}}{\text{a}-1}$ is the inverse of $\text{a}\in\text{R}-\{1\}$.

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Question 111 Mark

Let * be a binary operation on R defined by a * b = ab + 1. Then, * is:

Commutative but not associative.
Associative but not commutative.
Neither commutative nor associative.
Both commutative and associative.

Answer

Commutative but not associative.

Solution:
Commutativity:
Let $\text{a, b}\in\text{R}$
a * b = ab + 1
= ba + 1
= b * a
Therefore,
a * b = b * a, $\forall\text{ a, b}\in\text{R}$
Therefore, * is commutative on R.
Associativity:
Let $\text{ a, b, c}\in\text{R}$
a * (b * c) = a * (bc + 1)
= a(bc + 1) + 1
= abc + a + 1
(a * b) * c = (ab + 1) * c
= (ab + 1)c + 1
= abc + c + 1
$\therefore$ a * (b * c) $\neq$ (a * b) * c
For example: a = 1, b = 2 and c = 3 [which belong to R]
Now,
1 * (2 * 3) = 1 * (6 + 1)
= 1 * 7
= 7 + 1
= 8
(1 * 2) * 3 = (2 + 1) * 3
= 3 * 3
= 9 + 1
= 10
⇒ 1 * (2 * 3) $\neq$ (1 * 2) * 3
Therefore, $\exists$ a = 1, b = 2 and c = 3 which belong to R such that
a * (b * c) $\neq$ (a * b) * c
Hence, * is not associative on R.

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MCQ 121 Mark

$Q^+$ is the set of all positive rational numbers with the binary operation $^*$ defined by $\text{a}*\text{b}=\frac{\text{ab}}2\ \forall\text{ a, b}\in\text{Q}^+$. The inverse of an element $\text{a}\in\text{Q}^+$ is:

A
$\text{a}$
B
$\frac{1}{\text{a}}$
C
$\frac{2}{\text{a}}$
✓
$\frac{4}{\text{a}}$

Answer

Correct option: D.

$\frac{4}{\text{a}}$

Let $e$ be the identity element in $Q^+$ with respect to $^*$ such thata $^* e = a = e ^* a, \forall\text{ a}\in\text{Q}^+$
$a ^* e = a$ and $e ^* a = a, \forall\text{ a}\in\text{Q}^+$
$\frac{\text{ae}}2=\text{a}$ and $\frac{\text{ea}}2=\text{a}$,
$\forall\text{ a}\in\text{Q}^+$
$\text{e}=2\in\text{Q}^+, \forall\text{ a}\in\text{Q}^+$
Thus$, 2$ is the identity element in $Q^+$ with respect to$^ *.$
Let $\text{ a}\in\text{Q}^+$ and $\text{ b}\in\text{Q}^+$ be the inverse of $a$.
Then,
$a ^* e = a = e ^* a$
$a ^* b = e$ and $b ^* a = e$
$\frac{\text{ab}}2=2$ and $\frac{\text{ba}}2=2$
$\text{b}=\frac{4}{\text{a}}\in\text{Q}^+$
Thus, $\frac{4}{\text{a}}$ is the inverse of $\text{ a}\in\text{Q}^+$.

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Question 131 Mark

An operation * is defined on the set Z of non-zero integers by a * b = ab for all a, b ∈ Z. Then the property satisfied is:

Closure.
Commutative.
Associative.
None of these.

Answer

None of these.

Solution:
* is not clouser because when a = 1 and b = 2,
$\text{a}*\text{b}=\frac{\text{a}}{\text{b}}=\frac{1}{2}\in\text{Z}$
* is not commutative because when a = 1, b = 2 and c = 3,
$1*(2*3)=1*\Big(\frac{2}3\Big)$
$=\frac{1}{\big(\frac{2}{3}\big)}$
$=\frac{3}2$
$(1*2)*3=\frac{1}2*3$
$=\frac{\big(\frac{1}2\big)}{3}$
$=\frac{1}6$
Thus,
$1*(2*3)\neq(1*2)*3$

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Question 141 Mark

The law a + b = b + a is called:

Closure law.
Associative law.
Commutative law.
Distributive law.

Answer

Commutative law.

Solution:
The law a + b = b + a is commutative.

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Question 151 Mark

A binary operation * on Z defined by a * b = 3a + b for all a, b ∈ Z, is:

Commutative.
Associative.
Not commutative.
Commutative and associative.

Answer

Not commutative.

Solution:
Let $\text{a, b}\in\text{Z}$
a * b = 3a + b
b * a = 3b + a
Thus, a * b $\neq$ b * a
If a = 1 and b = 2,
1 * 2 = 3(1) + 2
= 5
2 * 1 = 3(2) + 1
= 7
1 * 2 $\neq$ 2 * 1
Thus, * is not commutative on Z.

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MCQ 161 Mark

If the binary operation $\odot$ is defined on the set $Q^+$ of all positive rational numbers by $\text{a}\odot\text{b}=\frac{\text{ab}}4$. Then, $3\odot\Big(\frac{1}5\odot\frac{1}2\Big)$ is equal to:

✓
$\frac{3}{160}$
B
$\frac{5}{160}$
C
$\frac{3}{10}$
D
$\frac{3}{40}$

Answer

Correct option: A.

$\frac{3}{160}$

Given $\text{a}\odot\text{b}=\frac{\text{ab}}4$
$\Rightarrow\Big(\frac{1}5\odot\frac{1}2\Big)$
$=\frac{\frac{1}5.\frac{1}2}{4}$
$=\frac{1}{40}$
$3\odot\Big(\frac{1}5\odot\frac{1}2\Big)$
$=3\odot\frac{1}{40}$
$=\frac{\frac{1}{40}.3}{4}$
$=\frac{3}{160}$

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Question 171 Mark

The number of binary operation that can be defined on a set of 2 elements is:

Answer

Solution:
Total number of binary operations on a set containing n elements is
$\text{(n)}^{\text{n}^2}$ so for n = 2 we have $(2)^{2^2}=2^4=16$

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MCQ 181 Mark

Let * be a binary operation on $Q^+$ defined by $\text{a}*\text{b}=\frac{\text{ab}}{100}\forall\text{ a, b}\in\text{Q}^+$. The inverse of $0.1$ is:

✓
$10^5$
B
$10^4$
C
$10^6$
D
None of these.

Answer

Correct option: A.

$10^5$

Let $e$ be the identity element in $Q^+$ with respect to * such thata *$ e = a = e * a, \forall\text{ a}\in\text{Q}^+$
$a * e = a$ and e$ * a = a, \forall\text{ a}\in\text{Q}^+$
$\frac{\text{ae}}{100}=\text{a}\text{ and }\frac{\text{ea}}{100}=\text{a},\forall\text{ a}\in\text{Q}^+$
$\text{e}=100,\forall\text{ a}\in\text{Q}^+$
Thus, $100$ is the identity element in $Q^+$ with repect to *.
$0.1 * b = e = b * 0.1$
$0.1 * b = e$ and $b * 0.1 = e$
$\frac{(0.1)\text{b}}{100}=100\text{ and }\frac{\text{b}(0.1)}{100}=100$
$\text{b}=\frac{100\times100}{0.1}$
$=10^5\in\text{Q}^+$
Thus$, 10^5$ is the inverse of $0.1.$

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Question 191 Mark

If a * b denote the bigger among a and b and if ab = (a * b) + 3, then 4.7 =

Answer

Solution:
4.7 = (4 * 7) + 3
= 7 + 3
= 10

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MCQ 201 Mark

$Q^+$ denote the set of all positive rational numbers. If the binary operation $\text{a }\odot$ on $Q^+$ is defined as:
$\text{a }\odot=\frac{\text{ab}}{2}$, then the inverse of $3$ is:

✓
$\frac{4}{3}$
B
$2$
C
$\frac{1}3$
D
$\frac{2}3$

Answer

Correct option: A.

$\frac{4}{3}$

Let us first find the identity element.
We know that if e is the identity element then,
$\text{a}\odot\text{e}=\text{e}$
Given $\text{a}\odot\text{e}=\frac{\text{ae}}2$
$\Rightarrow\text{a}=\frac{\text{ae}}2$
$\Rightarrow\text{e}=2$
Let $b$ be the inverse of $3,$ then
$3\odot\text{b}=\text{e}$
$\Rightarrow\frac{3\text{b}}2=2$
$\Rightarrow\text{b}=\frac{4}3$

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Question 211 Mark

Which of the following is true?

* defined by $\text{a}*\text{b}=\frac{\text{a + b}}2$ is a binary operation on Z.
* defined by $\text{a}*\text{b}=\frac{\text{a + b}}2$ is a binary operation on Q.
All binary commutative operations are associative.
Subtraction is a binary operation on N.

Answer

* defined by $\text{a}*\text{b}=\frac{\text{a + b}}2$ is a binary operation on Q.

Solution:
For option a, if we take 3 and 2 then
$3*2=\frac{5}2\in\text{Z}$. So, option a is not true.
For option b, if we take any two numbers a and b
then $\frac{\text{a + b}}2$ belongs to Q for $\text{a, b}\in\text{Q}$.
So, option b is correct.
For option d, if we take 2, 3 then $2-3=-1\in\text{N}$.
So, option d is not true.
Option c is not true.

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Question 221 Mark

Subtraction of integers is:

Commutative but no associative.
Commutative and associative.
Associative but not commutative.
Neither commutative nor associative.

Answer

Neither commutative nor associative.

Solution:
Let $\text{a, b}\in\text{Z}$, then
a * b = a - b
b * a = b - a
⇒ a * b $\neq$ b * a
Substraction is not commutative.
(a * b) * c
= (a - b) * c
= a - b - c
a * (b * c)
= a * (b - c)
= a - b + c
⇒ (a * b) * c $\neq$ a * (b * c)
Substraction is not associative.

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Question 231 Mark

The binary operation * defined on N by a * b = a + b + ab for all a, b ∈ N is:

Commutative only.
Associative only.
Commutative and associative both.
None of these.

Answer

Commutative and associative both.

Solution:
a * b = a + b + ab
b * a = b + a + ba
⇒ a * b = b * a
So * is commutative.
Now,
(a * b) * c
= (a + b + ab) * c
= a + b + ab + c + ca + cb + abc
a * (b * c)
= a * (b + c + bc)
= a + b + c + bc + ab + ac + abc
⇒ (a * b) * c = a * (b * c)
So * is associative.

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MCQ 241 Mark

On $Z$ an operation $*$ is defined by $a * b = a^2 + b^2$ for all $a, b \in Z.$ The operation $*$ on $Z$ is:

A
Commutative and associative.
B
Associative but not commutative.
✓
Not associative.
D
Not a binary operation.

Answer

Correct option: C.

Not associative.

$a * b = a^2 + b^2$
$b * a = b^2 + a^2$
$\Rightarrow a * b = b * a$
So $*$ is commutative.
Now
$(a * b) * c$
$= (a^2 + b^2) * c$
$= (a^2 + b^2)^2 + c^2$
$a * (b * c)$
$= a * (b^2 + c^2)$
$= a^2 + (b^2 +c^2)^2$
$\Rightarrow (a * b) * c \neq a * (b * c)$
So $*$ is not associative.

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Question 251 Mark

Let * be a binary operation on N defined by a * b = a + b + 10 for all a, b ∈ N. The identity element for * in N is:

−10
0
10
Non-existent.

Answer

Non-existent.

Solution:
Given a * b = a + b + 10
Let the identity element be e, then
a * e = a
⇒ a + e + 10 = a
⇒ e = -10
But the operation is defined on the set of natural numbers.
So, the identity element doesn't exist.

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Question 261 Mark

Consider the binary operation * defined on Q − {1} by the rule a * b = a + b − ab for all a, b ∈ Q − {1}. The identity element in Q − {1} is:

$0$
$1$
$\frac{1}2$
$-1$

Answer

Solution:
Let e be the identity element in Q - {1} with respect to * such that
a * e = a = e * a, $\forall\text{ a}\in\text{Q}-\{-1\}$
a * e = a and e * a = a, $\forall\text{ a}\in\text{Q}-\{-1\}$
Then,
a + e - ae = a and e + a - ea = a, $\forall\text{ a}\in\text{Q}-\{-1\}$
e(1 - a) = 0, $\forall\text{ a}\in\text{Q}-\{-1\}$
$\text{e}=0\in\text{Q}-\{-1\}$ $[\because\text{ a}\neq1]$
Thus, 0 is the identity element in Q - {1} with respect to *.

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MCQ 271 Mark

If the binary operation $*$ on $Z$ is defined by $a * b = a^2 − b^2 + ab + 4,$ then value of $(2 * 3) * 4$ is:

A
$233$
✓
$33$
C
$55$
D
$−55$

Answer

Correct option: B.

$33$

Given that $a * b = a^2 - b^2 + ab + 4$
So,
$2 * 3$
$= 2^2 - 3^2 + 2.3 + 4$
$= 4 - 9 + 6 + 4$
$= 5$
Now,
$(2 * 3) * 4$
$= 5 * 4$
$= 5^2 - 4^2 + 5.4 + 4$
$= 25- 16 + 20 + 4$
$= 33$

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MCQ 281 Mark

The binary operation $^*$ is defined by $a ^* b = a^2 + b^2 + ab + 1,$ then $(2 ^* 3) ^* 2$ is equal to$:$

A
$20$
B
$40$
C
$400$
✓
$445$

Answer

Correct option: D.

$445$

Given$: a ^* b = a^2 + b^2 + ab + 1$
$2 ^* 3 = 2^2 + 3^2 + 2 \times 3 + 1$
$= 4 + 9 + 6 + 1$
$= 20$
$(2 ^* 3) ^* 2 = 20 ^* 2$
$= 20^2 + 2^2 + 20 \times 2 + 1$
$= 400 + 4 + 40 + 1$
$= 445$

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MCQ 291 Mark

If $a ^* b = a^2 + b^2,$ then the value of $(4 ^* 5) ^* 3$ is:

A
$(4^2 + 5^2) + 3^2$
B
$(4 + 5)^2 + 3^2$
✓
$41^2 + 3^2$
D
$(4 + 5 + 3)^2$

Answer

Correct option: C.

$41^2 + 3^2$

Given $a ^* b = a^2 + b^2$
So, $4 ^* 5 = 4^2 + 5^2$
Now,
$(4 ^* 5) ^* 3 = (4 ^* 5)^2 + 3^2$
$= (4^2 + 5^2)^2 + 3^2$
$= 41^2 + 3^2$

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