M.C.Q (1 Marks) - Maths STD 12 Science Questions - Vidyadip

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M.C.Q (1 Marks)

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29 questions · auto-graded multiple-choice test.

MCQ 11 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 $\triangle$

A
Commutative and associative without an identity.
B
Commutative but not associative with an identity.
C
Associative but not commutative without an identity.
✓
Associative and commutative with an identity.

Answer

Correct option: D.

Associative and commutative with an identity.

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 21 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 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,
$\frac{\text{ae}}{3}=\text{a}$ 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 $b^ * a = e$
$\therefore\ \frac{\text{ab}}3=3$ 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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MCQ 31 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:

A
$1$
B
$\frac{\text{a}-1}{\text{a}}$
C
$\frac{\text{a}}{\text{a}-1}$
✓
$0$

Answer

Correct option: D.

$0$

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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MCQ 41 Mark

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

A
8
B
6
C
4
✓
2

Answer

Correct option: D.

2

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 51 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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MCQ 61 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:

A
$\begin{bmatrix}-2&3\\-3&-2\end{bmatrix}$
B
$\begin{bmatrix}2&3\\-3&2\end{bmatrix}$
✓
$\begin{bmatrix}\frac{2}{13}&\frac{-3}{13}\\\frac{3}{13}&\frac{2}{13}\end{bmatrix}$
D
$\begin{bmatrix}1&0\\0&1\end{bmatrix}$

Answer

Correct option: C.

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

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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MCQ 71 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:

A
$\begin{bmatrix}1&1\\1&1\end{bmatrix}$
B
$\begin{bmatrix}-\frac{1}2&-\frac{1}2\\-\frac{1}2&-\frac{1}2\end{bmatrix}$
✓
$\begin{bmatrix}\frac{1}2&\frac{1}2\\\frac{1}2&\frac{1}2\end{bmatrix}$
D
$\begin{bmatrix}-1&-1\\-1&-1\end{bmatrix}$

Answer

Correct option: C.

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

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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MCQ 81 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:

A
$-\text{a}$
✓
$-\frac{\text{a}}{\text{a}-1}$
C
$\frac{1}{\text{a}}$
D
$\text{a}^2$

Answer

Correct option: B.

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

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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MCQ 91 Mark

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

✓
Commutative but not associative.
B
Associative but not commutative.
C
Neither commutative nor associative.
D
Both commutative and associative.

Answer

Correct option: A.

Commutative but not associative.

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 101 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 111 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 that
$a^ * e = a = e^ * a, \forall\text{ a}\in\text{Q}^+$
$a^ * e = a$ and $e^ * a = a, \forall\text{ a}\in\text{Q}^+$
$\frac{\text{ae}}2=\text{a}$ 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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MCQ 121 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:

A
Closure.
B
Commutative.
C
Associative.
✓
None of these.

Answer

Correct option: D.

None of these.

* 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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MCQ 131 Mark

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

A
Closure law.
B
Associative law.
✓
Commutative law.
D
Distributive law.

Answer

Correct option: C.

Commutative law.

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

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MCQ 141 Mark

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

A
Commutative.
B
Associative.
✓
Not commutative.
D
Commutative and associative.

Answer

Correct option: C.

Not commutative.

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 151 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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MCQ 161 Mark

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

A
8
B
4
✓
16
D
64

Answer

Correct option: C.

16

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 171 Mark

Subtraction of integers is:

✓
Commutative but no associative.
B
Commutative and associative.
C
Associative but not commutative.
D
Neither commutative nor associative.

Answer

Correct option: A.

Commutative but no associative.

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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MCQ 181 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:

A
−10
B
0
C
10
✓
Non-existent.

Answer

Correct option: D.

Non-existent.

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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MCQ 191 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 201 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 that
$a ^* e = a = e ^* a,$ $\forall\text{ a}\in\text{Q}^+$
$a ^* e = a$ and $e ^* a = a$, $\forall\text{ a}\in\text{Q}^+$
$\frac{\text{ae}}{100}=\text{a}\text{ and }\frac{\text{ea}}{100}=\text{a},\forall\text{ a}\in\text{Q}^+$
$\text{e}=100,\forall\text{ a}\in\text{Q}^+$
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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MCQ 211 Mark

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

A
14
B
31
✓
10
D
8

Answer

Correct option: C.

10

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

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MCQ 221 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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MCQ 231 Mark

Which of the following is true?

A
* 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.
C
All binary commutative operations are associative.
D
Subtraction is a binary operation on N.

Answer

Correct option: B.

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

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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MCQ 241 Mark

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

A
Commutative only.
B
Associative only.
✓
Commutative and associative both.
D
None of these.

Answer

Correct option: C.

Commutative and associative both.

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 251 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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MCQ 261 Mark

On $Z$ an operation $^*$ is defined by $a ^* b = a^2 + b^2$ for all $a, b ∈ Z$. 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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MCQ 271 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:

A
2
✓
3
C
4
D
5

Answer

Correct option: B.

3

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

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MCQ 281 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:

A
0
✓
-1
C
1
D
2

Answer

Correct option: B.

-1

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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MCQ 291 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:

✓
$0$
B
$1$
C
$\frac{1}2$
D
$-1$

Answer

Correct option: A.

$0$

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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