MCQ 11 Mark
The range of the function $\text{f(x)}=^{7-\text{x}}\text{P}_{\text{x}-3}$ is:
- A
$\{1, 2, 3, 4, 5\}$
- B
$\{1, 2, 3, 4, 5, 6\}$
- C
$\{1, 2, 3, 4\}$
- ✓
$\{1, 2, 3\}$
AnswerCorrect option: D. $\{1, 2, 3\}$
We know that
$7-\text{x}>0;\ \text{x}-3\geq0$ and $7-\text{x}\geq\text{x}-3$
$\Rightarrow\ \text{x}<7;\ \text{x}\geq3$ and $2\text{x}\leq10$
$\Rightarrow\ \text{x}<7;\ \text{x}\geq3$ and $\text{x}\leq5$
Therefore, x = 3, 4, 5
Range of $\text{f}=\Big\{^{(7-3)}\text{P}_{(3-3)},\ ^{(7-4)}\text{P}_{(4-3)},\ ^{(5-3)}\text{P}_{(7-5)}\Big\}$
$=\left\{4 \mathrm{P}_0, 3 \mathrm{P}_1, 2 \mathrm{P}_2\right\}$
$= {1, 3, 2}$
$= {1, 2, 3}$
View full question & answer→MCQ 21 Mark
If $f : R \rightarrow R, g : R \rightarrow R$ and $h : R \rightarrow R$ are such that $f(x) = x^2$, $\text{g(x)}=\tan\text{x}$ and $\text{h(x)}=\log\text{x},$ then the value of $(go(foh)) (x),$ if $x = 1$ will be:
View full question & answer→MCQ 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 $\triangle$
- 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.
AnswerCorrect 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,
$\phi\triangle\text{A}=\big(\overline{\phi}\cap\text{A}\big)\cup\big(\phi\cap\overline{\text{A}}\big)$
$=\text{A}\cup\phi$
$=\text{A}$
View full question & answer→MCQ 41 Mark
The relation S defined on the set R of all real number by the rule aSb iff a ≥ b 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.
Solution:
The relation S is reflexive, since for any $(\text{a, a})\in\text{S}$ the condition a2b holds,
The relation S is not symmetric since, for any $(\text{a, b}]\in\text{S}$ but $(\text{b, a})\notin\text{S}$
The relation S is transitive since, for any $(\text{a, b}]\in\text{S}$ and $(\text{b, c})\in\text{S}$
Therefore, $(\text{a, c})\notin\text{S}$
View full question & answer→MCQ 51 Mark
$S$ is a relation over the set $R$ of all real numbers and it is given by $(\text{a, b})\in\text{S}\Leftrightarrow\text{ab}\geq0.$ Then, $S$ is:
- A
Symmetric and transitive only.
- B
Reflexive and symmetric only.
- C
- ✓
AnswerReflexivity: Let $\text{a}\in\text{R}$
Then,
$aa = a^2 > 0 \Rightarrow\ \text{a, }\forall$
So, $S$ is reflexive on $R.$
Symmetry: Let $(\text{a, b})\in\text{S}$
Then,
$\text{a, b}\in\text{S}\Rightarrow\ \text{ab}\geq0$
$\Rightarrow\ \text{ba}\geq0 $
$\Rightarrow\ \text{ba}\geq0$
$\Rightarrow\ \text{b, a}\in\text{S}\ \forall\ \text{a, b}\in\text{R}$
So, S is symmetric on $R.$
Transitivity: If $\text{a, b, b, c}\in\text{S}$
$\Rightarrow\ \text{ab}\geq0$ and $\text{bc}\geq0$
$\Rightarrow\ \text{ab}\times\text{bc}\geq0$
$\Rightarrow\ \text{ac}\geq0$
$\text{b}^2\geq0\Rightarrow\ \text{a, c}\in\text{S}$ for all $\text{a, b, c}\in\text{set R}$
Hence, $S$ is an equivalence relation on $R.$
View full question & answer→MCQ 61 Mark
Choose the correct answer from the given four options.
Let $\text{f}:\text{R}-\Big\{\frac{3}{5}\Big\}\rightarrow\ \text{R}$ be defined by $\text{f}(\text{x})=\frac{3\text{x}+2}{5\text{x}-3}.$ Then,
- ✓
$\text{f}^{-1}(\text{x})=\text{f}(\text{x})$
- B
$\text{f}^{-1}(\text{x})=-\text{f}(\text{x})$
- C
$(\text{fof})\text{x}=-\text{x}$
- D
$\text{f}^{-1}\text{x}=\frac{1}{19}\text{f}(\text{x})$
AnswerCorrect option: A. $\text{f}^{-1}(\text{x})=\text{f}(\text{x})$
Solution:
We have, $\text{f}(\text{x})=\frac{3\text{x}+2}{5\text{x}-3}=\text{y}\ (\text{let})$
$\Rightarrow\ 3\text{x}+2=5\text{xy}-3\text{y}$
$\Rightarrow\ \text{x}(3-5\text{y})=-3\text{y}-2$
$\Rightarrow\ \text{x}=\frac{3\text{y}+2}{5\text{y}-3}$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{3\text{x}+2}{5\text{x}-3}$
$\therefore\ \text{ f}^{-1}\text{x}=\text{f}(\text{x})$
View full question & answer→MCQ 71 Mark
The relation R = {(1, 1), (2, 2), (3, 3)} on the set {1, 2, 3} is:
AnswerSolution:
R = {(a, b): a = b and a, b $\in\text{A}$}
Reflexivity: Let $\text{a}\in\text{A}$
Here,
a = a
$\Rightarrow\ (\text{a, a})\in\text{R}$ for all $\text{a}\in\text{A}$
So, R is reflexive on A.
Symmetry: Let $\text{a, b}\in\text{A}$ such that $ (\text{a, b})\in\text{R}.$ Then,
$ (\text{a, b})\in\text{R}$
$\Rightarrow\ \text{a}=\text{b}$
$\Rightarrow\ \text{b}=\text{a}$
$\Rightarrow\ (\text{b, a})\in\text{R}$ for all $\text{a}\in\text{A}$
So, R is symmetric on A.
Transitive: Let $\text{a, b, c}\in\text{A}$ such that $ (\text{a, b})\in\text{R}$ and $ (\text{b, c})\in\text{R}.$ Then,
$ (\text{a, b})\in\text{R}\Rightarrow\ \text{a}=\text{b}$
and $ (\text{b, c})\in\text{R}\Rightarrow\ \text{b}=\text{c}$
$\Rightarrow\ \text{a}=\text{c}$
$\Rightarrow\ (\text{a, c})\in\text{R}$ for all $\text{a}\in\text{A}$
So, R is transitive on A.
Hence, R is an equivalence relation on A.
View full question & answer→MCQ 81 Mark
Choose the correct answer from the given four options.
Let f : R → R be defined by $\text{f}(\text{x})=\frac{1}{\text{x}}\ \forall\ \text{x}\in\text{R}.$ Then f is:
AnswerGiven that, $\text{f}(\text{x})=\frac{1}{\text{x}}\ \forall\ \text{x}\in\text{R}.$
For x = 0, f(x) is not defined.
Hence, f(x) is a not defined function.
View full question & answer→MCQ 91 Mark
Let $f : R \rightarrow R$ be a function defined by $\text{f(x)}=\frac{\text{e}^{|\text{x}|}-\text{e}^{-\text{x}}}{\text{e}^\text{x}+\text{e}^{-\text{x}}}$ then $f(x)$ is:
- A
One$-$one onto.
- B
One$-$one but not onto.
- C
Onto but not one$-$one.
- ✓
View full question & answer→MCQ 101 Mark
Let $\text{A}=\{\text{x}\in\text{R}:-1\leq\text{x}\leq1\}=\text{B}.$ Then, the mapping $f : A \rightarrow B$ given by $f(x) = x|x|$ is:
- A
Injective but not surjective.
- B
Surjective but not injective.
- ✓
- D
AnswerGiven function is $\text{A}=\{\text{x}:-1\leq\text{x}\leq1\}$ and $f : A \rightarrow A$ such that $f(x) = x|x|$
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 111 Mark
Choose the correct answer from the given four options. 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. $(\text{x}-5)^\frac{1}{3}$
we are given that, $\text{f}(\text{x})=\text{x}^3 +5$
Let us suppose, $\text{y}=\text{x}^3+5$
$\Rightarrow\ \text{x}^3=\text{y}-5$
$\Rightarrow\text{x}=(\text{y}-5)^{\frac{1}{3}}$
$\begin{bmatrix}\because\text{f}(\text{x})=\text{y}\\\Rightarrow\text{x}=\text{f}^{-1}(\text{y})\end{bmatrix}$
$\Rightarrow\text{f}^{-3}(\text{y})=(\text{y}-5)^{\frac{1}{3}}$
$\Rightarrow\text{f}^{-1}(\text{x})=(\text{x}-5)^{\frac{1}{3}}$
View full question & answer→MCQ 121 Mark
Let R be the relation in the set $\{1, 2, 3, 4\}$ given by $R = \{(1, 2), (2, 2), (1, 1), (4,4), (1, 3), (3, 3), (3, 2)\}.$Choose the correct answer.
- A
$R$ is reflexive and symmetric but not transitive.
- ✓
$R$ is reflexive and transitive but not symmetric.
- C
$R$ is symmetric and transitive but not reflexive.
- D
$R$ is an equivalence relation.
AnswerCorrect option: B. $R$ is reflexive and transitive but not symmetric.
Let $R$ be the relation in the set $\{1, 2, 3, 4\}$ is given by
$R = \{(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)\}$
| (a) |
$(1,1),(2,2),(3,3),(4,4)\in\text{R}$ |
$\therefore$ |
$R$ is reflexive |
| (b) |
$(1,2)\in\text{R}$ but $(2,1)\notin\text{R}$ |
$\therefore$ |
$R$ is not symmetric |
| (c) |
$\text{If }(1,3)\in\text{R}$ and $(3,2)\in\text{R}$ then $(1,2)\in\text{R}$ |
$\therefore$ |
$R$ is transitive |
Therefore, option $(B)$ is correct. View full question & answer→MCQ 131 Mark
Let L denote the set of all straight lines in a plane. Let a relation R be defined by lRm if l is perpendicular to m for all l, m ∈ L. Then, R is:
AnswerGiven that L denote the set of all straight lines in a plane.
A relation R be defined by lRm if is perpendicular to m for all l, m ∈ L.
R is not reflexive. R is symmetric as we can say $\text{l}\bot\text{m}$ or $\text{m}\bot\text{l}.$
View full question & answer→MCQ 141 Mark
In the set Z of all integers, which of the following relation R is not an equivalence relation?
AnswerCorrect option: A. xRy : if $\text{x}\leq\text{y}$
In the set of Z of all integers xRy : if $\text{x}\leq\text{y}$ is not an equivalence relation.
For the relation $\text{x}\leq\text{y}(\text{x, y})\in\text{R}$ but (y, x) not belongs to y as $\text{y}\geq\text{x}$ given.
Hence, it is not an equivalence relation.
View full question & answer→MCQ 151 Mark
If $f :R \rightarrow R$ is given by $f(x) = x^3+ 3,$ then $f^{-1}(x)$ is equal to:
AnswerCorrect option: C. $(\text{x}-3)^\frac{1}{3}$
$ \text { Let } f^{-1}(x)=y $
$ f(y)=x $
$ \Rightarrow y^3+3=x $
$ \Rightarrow y^3=x-3 $
$ \Rightarrow y=(x-3)^3 $
$ \Rightarrow y=(x-3)^{\frac{1}{3}}$
View full question & answer→MCQ 161 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
AnswerCorrect 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,
$\frac{\text{ae}}{3}=\text{a}\text{ and }\frac{\text{ea}}{3}=\text{a},\forall\text{ a}\in\text{Q}^+$
$e = 3, \forall\text{ a}\in\text{Q}^+$
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,
$\text{a}^{-1}=\frac{9}{\text{a}}$
$(4*6)^{-1}=8^{-1}$
$=\frac{9}8$
View full question & answer→MCQ 171 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:
AnswerLet 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 *.
View full question & answer→MCQ 181 Mark
Let $\text{f(x)}=\frac{1}{1-\text{x}}.$ Then, {fo(fof)}(x):
- A
x for all $\text{x}\in\text{R}$
- B
x for all $\text{x}\in\text{R}-\{1\}$
- ✓
x for all $\text{x}\in\text{R}-\{0,1\}$
- D
AnswerCorrect option: C. x for all $\text{x}\in\text{R}-\{0,1\}$
Domain of f: $1-\text{x}\neq0$
$\Rightarrow\ \text{x}\neq1$
Domain of f = R - {1}
Range of f: $\text{y}=\frac{1}{1-\text{x}}$
$\Rightarrow\ 1-\text{x}=\frac{1}{\text{y}}$
$\Rightarrow\ \text{x}=1-\frac{1}{\text{y}}$
$\Rightarrow\ \text{y}\neq0$
Range of f = R - {0}
So, f : R - {1} → R - {0} and f : R - {1} → R - {0}
Range of f is not a subset of the domain of f.
Domain (fof) = {x : $\text{x}\in$ domain of f and $\text{f(x)}\in$ domain of f}
Domain (fof) $=\big\{\text{x}:\text{x}\in\text{R}-\{1\}\text{ and }\frac{1}{1-\text{x}}\in\text{R}-\{1\}\big\}$
Domain (fof) $=\big\{\text{x}:\text{x}\neq1\text{ and }\frac{1}{1-\text{x}}\neq1\big\}$
Domain (fof) $=\{\text{x}:\text{x}\neq1\text{ and }1-\text{x}\neq1\}$
Domain (fof) $=\{\text{x}:\text{x}\neq1\text{ and }\text{x}\neq0\}$
Domain (fof) = R - {0, 1}
(fof)(x) = f(f(x))
$=\text{f}\Big(\frac{1}{1-\text{x}}\Big)=\frac{1}{1-\frac{1}{1-\text{x}}}=\frac{1-\text{x}}{1-\text{x}-1}$
$=\frac{1-\text{x}}{-\text{x}}=\frac{\text{x}-1}{\text{x}}$
For range of fof, $\text{x}\neq0$
Now, fof : R → {0, 1} → R - {0} and f : R - {1} → R - {0}
Range of fof is not a subset of domain of f.
Domain (fo(fof)) $=\{\text{x}:\text{x}\in$ domain of fof and (fof)(x) $\in$ domain of f$\}$
Domain (fo(fof)) $=\Big\{\text{x}:\text{x}\in\text{R}-\{0,1\}\text{ and }\frac{\text{x}-1}{\text{x}}\in\text{R}-\{1\}\Big\}$
Domain (fo(fof)) $=\Big\{\text{x}:\text{x}\neq0,1\text{ and }\frac{\text{x}-1}{\text{x}}\neq1\Big\}$
Domain (fo(fof)) $=\{\text{x}:\text{x}\neq0,1\text{ and }\text{x}-1\neq\text{x}\}$
Domain (fo(fof)) $=\{\text{x}:\text{x}\neq0,1\text{ and }\text{x}\in\text{R}\}$
Domain (fo(fof)) = R - {0, 1}
Domain (fo(fof)) = f((fof)(x))
$=\text{f}\Big(\frac{\text{x}-1}{\text{x}}\Big)$
$=\frac{1}{1-\frac{\text{x}-1}{\text{x}}}$
$=\frac{\text{x}}{\text{x}-\text{x}+1}$
$=\text{x}$
So, (fo(fof))(x) = x, where $\text{x}\neq0,1$
View full question & answer→MCQ 191 Mark
Choose the correct answer out of the given four options.Let * be binary operation defined on R by a * b = 1 + ab ∀ a, b ∈ R. Then the operation * is:
- ✓
Commutative but not associative.
- B
Associative but not commutative.
- C
Neither commutative nor associative.
- D
Both commutative and associative.
AnswerCorrect option: A. Commutative but not associative.
We are given that, a * b = 1 + ab ∀ a, b ∈ R
Consider, a * b = ab + 1
= ba + 1
= b * a
Hence, * is a communicative binary operation.
Also, a * (b * c) = a * (bc + 1) $[\because$ b * c = bc + 1$]$
= a(bc + 1) + 1
= a + abc + 1
Now, (a * b) * c = (ab + 1) * c $[\because$ a * b = ab + 1$]$
= (1 + ab)c + 1
= c + abc +1
Now, $\text{a}+\text{abc}+1\neq\text{c}+\text{abc}+1$
$\Rightarrow\ \text{a}\ ^*\ (\text{b}\ ^* \ \text{c})\neq(\text{a}\ ^* \ \text{b})\ ^* \ \text{c}$
Therefore, * is not associative.
Hence, * is communicative but not associative.
View full question & answer→MCQ 201 Mark
Let A = N × N and × be the binary operation on A defined by (a, b) × (c, d) = (a + c, b + d). Then × is:
MCQ 211 Mark
The relation $R$ is defined on the set of natural numbers as $\{(a, b) : a = 2b\}$. Then, $R^{-1}$ is given by:
- A
$\{(2, 1), (4, 2), (6, 3), ….\}$
- ✓
$\{(1, 2), (2, 4), (3, 6), …….. \}$
- C
$R^{-1}$ is not defiend.
- D
AnswerCorrect option: B. $\{(1, 2), (2, 4), (3, 6), …….. \}$
$\{(1, 2), (2, 4), (3, 6), ……..\}$
View full question & answer→MCQ 221 Mark
The number of commutative binary operation that can be defined on a set of 2 elements is:
AnswerThe 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 $=2\frac{2(2-1)}{2}=2^1$
$=2$
View full question & answer→MCQ 231 Mark
The domain of the function $\text{f(x)}=\frac{1}{\sqrt{\{\sin\text{x}\}+\{\sin(\pi+\text{x})}\}}$ where $\{.\}$ denotes fractional part, is:
View full question & answer→MCQ 241 Mark
Choose the correct answer from the given four options. The maximum number of equivalence relations on the set $A = \{1, 2, 3\}$ are:
AnswerGiven that, $A = \{1, 2, 3\}$
Now, number of equivalence relations as follows
$R_1= \{(1, 1), (2, 2), (3, 3)\}$
$R_2= \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$
$R_3= \{(1, 1), (2, 2), (3, 3), (1, 3), (3, 1)\}$
$R_4= \{(1, 1), (2, 2), (3, 3), (2, 3), (3, 2)\}$
$R_5= \{(1, 2, 3) \Leftrightarrow A \times A = A^2\}$
$\therefore$ Maximum number of equivalence relation is $'5\ '.$
View full question & answer→MCQ 251 Mark
Choose the correct answer from the given four options. Let $f : R \rightarrow R$ be given by$ f(x) = \tan x.$ Then $f^{-1}(1)$ is:
AnswerCorrect option: A. $\frac{\pi}{4}$
Given that, $f(x) = \tan x$
Let $y =\tan x \Rightarrow x = \tan^{-1}y$
$\Rightarrow f^{-1}(x) = \tan^{-1}x \Rightarrow f^{-1}(1) = \tan^{-1}1$
$\Rightarrow\ \tan^{-1}\tan\frac{\pi}{4}=\frac{\pi}{4}\ \Big[\because\ \tan\frac{\pi}{4}=1\Big]$
View full question & answer→MCQ 261 Mark
Choose the correct answer from the given four options. Let $f : [2, \infty ) \rightarrow R$ be the function defined by $f(x) = x^2– 4x + 5$, then the range of $f$ is:
- A
$\text{R}$
- ✓
$[1,\infty)$
- C
$[4,\infty)$
- D
$[5,\infty)$
AnswerCorrect option: B. $[1,\infty)$
Given that, $\text{f}(\text{x})=\text{x}^2-4\text{x}+5,$
Let $\text{y}=\text{x}^2-4\text{x}+5$
$\Rightarrow\ \text{y}=\text{x}^2-4\text{x}+4+1$
$=(\text{x}-2)^2+1$
$\Rightarrow\ (\text{x}-2)^2=\text{y}-1$
$\Rightarrow\ \text{x}-2=\sqrt{\text{y}-1}$
$\Rightarrow\ \text{x}=2+\sqrt{\text{y}-1}$
$\therefore\ \text{y}-1\geq0,\ \text{y}\geq1$
Range $=[1,\infty)$
View full question & answer→MCQ 271 Mark
Let $f : R \rightarrow R$ be defind by $\text{f(x)}=\frac{1}{\text{x}}\forall\times\in\text{ R}.$ Then $f$ is:
- A
One$-$one.
- B
- C
- ✓
$F$ is not defined.
AnswerCorrect option: D. $F$ is not defined.
View full question & answer→MCQ 281 Mark
Choose the correct answer from the given four options.
The identity element for the binary operation * defined on $\text{Q}\sim\{0\}$ as $\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{2}\ \forall\ \text{a, b}\in\text{Q}\sim\{0\}$ is:
AnswerGiven that, $\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{2}\ \forall\ \text{a, b}\in\text{Q}\sim\{0\}$
Let e be the identity element for *
$\therefore\ \text{a}\ ^*\ \text{e}=\frac{\text{ae}}{2}(\text{a}\ *\ \text{e}=\text{e}\ *\ \text{a}=\text{a})$
$\Rightarrow\ \text{a}=\frac{\text{ae}}{2}$
$\Rightarrow\ \text{e}=2$
View full question & answer→MCQ 291 Mark
The identity element for the binary operation $\times $ defined on $Q - \{0\}$ as $\text{a}\times\text{b}=\frac{\text{ab}}{2}\ \forall a, b \in Q - \{0\}$ is:
View full question & answer→MCQ 301 Mark
If $g(x) = x^2+ x - 2$ and $\frac{1}{2}\text{gof(x)}=2\text{x}^2-5\text{x}+2,$ then $f(x)$ is equal to:
- ✓
$2x - 3$
- B
$2x + 3$
- C
$2x^2 + 3x + 1$
- D
$2x^2 - 3x - 1$
AnswerCorrect option: A. $2x - 3$
We will solve this problem by the trial $-$ and $-$ error method.
Let us check option $(a)$ first.
If $f(x) = 2x - 3$
$\frac{1}{2}(\text{gof})(x)=\text{g(f(x))}$
$=\frac{1}{2}\text{g}(2\text{x}-3)$
$=\frac{1}{2}\big[(2\text{x}-3)^2+(2\text{x}-3)-2\big]$
$=\frac{1}{2}[4\text{x}^2+9-12\text{x}+2\text{x}-3-2]$
$=\frac{1}{2}[4\text{x}^2-10\text{x}+4]$
$=2\text{x}^2-5\text{x}+2$
The given condition is satisfied by $(a).$
View full question & answer→MCQ 311 Mark
The distinct linear functions that map $[-1, 1]$ onto $[0, 2]$ are:
- A
$f(x) = x + 1, g(x) = -x + 1$
- B
$f(x) = x - 1, g(x) = x + 1$
- ✓
$f(x) = -x - 1, g(x) = x - 1$
- D
AnswerCorrect option: C. $f(x) = -x - 1, g(x) = x - 1$
Since $f$ is invertible, range of $f =$ co$-$domain of $f = x$
So, we need to find the range of $f$ to find $X.$
For finding the range, let $f(x) = y$
$\Rightarrow 4x - x^2 = y$
$\Rightarrow x^2 - 4x = -y$
$\Rightarrow x^2 - 4x + 4 = 4 - y$
$\Rightarrow (x - 2)^2 = 4 - y$
$\Rightarrow\ \text{x}-2=\pm4-\text{y}$
$\Rightarrow\ \text{x}=2\pm4-\text{y}$
This is defined only when $4-\text{y}\geq0$
$\Rightarrow\ \text{y}\leq4,$
$X =$ Range of $f=(-\infty,4]$
View full question & answer→MCQ 321 Mark
Choose the correct answer from the given four options. Let $A = \{1, 2, 3, ...n\}$ and $B = \{a, b\}.$ Then the number of surjections from $A$ into $B$ is:
- A
$^nP_2$
- ✓
$2^n – 2$
- C
$2^n – 1$
- D
AnswerCorrect option: B. $2^n – 2$
Given that, $A = \{1, 2, 3, ..... n\}$ and $B = \{a, b\}$
If function is subjective then its range must be set $B = \{a, b\}$
Now number of onto functions $=$ Number of ways $'n\ '$ distinct objects can be distributed in two boxes $'a\ '$ and $'b\ '$ in such a way that no box remains empty.
Now for each object there are two options, either it is put in box $'a\ '$ or in box $'b\ '$
So total number of ways of $'n\ '$ different objects $ = 2 \times 2 \times 2 .... n$ times $= 2^n$
But in one case all the objects are put box $'a\ '$ and in one case all the objects are put in box $'b\ '$
So, number of subjective functions $= 2^n - 2$
View full question & answer→MCQ 331 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
$4$
AnswerCorrect 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 $8.$
View full question & answer→MCQ 341 Mark
The function $\text{f}:[0,\infty)\rightarrow\ \text{R}$ given by $\text{f(x)}=\frac{\text{x}}{\text{x}+1}$ is:
AnswerGiven function is $\text{f(x)}=\frac{\text{x}}{\text{x}+1}$ on $\text{f}:[0,\infty)\rightarrow\ \text{R}$
If f(x) = f(y)
$\Rightarrow\ \frac{\text{x}}{\text{x}+1}=\frac{\text{y}}{\text{y}+1}$
⇒ xy + x = xy + y
⇒ x = y
Hence, f is one-one.
If y = f(x)
$\text{y}=\frac{\text{x}}{\text{x}+1}$
⇒ xy + y = x
⇒ xy - x = -y
x(y - 1) = -y
$\text{x}=\frac{-\text{y}}{\text{y}-1}\neq\text{f(x)}$
It is not onto.
View full question & answer→MCQ 351 Mark
If $f : R \rightarrow R$ defined by $\text{f(x)}=\frac{3\text{x}+5}{2}$ is an invertible function, then find $f^{-1}.$
- ✓
$\frac{2\text{x}-5}{3}$
- B
$\frac{\text{x}-5}{3}$
- C
$\frac{5\text{x}-2}{3}$
- D
$\frac{\text{x}-2}{3}$
AnswerCorrect option: A. $\frac{2\text{x}-5}{3}$
$\frac{2\text{x}-5}{3}$
View full question & answer→MCQ 361 Mark
Let $f : R \rightarrow R$ be defined as $\text{f(x)}=\begin{cases}2\text{x},&\text{if x}>3\\\text{x}^2,&\text{if }1<\text{x}\leq3\\3\text{x},&\text{if x}\leq1\end{cases}.$ Then, find $f(-1) + f(2) + f(4):$
AnswerWe have,
$\text{f(x)}=\begin{cases}2\text{x},&\text{if x}>3\\\text{x}^2,&\text{if }1<\text{x}\leq3\\3\text{x},&\text{if x}\leq1\end{cases}$
Now,
$f(-1) + f(2) + f(4)$
$= 3(-1) + 2^2 + 2(4)$
$= -3 + 4 + 8$
$= 9$
View full question & answer→MCQ 371 Mark
If $\times$ is a binary operation on set of integers I defined by $a \times b = 3a + 4b - 2,$ then find the value of $4 \times 5.$
View full question & answer→MCQ 381 Mark
Which one of the following function is not invertible?
- A
$\text{f} : \text{R} \rightarrow \text{R}, \text{f(x)} = 3\text{x} + 1$
- B
$\text{f} : \text{R} \rightarrow [0,\infty), \text{f(x)} = \text{x}^2$
- C
$\text{f} : \text{R}^+\rightarrow\text{R}^+, \text{f(x)} =\frac{1}{\text{x}^3}$
- ✓
View full question & answer→MCQ 391 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
$\begin{bmatrix}1&0\\0&1\end{bmatrix}$
AnswerCorrect 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 $\begin{bmatrix}\frac{2}{13}&\frac{-3}{13}\\\frac{3}{13}&\frac{2}{13}\end{bmatrix}$
View full question & answer→MCQ 401 Mark
If $f : R \rightarrow R, g : R \rightarrow R$ and $h : R \rightarrow R$ is such that $f(x) = x^2, g(x) = \tan x$ and $h(x) = \log bx$, then the value of $[ho(gof)](x),$ if $\text{x}=\frac{\sqrt{\pi}}{2}$ will be:
View full question & answer→MCQ 411 Mark
Choose the correct answer from the given four options.
If a relation R on the set {1, 2, 3} be defined by R = {(1, 2)}, then R is:
AnswerR on the set {1, 2, 3} be defined by R = {(1, 2)}
It is clear that R is transitive.
View full question & answer→MCQ 421 Mark
Let $S = \{1, 2, 3, 4, 5\}$ and let $A = S \times S.$ Define the relation $R$ on $A$ as follows$:\ (a, b) R (c, d)$ if $ad = cb.$ Then$, R$ is;
View full question & answer→MCQ 431 Mark
If A = {1, 2, 3}, then a relation R = {(2, 3)} on A is:
- A
Symmetric and transitive only.
- B
- ✓
- D
AnswerThe relation R is not reflexive because every element of A is not related to itself. Also, R is not symmetric since on interchanging the elements, the ordered pair in R is not contained in it.
R is transitive by default because there is only one element in it.
View full question & answer→MCQ 441 Mark
$R$ is a relation from $\{11, 12, 13\}$ to $\{8, 10, 12\}$ defined by $y = x - 3$. Then, $R^{-1}$ is:
- ✓
$\{(8, 11), (10, 13)\}$
- B
$\{(11, 8), (13, 10)\}$
- C
$\{(10, 13), (8, 11)\}$
- D
AnswerCorrect option: A. $\{(8, 11), (10, 13)\}$
Given that $R$ is a relation from $\{11, 12, 13\}$ to $\{8, 10, 12\}$ defined by $y = x - 3.$
$R = \{(8, 11), (10, 13)\}$
$R^{-1} =\{(8, 11), (10, 13)\}$
As inverse function of $R$ is,
$y + 3 = x$
$⇒ y = x + 3$
View full question & answer→MCQ 451 Mark
Let $f : R \rightarrow R$ be given by $f(x) = [x^2] + [x + 1] - 3$ where $[x]$ denotes the greatest integer less than or equal to $x.$ Then, $f(x)$ is:
- A
Many $-$ one and onto.
- ✓
Many $-$ one and into.
- C
One $-$ one and into.
- D
One $-$ one and onto.
AnswerCorrect option: B. Many $-$ one and into.
$f : R \rightarrow R$
$= [x^2] + [x + 1] - 3$
It is many one function because in this case for two different values of $x$ we would get the same value of $f(x).$
For $\text{x}=1.1,\ 1.2\in\text{R}$
$f(1.1) = (1.1)^2 + [1.1 + 1] - 3$
$= [1.21] + [2.1] - 3$
$= 1 + 2 + 3 = 0$
$f(1.1) = [1.2]^2 + [1.2 + 1] - 3$
$= [1.44] + [2.2] - 3$
$= 1 + 2 - 3$
$= 0$
It is into function because for the given domain we would only get the integral values of $f(x).$
But $R$ is the co $-$ domain of the given function.
That means, $\text{Co-domain}\neq\text{Range}$
Hence, the given function is into function.
Therefore, $f(x)$ is many one and into.
View full question & answer→MCQ 461 Mark
Choose the correct answer from the given four options.
Consider the 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.
- ✓
Transitive but not symmetric.
- C
Neither symmetric nor transitive.
- D
Both symmetric and transitive.
AnswerCorrect option: B. Transitive but not symmetric.
We are given that a relation R defined aRb ⇒ a is brother of b.
aRa ⇒ a is brother of a, which is not true.
Hence, R is not reflexive.
aRb ⇒ a is brother of b.
This does not mean b is also a brother of a and b can be a sister of a.
Hence, it is not symmetric.
aRb ⇒ a is brother of b
and bRc ⇒ b is a brother of c.
So, a is brother of c.
Hence, R is transitive.
View full question & answer→MCQ 471 Mark
The function f : R → R defined by f(x) = (x - 1)(x - 2)(x - 3) is:
- A
- ✓
- C
- D
Neither one-one nor onto.
AnswerGiven function is f(x) = (x - 1)(x - 2)(x - 3)
If f(x) = f(y) then
(x - 1)(x - 2)(x - 3) = (y - 1)(y - 2)(y - 3)
⇒ f(1) = f(2) = f(3) = 0
It is not one-one.
y = f(x)
$\text{x}\in\text{R}$ also $\text{y}\in\text{R}$ hence f is onto.
View full question & answer→MCQ 481 Mark
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.
View full question & answer→MCQ 491 Mark
The inverse of the function $\text{y}=\frac{10^\text{x}-10^{-\text{x}}}{10^\text{x}+10^{-\text{x}}}$ is:
- A
$\log_{10}(2-\text{x})$
- ✓
$\frac{1}{2}\log_{10}\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$
- C
$\frac{1}{2}\log_{10}(2\text{x}-1)$
- D
$\frac{1}{4}\log\big(\frac{2\text{x}}{2-\text{x}}\Big)$
AnswerCorrect option: B. $\frac{1}{2}\log_{10}\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$
View full question & answer→MCQ 501 Mark
$f : R \rightarrow R$ is defined by $\text{f(x)}=\frac{\text{e}^{\text{x}^2}-\text{e}^{-\text{x}^2}}{\text{e}^{\text{x}^2}+\text{e}^{-\text{x}^2}}$ is:
AnswerCorrect option: D. Neither one$-$one nor onto.
We have,
$\text{f(x)}=\frac{\text{e}^{\text{x}^2}-\text{e}^{-\text{x}^2}}{\text{e}^{\text{x}^2}+\text{e}^{-\text{x}^2}}$
Here, $-2,2\in\text{R}$
Now, $2\neq-2$
But$, f(2) = f(-2)$
Therefore, function is not one$-$one.
And,
The minimum value of the function is $0$ and maximum value is $1.$
That is range of the function is $[0, 1]$ but the $co-$domain of the function is given $R.$
Therefore, function is not onto.
$\therefore$ function is neither one$-$one nor onto.
View full question & answer→MCQ 511 Mark
Let $\text{A}=\{\text{x}\in\text{R}:\text{x}\geq1\}.$ The inverse of the function f : A → A given by $\text{f(x)}=2^{\text{x}(\text{x}-1)},$ is:
- A
$\big(\frac{1}{2}\big)^{\text{x}(\text{x}-1)}$
- ✓
$\frac{1}{2}\big\{1+\sqrt{1+4\log_2\text{x}}\big\}$
- C
$\frac{1}{2}\big\{1-\sqrt{1+4\log_2\text{x}}\big\}$
- D
$\text{Not defined}$
AnswerCorrect option: B. $\frac{1}{2}\big\{1+\sqrt{1+4\log_2\text{x}}\big\}$
Given function is $\text{A}=\{\text{x}\in\text{R}:\text{x}\geq1\}.$
The inverse of the function f : A → A given by $\text{f(x)}=2^{\text{x}(\text{x}-1)}$
$\text{f(x)}=\text{y}$
$2^{\text{x}(\text{x}-1)}=\text{y}$
$\text{x}(\text{x}-1)=\log_2\text{y}$
$\text{x}^2+\text{x}=\log_2\text{y}$
$\text{x}^2+\text{x}+\frac{1}{4}=\log_2\text{y}+\frac{1}{4}$
$\Big(\text{x}-\frac{1}{2}\Big)^2=\frac{4\log_2\text{y}+1}{4}$
$\text{x}-\frac{1}{2}=\pm\sqrt{\frac{4\log_2\text{y}+1}{4}}$
$\text{x}=\frac{1}{2}\pm\sqrt{\frac{4\log_2\text{y}+1}{4}}$
$\text{x}=\frac{1}{2}+\sqrt{\frac{4\log_2\text{y}+1}{4}}$
$\text{f}^{-1}(\text{x})=\frac{1+\sqrt{4\log_2\text{y}+1}}{2}$
View full question & answer→MCQ 521 Mark
Total number of equivalence relations defined in the set $S = \{a, b, c\}$ is:
View full question & answer→MCQ 531 Mark
If f : A → B is surjective then:
- A
no two elements of A have the same image in B
- B
every element of A has an image in B
- ✓
every element of B has at least one pre-image in A
- D
A and B are finite non empty sets
AnswerCorrect option: C. every element of B has at least one pre-image in A
Surjective means onto function.co domain = Range
So every element of B has at least one pre-image in A.
View full question & answer→MCQ 541 Mark
Choose the correct answer from the given four options. Which of the following functions from $Z$ into $Z$ are bijections?
- A
$f(x) = x^3$
- ✓
$f(x) = x + 2$
- C
$f(x) = 2x + 1$
- D
$f(x) = x^2 + 1$
AnswerCorrect option: B. $f(x) = x + 2$
Consider, the second option i.e., $f(x) = x + 2$
Now, $f(x_1) = f(x_2)$
$\Rightarrow x_1 + 2 = x_2 + 2$
$\Rightarrow x_1= x_2$
Hence, $f(x) = x + 2$ is one$-$one function.
Now, let us suppose, $y = x + 2$
$\text{x}=\text{y}-2\in\text{Z},\ \forall\ \text{y}\in\text{x}$
Hence, $f(x)$ is one$-$one and onto.
View full question & answer→MCQ 551 Mark
If R is a relation on the set A = {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3)}, then R is:
AnswerR = a, b : a = b and $\text{a, b}\in\text{A}$
Reflexivity: Let $\text{a}\in\text{A}$
Then, a = a
$\Rightarrow\ \text{a, a}\in\text{R}$ for all $\text{a}\in\text{A}$
So, R is reflexive on A.
Symmetry: Let $\text{a, b}\in\text{A}$ such that $\text{a, b}\in\text{R.}$
Then, $\text{a, b}\in\text{R}$
⇒ a = b ⇒ b = a ⇒ b, $\text{a}\in\text{R}$ for all $\text{a}\in\text{A}$
So, R is symmetric on A.
View full question & answer→MCQ 561 Mark
Let us define a relation $R$ in $R$ as $aRb$ if $a \geq b.$ 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.
View full question & answer→MCQ 571 Mark
Range of $\text{f(x)}=\sqrt{(1-\cos\text{x})\sqrt{(1-\cos\text{x})\sqrt{(1-\cos\text{x}).....\infty}}}$
- A
$[0, 1]$
- B
$(0, 1)$
- ✓
$[0, 2]$
- D
$(0, 2)$
AnswerCorrect option: C. $[0, 2]$
View full question & answer→MCQ 581 Mark
If $f: R \rightarrow R$ be given by $f(\text{x})=(3-\text{x}^3)^{\frac{1}{3}},$ then $fof(x)$ is:
- A
$\text{x}^{\frac{1}{3}}$
- B
$x^3$
- ✓
$x$
- D
$(3 - x^3).$
Answer$f: R \rightarrow R$ and $f(\text{x})=(3-\text{x}^3)^{\frac{1}{3}}$
$\Rightarrow\ \ f[f(\text{x})]=\Big[3-[f(\text{x})^3\Big]^{\frac{1}{3}}$
$=\left[3-\Big\{(3-\text{x}^3)^{\frac{1}{3}}\Big\}^3\right]^{\frac{1}{3}}$
$=[3-(3-\text{x}^3)]^{\frac{1}{3}}$
$=(3-3+\text{x}^3)^{\frac{1}{3}}$
$=\text{x}$
Therefore, option $(C)$ is correct
View full question & answer→MCQ 591 Mark
Let $f : [0, \infty) \rightarrow [0, 2]$ be defined by $\text{f(x)}=\frac{2\text{x}}{1+\text{x}},$ then $f$ is:
AnswerCorrect option: A. One$-$one but not onto.
View full question & answer→MCQ 601 Mark
Let $A = \{1, 2, 3, …. n\}$ and $B = \{a, b\}.$ Then the number of surjections from $A$ into $B$ is:
- A
$^\text{n}\text{P}_2$
- ✓
$2^n - 2$
- C
$2^n - 1$
- D
AnswerCorrect option: B. $2^n - 2$
$2^n - 2$
View full question & answer→MCQ 611 Mark
If X is brother of the son of Y's son. How is X related to Y?
AnswerSon of Y's Son- Grandson, Brother of Y's Grandson- Y's Grandson
Option D is correct.
View full question & answer→MCQ 621 Mark
Which of the following functions form $Z$ to itself are bijections?
- A
$f(x) = x^3$
- ✓
$f(x) = x + 2$
- C
$f(x) = 2x + 1$
- D
$f(x) = x^2 + x$
AnswerCorrect option: B. $f(x) = x + 2$
- $f$ is not because for $\text{y}=3\in\text{Co-domain (Z)},$ there is no value of $\text{x}\in\text{Domain (Z)}$ $\text{x}^3=3$
$\Rightarrow\ \text{x}=\sqrt[3]{3}\notin\text{Z}$
$\Rightarrow f$ is not onto.
So, $f$ is not a bijection.
- Injectivity: Let $x$ and $y$ be two elements of the domain $(Z),$ such that
$x + 2 = y + 2$
$\Rightarrow x = y$
So, $f$ is one$-$one.
Surjectivity: Let $y$ be an element in the co$-$domain $(Z),$ such that
$y = f(x)$
$\Rightarrow y = x + 2$
$\Rightarrow\ \text{x}=\text{y}-2\in\text{Z} ($Domain$)$
$\Rightarrow f$ is onto.
So, $f$ is a bijection.
- $f(x) = 2x + 1$ is not onto because if we take $4\in\text{Z}$ (co domain), then $4 = f(x)$
$\Rightarrow4 = 2\text{x} + 1$
$\Rightarrow 2\text{x} = 3$
$\Rightarrow\ \text{x}=\frac{3}{2}\notin\text{Z}$
So, $f$ is not a bijection.
- $f(0) = 0^2 + 0 = 0$
$\Rightarrow $ and $f(-1) = (-1)^2 + (-1) = 1 - 1 = 0$
$\Rightarrow 0$ and $-1$ have the same image.
$\Rightarrow f$ is not one$-$one.
So, $f$ is not a bijection. View full question & answer→MCQ 631 Mark
If the function $f : R \rightarrow R$ be such that $f(x) = x - [x],$ where $[x]$ denotes the greatest integer less than or equal to $x,$ then $f^{-1}(x)$ is:
AnswerGiven function is $f(x) = x - [x]$
$[x]$ is a greatest integer function.
Hence, we will have same values of the function for the different values of $x.$
As we are considering integer only not fraction part.
Hence, it is not defined.
View full question & answer→MCQ 641 Mark
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:
AnswerAs, the number of bijection from A into B can only be possible when provided $\frac{7}{(\text{A})}>\frac{7}{(\text{B})}$
But here n(A) < n(B)
So, the number of bijection.
i.e. one-one and onto mapping from A to B.
View full question & answer→MCQ 651 Mark
Let g(x) = 1 + x - [x] and $\text{f(x)}=\begin{cases}-1,&\text{x}<0\\0,&\text{x}=0\\1,&\text{x}>0\end{cases}$ where [x] denotes the greatest integer less than or equal to x. Then for all x, f(g(x)) is equal to:
AnswerWhen, -1 < x < 0
Then, g(x) = 1 + x - [x]
= 1 + x - (-1) = 2 + x
$\therefore$ f(g(x)) = 1
When, x = 0
Then, g(x) = 1 + x - [x]
= 1 + x - 0 = 1 + x
$\therefore$ f(g(x)) = 1
When, x > 1
Then, g(x) = 1 + x - [x]
= 1 + x - 1 = x
$\therefore$ f(g(x)) = 1
Therefore, for each interval f(g(x)) = 1
View full question & answer→MCQ 661 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
$\begin{bmatrix}-1&-1\\-1&-1\end{bmatrix}$
AnswerCorrect 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.
View full question & answer→MCQ 671 Mark
If $f : R \rightarrow (-1, 1)$ is defined by $\text{f(x)}=\frac{-\text{x}|\text{x}|}{1+\text{x}^2},$ then $f^{-1}(x)$ equals,
- A
$\sqrt{\frac{|\text{x}|}{1-|\text{x}|}}$
- ✓
$-\text{Sgn (x)}\sqrt{\frac{|\text{x}|}{1-|\text{x}|}}$
- C
$-\sqrt{\frac{\text{x}}{1-\text{x}}}$
- D
$\text{None of these}$
AnswerCorrect option: B. $-\text{Sgn (x)}\sqrt{\frac{|\text{x}|}{1-|\text{x}|}}$
Given function is $f : R \rightarrow (-1, 1)$ is defined by $\text{f(x)}=\frac{-\text{x}|\text{x}|}{1+\text{x}^2}$
Here, for mod function we will have to consider three cases as,
$x < 0, x = 0, x > 0$
$x < 0 $
$\Rightarrow |x| = -x$
$\text{f(|x|)}=\frac{-\text{x}(-\text{x})}{1+\text{x}^2}$
$\text{y}=\frac{\text{x}^2}{1+\text{x}^2}$
$\text{y}(1+\text{x}^2)=\text{x}^2$
$\text{y}+\text{yx}^2=\text{x}^2$
$\text{y}=\text{x}^2-\text{yx}^2$
$\text{y}=(1-\text{y})\text{x}^2$
$\text{x}^2=\frac{\text{y}}{1-\text{y}}$
$\text{x}=-\sqrt{\frac{\text{y}}{1-\text{y}}}$
$\Rightarrow\ \text{x}=-\sqrt{\frac{|\text{y}|}{1-|\text{y}|}}\ \text{x} < 0$
Also you can check for the cases $x = 0$ and $x > 0$ that $\text{x}=-\sqrt{\frac{|\text{y}|}{1-|\text{y}|}}$
$-\text{Sgn (x)}\sqrt{\frac{|\text{x}|}{1-|\text{x}|}}$
View full question & answer→MCQ 681 Mark
A constant function f : A → B will be one-one if:
AnswerGiven f is a constant functions.
⇒ range of f is {c}(say)
Since f is one-one
⇒ domain of A should also contain
one element.
$\therefore\text{n(A)}=1$
View full question & answer→MCQ 691 Mark
If $f(x)=1-\frac{1}{\text{x}},$ then $\text{f}(\text{f}(\frac{1}{\text{x}}))$
AnswerCorrect option: C. $\frac{\text{x}}{\text{x}-1}$
View full question & answer→MCQ 701 Mark
Let R be the relation in the set N given by R = {(a, b) : a = b – 2, b > 6}. Choose
the correct answer.
- A
$(2,4)\in\text{R}$
- B
$(3,8)\in\text{R}$
- ✓
$(6,8)\in\text{R}$
- D
$(8,7)\in\text{R}.$
AnswerCorrect option: C. $(6,8)\in\text{R}$
Given: a = b − 2, b > 6
(A) a = 2, b = 4 , Here b > 6 is not true, therefore, this option is incorrect
| (B) |
a = 3, b = 8 and a = b - 2 ⇒ 3 = 8-2 |
⇒ |
3 = 6, which is false. |
| |
Therefore, this option is incorrect |
|
|
| (C) |
a = 6, b = 8 and b = b - 2 ⇒ 6 = 8 - 2 |
⇒ |
6 = 6, which is true. |
| |
Therefore, this option is correct |
|
|
| (D) |
a = 8, b = 7 and a = b - 2 ⇒ 8 = 7 - 2 |
⇒ |
8 = 5, which is false. |
View full question & answer→MCQ 711 Mark
Let $f :\text{R}-\{\frac{3}{5}\}\rightarrow\text{R}$ be defined by $\text{f(x)}=\frac{3\text{x}+2}{5\text{x}-2}.$ Then:
AnswerCorrect option: A. $f^{-1}(x) = f(x)$
$f^{-1}(x) = f(x)$
View full question & answer→MCQ 721 Mark
Let $f: R \rightarrow R$ be defined as $f(x) = x^4.$ Choose the correct answer.
AnswerCorrect option: D. $f$ is neither one $-$ one nor onto.
$f: R \rightarrow R$ is defined as $f(x) = x^4.$
Let $\text{x},\text{y}\in\text{R}$ such that $f(x) = f(y).$
$\Rightarrow x^4 = y^4$
$\Rightarrow\text{x}=\pm\text{y}$
$\therefore f(x_1) = f(x_2)$ does not imply that $x_1 = x_2$
For instance,
$f(1) = f(-1) = 1$
$\therefore f $ is not one $-$ one.
Consider an element $2$ in co $-$ domain $R$. It is clear that there does not exist any $x$ in domain $R$ such that $f(x) = 2.$
$\therefore$ f is not onto.
Hence, function $f$ is neither one $-$ one nor onto.
The correct answer is $D.$
View full question & answer→MCQ 731 Mark
Let $R$ be the relation “is congruent to” on the set of all triangles in a plane is:
View full question & answer→MCQ 741 Mark
If $A = \{a, b, c, d\},$ then a relation $R = \{(a, b), (b, a), (a, a)\}$ on $A$ is:
- ✓
Symmetric and transitive only.
- B
Reflexive and transitive only.
- C
- D
AnswerCorrect option: A. Symmetric and transitive only.
Given that $A = \{a, b, c, d\}$ then a relation $R = \{(a, b), (b, a), (a, a)\}$ on $A.$
$(a, b), (b, a) \in\text{R}$
$\Rightarrow R$ is symmetric.
Also for $(a, a) R$ is symmetric.
View full question & answer→MCQ 751 Mark
If R is the largest equivalence relation on a set A and S is any relation on A, then:
AnswerCorrect option: B. $\text{S}\subset\text{R}$
Given that R is the largest relation on A and S is any relation on A.
We know that R is always subset of A × A.
Hence, $\text{S}\subset\text{R}.$
View full question & answer→MCQ 761 Mark
Choose the correct answer from the given four options.Let f : N → R be the function defined by $\text{f}(\text{x})=\frac{2\text{x}-1}{2}$ and g : Q → R be another function defined by g(x) = x + 2. Then $(\text{gof})\frac{3}{2}$ is:
- A
$1$
- B
$1$
- C
$\frac{7}{2}$
- ✓
$\text{None of these}.$
AnswerCorrect option: D. $\text{None of these}.$
We have $\text{f}(\text{x})=\frac{2\text{x}-1}{2}$ and g(x) = x + 2
$\text{gof}\Big(\frac{3}{2}\Big)=\text{g}\Big(\text{f}\Big(\frac{3}{2}\Big)\Big)$
$=\text{g}\bigg(\frac{2\times\frac{3}{2}-1}{2}\bigg)$
$=\text{g}(1)=1+2=3$
View full question & answer→MCQ 771 Mark
Let $\text{f(x)}=\frac{\alpha\text{x}}{\text{x}+1},\ \text{x}\neq-1.$ Then, for what value of $\alpha$ is f(f(x)) = x?
- A
$\sqrt{2}$
- B
$-\sqrt{2}$
- C
- ✓
Answer Given function is $\text{f(x)}=\frac{\alpha\text{x}}{\text{x}+1},\ \text{x}\neq-1$
Also f(f(x)) = x
$\text{f}\Big(\frac{\alpha\text{x}}{\text{x}+1}\Big)=\text{x}$
$\frac{\alpha\big(\frac{\alpha\text{x}}{\text{x}+1}\big)}{\frac{\alpha\text{x}}{\text{x}+1}+1}=\text{x}$
$\frac{\alpha^2\text{x}}{\alpha\text{x}+\text{x}+1}=\text{x}$
$\alpha^2=\alpha\text{x}+\text{x}+1$
$\alpha^2=(\alpha+1)\text{x}+1$
Comparing on both sides,
$\alpha+1=0\Rightarrow\ \alpha=-1$
View full question & answer→MCQ 781 Mark
Let $\times$ be a binary operation on set $Q$ of rational numbers defined as $\text{a}\times\text{b}=\frac{\text{ab}}{5}.$ Write the identity for $\times .$
View full question & answer→MCQ 791 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:
AnswerCorrect 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 $\text{a}\in\text{R}-\{1\}$.
View full question & answer→MCQ 801 Mark
Let $A = \{1, 2, ......., n\}$ and $B = \{a, b\}.$ Then the number of subjections from $A$ into $B$ is:
- A
$^{\text{n}}\text{P}_2$
- ✓
$2^\text{n}-2$
- C
$2^\text{n}-1$
- D
$^{\text{n}}\text{C}_2$
AnswerCorrect option: B. $2^\text{n}-2$
The number of functions from a set with $n$ number of elements into a set with $2$ number of elements $= 2^n$
But two functions can be many$-$one into function.
View full question & answer→MCQ 811 Mark
The function $f : R \rightarrow R$ defined by $f(x) = 6^x + 6^{|x|}$ is:
- A
One $-$ one and onto.
- B
- C
One $-$ one and into.
- ✓
AnswerGraph of the given function is as follows:
A line parallel to $X-$ axis is cutting the graph at two different values.
Therefore, for two different values of $x$ we are getting the same value of $y.$
That means it is many one function.
From the given graph we can see that the range is $[2,\infty]$ and $R$ is the co $-$ domain of the given function.
Hence, Co $-$ dornain $=$ Range
Therefore, the given function is into. View full question & answer→MCQ 821 Mark
Let $f, g : R \rightarrow R$ be defined by $f(x) = 3x + 1$ and $g(x) = x^2 - 2, \forall x \in R,$ respectively. Then, $fog$ is:
- ✓
$3x^2 - 5$
- B
$9x^2 + 6x - 1$
- C
$3x^2$
- D
$9x^2 - 6x - 3$
AnswerCorrect option: A. $3x^2 - 5$
Given,
$f(x) = 3x + 1$
$g(x) = x^2 - 2$
$f o g = f[g(x)]$
$= f(x^2 - 2)$
$= 3(x^2 - 2) + 1$
$= 3x^2 - 6 + 1$
$= 3x^2 - 5$
View full question & answer→MCQ 831 Mark
If $N$ be the set of all$-$natural numbers, consider $f : N \rightarrow N$ such that $f(x)=2 x, \forall \ x \in N$, then $f$ is:
- A
One$-$one onto.
- ✓
One$-$one into.
- C
Many$-$one onto.
- D
AnswerCorrect option: B. One$-$one into.
View full question & answer→MCQ 841 Mark
Let f : R → R be a function defined by $\text{f(x)}=\frac{\text{e}^{|\text{x}|}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}.$ Then,
- A
- B
- C
- ✓
f is neither an injection nor a surjection.
AnswerCorrect option: D. f is neither an injection nor a surjection.
f : R → R
$\text{f(x)}=\frac{\text{e}^{|\text{x}|}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$
For x = -2 and -3 $\in\text{R}$
$\text{f(-2)}=\frac{\text{e}^{|-2|}-\text{e}^2}{\text{e}^{-2}+\text{e}^2}$
$=\frac{\text{e}^2-\text{e}^2}{\text{e}^{-2}+\text{e}^2}$
$=0$
Hence, for different values of x we are getting same values of f(x)
That means, the given function is many one.
Therefore, this function is not injective.
For x < 0
f(x) = 0
For x > 0
$\text{f(x)}=\frac{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$
$=\frac{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{}e^{-\text{x}}}-\frac{2\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$
$=1-\frac{2\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$
The value of $\frac{2\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$ is always positive.
Therefore, the value of f(x) is always less than 1.
Numbers more than 1 are not included in the range but they are included in co-domain.
As the codomain is R.
$\therefore\ \text{Co-domain}\neq\text{Range}$
Hence, the given function is not onto.
Therefore, this function is not surjective.
View full question & answer→MCQ 851 Mark
If a relation $R$ is defined on the set $Z$ of integers as follows: $(a, b) \in R \Leftrightarrow a^2 + b^2 = 25.$ Then, domain $(R)$ is:
- A
$\{3, 4, 5\}$
- B
$\{0, 3, 4, 5\}$
- ✓
$\{0,\pm3,\pm4,\pm5\}$
- D
AnswerCorrect option: C. $\{0,\pm3,\pm4,\pm5\}$
As $ aRb \Leftrightarrow a < b$
does not satisfy reflexive and symmetric relation.
View full question & answer→MCQ 861 Mark
The function $f : A \rightarrow B$ defined by $f(x) = -x^2 + 6x- 8$ is a bijection if,
- ✓
$\text{A}=(-\infty,3]$ and $\text{B}=(-\infty,1]$
- B
$\text{A}=[-3,\infty)$ and $\text{B}=(-\infty,1]$
- C
$\text{A}=(-\infty,3]$ and $\text{B}=[1,\infty)$
- D
$\text{A}=[3,\infty)$ and $\text{B}=[1,\infty)$
AnswerCorrect option: A. $\text{A}=(-\infty,3]$ and $\text{B}=(-\infty,1]$
$f(x) = -x^2 + 6x - 8,$ is a polynomial function and the domain of polynomial function is real number.
$\therefore\ \text{x}\in\text{R}$
$f(x) = -x^2 + 6x - 8$
$= -(x^2 - 6x + 8)$
$= -(x^2 - 6x + 9 - 1)$
$= -(x - 3)^2 + 1$
Maximum value of $-(x - 3)^2$ woud be $0$
$\therefore$ Maximum value of $-(x - 3)^2 + 1$ woud be $1$
$\therefore\ \text{f(x)}\in(-\infty,1]$
We can see from the given graph that function is symmetrical about $x = 3$ and the given function is bijective.
So, $x$ would be either $(-\infty,3]\text{ or }[3,\infty)$
The correct option which satisfy $A$ and $B$ both is:
$\text{A}=(-\infty,3]$ and $\text{B}=(-\infty,1]$ View full question & answer→MCQ 871 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.
AnswerCorrect 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.
View full question & answer→MCQ 881 Mark
If A = {1, 2, 3}, B = {1, 4, 6, 9} and R is a relation from A to B defined by 'x is greater than y'. The range of R is:
AnswerHere, $\text{R}=\text{x, y}:\text{x}\in\text{A}$ and $\text{y}\in\text{B}:\text{x}>\text{y}$ ⇒ R = 2, 1, 3, 1
Thus, Range of R = {1}
View full question & answer→MCQ 891 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:
AnswerGiven 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,
$(2 ^* 3) ^* 4$
$= 5 ^* 4$
$= 5^2 - 4^2 + 5.4 + 4$
$= 25- 16 + 20 + 4$
$= 33$
View full question & answer→MCQ 901 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
$\frac{2}{\text{a}}$
- ✓
$\frac{4}{\text{a}}$
AnswerCorrect 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 $\text{ a}\in\text{Q}^+$.
View full question & answer→MCQ 911 Mark
Let $f(x) = x^2$ and $g(x) = 2^x.$ Then, the solution set of the equation $fog(x) = gof(x)$ is:
AnswerCorrect option: C. $\{0, 2\}$
Since $(fog)(x) = (gof)(x),$
$f(g(x)) = g(f(x))$
$\Rightarrow\ \text{f}(2^\text{x})=\text{g}(\text{x}^2)$
$\Rightarrow\ \big(2^{\text{x}}\big)^{2}=2^{\text{x}^2}$
$\Rightarrow\ 2^{2\text{x}}=2^{\text{x}^2}$
$\Rightarrow\ \text{x}^2=2\text{x}$
$\Rightarrow\ \text{x}^2-2\text{x}=0$
$\Rightarrow\ \text{x}(\text{x}-2)=0$
$\Rightarrow\ \text{x}=0, 2$
$\Rightarrow\ \text{x}\in\{0,2\}$
View full question & answer→MCQ 921 Mark
Let $\times$ be a binary operation on set $Q - \{1\}$ defind by $a \times b = a + b - ab : a, b \in Q - {1}.$ Then $\times$ is:
- A
- B
- ✓
Both $(a)$ and $(b).$
- D
AnswerCorrect option: C. Both $(a)$ and $(b).$
View full question & answer→MCQ 931 Mark
Let $f : R \rightarrow R$ be given by $\text{f(x)}=\tan\text{x}.$ Then, $f^{-1}(1)$ is:
AnswerCorrect option: B. $\big\{\text{n}\pi+\frac{\pi}{4}:\text{n}\in\text{Z}\big\}$
We have, $f : R \rightarrow R$ is given by
$\text{f(x)}=\tan\text{x}$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\tan^{-1}\text{x}$
$\therefore\ \text{f}^{-1}(1)=\tan^{-1}1$
$=\big\{\text{n}\pi+\frac{\pi}{4}:\text{n}\in\text{Z}\big\}$
View full question & answer→MCQ 941 Mark
The maximum number of equivalence relations on the set $A = \{1, 2, 3\}$ is:
AnswerThe maximum number of equivalence relations on the set $A = \{1, 2, 3\}$ is,
$R_1 = \{(1, 1)\}$
$R_2 = \{(2, 2)\}$
$R_3 = \{(3, 3)\}$
$R_4 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$
$R_5 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)\}$
The maximum number of equivalence relations on the set $A = \{1, 2, 3\}$ is $5.$
View full question & answer→MCQ 951 Mark
Let R be the relation over the set of all straight lines in a plane such that $\text{l}_1\text{Rl}_2\Leftrightarrow\text{l}_1\bot\text{l}_2.$ Then, R is:
AnswerGiven R is the relation over the set of all straight lines in a plane such that $\text{l}_1\text{Rl}_2\Leftrightarrow\text{l}_1\bot\text{l}_2.$
It is symmetric relation as we can say either or $\text{l}_2\bot\text{l}_1.$
View full question & answer→MCQ 961 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:
Answer* 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,
$1*(2*3)\neq(1*2)*3$
View full question & answer→MCQ 971 Mark
Let $f(x) = x^3$ be a function with domain $\{0, 1, 2, 3\}$. Then domain of $f^{-1}$ is:
- A
$\{3, 2, 1, 0\}$
- B
$\{0, -1, -2, -3\}$
- ✓
$\{0, 1, 8, 27\}$
- D
$\{0, -1, -8, -27\}$
AnswerCorrect option: C. $\{0, 1, 8, 27\}$
Given function is $f(x) = x^3$ be a function with domain $\{0, 1, 2, 3\}.$
Range $= \{0, 1^3, 2^3, 3^3\} = \{0, 1, 8, 27\}$
$f$ can be written as
$\{(0, 0), (1, 1), (2, 8), (3, 27)\}$
Hence, $f^{-1}$ can be written as
$\{(0, 0), (1, 1), (8, 2), (27, 3)\}$
Domain of $f^{-1}$ is $\{0, 1, 8, 27\}$
View full question & answer→MCQ 981 Mark
The smallest integer function $f(x) = [x]$ is:
- A
One$-$one.
- ✓
Many$-$one.
- C
Both $(a)$ and $(b).$
- D
AnswerCorrect option: B. Many$-$one.
View full question & answer→MCQ 991 Mark
The number of binary operations that can be defined on a set of $2$ elements is:
View full question & answer→MCQ 1001 Mark
Let R be the relation on the set A = {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Then,
- A
R is reflexive and symmetric but not transitive.
- ✓
R is reflexive and transitive but not symmetric.
- C
R is symmetric and transitive but not reflexive.
- D
R is an equivalence relation.
AnswerCorrect option: B. R is reflexive and transitive but not symmetric.
Reflexivity: Clearly, $(\text{a, a})\in\text{R}\ \forall\ \text{a}\in\text{A}$
So, R is reflexive on A.
Symmetry: Since, $1,2\in\text{R},$ but $2,1\notin\text{R,}$ R is not symmetric on A.
Transitivity: Since, $1,3,3,2\in\text{R}$ and $1,2\in\text{R},$ R is transitive on A.
View full question & answer→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→MCQ 1511 Mark
If $f : R \rightarrow R$ and $g : R \rightarrow R$ defined by $f(x) = 2x + 3$ and $g(x) = x^2 + 7,$ then the value of $x$ for which $f(g(x)) = 25$ is:
- A
$\pm1$
- ✓
$\pm2$
- C
$\pm3$
- D
$\pm4$
AnswerCorrect option: B. $\pm2$
$\pm2$
View full question & answer→MCQ 1521 Mark
If $f : R \rightarrow R$ defind by $\text{f(x)}=\frac{2\text{x}-7}{4}$ is an invertible function, then find $f^{-1}.$
- A
$\frac{4\text{x}+5}{2}$
- ✓
$\frac{4\text{x}+7}{2}$
- C
$\frac{3\text{x}+2}{2}$
- D
$\frac{9\text{x}+3}{5}$
AnswerCorrect option: B. $\frac{4\text{x}+7}{2}$
$\frac{4\text{x}+7}{2}$
View full question & answer→MCQ 1531 Mark
Let $f(x) = x^2 – x + 1, \text{x}\geq\frac{1}{2},$ then the solution of the equation $f(x) = f^{-1}(x)$ is:
- ✓
$x = 1$
- B
$x = 2$
- C
$\text{x}=\frac{1}{2}$
- D
AnswerCorrect option: A. $x = 1$
$x = 1$
View full question & answer→MCQ 1541 Mark
Let $A = \{x : -1 \leq x \leq 1\}$ and $f : A \rightarrow A$ is a function defined by $f(x) = x |x|$ then $f$ is:
- ✓
- B
Injection but not surjection.
- C
Surjection but not injection.
- D
Neither injection nor surjection.
View full question & answer→MCQ 1551 Mark
If a function $\text{f}:[2,\infty)\rightarrow\ \text{B}$ defined by $f(x) = x^2 - 4x + 5$ is a bijection, then $B =$
- A
$\text{R}$
- ✓
$[1,\infty)$
- C
$[4,\infty)$
- D
$[5,\infty)$
AnswerCorrect option: B. $[1,\infty)$
Since $f$ is a bijection, co$-$domain of $f =$ range of $f$
$\Rightarrow B =$ range of $f$
Given: $f(x) = x^2 - 4x + 5$
Let $f(x) = y$
$\Rightarrow y = x^2 - 4x + 5$
$\Rightarrow x^2 - 4x + (5 - y) = 0$
$\because$ Discrimant, $\text{D}=\text{b}^2-4\text{ac}\geq0,$
$(-4)^2-4\times1\times(5-\text{y})\geq0$
$\Rightarrow 16-20+4\text{y}\geq0$
$\Rightarrow 4\text{y}\geq4$
$\Rightarrow \text{y}\geq1$
$\Rightarrow \text{y}\in[1,\infty)$
$\Rightarrow $ Range of $\text{f}=[1,\infty)$
$\Rightarrow \text{B}=[1,\infty)$
View full question & answer→MCQ 1561 Mark
Set $A$ has $3$ elements, and set $B$ has $4$ elements. Then the number of injective mappings that can be defined from $A$ to $B$ is:
AnswerThe total number of injective mappings from the set containing $3$ elements into the set containing $4$ elements is $^4P_3 = 4! = 4 \times 3 \times 2 \times 1 = 24.$
View full question & answer→MCQ 1571 Mark
If a * b denote the bigger among a and b and if ab = (a * b) + 3, then 4.7 =
Answer4.7 = (4 * 7) + 3
= 7 + 3
= 10
View full question & answer→MCQ 1581 Mark
Which of the following functions from $\text{A}=\{\text{x}\in\text{R}:-1\leq\text{x}\leq1\}$ to itself are bijections?
AnswerCorrect option: B. $\text{f(x)}=\sin\frac{\pi\text{x}}{2}$
It is clear that f(x) is one-one.
Range of $\text{f}=\Big[\sin\frac{\pi(-1)}{2},\sin\frac{\pi(1)}{2}\Big]=\Big[\sin\frac{-\pi}{2},\sin\frac{\pi}{2}\Big]$
= A = Co-domain of f
⇒ f is onto.
So, f is a bijection.
View full question & answer→MCQ 1591 Mark
Let $R = \{(a, a), (b, b), (c, c), (a, b)\}$ be a relation on set $A = a, b, c.$ Then$, R$ is:
AnswerReflexivity: Since $(\text{a, a})\in\text{R}\ \forall\ \text{a}\in\text{A}, R$ is reflexive on $A.$
Symmetry: Since $(\text{a, b})\in\text{R}$ but $(\text{b, a})\notin\text{R,}$ is not symmetric on $A.$
$\Rightarrow R$ is not antisymmetric on $A.$
Also$, R$ is not an identity relation on $A.$
View full question & answer→MCQ 1601 Mark
Number of binary operations on the set $\{a, b\}$ are:
AnswerA binary operation $^*$ on $\{a, b\}$ is a function from $\{a, b\} \times \{a, b\} \rightarrow \{a, b\}$
i.e., $^*$ is a function from $\{(a, a), (a, b), (b, a), (b, b)\}.$
Hence, the total number of binary operations on the set $\{a, b\}$ is $2^4$ i.e., $16.$
The correct answer is $B.$
View full question & answer→MCQ 1611 Mark
If the set A contains 7 elements and the set B contains 10 elements, then the number one-one functions from A to B is:
AnswerCorrect option: B. $^{10}\text{C}_7\times7!$
As, the number of one-one functions from A to B with m and n elements, respectively $= \ ^{\text{n}}\text{P}_\text{m}=\ ^{\text{n}}\text{C}_\text{m}\times\text{m}!$
So, the number of one-one functions from A to B with 7 and 10 elements, respectively $=\ ^{10}\text{P}_7=\ ^{10}\text{C}_7\times7!$
View full question & answer→MCQ 1621 Mark
The function $f : R \rightarrow R$ given by $f(x) = x^3 – 1$ is:
View full question & answer→MCQ 1631 Mark
If $f(x) = \sin^2 x$ and the composite function $\text{g(f(x))} = |\sin\text{x}|,$ then $g(x)$ is equal to:
- A
$\sqrt{\text{x}-1}$
- ✓
$\sqrt{\text{x}}$
- C
$\sqrt{\text{x}+1}$
- D
$-\sqrt{\text{x}}$
AnswerCorrect option: B. $\sqrt{\text{x}}$
Given that $\text{f(x)}=\sin^2\text{x}$ and the composite function $\text{g(f(x))}=|\sin\text{x}|$
We will do it using trial and error method.
If we take $\text{g(x)}=-\sqrt{\text{x}}$ and $\text{f(x)}=\sin^2\text{x}$
$\text{g(f(x))}=\text{g}(\sin^2\text{x})$
$=-\sin\text{x}$
Which contradicts to the $\text{g(f(x))}=|\sin\text{x}|$
Hence, we take $\text{g(x)}=\sqrt{\text{x}}$
$\text{g(f(x))}=\text{g}(\sin^2\text{x})$
$=\sqrt{\sin^2\text{x}}$
$=|\sin\text{x}|$
View full question & answer→MCQ 1641 Mark
Let $\text{f(x)}=\frac{\text{x}-1}{\text{x}+1},$ then $f(f(x))$ is:
- A
$\frac{1}{\text{x}}$
- ✓
$-\frac{1}{\text{x}}$
- C
$\frac{1}{\text{x}+1}$
- D
$\frac{1}{\text{x}-1}$
AnswerCorrect option: B. $-\frac{1}{\text{x}}$
View full question & answer→MCQ 1651 Mark
Choose the correct answer from the given four options. Let $f : R \rightarrow R$ be defined by $\text{f}(\text{x})=\begin{cases}2\text{x}:\text{x}>3\\\text{x}^2:1<\text{x}\leq3\\3\text{x}:\text{x}\leq1\end{cases}$ Then $f(-1) + f(2) + f(4)$ is:
AnswerWe are given that, $\text{f}(\text{x})=\begin{cases}2\text{x}:\text{x}>3\\\text{x}^2:1<\text{x}\leq3\\3\text{x}:\text{x}\leq1\end{cases}$
Now, $f(-1) + f(2) + f(4)$
$ = 3(-1) + (2)^2 + 2 \times 4$
$= -3 + 4 + 8$
$= 9$
View full question & answer→MCQ 1661 Mark
Choose the correct answer from the given four options. Let $f : R \rightarrow R$ be defined by $f(x) = 3x^2 – 5$ and $g : R \rightarrow R$ by $\text{g}(\text{x})=\frac{\text{x}}{\text{x}^2+1}.$ Then $gof$ is:
- ✓
$\frac{3\text{x}^2-5}{9\text{x}^4-30\text{x}^2+26}$
- B
$\frac{3\text{x}^2-5}{9\text{x}^4-6\text{x}^2+26}$
- C
$\frac{3\text{x}^2}{\text{x}^4+2\text{x}^2-4}$
- D
$\frac{3\text{x}^2}{9\text{x}^4+30\text{x}^2-2}$
AnswerCorrect option: A. $\frac{3\text{x}^2-5}{9\text{x}^4-30\text{x}^2+26}$
Given that, $f(x) = 3x^2 - 5$ and $\text{g}(\text{x})=\frac{\text{x}}{\text{x}^2+1}$
$gof(x) = g(f(x))$
$= g(3x^2 - 5)$
$=\frac{3\text{x}^2-5}{(3\text{x}^2-5)^2+1}$
$=\frac{3\text{x}^2-5}{9\text{x}^4-30\text{x}^2+25+1}$
$=\frac{3\text{x}^2+5}{9\text{x}^4-30\text{x}^2+26}$
View full question & answer→MCQ 1671 Mark
A function f from the set of natural numbers to integers defined by $\text{f(n)}=\begin{cases}\frac{\text{n}-1}{2},&\text{when n is odd}\\-\frac{\text{n}}{2},&\text{when n is even}\end{cases}$
- A
Neither one-one nor onto.
- B
- C
- ✓
AnswerInjectivity: Let x and y be any two elements in the domain (N).
Case-1: Both x and y are even.
Let f(x) = f(y)
$\Rightarrow\ \frac{-\text{x}}{2}=\frac{-\text{y}}{2}$
$\Rightarrow-\text{x}=-\text{y}$
$\Rightarrow\ \text{x}=\text{y}$
Case-2: Both x and y are odd.
Let f(x) = f(y)
$\Rightarrow\ \frac{\text{x}-1}{2}=\frac{\text{y}-1}{2}$
$\Rightarrow\ \text{x}-1=\text{y}-1$
$\Rightarrow\ \text{x}=\text{y}$
Case-3: Let x be even and y be odd.
Then, $\text{f(x)}=\frac{-\text{x}}{2}$ and $\text{f(y)}=\frac{\text{y}-1}{2}$
Then, clearly
$\text{x}\neq\text{y}$
$\Rightarrow\ \text{f(x)}\neq\text{f(y)}$
From all the cases, f is one-one.
Surjectivity: Co-domain of f = Z = {......, -3, -2, -1, 0, 1, 2, 3, ......}
Range of $\text{f}=\Big\{....,\ \frac{-3-1}{2},\ \frac{-(-2)}{2},\ \frac{-1-1}{2},\ \frac{0}{2},\ \frac{1-1}{2},\ \frac{-2}{2},\ \frac{3-1}{2},\ ....\Big\}$
⇒ Range of f = {....., -2, 1, -1, 0, 0, -1, 1, .....}
⇒ Range of f = {....., -2, -1, 0, 1, 2, ......}
⇒ Co-domain of f = Range of f
⇒ f is onto.
View full question & answer→MCQ 1681 Mark
Let M be the set of all 2 × 2 matrices with entries from the set R of real numbers. Then, the function f : M→ R defined by f(A) = |A| for every A ∈ M, is:
- A
- B
Neither one-one nor onto.
- C
- ✓
Answer$\text{M}=\begin{Bmatrix}\text{A}=\begin{bmatrix}\text{a}&\text{b}\\\text{c}&\text{d}\end{bmatrix}:\text{a, b, c, d}\in\text{R}\end{Bmatrix}$
f : M → R is given by f(A) = |A|
Injectivity: $\text{f}\begin{pmatrix}\begin{bmatrix}0&0\\0&0\end{bmatrix}\end{pmatrix}=\begin{vmatrix}0&0\\0&0\end{vmatrix}=0$
and $\text{f}\begin{pmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix}\end{pmatrix}=\begin{vmatrix}1&0\\0&0\end{vmatrix}=0$
$\Rightarrow\ \text{f}\begin{pmatrix}\begin{bmatrix}0&0\\0&0\end{bmatrix}\end{pmatrix}=\text{f}\begin{pmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix}\end{pmatrix}=0$
So, f is not one-one.
Surjectivity: Let y be an element of the co-domain, such that
$\text{f(A)}=-\text{y},\ \text{A}=\begin{bmatrix}\text{a} & \text{b} \\\text{c} & \text{d} \end{bmatrix}$
$\Rightarrow\ \begin{vmatrix}\text{a}&\text{b}\\\text{c}&\text{d}\end{vmatrix}=\text{y}$
$\Rightarrow\ \text{ad}-\text{bc}=\text{y}$
$\Rightarrow\ \text{a, b, c, d}\in\text{R}$
$\Rightarrow\ \text{A}=\begin{bmatrix}\text{a} & \text{b} \\\text{c} & \text{d} \end{bmatrix}\in\text{M}$
⇒ f is onto.
View full question & answer→MCQ 1691 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
$\frac{2}3$
AnswerCorrect 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
$3\odot\text{b}=\text{e}$
$\Rightarrow\frac{3\text{b}}2=2$
$\Rightarrow\text{b}=\frac{4}3$
View full question & answer→MCQ 1701 Mark
Which of the following is true?
AnswerCorrect 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.
View full question & answer→MCQ 1711 Mark
Let $\times$ be a binary operation on set of integers I, defined by $a \times b = a + b - 3,$ then find the value of $3 \times 4.$
View full question & answer→MCQ 1721 Mark
Let A = {1, 2, 3}. Then the number of relations containing (1, 2) and (1, 3), which are reflexive and symmetric but not transitive is:
AnswerRelation R is reflexive as (1, 1), (2, 2), (3, 3) ∈ R.
Relation R is symmetric since (1, 2), (2, 1) ∈ R and (1, 3), (3, 1) ∈ R.
But relation R is not transitive as (3, 1), (1, 2) ∈ R but (3, 2)∉ R.
When we add any one of the two pairs, i.e. (3, 2) and (2, 3) or both, to relation R, it will become transitive.
Hence, the total number of desired relations is 1.
View full question & answer→MCQ 1731 Mark
Let $R$ be a relation on the set $N$ of natural numbers denoted by $nRm \Leftrightarrow n$ is a factor of $m ($i.e. $n | m).$ Then$, R$ is:
- A
- B
Transitive and symmetric.
- C
- ✓
Reflexive, transitive but not symmetric.
AnswerCorrect option: D. Reflexive, transitive but not symmetric.
View full question & answer→MCQ 1741 Mark
A function f from the set of natural numbers to the set of integers defined by $\text{f(n)}\begin{cases}\frac{\text{n}-1}{2},&\text{when n is odd}\\-\frac{\text{n}}{2},&\text{when n is even}\end{cases}$ is:
- A
Neither one-one nor onto.
- B
- C
- ✓
AnswerGiven function is,
$\text{f(n)}=\frac{\text{n}-1}{2}$ for n is odd
$=-\frac{\text{n}}{2}$ for n is even
For n is odd,
If f(n) = f(m) then
$\frac{\text{n}-1}{2}=\frac{\text{m}-1}{2}$
⇒ n = m
Also, for n is even if f(n) = f(m) then n = m
Hence, f is one-one.
Also, each element of y is associated with at least one element of x,
f is onto.
View full question & answer→MCQ 1751 Mark
The binary operation * defined on N by a * b = a + b + ab for all a, b ∈ N is:
- A
- B
- ✓
Commutative and associative both.
- D
AnswerCorrect 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.
View full question & answer→MCQ 1761 Mark
The function $\text{f}:\Big[\frac{-1}{2},\frac{1}{2},\frac{1}{2}\Big]\rightarrow\ \Big[\frac{-\pi}{2},\frac{\pi}{2}\Big],$ defined by $\text{f(x)}=\sin^{-1}(3\text{x}-4\text{x}^3),$ is:
- ✓
- B
Injection but not a surjection.
- C
Surjection but not an injection.
- D
Neither an injection nor a surjection.
Answer$\text{f(x)}=\sin^{-1}(3\text{x}-4\text{x}^3)$
$\Rightarrow\ \text{f(x)}=3\sin^{-1}\text{x}$
Injectivity: Let $x$ and $y$ be two elements in the domain $\Big[\frac{-1}{2},\frac{1}{2}\Big],$ such that
$f(x) = f(y)$
$\Rightarrow\ 3\sin^{-1}\text{x}=3\sin^{-1}\text{y}$
$\Rightarrow\ \sin^{-1}\text{x}=\sin^{-1}\text{y}$
$\Rightarrow\ \text{x}=\text{y}$
So,$ f$ is one$-$one.
Surjectivity: Let $y$ be any element in the $co-$domain, such that
$f(x) = y$
$\Rightarrow\ 3\sin^{-1}\text{x}=\text{y}$
$\Rightarrow\ \sin^{-1}\text{x}=\frac{\text{y}}{3}$
$\Rightarrow\ \text{x}=\sin\frac{\text{y}}{3}\in\Big[\frac{-1}{2},\frac{1}{2}\Big]$
$\Rightarrow f$ is onto.
$\Rightarrow f$ is a bijection.
View full question & answer→MCQ 1771 Mark
Let $f$ be an injective map with domain $\{x, y, z\}$ and range $\{1, 2, 3\},$ such that exactly one of the following statements is correct and the remaining are false.$\text{f(x)}=1,\ \text{f(y)}\neq1,\ \text{f(z)}\neq2.$ The value of $f^{-1}(1)$ is:
AnswerGiven that $f$ be an injective map with domain $\{x, y, z\}$ and range $\{1, 2, 3\}$
$f(x) = 1, \text{f(y)}\neq1,\ \text{f(z)}\neq2$
As $f(x) = 1 $
$\Rightarrow f^{-1}(1) = y$
View full question & answer→MCQ 1781 Mark
Let $\text{A}=\{\text{x}\in\text{R}:\text{x}\leq1\}$ and $f : A \rightarrow A$ be defined as $f(x) = x(2 - x).$ Then $f^{-1}(x)$ is:
- A
$1+\sqrt{1-\text{x}}$
- ✓
$1-\sqrt{1-\text{x}}$
- C
$\sqrt{1-\text{x}}$
- D
$1\pm\sqrt{1-\text{x}}$
AnswerCorrect option: B. $1-\sqrt{1-\text{x}}$
Let $y$ be the element in the co$-$domain $R$ such that $f^{-1}\ (x) = y ......(1)$
$\Rightarrow f(y) = x$ and $\text{y}\leq1$
$\Rightarrow y(2 - y) = x$
$\Rightarrow 2y - y^2 = x$
$\Rightarrow y^2 - 2y + x = 0$
$\Rightarrow y^2 - 2y = -x $
$\Rightarrow y^2 - 2y + 1 = 1 - x$
$\Rightarrow\ (\text{y}-1)^2=\sqrt{1-\text{x}}$
$\Rightarrow\ \text{y}-1=\pm\sqrt{1-\text{x}}$
$\Rightarrow\ \text{y}=1\pm\sqrt{1-\text{x}}$
$\Rightarrow\ \text{y}=1-\sqrt{1-\text{x}}$ $(\because\ \text{y}\leq1)$
View full question & answer→MCQ 1791 Mark
The mapping $f : N \rightarrow N$ is given by $f(n) = 1 + n^2, n \in N$ when $N$ is the set of natural numbers is:
AnswerCorrect option: C. One$-$one but not onto.
One$-$one but not onto.
View full question & answer→MCQ 1801 Mark
If R is a relation on the set A = {1, 2, 3, 4, 5, 6, 7, 8, 9} given by xRy ⇔ y = 3x, then R =
- A
{(3, 1), (6, 2), (8, 2), (9, 3)}
- B
- C
- ✓
AnswerThe relation R is defined as,
$\text{R}=\{(\text{x, y}):\ \text{x, y}\in\text{A}:\text{y}=3\text{x}\}$
⇒ R = {(1, 3), (2, 6), (3, 9)}
View full question & answer→MCQ 1811 Mark
Consider the binary operation $\times$ on $Q$ defind by $a \times b = a + 12b + ab$ for $a, b \in Q.$ Find $2\times\frac{1}{3}$
- ✓
$\frac{20}{3}$
- B
$4$
- C
$18$
- D
$\frac{16}{3}$
AnswerCorrect option: A. $\frac{20}{3}$
View full question & answer→MCQ 1821 Mark
A relation $\phi$ from C to R is defined by $\text{x }\phi\text{ y}\Leftrightarrow|\text{x}|=\text{y.}$ Which one is correct?
- A
$(2+3\text{i})\phi13$
- B
$3\phi(-3)$
- C
$(1+\text{i})\phi2$
- ✓
$\text{i}\phi1$
AnswerCorrect option: D. $\text{i}\phi1$
$\because\ |2+3\text{i}|=\sqrt{13}\neq13$
$|3|\neq-3$
$|1+\text{i}|=\sqrt{2}\neq2$
and $|\text{i}|=1$
So, $(\text{i, }1)\in\phi$
View full question & answer→MCQ 1831 Mark
Let $f : R \rightarrow R$ be given by $f(x) = x^2 - 3.$ Then, $f^{-1}$ is given by:
- A
$\sqrt{\text{x}+3}$
- B
$\sqrt{\text{x}}+3$
- C
$\text{x}+\sqrt{3}$
- ✓
$\text{None of these}$
AnswerCorrect option: D. $\text{None of these}$
Given function is $f : R \rightarrow R$ be given by $f(x) = x^2 - 3.$
$\text{y} = \text{x}^2 - 3$
$\text{y} + 3 = \text{x}^2$
$\text{x}=\pm\sqrt{\text{y}+3}$
$\Rightarrow\ \text{y}=\pm\sqrt{\text{x}+3}$
View full question & answer→MCQ 1841 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
$(4 + 5 + 3)^2$
AnswerCorrect option: C. $41^2 + 3^2$
Given $a^* b = a^2 + b^2$
So,$ 4^* 5 = 42 + 52$
Now,
$(4^* 5)^* 3 = (4^* 5)2 + 32$
$= (42 + 52)2 + 32$
$= 412 + 32$
View full question & answer→MCQ 1851 Mark
Let the function f : R - {-b} → R - {1} be defined by $\text{f(x)}=\frac{\text{x}+\text{a}}{\text{x}+\text{b}},\ \text{a}\neq\text{b}.$ Then,
- A
f is one-one but not onto.
- B
f is onto but not one-one.
- ✓
f is both one-one and onto.
- D
AnswerCorrect option: C. f is both one-one and onto.
Injectivity: Let x and y be two elements in the domain R - {-b}, such that
f(x) = f(y) ⇒ x + ax + b = y + ay + b
⇒ x + ay + b = x + by + a
⇒ xy + bx + ay + ab = xy + ax + by + ab
⇒ bx + ay = ax + by
⇒ a - bx = a - by
⇒ x = y
So, f is one-one.
Surjectivity: Let y be an element in the co-domain of f,
i.e., R - {1}, such that f(x) = y
⇒ x + ax + b = y
⇒ x + a ⇒ x = -a
So, f is onto.
View full question & answer→MCQ 1861 Mark
Let $f : N \rightarrow R : \text{f}(\text{x})=\frac{(2\text{x}-1)}{2}$ and $g : Q \rightarrow R : g(x) = x + 2$ be two functions. Then, $(gof) (\frac{3}{2})$ is:
View full question & answer→MCQ 1871 Mark
The inverse of the function $\text{f}:\text{R}\rightarrow\{\text{x}\in\text{R}:\text{x}<1\}$ given by $\text{f(x)}=\frac{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}{\text{e}^\text{x}+\text{e}^{-\text{x}}}$ is:
- ✓
$\frac{1}{2}\log\frac{1+\text{x}}{1-\text{x}}$
- B
$\frac{1}{2}\log\frac{2+\text{x}}{2-\text{x}}$
- C
$\frac{1}{2}\log\frac{1-\text{x}}{1+\text{x}}$
- D
AnswerCorrect option: A. $\frac{1}{2}\log\frac{1+\text{x}}{1-\text{x}}$
Let $f^{-1}(x) = y .....(1)$
$\Rightarrow \text{f(y)}=\text{x}$
$\Rightarrow \frac{\text{e}^{\text{y}}-\text{e}^{-\text{y}}}{\text{e}^{\text{y}}+\text{e}^{-\text{y}}}=\text{x}$
$\Rightarrow \frac{\text{e}^{-\text{y}}(\text{e}^{2\text{y}}-1)}{\text{e}^{-\text{y}}(\text{e}^{2\text{y}}+1)}=\text{x}$
$\Rightarrow (\text{e}^{2\text{y}}-1)=\text{x}(\text{e}^{2\text{y}}+1)$
$\Rightarrow \text{e}^{2\text{y}}-1=\text{xe}^{2\text{y}}+\text{x}$
$\Rightarrow \text{e}^{2\text{y}}=\frac{1+\text{x}}{1-\text{x}}$
$\Rightarrow 2\text{y}=\log_\text{e}\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$
$\Rightarrow \text{y}=\frac{1}{2}\log_\text{e}\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$
$\Rightarrow \text{f}^{-1}(\text{x})=\frac{1}{2}\log_\text{e}\Big(\frac{1+\text{x}}{1-\text{x}}\Big) [$From $(1)]$
View full question & answer→MCQ 1881 Mark
Let $A = \{1, 2, 3\}.$ Then number of equivalence relations containing $(1, 2)$ is:
AnswerThe given set is $A = \{1, 2, 3\}.$
The smallest equivalence relation containing $(1, 2)$ is given by,
$R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$
Now, we are left with only four pairs i.e., $(2, 3), (3, 2), (1, 3),$ and $(3, 1).$
If we add any one pair $[$say $(2, 3)]$ to $R_1,$ then for symmetry we must add $(3, 2)$.
Also, for transitivity we required to add $(1, 3)$ and $(3, 1).$
Hence, the only equivalence relation $($bigger than $R_1)$ is the universal relation.
This shows that the total number of equivalence relations containing $(1, 2)$ is two
View full question & answer→MCQ 1891 Mark
Let $f : R \rightarrow R$ be a function defined by $f(x) = x^3 + 4,$ then f is:
View full question & answer→MCQ 1901 Mark
The relation $R$ defined on the set $A = \{1, 2, 3, 4, 5\}$ by $R = \{(a, b): |a^2 - b^2| < 16\}$ is given by:
- A
$\{(1, 1), (2, 1), (3, 1), (4, 1), (2, 3)\}$
- B
$\{(2, 2), (3, 2), (4, 2), (2, 4)\}$
- C
$\{(3, 3), (4, 3), (5, 4), (3, 4)\}$
- ✓
Answer$R$ is given by $\{(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3), (4, 5), (5, 4),(1, 3), (3, 1), (1, 4), (4, 1), (2, 4), (4, 2)\}$ which is not mentioned in $(a), (b)$ or $(c).$
View full question & answer→MCQ 1911 Mark
If $f : [1, \infty ) \rightarrow [2, \infty)$ is given by $\text{f(x)}=\text{x}+\frac{1}{\text{x}},$ then $f^{-1}$ equals to:
- ✓
$\frac{\text{x}+\sqrt{\text{x}^2-4}}{2}$
- 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. $\frac{\text{x}+\sqrt{\text{x}^2-4}}{2}$
$\frac{\text{x}+\sqrt{\text{x}^2-4}}{2}$
View full question & answer→MCQ 1921 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.
- ✓
- D
Answer$a ^* b = a^2 + b^2$
$b ^* a = b^2 + a^2$
$\Rightarrow a ^* b = b ^* a$
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.
View full question & answer→MCQ 1931 Mark
Which of the following functions from $Z$ into $Z$ are bijective$?$
- A
$f(x) = x^3$
- ✓
$f(x) = x + 2$
- C
$f(x) = 2x + 1$
- D
$f(x) = x^2 + 1$
AnswerCorrect option: B. $f(x) = x + 2$
$f(x) = x + 2$
View full question & answer→MCQ 1941 Mark
Let $T$ be the set of all triangles in the Euclidean plane, and let a relation $R$ on $T$ be defined as $\text{aRb}$ if a is congruent to $b \forall a, b \in T.$ Then $R$ is:
- A
Reflexive but not transitive.
- B
Transitive but not symmetric.
- ✓
- D
View full question & answer→MCQ 1951 Mark
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.
We have,
R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}
As, $(\text{a, a})\in\text{R}\ \forall\ \text{a}\in\text{A}$
So, R is reflexive relation.
Also, $(1,2)\in\text{R}$ but $(2,1)\notin\text{R}$
So, R is not symmetric relation.
And, $(1,2)\in\text{R},\ (2,3)\in\text{R}$ and $(1,3)\in\text{R}$
So, R is transitive relation.
View full question & answer→MCQ 1961 Mark
Number of binary operations on the set {a, b} are:
AnswerLet the given set be A = {a, b}
n(A) = 2
Total number of binary operations = 2(2 × Number of elements in the set)
= 2(2 × 2)
= 24
= 16
Therefore, the number of binary operations on the set {a, b} are 16.
View full question & answer→MCQ 1971 Mark
If 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:
AnswerGiven,
n(A) = 5
n(B) = 6
Each element in set B is assigned to only one element in set A for the one-one function.
Here, only ‘5’ elements of set B are assigned to ‘5’ elements of set ‘A’ and one element will be left in set B.
The range of the function must be equal to B.
However, for the given sets, it is not possible.
Thus, the range of functions does not contain all ‘6’ elements of set ‘ B’.
Therefore, if the function is one-one it cannot be onto.
Hence, the number of one-one and onto mappings from A to B is 0.
View full question & answer→MCQ 1981 Mark
Let $X = \{-1, 0, 1\}, Y = \{0, 2\}$ and a function $f : X \rightarrow Y$ defiend by $y = 2x^4,$ is:
- A
One$-$one onto.
- B
One$-$one into.
- ✓
Many$-$one onto.
- D
Many$-$one into.
AnswerCorrect option: C. Many$-$one onto.
Many$-$one onto.
View full question & answer→MCQ 1991 Mark
If $f : R \rightarrow R$ is given by $f(x) = 3x - 5,$ then $f^{-1}(x)$
- A
is given by $\frac{1}{3\text{x}-5}$
- ✓
is given by $\frac{\text{x}+5}{3}$
- C
does not exist because $f$ is not one$-$one.
- D
does not exist because $f$ is not onto.
AnswerCorrect option: B. is given by $\frac{\text{x}+5}{3}$
Given function is $f : R \rightarrow R$ is given by $f(x) = 3x - 5$
To find $f^{-1}(x)$
$y = f(x)$
$\Rightarrow y = 3x - 5$
$\Rightarrow y + 5 = 3x$
$\Rightarrow\ \text{y}=\frac{\text{y}+5}{3}$
Hence, $\text{f}^{-1}(\text{x})=\frac{\text{x}+5}{3}$
View full question & answer→MCQ 2001 Mark
If the function f : R → A given by $\text{f(x)}=\frac{\text{x}^2}{\text{x}^2+1}$ is a surjection, then A =
AnswerAs f is surjective, range of f = co-domain of f
⇒ A = range of f
$=\frac{\text{x}^2}{\text{x}^2+1},$
$\text{y}=\frac{\text{x}^2}{\text{x}^2+1}$
$\Rightarrow\ \text{y}(\text{x}^2+1)$
$\Rightarrow\ \text{x}^2=\frac{-\text{y}}{(\text{y}-1)}$
$\Rightarrow\ \text{x}=\sqrt{\frac{\text{y}}{(1-\text{y})}}$
$\Rightarrow\ \frac{\text{y}}{(1-\text{y})}\geq0$
$\Rightarrow\ \text{y}\in[0,1)$
⇒ Range of f = [0, 1)
⇒ A = [0, 1)
View full question & answer→MCQ 2011 Mark
Let R be a relation on the set N of natural numbers defined by nRm if n divides m. Then, R is:
- A
- B
Transitive and symmetric.
- C
- ✓
Reflexive, transitive but not symmetric.
AnswerCorrect option: D. Reflexive, transitive but not symmetric.
We have,
R = {(m, n): n divides m; m, n ∈ N}
As, m divides m
$\Rightarrow\ (\text{m, m})\in\text{R}\ \forall\ \text{m}\in\text{N}$
So, R is reflexive.
Since, $(2,1)\in\text{R}$ i.e. 1 divides 2
but 2 cannot divide 1 i.e. $(2,1)\notin\text{R}$
So, R is not symmetric.
Let $(\text{m, n})\in\text{R}$ and $(\text{n, p})\in\text{R.}$ Then,
n divides m and p divides n
⇒ p divides m
$\Rightarrow\ (\text{m, p})\in\text{R}$
So, R is transitive.
View full question & answer→MCQ 2021 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:
AnswerGiven: a * b = 3a - b
2 * 3 = 3 (2) - 3
= 6 - 3
= 3
(2 * 3) * 4 = 3 * 4
= 3(3) - 4
= 9 - 4
= 5
View full question & answer→MCQ 2031 Mark
Let $\text{A}=\{\text{x}\in\text{R}:-1\leq\text{x}\leq1\}=\text{B}$ and $\text{C}=\{\text{x}\in\text{R}:\text{x}\geq0\}$ and let $\text{S}=\{(\text{x, y})\in\text{A}\times\text{B}:\text{x}^2+\text{y}^2=1\}$ and $\text{S}_0=\{(\text{x, y})\in\text{A}\times\text{C}:\text{x}^2+\text{y}^2=1\}.$ Then,
- ✓
$S$ defines a function from $A$ to $B.$
- B
$S_0$ defines a function from $A$ to $C.$
- C
$S_0$ defines a function from $A$ to $B$.
- D
$S$ defines a function from $A$ to $C$.
AnswerCorrect option: A. $S$ defines a function from $A$ to $B.$
Given that $\text{A}=\{\text{x}\in\text{R}:-1\leq\text{x}\leq1\}=\text{B}$ and $\text{C}=\{\text{x}\in\text{R}:\text{x}\geq0\}$ and $\text{S}=\{(\text{x, y})\in\text{A}\times\text{B}:\text{x}^2+\text{y}^2=1\}$ and $\text{S}_0=\{(\text{x, y})\in\text{A}\times\text{C}:\text{x}^2+\text{y}^2=1\}$
$\text{x}^2+\text{y}^2=1$
$\Rightarrow\ \text{y}^2=1-\text{x}^2$
$\Rightarrow\ \text{y}=\sqrt{1-\text{x}^2}$
$\text{y}\in\text{B}$
Hence, $S$ defines a function from $A$ to $B.$
View full question & answer→MCQ 2041 Mark
Let $\text{f}:\text{R}-\Big\{\frac{3}{5}\Big\}\rightarrow\ \text{R}$ be defined by $\text{f(x)}=\frac{3\text{x}+2}{5\text{x}-3}.$ Then,
AnswerCorrect option: A. $f^{-1}(x) = f(x)$
Given function is $\text{f}:\text{R}-\Big\{\frac{3}{5}\Big\}\rightarrow\ \text{R}$ be defined by $\text{f(x)}=\frac{3\text{x}+2}{5\text{x}-3}$
$fof(x) = f(f(x))$
$=\text{f}\Big(\frac{3\text{x}+2}{5\text{x}-3}\Big)$
$=\frac{3\big(\frac{3\text{x}+2}{5\text{x}-3}\big)+2}{5\big(\frac{3\text{x}+2}{5\text{x}-3}\big)-3}$
After solving you will get
$f(f(x)) = x$
Also, $f^{-1}(x) = f(x)$ you can check.
View full question & answer→MCQ 2051 Mark
Choose the correct answer from the given four options.
Let f : [0, 1] → [0, 1] be defined by $\text{f}(\text{x})=\begin{cases}\text{x, if x is rational}\\1-\text{x, if x is irrational}\end{cases}$ Then (fof)x is:
AnswerWe are given that, f : [0, 1] → [0, 1] be defined by $\text{f}(\text{x})=\begin{cases}\text{x, if x is rational}\\1-\text{x, if x is irrational}\end{cases}$
Now, $(\text{fof})\text{x} = \text{f}(\text{f(x)})$
$=\text{x}$
View full question & answer→MCQ 2061 Mark
If $f(x) = (ax^2 – b)^3,$ then the function g such that $f\{g(x)\} = g\{f(x)\}$ is given by:
- A
$\text{g(x)}=\Big(\frac{\text{b}-\text{x}^\frac{1}{3}}{\text{a}}\Big)^\frac{1}{2}$
- B
$\text{g(x)}=\frac{1}{(\text{ax}^2+\text{b})^3}$
- C
$\text{g(x)}=(\text{ax}^2+\text{b})^\frac{1}{3}$
- ✓
$\text{g(x)}=\Big(\frac{\text{x}^\frac{1}{3}+\text{b}}{\text{a}}\Big)^\frac{1}{2}$
AnswerCorrect option: D. $\text{g(x)}=\Big(\frac{\text{x}^\frac{1}{3}+\text{b}}{\text{a}}\Big)^\frac{1}{2}$
$\text{g(x)}=\Big(\frac{\text{x}^\frac{1}{3}+\text{b}}{\text{a}}\Big)^\frac{1}{2}$
View full question & answer→MCQ 2071 Mark
Find the identity element in the set $I^+$ of all positive integers defined by $a \times b = a + b$ for all $a, b \in I^+.$
View full question & answer→MCQ 2081 Mark
The function $f : R \rightarrow R$ defined by $f(x) = 3 – 4x$ is:
- ✓
- B
- C
None one$-$one.
- D
None one$-$into.
View full question & answer→MCQ 2091 Mark
If $f : A \rightarrow B$ given by $3^{f(x)} + 2^{-x} = 4$ is a bijection, then
- A
$\text{A}=\{\text{x}\in\text{R}:-1<\text{x}<\infty\},$ $\text{B}=\{\text{x}\in\text{R}:2<\text{x}<4\}$
- B
$\text{A}=\{\text{x}\in\text{R}:-3<\text{x}<\infty\},$ $\text{B}=\{\text{x}\in\text{R}:2<\text{x}<4\}$
- C
$\text{A}=\{\text{x}\in\text{R}:-2<\text{x}<\infty\},$ $\text{B}=\{\text{x}\in\text{R}:2<\text{x}<4\}$
- ✓
Answer$\text{f}:\text{A}\Rightarrow \text{B}$
$3^\text{f(x)}+2^{-\text{x}}=4$
$\Rightarrow 3^{\text{f(x)}}=4-2^{-\text{x}}$
Taking $\log$ on both the sides,
$\text{f(x)}\log3=\log(4-2^{-\text{x}})$
$\Rightarrow \text{f(x)}=\frac{\log(4-2^{-\text{x}})}{\log3}$
Logaritmic function will only be defined if $4-2^{-\text{x}}>0$
$\Rightarrow 4>2^{-\text{x}}$
$\Rightarrow 2^2>2^{-\text{x}}$
$\Rightarrow 2>-\text{x}$
$\Rightarrow-2<\text{x}$
$\Rightarrow \text{x}\in(-2,\infty)$
That means $\text{A}=\{\text{x}\in\text{R}:-2<\text{x}<\infty\}$
As we know that, $\text{f(x)}=\frac{\log(4-2^{-\text{x}})}{\log3}$
We take $\text{x}=0\in(-2,\infty)$
$\Rightarrow f(x) = 1$ which does not belong to any of the options.
View full question & answer→MCQ 2101 Mark
If $\text{g(f(x))}=|\sin\text{x}|$ and $\text{f(g(x))}=(\sin\sqrt{\text{x}})^2,$ then
- ✓
$\text{f(x)}=\sin^2\text{x},\ \text{g(x)}=\sqrt{\text{x}}$
- B
$\text{f(x)}=\sin\text{x},\ \text{g(x)}=|\text{x}|$
- C
$\text{f(x)}=\text{x}^2,\ \text{g(x)}=\sin\sqrt{\text{x}}$
- D
$\text{f and g cannot be determined.}$
AnswerCorrect option: A. $\text{f(x)}=\sin^2\text{x},\ \text{g(x)}=\sqrt{\text{x}}$
If we solve it by the trial-and-error method, we can see that (a) satisfies the given condition.
From (a):
$\text{f(x)}=\sin^2\text{x}$ and $\text{g(x)}=\sqrt{\text{x}}$
$\Rightarrow\ \text{f(g(x))}=\text{f}(\sqrt{\text{x}})=\sin^2\sqrt{\text{x}}$
$=(\sin\sqrt{\text{x}})^2$
View full question & answer→MCQ 2111 Mark
Consider the function $f$ in $\text{A}=\text{R}-\{\frac{2}{3}\}$ defiend as $\text{f(x)}=\frac{4\text{x}+3}{6\text{x}-4}.$ Find $f^{-1}$.
- ✓
$\frac{3+4\text{x}}{6\text{x}-4}$
- B
$\frac{6\text{x}-4}{3+4\text{x}}$
- C
$\frac{3-4\text{x}}{6\text{x}-4}$
- D
$\frac{9+2\text{x}}{6\text{x}-4}$
AnswerCorrect option: A. $\frac{3+4\text{x}}{6\text{x}-4}$
$\frac{3+4\text{x}}{6\text{x}-4}$
View full question & answer→MCQ 2121 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:
AnswerWe have,
$a ^* b = a + b + 1$
Let e be the identity element of $*.$ Then,
$a^ * e = a = e ^* a$
$a + e + 1 = a$
$e = a - a - 1$
$e = -1$
View full question & answer→MCQ 2131 Mark
Find which of the binary operations are commutative and which are associative. Consider a binary operation $*$ on $N$ defined as $a * b = a^3 + b^3.$ Choose the correct answer.
- A
Is $*$ both associative and commutative?
- ✓
Is $*$ commutative but not associative?
- C
Is $*$ commutative but not associative?
- D
Is $*$ neither commutative nor associative?
AnswerCorrect option: B. Is $*$ commutative but not associative?
$a * b = a^3 + b^3 = b^3 + a^3 = b * a$
$\therefore$ The operation is commutative.
Again, $(a * b) * c = a * (a^3 + b^3) = a^3(a^3 + b^3)^3$
And $(a * b) * c= (a^3 + b^3) * c = (a^3 + b^3)^3 + c^3 \neq\text{a}*(\text{b}*\text{c})$
$\therefore$ The operation $*$ is not associative.
Therefore, option $(B)$ is correct.
Is $*$ commutative but not associative?
View full question & answer→MCQ 2141 Mark
Let A = {2, 3, 4, 5, ..., 17, 18}. Let $'\simeq'$ be the equivalence relation on A × A, cartesian product of A with itself, defined by $(\text{a, b})\simeq(\text{c, d)}$ if ad = bc. Then, the number of ordered pairs of the equivalence class of (3, 2) is:
AnswerThe ordered pairs of the equivalence class of (3, 2) are {(3, 2), (6, 4), (9, 6), (12, 8), (15, 10), (18, 12)}.
We observe that these are 6 pairs.
View full question & answer→MCQ 2151 Mark
The number of bijective functions from set $A$ to itself when $A$ contains $106$ elements is:
- A
$106$
- B
$(106)^2$
- ✓
$106!$
- D
$2^{106}$
AnswerCorrect option: C. $106!$
$106!$
View full question & answer→MCQ 2161 Mark
Let $\times$ be the binary operation on $N$ given by $a \times b = \text{HCF} (a, b)$ where$, a, b \in N.$ Find the value of $22 \times 4.$
View full question & answer→MCQ 2171 Mark
A relation R is defined from {2, 3, 4, 5} to {3, 6, 7, 10} by : xRy ⇔ x is relatively prime to y. Then, domain of R is:
AnswerGiven that relation R is defined from {2, 3, 4, 5} to {3, 6, 7, 10} by : xRy ⇔ x is relatively prime to y. R can be written as,
{(2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3), (5, 6), (5, 7)}
Here we can see that domain means x element which is $2\leq\text{x}\leq5.$
Hence, {2, 3, 4, 5}
View full question & answer→MCQ 2181 Mark
Let A = {1, 2, 3}. Then, the number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is:
AnswerGiven that A = {1, 2, 3}
To find the number of relations containing (1, 2) and (1, 3) then R can be written as {(1, 2), (1, 3), (1, 1), (2, 2), (3, 3), (2, 1), (3, 1)}
Here, we can see that
(3, 1) and (1, 2) ⇒ (3, 2) which is not belongs to R.
The number of relations containing (1, 2) and (1, 3)
Which are reflexive and symmetric but not transitive is 1.
View full question & answer→MCQ 2191 Mark
Let $\text{f}:[2,\infty)\rightarrow\ \text{X}$ be defined by $f(x) = 4x - x^2$. Then, $f$ is invertible if $X =$
- A
$[2,\infty)$
- ✓
$(-\infty,2]$
- C
$(-\infty,4]$
- D
$[4,\infty)$
AnswerCorrect option: B. $(-\infty,2]$
Since $f$ is invertible, range of $f =$ co$-$domain of $f = X$
So, we need to find the range of $f$ to find $X$.
For finding the range, let
$f(x) = y$
$\Rightarrow 4x - x^2 = y$
$\Rightarrow x^2 - 4x = -y$
$\Rightarrow x^2 - 4x + 4 = 4 - y$
$\Rightarrow (x - 2)^2 = 4 - y$
$\Rightarrow \text{x}-2=\pm\sqrt{4-\text{y}}$
$\Rightarrow \text{x}=2\pm\sqrt{4-\text{y}}$
This is defined only when
$4-\text{y}\geq0$
$\Rightarrow \text{y}\leq4$
$X =$ Range of $\text{f}=(-\infty,4]$
View full question & answer→MCQ 2201 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
$-1$
AnswerLet 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 *.
View full question & answer→