MCQ 1011 Mark
The law a + b = b + a is called:
AnswerThe law a + b = b + a is commutative.
View full question & answer→MCQ 1021 Mark
Let R be a relation on the set N given by R = {(a, b): a = b - 2, b > 6}. Then,
Answera = b - 2 ⇒ 6 = 8 - 2 and b = 8 > 6
Hence, (6, 8) ∈ R
View full question & answer→MCQ 1031 Mark
Consider a non-empty set consisting of children in a family and a relation R defined as aRb if a is brother of b. Then, R is:
- A
Symmetric but not transitive.
- B
Transitive but not symmetric.
- C
Neither symmetric nor transitive.
- ✓
Both symmetric and transitive.
AnswerCorrect option: D. Both symmetric and transitive.
We have,
R = {(a, b): a is brother of b}
Let $(\text{a, b})\in\text{R}.$ Then,
a is brother of b.
but b is not necessary brother of a (As, b can be sister of a)
$\Rightarrow\ (\text{b, a})\notin\text{R}$
So, R is not symmetric.
Also,
Let $(\text{a, b})\in\text{R}$ and $(\text{b, c})\in\text{R}$
⇒ a is brother of b and b is brother of c
⇒ a is brother of c
$\Rightarrow\ (\text{a, c})\in\text{R}$
So, R is transitive.
View full question & answer→MCQ 1041 Mark
Let $f : R \rightarrow R$ be given by $f(x) = \tan x$. Then $f^{-1}(1)$ is:
AnswerCorrect option: B. $\{\text{n}\pi+\frac{\pi}{4};\text{n }\in\text{ Z}\}$
$\{\text{n}\pi+\frac{\pi}{4};\text{n }\in\text{ Z}\}$
View full question & answer→MCQ 1051 Mark
Let R be a relation on N defined by x + 2y = 8. The domain of R is:
AnswerThe relation R is defined as R = x, y: $\text{x, y}\in\text{N}$ and x + 2y = 8
⇒ R = x, y: $\text{x, y}\in\text{N}$ and $\text{y}=\frac{8-\text{x}}{2}$
Domain of R is all values of $\text{x}\in\text{N}$ satisfying the relation R.
Also, there are only three values of x that result in y, which is a natural number.
These are {2, 6, 4}.
View full question & answer→MCQ 1061 Mark
Let f : Z → Z be given by $\text{f(x)}=\begin{cases}\frac{\text{x}}{2},&\text{if x is even}\\0,&\text{if x is odd}\end{cases}.$ Then, f is:
- ✓
- B
- C
- D
Neither one-one nor onto.
AnswerGiven function is
$\text{f(x)}=\frac{\text{x}}{2}$ if x is even
= 0 if x is odd
For f(3) = 0 and f(4) = 0
⇒ f(3) = f(4)
But, $3\neq4$
Hence, it is not one-one.
$\text{x}\in\text{R}\Rightarrow\ \text{y}\in\text{R}$
Here, Domain = range of f
Hence, it is onto.
View full question & answer→MCQ 1071 Mark
Let $\text{f}:\text{R}-\Big\{-\frac{4}{3}\Big\}\rightarrow\text{R}$ be a function defined as $f(\text{x})=\frac{4\text{x}}{3\text{x}+4}.$ The inverse of f is the map g: Range $\text{f}:\text{R}-\Big\{-\frac{4}{3}\Big\}\rightarrow\text{R}$ given by:
- A
$\text{g}(\text{y})=\frac{3\text{y}}{3-4\text{y}}$
- ✓
$\text{g}(\text{y})=\frac{4\text{y}}{4-3\text{y}}$
- C
$\text{g}(\text{y})=\frac{4\text{y}}{3-4\text{y}}$
- D
$\text{g}(\text{y})=\frac{3\text{y}}{4-3\text{y}}.$
AnswerCorrect option: B. $\text{g}(\text{y})=\frac{4\text{y}}{4-3\text{y}}$
Given: $\text{f}:\text{R}-\Big\{-\frac{4}{3}\Big\}\rightarrow\text{R}\ \text{and}\ f(\text{x})=\frac{4\text{x}}{3\text{x}+4}$
Now, Range of $\text{f}\rightarrow\text{R}-\Big\{-\frac{4}{3}\Big\}$
Let $\text{y}=f(\text{x})\ \ \ \therefore\ \text{y}=\frac{4\text{x}}{3\text{x}+4}\ \ \Rightarrow\ 3\text{xy}+4\text{y}=4\text{x}$
$\Rightarrow\ \ \ \text{x}(4-3\text{y})=4\text{y}\ \ \Rightarrow\ \text{x}=\frac{4\text{y}}{4-3\text{y}}$
$\therefore\ \ f^{-1}(\text{y})=\text{g(y)}=\frac{4\text{y}}{3-4\text{y}}$
Therefore, option (B) is correct.
View full question & answer→MCQ 1081 Mark
The binary operation $\times$ defined on $N$ by $a \times b = a + b + ab$ for all $a, b \in N$ is:
- A
- B
- ✓
Both commutative and associative.
- D
AnswerCorrect option: C. Both commutative and associative.
View full question & answer→MCQ 1091 Mark
The function $f : R \rightarrow R, f(x) = x^2$ is:
- A
Injective but not surjective.
- B
Surjective but not injective.
- C
Injective as well as surjective.
- ✓
Neither injective nor surjective.
AnswerCorrect option: D. Neither injective nor surjective.
Given function is $f : R \rightarrow R, f(x) = x^2$
If $f(x) = f(y)$ then
$x^2 = y^2$
$\Rightarrow\ \text{x}\pm\text{y}$
Hence, it is not one$-$one or injective.
$f(x) = y$
$y = x^2$
$\text{x}=\pm\sqrt{\text{y}}$
But co$-$domain is $R.$
Hence, it is not onto or surjective.
View full question & answer→MCQ 1101 Mark
Let $f : R \rightarrow R, g : R \rightarrow R$ be two functions such that $f(x) = 2x – 3, g(x) = x^3 + 5.$ The function $(fog)^{-1} (x)$ is:
- A
$\Big(\frac{\text{x}+7}{2}\Big)^\frac{1}{3}$
- B
$\Big(\text{x}-\frac{7}{2}\Big)^\frac{1}{3}$
- C
$\Big(\frac{\text{x}-2}{7}\Big)^\frac{1}{3}$
- ✓
$\Big(\frac{\text{x}-7}{2}\Big)^\frac{1}{3}$
AnswerCorrect option: D. $\Big(\frac{\text{x}-7}{2}\Big)^\frac{1}{3}$
$\Big(\frac{\text{x}-7}{2}\Big)^\frac{1}{3}$
View full question & answer→MCQ 1111 Mark
The relation 'R' in N × N such that (a, b)R(c, d) ⇔ a + d = b + c is:
- A
Reflexive but not symmetric.
- B
Reflexive and transitive but not symmetric.
- ✓
- D
AnswerWe observe the following properties of relation R.
Reflexivity: Let $(\text{a, b})\in\text{N}\times\text{N}$
$\Rightarrow\ \text{a, b}\in\text{N}$
$\Rightarrow\ \text{a}+\text{b}=\text{b}+\text{a}$
$\Rightarrow\ (\text{a, b})\in\text{R}$
So, R is reflexive on N × N.
Symmetry: Let $(\text{a, b}),\ (\text{c, d})\in\text{N}\times\text{N}$ such that (a, b)R(c, d)
$\Rightarrow\ \text{a}+\text{d}=\text{b}+\text{c}$
$\Rightarrow\ \text{d}+\text{a}=\text{c}+\text{b}$
$\Rightarrow\ (\text{d, c}),\ (\text{b, a})\in\text{R}$
So, R is symmetric on N × N.
Transitivity: Let $(\text{a, b}),\ (\text{c, d}),\ (\text{e, f})\in\text{N}\times\text{N}$ such that (a, b)R(c, d) and (c, d)R(e, f)
⇒ a + d = b + c and c + f = d + e
⇒ a + d + c + f = b + c + d + e
⇒ a + f = b + e
⇒ (a, b)R(e, f)
So, R is transitive on N × N.
Hence, R is an equivalence relation on N.
View full question & answer→MCQ 1121 Mark
If $f$ is an invertible function defined as $\text{f(x)}=\frac{3\text{x}-4}{5},$ then $f^{-1}(x)$ is:
- A
$5x + 3$
- B
$5x + 3$
- ✓
$\frac{5\text{x}+4}{3}$
- D
$\frac{3\text{x}+2}{3}$
AnswerCorrect option: C. $\frac{5\text{x}+4}{3}$
$\frac{5\text{x}+4}{3}$
View full question & answer→MCQ 1131 Mark
Let f: R → R be defined as f(x) = 3x. Choose the correct answer.
- ✓
- B
- C
f is one-one but not onto
- D
f is neither one-one nor onto.
Answerf: R → R is defined as f(x) = 3x. Let $\text{x},\text{y}\in\text{R}$ such that f(x) = f(y). ⇒ 3x = 3y ⇒ x = y $\therefore$ f is one-one.Also, for any real number (y) in co-domain R, there exists $\frac{\text{y}}{3}$ in R such that $f\Big(\frac{\text{y}}{3}\Big)=3\Big(\frac{\text{y}}{3}\Big)=\text{y}$
$\therefore$ f is onto. Hence, function f is one-one and onto. The correct answer is A
View full question & answer→MCQ 1141 Mark
The number of commutative binary operation that can be defined on a set of $2$ elements is:
View full question & answer→MCQ 1151 Mark
Let $A = R – \{3\}, B = R – \{1\}.$ Let $f : A \rightarrow B$ be defined by $\text{f(x)}=\frac{\text{x}-2}{\text{x}-3}.$ Then,
AnswerCorrect option: A. $F$ is bijective.
View full question & answer→MCQ 1161 Mark
If the binary operation $\times$ is defind on the set $Q +$ of all positive rational numbers by $\text{a}\times\text{b}=\frac{\text{ab}}{4}.$ Then, $3\times\Big(\frac{1}{5}\times\frac{1}{2}\Big)$ is equal to:
- ✓
$\frac{3}{160}$
- B
$\frac{5}{160}$
- C
$\frac{3}{10}$
- D
$\frac{3}{40}$
AnswerCorrect option: A. $\frac{3}{160}$
View full question & answer→MCQ 1171 Mark
If $f(x) = (ax^2 + b)^3,$ then the function $g$ such that $f(g(x)) = g(f(x))$ is given by:
- A
$\text{g}(\text{x})=\Big(\frac{\text{b}-\text{x}^\frac{1}{3}}{\text{a}}\Big)$
- B
$\text{g}(\text{x})=\frac{1}{(\text{ax}^2+\text{b})^3}$
- C
$\text{g}(\text{x})=(\text{ax}^2+\text{b})^\frac{1}{3}$
- ✓
$\text{g}(\text{x})=\Big(\frac{\text{x}^\frac{1}{3}-\text{b}}{\text{a}}\Big)^\frac{1}{2}$
AnswerCorrect option: D. $\text{g}(\text{x})=\Big(\frac{\text{x}^\frac{1}{3}-\text{b}}{\text{a}}\Big)^\frac{1}{2}$
$\text{g}(\text{x})=\Big(\frac{\text{x}^\frac{1}{3}-\text{b}}{\text{a}}\Big)^\frac{1}{2}$
View full question & answer→MCQ 1181 Mark
A binary operation * on Z defined by a * b = 3a + b for all a, b ∈ Z, is:
- A
- B
- ✓
- D
Commutative and associative.
AnswerLet $\text{a, b}\in\text{Z}$
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.
View full question & answer→MCQ 1191 Mark
Choose the correct answer from the given four options.
Let A = {1, 2, 3} and consider the relation R = {1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}. Then R is:
- ✓
Reflexive but not symmetric.
- B
Reflexive but not transitive.
- C
Symmetric and transitive.
- D
Neither symmetric, nor transitive.
AnswerCorrect option: A. Reflexive but not symmetric.
Given that, A = {1, 2, 3}
and R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}
$\because\ (1,1), (2,2),(3,3)\in\text{R}$
Hence, R is reflexive.
$(1,2)\in\text{R}$ but $(2,1)\notin\text{R}$
Hence, R is not symmetric.
$(1,2)\in\text{R}$ and $(2,3)\in\text{R}$
$\Rightarrow\ (1,3)\in\text{R}$
Hence, R is transitive.
View full question & answer→MCQ 1201 Mark
Which of the following functions from $\text{A}=\{\text{x}:-1\leq\text{x}\leq1\}$ to itself are bijections?
AnswerCorrect option: B. $\text{g(x)}=\sin\big(\frac{\pi\text{x}}{2}\big)$
- Range of $\text{f}=\Big[\frac{-1}{2},\frac{1}{2}\Big]\neq\text{A}$
- So$, f$ is not a bijection.
- Range $=\Big[\sin\Big(\frac{-\pi}{2}\Big),\ \sin\Big(\frac{\pi}{2}\Big)\Big]=[-1,1]=\text{A}$
- So, $g$ is a bijection.
- $h(-1) = |-1| = 1$
- And $h(1) = |1| = 1$
$\Rightarrow -1$ and $1$ have the same images.
So, $h$ is not a bijection.
- $k(-1) = (-1)^2 = 1$
- And $k(1) = (1)^2 = 1$
$\Rightarrow -1$ and $1$ have the same images.
So, $k$ is not a bijection.
View full question & answer→MCQ 1211 Mark
The function $f : R \rightarrow R$ defined by $f(x) = 2^x + 2^{|x|}$ is:
- A
One $-$ one and onto.
- B
Many $-$ one and onto.
- ✓
One $-$ one and into.
- D
Many $-$ one and into.
AnswerCorrect option: C. One $-$ one and into.
The function $f : R \rightarrow R$ defined by $f(x) = 2^x + 2^{|x|}$
Here, for each value of $x$ we will get different values of $f(x).$
Hence, it is one $-$ one function.
Also, each element of codomain is mapped to at most one element of the domain.
Function is one $-$ one and into.
View full question & answer→MCQ 1221 Mark
Choose the correct answer from the given four options.
Let us define a relation R in R as aRb if a ≥ b. Then R is:
- A
- ✓
Reflexive, transitive but not symmetric.
- C
Symmetric, transitive but not reflexive.
- D
Neither transitive nor reflexive but symmetric.
AnswerCorrect option: B. Reflexive, transitive but not symmetric.
We are given that, aRb if a ≥ b
⇒ aRa ⇒ a ≥ a which is true.
For relation aRb to be symmetric, we must have a ≥ b and b ≥ a which can’t be possible.
Hence, R is not symmetric.
For relation aRb to be transitive, we must have aRb and bRc.
⇒ a ≥ b and b ≥ c
⇒ a ≥ c
Hence, R is transitive.
View full question & answer→MCQ 1231 Mark
Choose the correct answer from the given four options.
If the set A contains 5 elements and the set B contains 6 elements, then the number of one-one and onto mappings from A to B is:
AnswerSince, the number of elements in B is more than A.
Hence, there cannot be any one-one and onto mapping from A to B.
View full question & answer→MCQ 1241 Mark
Let $f : R \rightarrow R$ be the functions defined by $f(x) = x^3 + 5.$ Then $f^{-1}(x)$ is:
- A
$(\text{x}+5)^\frac{1}{3}$
- ✓
$(\text{x}-5)^\frac{1}{3}$
- C
$(5-\text{x})^\frac{1}{3}$
- D
$5-\text{x}$
AnswerCorrect option: B. $(\text{x}-5)^\frac{1}{3}$
$(\text{x}-5)^\frac{1}{3}$
View full question & answer→MCQ 1251 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
$\frac{3}{40}$
AnswerCorrect option: A. $\frac{3}{160}$
Given $\text{a}\odot\text{b}=\frac{\text{ab}}4$
$\Rightarrow\Big(\frac{1}5\odot\frac{1}2\Big)$
$=\frac{\frac{1}5.\frac{1}2}{4}$
$=\frac{1}{40}$
$3\odot\Big(\frac{1}5\odot\frac{1}2\Big)$
$=3\odot\frac{1}{40}$
$=\frac{\frac{1}{40}.3}{4}$
$=\frac{3}{160}$
View full question & answer→MCQ 1261 Mark
Let A = {1, 2, 3} and B = {(1, 2), (2, 3), (1, 3)} be a relation on A. Then, R is:
- A
Neither reflexive nor transitive.
- B
Neither symmetric nor transitive.
- ✓
- D
AnswerReflexivity: Since $(1, 1)\notin\text{B,}$ B is not reflexive on A.
Symmetry: Since $1,2\in\text{B}$ but $2,1\notin\text{B,}$ B is not symmetric on A.
Transitivity: Since $1,2\in\text{B},\ 2,3\in\text{B}$ and $1,3\in\text{B,}$ B is transitive on A.
View full question & answer→MCQ 1271 Mark
Consider a binary operation $∗$ on $N$ defined as $a^∗ b = a^3 + b^3.$
- A
$∗$ is both associative and commutative.
- ✓
$∗$ is commutative but not associative.
- C
$∗$ is neither commutative nor associative.
- D
$∗$ is associative but not commutative.
AnswerCorrect option: B. $∗$ is commutative but not associative.
Given that the binary operation $∗$ on $N$ is defined as $a∗b = a^3 + b^3.$
Apply the given binary operation on $b∗a$.
$b∗a = b^3 + a^3 = a^3 + b^3$
It shows that the value of $a∗b$ is equal to that of $b∗a.$
So, the operation is commutative.
Consider different values of the variable as $a = 1, b = 2$ and $c = 3.$
Apply the given binary operation on $(a∗b)∗c.$
$(a∗b)∗c = (1∗2)∗3 = (1^3 + 2^3)∗3 = 9^3 + 3^3 = 729 + 27 = 756$
Apply the given binary operation on $a∗(b∗c)$.
$(a∗b)∗c = 1∗(2∗3) = 1∗(2^3 + 3^3) = 1^3 + 35^3 = 42876$
$(a∗b)∗c \neq a∗(b∗c)$
So the operation is not associative.
Therefore, the given operation is commutative but not associative.
View full question & answer→MCQ 1281 Mark
For binary operation $\times$ defind on $R – \{1\}$ such that $\text{a}\times\text{b}=\frac{\text{a}}{\text{b}+1}$ is:
- A
- B
- C
- ✓
Both $(a)$ and $(b).$
AnswerCorrect option: D. Both $(a)$ and $(b).$
View full question & answer→MCQ 1291 Mark
R is a relation on the set Z of integers and it is given by (x, y) ∈ R ⇔ | x - y | ≤ 1. Then, R is:
- A
Reflexive and transitive.
- ✓
- C
Symmetric and transitive.
- D
AnswerReflexivity: Let $\text{x}\in\text{R.}$ Then,
$\text{x}-\text{x}=0<1$
$\Rightarrow\ |\text{x}-\text{x}|\leq1$
$\Rightarrow\ (\text{x, x})\in\text{R}$ for all $\text{x}\in\text{Z}$
So, R is reflexive on Z.
Symmetry: Let $\text{x, y}\in\text{R.}$ Then,
$|\text{x}-\text{y}|\leq0$
$\Rightarrow\ |-(\text{y}-\text{x})|\leq1$
$\Rightarrow\ |(\text{y}-\text{x})|\leq1$ [Since |x - y| = |y - x|]
$\Rightarrow\ (\text{y, x})\in\text{R}$ for all $\text{x, y}\in\text{Z}$
So, R is symmetric on Z.
Transitivity: Let $(\text{x, y})\in\text{R}$ and $(\text{y, z})\in\text{R}.$ Then,
$|\text{x}-\text{y}|\leq1$ and $|\text{y}-\text{z}|\leq1$
⇒ It is not always true that $|\text{x}-\text{y}|\leq1.$
$\Rightarrow\ (\text{x, z})\notin\text{R}$
So, R is not transitive on Z.
View full question & answer→MCQ 1301 Mark
The number of binary operation that can be defined on a set of 2 elements is:
AnswerTotal number of binary operations on a set containing n elements is
$\text{(n)}^{\text{n}^2}$ so for n = 2 we have $(2)^{2^2}=2^4=16$
View full question & answer→MCQ 1311 Mark
If $\text{F}:[1,\infty)\rightarrow[2,\infty)$ is given by $\text{f(x)}=\text{x}+\frac{1}{\text{x}},$ then $f^{-1}(x)$ equals:
- ✓
$\frac{\text{x}+\sqrt{\text{x}^2-4}}{2}$
- B
$\frac{\text{x}}{1+\text{x}^2}$
- C
$\frac{\text{x}-\sqrt{\text{x}^2-4}}{2}$
- D
$1+\sqrt{\text{x}^2-4}$
AnswerCorrect option: A. $\frac{\text{x}+\sqrt{\text{x}^2-4}}{2}$
Let $f^{-1}(x) = y$
$\Rightarrow\ \text{f(y)} = \text{x}$
$\Rightarrow\ \text{y}+\frac{1}{\text{y}}=\text{x}$
$\Rightarrow\ \text{y}^2 + 1 = \text{xy}$
$\Rightarrow\ \text{y}^2 - \text{xy} + 1 = 0$
$\Rightarrow\ \text{y}^2-2\times\text{y}\times\frac{\text{x}}{2}+\big(\frac{\text{x}}{2}\big)^2-\big(\frac{\text{x}}{2}\big)^2+1=0$
$\Rightarrow\ \text{y}^2-2\times\text{y}\times\frac{\text{x}}{2}+\big(\frac{\text{x}}{2}\big)^2=\frac{\text{x}^2-1}{4}$
$\Rightarrow\ \Big(\text{y}-\frac{\text{x}}{2}\Big)^2=\frac{\text{x}^2-1}{4}$
$\Rightarrow\ \text{y}-\frac{\text{x}}{2}=\frac{\sqrt{\text{x}^2-4}}{2}$
$\Rightarrow\ \text{y}=\frac{\text{x}}{2}+\frac{\sqrt{\text{x}^2-4}}{2}$
$\Rightarrow\ \text{y}=\frac{\text{x}+\sqrt{\text{x}^2-4}}{2}$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{\text{x}+\sqrt{\text{x}^2-4}}{2}$
View full question & answer→MCQ 1321 Mark
If $f(x) =\frac{3\text{x}+2}{5\text{x}-3}$ then $\text{(fof)(x)}$ is:
View full question & answer→MCQ 1331 Mark
Let $f : R \rightarrow R$ be a function defined by $\text{f(x)}=\frac{\text{x}^2-8}{\text{x}^2+2}.$ Then, $f$ is:
- A
One $-$ one but not onto.
- B
One $-$ one and onto.
- C
Onto but not one $-$ one.
- ✓
Neither one $-$ one nor onto.
AnswerCorrect option: D. Neither one $-$ one nor onto.
Injectivity: Let $x$ and $y$ be two elements in the domain $(R),$ such that
$f(x) = f(y)$
$\frac{\text{x}^2-8}{\text{x}^2+2}=\frac{\text{y}^2-8}{\text{y}^2+2}$
$\Rightarrow (x^2 - 8)(y^2 + 2) = (y^2 - 8)(x^2 + 2)$
$\Rightarrow x^2y^2 + 2x^2 - 8y^2 - 16 $
$= x^2y^2 + 2y^2 - 8x^2 - 16$
$\Rightarrow 10x^2 = 10y^2$
$\Rightarrow x^2 = y^2$
$\Rightarrow\ \text{x}=\pm\text{y}$
So, $f$ is not one $-$ one.
Surjectivity: $\text{f}(-1)=\frac{(-1)^2-8}{(-1)^2+2}=\frac{1-8}{1+2}=\frac{-7}{3}$
and $\text{f(1)}=\frac{(1)^2-8}{(1)^2+2}=\frac{1-8}{1+2}=\frac{-7}{3}$
$\Rightarrow\ \text{f}(-1)=\text{f}(1)=\frac{-7}{3}$
$\Rightarrow f$ is not onto.
View full question & answer→MCQ 1341 Mark
Subtraction of integers is:
- ✓
Commutative but no associative.
- B
Commutative and associative.
- C
Associative but not commutative.
- D
Neither commutative nor associative.
AnswerCorrect 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.
View full question & answer→MCQ 1351 Mark
If A = {a, b, c}, then the relation R = {(b, c)} on A is:
- A
- B
- ✓
- D
Reflexive and transitive only.
AnswerThe relation R = {(b, c)} is neither reflexive nor symmetric because every element of A is not related to itself. Also, the ordered pair of R obtained by interchanging its elements is not contained in R.
We observe that R is transitive on A because there is only one pair.
View full question & answer→MCQ 1361 Mark
The binary operation $\times $ defind on set $R,$ given by $\text{a}\times\text{b}=\frac{\text{a}+\text{b}}{2}$ for all $a, b \in R$ is:
- ✓
- B
- C
Both $(a)$ and $(b).$
- D
View full question & answer→MCQ 1371 Mark
The maximum number of equivalence relations on the set $A = \{1, 2, 3\}$ are:
View full question & answer→MCQ 1381 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:
AnswerGiven 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.
View full question & answer→MCQ 1391 Mark
For real numbers x and y, define xRy if $\text{x}-\text{y}+\sqrt{2}$ is an irrational number. Then the relation R is:
AnswerWe have,
$\text{R} = \big\{(\text{x, y}):\text{x}-\text{y}+\sqrt{2}$ $$ is an irrational number, $\text{x, y}\in\text{R}\big\}$
As, $\text{x}-\text{x}+\sqrt{2}=\sqrt{2},$ which is an irrational number
$\Rightarrow\ (\text{x, x})\in\text{R}$
So, R is reflexive relation.
Since, $\Big(\sqrt{2},2\Big)\in\text{R}$
i.e. $\sqrt{2}-2+\sqrt{2}=2\sqrt{2}-2,$ which is an irrational number
but $2-\sqrt{2}+\sqrt{2}=2,$ which is a rational number
$\Rightarrow\ \Big(2,\sqrt{2}\Big)\notin\text{R}$
So, R is not symmetric relation.
Also, $\Big(\sqrt{2},2\Big)\in\text{R}$ and $\Big(2,2\sqrt{2}\Big)\in\text{R}$
$\Rightarrow\ \Big(\sqrt{2},2\sqrt{2}\Big)\notin\text{R}$
So, R is not transitive relation.
View full question & answer→MCQ 1401 Mark
Let $[x]$ denote the greatest integer less than or equal to $x.$ If $f(x) = \sin^{-1}x, g(x) = [x^2]$ and $\text{h(x)}=2\text{x},\frac{1}{2}\leq\text{x}\leq\frac{1}{\sqrt{2}},$ then
- A
$\text{fogoh(x)}=\frac{\pi}{2}$
- B
$\text{fogoh(x)}=\pi$
- ✓
$\text{hofog}=\text{hogof}$
- D
$\text{hofog}\neq\text{hogof}$
AnswerCorrect option: C. $\text{hofog}=\text{hogof}$
$gof(x) = h(f(g(x)))$
$= h(f([x]))$
$= h(\sin^{-1}[x])$
$= 2\sin^{-1}[x]$
$= 2 \times 0 = 0$
$f(x) = \sin^{-1}x$
$gof(x) = gof(x) = 0$
View full question & answer→MCQ 1411 Mark
The binary operation $^*$ is defined by $a ^* b = a^2 + b^2 + ab + 1,$ then $(2 ^* 3) ^* 2$ is equal to:
AnswerGiven: $a ^* b = a^2 + b^2 + ab + 1$
$2 ^* 3 = 2^2 + 3^2 + 2 \times 3 + 1$
$= 4 + 9 + 6 + 1$
$= 20$
$(2 ^* 3) ^* 2 = 20 ^* 2$
$= 20^2 + 2^2 + 20 \times 2 + 1$
$= 400 + 4 + 40 + 1$
$= 445$
View full question & answer→MCQ 1421 Mark
Let $\times$ be a binary operation on $Q,$ defined by $\text{a}\times\text{b}=\frac{3\text{ab}}{5}$ is:
- A
- B
- ✓
Both $(a)$ and $(b).$
- D
AnswerCorrect option: C. Both $(a)$ and $(b).$
View full question & answer→MCQ 1431 Mark
Let T be the set of all triangles in the Euclidean plane, and let a relation R on T be defined as aRb if a is congruent to b for all a, b ∈ T. Then, R is:
- A
Reflexive but not symmetric.
- B
Transitive but not symmetric.
- ✓
- D
AnswerGiven that R is T be the set of all triangle in the Euclidean plane, and a relation R on T be defined as aRb if a is congruent to b for all a, b ∈ T.
Here, congruency of triangles follows reflexive, symmetric and transitive property.
Hence, it is an equivalence relation.
View full question & answer→MCQ 1441 Mark
The function $f : A \rightarrow B$ defined by $f(x) = 4x + 7, x \in R$ is:
AnswerCorrect option: A. One$-$one
View full question & answer→MCQ 1451 Mark
f : R → R given by $\text{f(x)}=\text{x}+\sqrt{\text{x}^2}$ is:
Answer$\text{f(x)}=\text{x}+\sqrt{\text{x}^2}=\text{x}\pm\text{x}=0\text{ or }2\text{x}$
⇒ Each element of the domain has 2 images.
f is not a function.
View full question & answer→MCQ 1461 Mark
Let $g(x) = x^2 - 4x - 5,$ then:
AnswerCorrect option: B. $G$ is not one$-$one on $R.$
$G$ is not one$-$one on $R.$
View full question & answer→MCQ 1471 Mark
Choose the correct answer out of the given four options.Let T be the set of all triangles in the Euclidean plane and let a relation R on T be defined as aRb, if a is congruent to $\text{b}\ \forall\ \text{a},\ \text{b}\in\text{T}.$ Then, R is:
- A
Reflexive but not transitive.
- B
Transitive but not symmetric.
- ✓
- D
AnswerConsider that aRb, if a is congruent to b, $\forall\ \text{a, b}\in\text{T}$
Then, $\text{aRa}\Rightarrow\ \text{a}\cong\text{a},$
Which is true for all $\text{a}\in\text{T}$
So, R is reflexive, ....(i)
Let $\text{aRb}\Rightarrow\ \text{a}\cong\text{b}$
$\Rightarrow\ \text{b}\cong\text{a}\Rightarrow\ \text{b}\cong\text{a}$
$\Rightarrow\ \text{bRa}$
So, R is symmetric. ...(ii)
Let aRb and bRc
$\Rightarrow\ \text{a}\cong\text{b}\text{ and }\text{b}\cong\text{c}$
$\Rightarrow\ \text{a}\cong\text{c}\Rightarrow\ \text{aRc}$
So, R is transitive. .....(iii)
Hence, R is equivalence relation.
View full question & answer→MCQ 1481 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:
AnswerCorrect 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 $0.1.$
View full question & answer→MCQ 1491 Mark
Let $\text{A}=\{\text{x}:-1\leq\text{x}\leq1\}$ and $f : A \rightarrow A$ such that $\text{f(x)}=\text{x}|\text{x}|,$ then $f$ is:
- ✓
- B
Injective but not surjective.
- C
Surjective but not injective.
- D
Neither injective nor surjective.
AnswerGiven function is $\text{A}=\{\text{x}:-1\leq\text{x}\leq1\}$ and $f : A \rightarrow A$ such that $\text{f(x)}=\text{x}|\text{x}|$
For the mod function we have to check three cases as $x < 0, x = 0, x > 0.$
For example,
$x < 0$
$f(x) = x|x| < 0$
$|x| = -x$
$y = -x^2$
$\text{x}=-\sqrt{-\text{y}}$ which is not possible for $x > 0$
Hence, $f$ is onto.
$\Rightarrow f$ is bijection.
View full question & answer→MCQ 1501 Mark
Choose the correct answer from the given four options. Let $f : A \rightarrow B$ and $g : B \rightarrow C$ be the bijective functions. Then $(gof)^{-1}$ is:
- ✓
$f^{-1}og^{-1}$
- B
$fog$
- C
$g^{-1}of^{-1}$
- D
$gof$
AnswerCorrect option: A. $f^{-1}og^{-1}$
Given that, $f : A \rightarrow B$ and $g : B \rightarrow C$ be the bijective functions.
$(\text{f}^{-1}\text{o}\text{g}^{-1})\text{o}(\text{gof})=\text{f}^{-1}\text{o}(\text{g}^{-1}\text{gof})$
$=\text{f}^{-1}\text{o}(\text{g}^{-1}\text{og})\text{of} ($As composition of functions is associative$)$
$=(\text{f}^{-1}\text{o}\text{I}_{\text{B}}\text{of}) ($where $I_B$ is identity function on $B)$
$=(\text{f}^{-1}\text{o}\text{I}_{\text{B}})\text{of}$
$=\text{f}^{-1}\text{of}$
$=\text{I}_{\text{A}}$
Thus $(\text{gof})^{-1}=\text{f}^{-1}\text{og}^{-1}$
View full question & answer→