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
$\{1,2,3,4,3,6\}$
- 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 $\text{f}=\Big\{^{(7-3)}\text{P}_{(3-3)},\ ^{(7-4)}\text{P}_{(4-3)},\ ^{(5-3)}\text{P}_{(7-5)}\Big\}$
$=\left\{4 P_0, 3 P_1, 2 P_2\right\}$
$=\{1,3,2\}$
$=\{1,2,3\}$
View full question & answer→MCQ 21 Mark
If $f: R \rightarrow R, g: R \rightarrow R$ and $h: R \rightarrow R$ are such that $f(x)=x^2, g(x)=\tan x$ and $h(x)=\log x$, 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.
$\text{X}\triangle\text{Y}=(\overline{\text{X}}\cap\text{Y})\cup(\text{X}\cap\overline{\text{Y}})$
$=(\overline{\text{Y}}\cap\text{X})\cup(\text{Y}\cap\overline{\text{X}})$
$=\text{Y}\triangle\text{X}$
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.
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. $\text{f}^{-1}(\text{x})=\text{f}(\text{x})$
We have, $\text{f}(\text{x})=\frac{3\text{x}+2}{5\text{x}-3}=\text{y}\ ($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:
Answer$R = \{(a, b): a = b$ 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 $f : R \rightarrow R$ be defined by $\text{f}(\text{x})=\frac{1}{\text{x}}\ \forall\ \text{x}\in\text{R}.$ Then $f$ is:
- A
one$-$one.
- B
- C
- ✓
$f$ is not defined.
AnswerCorrect option: D. $f$ is not defined.
View full question & answer→MCQ 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 $A=\{x:-1 \leq x \leq 1\}$ 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, $x < 0$
$f(x)=x|x| < 0$
$|x|=-x$
$y=-x^2$
$x=-\sqrt{-y}$ 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 $(3,2)\in\text{R}\ \text{then}\ (1,2)\in\text{R}$ |
$\therefore$ |
$R$ 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 $\text{lRm}$ if l is perpendicular to $m$ for all $l, m \in L.$ Then$, R$ is:
AnswerGiven that $L$ denote the set of all straight lines in a plane.
A relation $R$ be defined by $\text{lRm}$ if is perpendicular to $m$ for all $l, m \in L.$
$R$ 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?
- ✓
$\text{xRy} :\ $ if $\text{x}\leq\text{y}$
- B
$\text{xRy} :\ $ if $x = y$
- C
$\text{xRy} :\ $ if $x - y$ is an even integer
- D
$\text{xRy} :\ $ if $\text{x}\equiv\text{y}\ ($mod $3)$
AnswerCorrect option: A. $\text{xRy} :\ $ if $\text{x}\leq\text{y}$
In the set of $Z$ of all integers $\text{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. $(\text{x}-3)^\frac{1}{3}$
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 $a^ * b=\frac{ ab }{3} \forall a , b \in Q ^{+}$. 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,
$\frac{\text{ae}}{3}=\text{a}$ and $\frac{\text{ea}}{3}=\text{a},\forall\text{ a}\in\text{Q}^+$
$e = 3, \forall\text{ a}\in\text{Q}^+$
Thus$, 3$ is the identity element in $Q^{+}$ with respect to $*.$
Let $\text{a}\in\text{Q}^+$ and $\text{b}\in\text{Q}^+$ be the inverse of $a.$ Then,
$a^ * b = e = b^ * a$
$a^ * b = e$ and $b^ * a = e$
$\therefore\ \frac{\text{ab}}3=3$ and $\frac{\text{ba}}3=3$
$\text{b}=\frac{9}{\text{a}}\in\text{Q}^+$
Thus, $\frac{9}{\text{a}}$ is the inverse of $\text{a}\in\text{Q}^+$.
Given: $\text{a}*\text{b}=\frac{\text{ab}}3$
$4^*6=\frac{4\times6}3=8$
Now,
$\text{a}^{-1}=\frac{9}{\text{a}}$
$(4^*6)^{-1}=8^{-1}$
$=\frac{9}8$
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$, \{fo(fof)\}(x):$
- ✓
$x$ for all $\text{x}\in\text{R}$
- B
$x$ for all $\text{x}\in\text{R}-\{1\}$
- C
$x$ for all $\text{x}\in\text{R}-\{0,1\}$
- D
AnswerCorrect option: A. $x$ for all $\text{x}\in\text{R}$
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$, f : R - \{1\} \rightarrow R - \{0\}$ and $f : R - \{1\} \rightarrow R - \{0\}$
Range of $f$ is not a subset of the domain of $f.$
Domain $\text{(fof)} = \{x : \text{x}\in$ domain of $f$ and $\text{f(x)}\in$ domain of $f\}$
Domain $\text{(fof)} =\big\{\text{x}:\text{x}\in\text{R}-\{1\}$ and $\frac{1}{1-\text{x}}\in\text{R}-\{1\}\big\}$
Domain $\text{(fof)} =\big\{\text{x}:\text{x}\neq1$ and $\frac{1}{1-\text{x}}\neq1\big\}$
Domain $\text{(fof)} =\{\text{x}:\text{x}\neq1$ and $1-\text{x}\neq1\}$
Domain $\text{(fof)} =\{\text{x}:\text{x}\neq1$ and $\text{x}\neq0\}$
Domain $\text{(fof)} = R - \{0, 1\}$
$\text{(fof)(x) = f(f(x))}$
$=\text{f}\Big(\frac{1}{1-\text{x}}\Big)=\frac{1}{1-\frac{1}{1-\text{x}}}=\frac{1-\text{x}}{1-\text{x}-1}$
$=\frac{1-\text{x}}{-\text{x}}=\frac{\text{x}-1}{\text{x}}$
For range of fof, $\text{x}\neq0$
Now$, \text{fof} : R \rightarrow \{0, 1\} \rightarrow R - \{0\}$ and $f : R - {1} \rightarrow R - \{0\}$
Range of fof is not a subset of domain of $f.$
Domain $\text{(fo(fof))} =\{\text{x}:\text{x}\in$ domain of $\text{fof}$ and $(fof)(x) \in$ domain of f$\}$
Domain $\text{(fo(fof))} =\Big\{\text{x}:\text{x}\in\text{R}-\{0,1\}$ and $\frac{\text{x}-1}{\text{x}}\in\text{R}-\{1\}\Big\}$
Domain $\text{(fo(fof))} =\Big\{\text{x}:\text{x}\neq0,1$ and $\frac{\text{x}-1}{\text{x}}\neq1\Big\}$
Domain $\text{(fo(fof))} =\{\text{x}:\text{x}\neq0,1$ and $\text{x}-1\neq\text{x}\}$
Domain $\text{(fo(fof))} =\{\text{x}:\text{x}\neq0,1$ and $\text{x}\in\text{R}\}$
Domain $\text{(fo(fof))} = R - \{0, 1\}$
Domain $\text{(fo(fof)) = f((fof)(x))}$
$=\text{f}\Big(\frac{\text{x}-1}{\text{x}}\Big)$
$=\frac{1}{1-\frac{\text{x}-1}{\text{x}}}$
$=\frac{\text{x}}{\text{x}-\text{x}+1}$
$=\text{x}$
So$, \text{(fo(fof))(x) = x},$ 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 $a * b = 1 + ab \forall a, b \in R.$ 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, $a^ * b = 1 + ab \forall a, b \in R$
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 $A = N \times N$ and $\times $ be the binary operation on $A$ defined by $(a, b) \times (c, d) = (a + c, b + d).$ Then $\times $ is:
- A
- B
- ✓
Both $(a)$ and $(b).$
- D
AnswerCorrect option: C. Both $(a)$ and $(b).$
View full question & answer→MCQ 211 Mark
The relation $R$ is defined on the set of natural numbers as $\{(a, b): a=2 b\}$. Then, $R^{-1}$ is given by:
- A
$\{(2,1),(4,2),(6,3), \ldots\}$
- ✓
$\{(1,2),(2,4),(3,6), \ldots \ldots .$.
- C
$R ^{-1}$ is not defiend.
- D
AnswerCorrect option: B. $\{(1,2),(2,4),(3,6), \ldots \ldots .$.
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
$R_1=\{(1,1),(2,2),(3,3)\}$
$R_2=\{(1,1),(2,2),(3,3),(1,2),(2,1)\}$
$R_3=\{(1,1),(2,2),(3,3),(1,3),(3,1)\}$
$R_4=\{(1,1),(2,2),(3,3),(2,3),(3,2)\}$
$R_5=\left\{(1,2,3) \Leftrightarrow A \times A=A^2\right\}$
$\therefore$ 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 $
$\Rightarrow f ^{-1}(1)=\tan ^{-1} y$
$\Rightarrow \tan ^{-1} \tan \frac{\pi}{4}=\frac{\pi}{4}\left[\because \tan \frac{\pi}{4}=1\right]$
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 $f(x)=x^2-4 x+ 5$, 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 $\frac{1}{2} \operatorname{gof}(x)=2 x^2-5 x+2$, then $f(x)$ is equal to:
- ✓
$2 x-3$
- B
$2x + 3$
- C
$2 x^2+3 x+1$
- D
$2 x^2-3 x-1$
AnswerCorrect option: A. $2 x-3$
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 $f(x) = y$
$\Rightarrow 4 x - x ^2= y$
$\Rightarrow x ^2-4 x =- y$
$\Rightarrow x ^2-4 x +4=4- y$
$\Rightarrow( x -2)^2=4- y$
$\Rightarrow\ \text{x}-2=\pm4-\text{y}$
$\Rightarrow\ \text{x}=2\pm4-\text{y}$
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
${ }^n P_2$
- ✓
$2^n-2$
- 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:
- A
One$-$one and onto.
- ✓
One$-$one but not onto.
- C
Onto but not one$-$one.
- D
Onto but not one$-$one.
AnswerCorrect option: B. One$-$one but not onto.
Given function is $\text{f(x)}=\frac{\text{x}}{\text{x}+1}$ on $\text{f}:[0,\infty)\rightarrow\ \text{R}$
If $f(x) = f(y)$
$\Rightarrow\ \frac{\text{x}}{\text{x}+1}=\frac{\text{y}}{\text{y}+1}$
$\Rightarrow xy + x = xy + y$
$\Rightarrow x = y$
Hence$, f$ is one$-$one.
If $y = f(x)$
$\text{y}=\frac{\text{x}}{\text{x}+1}$
$\Rightarrow xy + y = x$
$\Rightarrow xy - x = -y$
$x(y - 1) = -y$
$\text{x}=\frac{-\text{y}}{\text{y}-1}\neq\text{f(x)}$
It is not onto.
View full question & answer→MCQ 351 Mark
If $f: R \rightarrow R$ defined by $f(x)=\frac{3 x+5}{2}$ 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. $\frac{2\text{x}-5}{3}$
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,
$\text{f(x)}=\begin{cases}2\text{x}, \text{if x}>3\text{x}^2, \text{if }1<\text{x}\leq33\text{x}, \text{if x}\leq1\end{cases}\}$
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
$\begin{bmatrix}2&3\\-3&2\end{bmatrix}*\begin{bmatrix}\text{e}&\text{f}\\-\text{f}&\text{e}\end{bmatrix}=\begin{bmatrix}2&3\\-3&3\end{bmatrix}$
$\Rightarrow 2e - 3f = 2 \rightarrow (1)$
$2f + 3e = 3 \rightarrow (2)$
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
$\begin{bmatrix}2&3\\-3&2\end{bmatrix}*\begin{bmatrix}\text{a}&\text{b}\\-\text{b}&\text{a}\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow 2a - 3b = 1 \rightarrow (1)$
$2b + 3a = 0 \rightarrow (2)$
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 $h(x)=\log x$, then the value of $\text{[ho(gof)] (x)},$ if $x =\frac{\sqrt{\pi}}{2}$ 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:
Answer$R$ 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,
$y + 3 = x$
$\Rightarrow y = x + 3$
View full question & answer→MCQ 451 Mark
Let $f: R \rightarrow R$ be given by $f(x)=\left[x^2\right]+[x+1]-3$ 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.
$f: R \rightarrow R$
$=\left[x^2\right]+[x+1]-3$
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 $\text{aRb}$ 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 $\text{aRb} \Rightarrow a$ is brother of $b.$
$aRa \Rightarrow a$ is brother of $a,$ which is not true.
Hence$, R$ is not reflexive.
$aRb \Rightarrow a$ 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.
$aRb \Rightarrow a$ is brother of $b$
and $bRc \Rightarrow b$ is a brother of $c.$
So$, a$ is brother of $c.$
Hence$, R$ is transitive.
View full question & answer→MCQ 471 Mark
The function $f : R \rightarrow R$ defined by $f(x) = (x - 1)(x - 2)(x - 3)$ is:
AnswerCorrect option: B. Onto but not one$-$one.
Given function is $f(x) = (x - 1)(x - 2)(x - 3)$
If $f(x) = f(y)$ then
$(x - 1)(x - 2)(x - 3) = (y - 1)(y - 2)(y - 3)$
$\Rightarrow f(1) = f(2) = f(3) = 0$
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→