MCQ 1511 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. $\frac{4\text{x}+7}{2}$
View full question & answer→MCQ 1521 Mark
Let $f(x)=x^2-x+1, x \geq \frac{1}{2}$, 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. $x = 1$
View full question & answer→MCQ 1531 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 1541 Mark
If a function $f:\ (2, \infty) \rightarrow B$ defined by $f(x)=x^2-4 x+5$ 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: $f(x)=x^2-4 x+5$
Let $f(x)=y$
$\Rightarrow y=x^2-4 x+5$
$\Rightarrow x^2-4 x+(5-y)=0$
$\because$ 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 1551 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:
View full question & answer→MCQ 1561 Mark
If $a ^* b$ denote the bigger among $a$ and $b$ and if $ab = (a ^* b) + 3,$ then $4.7 =$
View full question & answer→MCQ 1571 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 $f$
$\Rightarrow f$ is onto.
So$, f$ is a bijection.
View full question & answer→MCQ 1581 Mark
Let $R =\{(a, a), (b, b), (c, c), (a, b)\}$ be a relation on set $A = \{a, b, c\}.$ Then, $R$ is:
View full question & answer→MCQ 1591 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. $^{10} \text{C} _7\times7!$
View full question & answer→MCQ 1601 Mark
The function $f: R \rightarrow R$ given by $f(x)=x^3-1$ is:
View full question & answer→MCQ 1611 Mark
If $f(x)=\sin ^2 x$ and the composite function $g(f(x))=|\sin x|$, 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 1621 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 1631 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 1641 Mark
Choose the correct answer from the given four options. Let $f: R \rightarrow R$ be defined by $f(x)=3 x^2-5$ and $g: R \rightarrow R$ by $g ( x )=\frac{ x }{ x ^2+1}$. 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, $f(x)=3 x^2-5$ and $\text{g}(\text{x})=\frac{\text{x}}{\text{x}^2+1}$
$g^ \circ f(x)=g(f(x))$
$=g\left(3 x^2-5\right)$
$=\frac{3\text{x}^2-5}{(3\text{x}^2-5)^2+1}$
$=\frac{3\text{x}^2-5}{9\text{x}^4-30\text{x}^2+25+1}$
$=\frac{3\text{x}^2+5}{9\text{x}^4-30\text{x}^2+26}$
View full question & answer→MCQ 1651 Mark
A function $f$ from the set of natural numbers to integers defined by $\text{f(n)}=\begin{cases}\frac{\text{n}-1}{2},&\text{when n is odd}\\-\frac{\text{n}}{2},&\text{when n is even}\end{cases}$
AnswerCorrect option: D. One$-$one and onto both.
Injectivity: 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 $\text{f}=\Big\{....,\ \frac{-3-1}{2},\ \frac{-(-2)}{2},\ \frac{-1-1}{2},\ \frac{0}{2},\ \frac{1-1}{2},\ \frac{-2}{2},\ \frac{3-1}{2},\ ....\Big\}$
$\Rightarrow$ Range of $f = \{....., -2, 1, -1, 0, 0, -1, 1, .....\}$
$\Rightarrow$ Range of $f = \{....., -2, -1, 0, 1, 2, ......\}$
$\Rightarrow Co-$domain of $f =$ Range of $f$
$\Rightarrow f$ is onto.
View full question & answer→MCQ 1661 Mark
Let $M$ be the set of all $2 \times 2$ matrices with entries from the set $R$ of real numbers. Then, the function $f : M\rightarrow R$ defined by $f(A) = |A|$ for every A \in M, is:
AnswerCorrect option: D. Onto but not one$-$one.
$\text{M}=\begin{Bmatrix}\text{A}=\begin{bmatrix}\text{a}&\text{b}\\\text{c}&\text{d}\end{bmatrix}:\text{a, b, c, d}\in\text{R}\end{Bmatrix}$
$f : M \rightarrow R$ 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 1671 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 1681 Mark
Which of the following is true?
- A
$*$ defined by $\text{a}^*\text{b}=\frac{\text{a + b}}2$ is a binary operation on $Z.$
- ✓
$*$ defined by $\text{a}^*\text{b}=\frac{\text{a + b}}2$ is a binary operation on $Q.$
- C
All binary commutative operations are associative.
- D
Subtraction is a binary operation on $N.$
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 1691 Mark
Let $\times$ be a binary operation on set of integers defined by $a \times b = a + b - 3,$ then find the value of $3 \times 4.$
View full question & answer→MCQ 1701 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 $(1, 1), (2, 2), (3, 3) \in R.$
Relation $R$ is symmetric since $(1, 2), (2, 1) \in R$ and $(1, 3), (3, 1) \in R.$
But relation $R$ is not transitive as $(3, 1), (1, 2) \in R$ but $(3, 2) \in R.$
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 1711 Mark
Let $R$ be a relation on the set $N$ of natural numbers denoted by $nRm \Rightarrow n$ 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 1721 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:
AnswerCorrect option: D. One$-$one and onto.
Given 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
$\frac{\text{n}-1}{2}=\frac{\text{m}-1}{2}$
$\Rightarrow n = m$
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 1731 Mark
The binary operation $*$ defined on $N$ by $a^ * b = a + b + ab$ for all $a, b \in N$ is:
- A
- B
- ✓
Commutative and associative both.
- D
AnswerCorrect option: C. Commutative and associative both.
$a^ * b = a + b + ab$
$b^ * a = b + a + ba$
$\Rightarrow a^ * b = b^ * a$
So $*$ is commutative.
Now,
$(a^ * b)^ * c$
$= (a + b + ab)^ * c$
$= a + b + ab + c + ca + cb + abc$
$a^ * (b^ * c)$
$= a^ * (b + c + bc)$
$= a + b + c + bc + ab + ac + abc$
$\Rightarrow (a^ * b)^ * c = a^ * (b^ * c)$
So $*$ is associative.
View full question & answer→MCQ 1741 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 1751 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 1761 Mark
Let $A=\{x \in R: x \leq 1\}$ 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 $f^{-1}(x)=y \ldots \ldots(1)$
$\Rightarrow f(y) = x$ and $\text{y}\leq1$
$\Rightarrow y(2 - y) = x$
$\Rightarrow 2 y-y^2=x$
$\Rightarrow y^2-2 y+x=0$
$\Rightarrow y^2-2 y=-x$
$\Rightarrow y^2-2 y+1=1-x$
$\Rightarrow (\text{y}-1)^2=\sqrt{1-\text{x}}$
$\Rightarrow \text{y}-1=\pm\sqrt{1-\text{x}}$
$\Rightarrow \text{y}=1\pm\sqrt{1-\text{x}}$
$\Rightarrow \text{y}=1-\sqrt{1-\text{x}}$ $(\because\ \text{y}\leq1)$
View full question & answer→MCQ 1771 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.
View full question & answer→MCQ 1781 Mark
If $R$ is a relation on the set $A = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ given by $xRy \Rightarrow y = 3x,$ then $R =$
- A
$\{(3, 1), (6, 2), (8, 2), (9, 3)\}$
- B
$\{(3, 1), (6, 2), (9, 3)\}$
- C
$\{(3, 1), (2, 6), (3, 9)\}$
- ✓
AnswerThe relation $R$ is defined as,
$\text{R}=\{(\text{x, y}):\ \text{x, y}\in\text{A}:\text{y}=3\text{x}\}$
$\Rightarrow R = \{(1, 3), (2, 6), (3, 9)\}$
View full question & answer→MCQ 1791 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 1801 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 1811 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
$\text{x}+\sqrt{3}$
- ✓
AnswerGiven 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 1821 Mark
If $a ^* b=a^2+b^2$, then the value of $(4 ^* 5) ^* 3$ is:
AnswerCorrect option: C. $41^2+3^2$
Given $a^* b=a^2+b^2$
So, $4 ^* 5=4^2+5^2$
Now,
$(4 ^* 5) ^* 3=(4 ^* 5)^2+3^2$
$=\left(4^2+5^2\right)^2+3^2$
$=41^2+3^2$
View full question & answer→MCQ 1831 Mark
Let the function $f :\ R - \{-b\} \rightarrow R - \{1\}$ 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
$f(x) = f(y)$
$ \Rightarrow x + ax + b = y + ay + b$
$\Rightarrow x + ay + b = x + by + a$
$\Rightarrow xy + bx + ay + ab = xy + ax + by + ab$
$\Rightarrow bx + ay = ax + by$
$\Rightarrow a - bx = a - by$
$\Rightarrow x = y$
So$, f$ is one$-$one.
Surjectivity: Let $y$ be an element in the $co-$domain of $f,$
i.e., $R - \{1\},$ such that $f(x) = y$
$\Rightarrow x + ax + b = y$
$\Rightarrow x + a $
$\Rightarrow x = -a$
So$, f$ is onto.
View full question & answer→MCQ 1841 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$, (gof) \Big(\frac{3}{2}\Big)$ is:
View full question & answer→MCQ 1851 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 1861 Mark
Let $A = {1, 2, 3}.$ Then number of equivalence relations containing $(1, 2)$ is:
AnswerSOLution The 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.
b. 2 .
View full question & answer→MCQ 1871 Mark
Let $f: R \rightarrow R$ be a function defined by $f(x)=x^3+4$, then $f$ is:
View full question & answer→MCQ 1881 Mark
The relation $R$ defined on the set $A=\{1,2,3,4,5\}$ by $R=\left\{(a, b):\left|a^2-b^2\right|<16\right\}$ 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 1891 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. $\frac{\text{x}+\sqrt{\text{x}^2-4}}{2}$
View full question & answer→MCQ 1901 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
$\left(a^* b\right)^* c$
$=\left(a^2+b^2\right)^* c$
$=\left(a^2+b^2\right)^2+c^2$
$a^*\left(b^* c\right)$
$=a^*\left(b^2+c^2\right)$
$=a^2+\left(b^2+c^2\right)^2$
$\Rightarrow\left(a^* b\right)^* c \neq a^*\left(b^* c\right)$
So $*$ is not associative.
View full question & answer→MCQ 1911 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. $f(x) = x + 2$
View full question & answer→MCQ 1921 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 1931 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 1941 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 $= 2(2 \times$ Number of elements in the set$)$
$=2^{(2 \times 2)}$
$=2^4$
$= 16$
Therefore, the number of binary operations on the set $\{a, b\}$ are $16.$
View full question & answer→MCQ 1951 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 1961 Mark
Let $X=\{-1,0,1\}, Y=\{0,2\}$ and a function $f: X \rightarrow Y$ defiend by $y=2 x^4$, is:
- A
One$-$one onto.
- B
One$-$one into.
- ✓
Many$-$one onto.
- D
Many$-$one into.
AnswerCorrect option: C. Many$-$one onto.
View full question & answer→MCQ 1971 Mark
If $f: R \rightarrow R$ is given by $f(x)=3 x-5$, 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 $f(x)=3 x-5$,
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 1981 Mark
If the function $f : R \rightarrow A$ given by $\text{f(x)}=\frac{\text{x}^2}{\text{x}^2+1}$ is a surjection, then $ A =$
- A
$R$
- B
$[0, 1]$
- C
$(0, 1)$
- ✓
$(0, 1)$
AnswerCorrect option: D. $(0, 1)$
As $f$ is surjective, range of $f = co-$domain of $f$
$\Rightarrow A =$ range of $f$
$=\frac{\text{x}^2}{\text{x}^2+1},$
$\text{y}=\frac{\text{x}^2}{\text{x}^2+1}$
$\Rightarrow\ \text{y}(\text{x}^2+1)$
$\Rightarrow\ \text{x}^2=\frac{-\text{y}}{(\text{y}-1)}$
$\Rightarrow\ \text{x}=\sqrt{\frac{\text{y}}{(1-\text{y})}}$
$\Rightarrow\ \frac{\text{y}}{(1-\text{y})}\geq0$
$\Rightarrow\ \text{y}\in[0,1)$
$\Rightarrow $ Range of $f = (0, 1)$
$\Rightarrow A = (0, 1)$
View full question & answer→MCQ 1991 Mark
Let $R$ be a relation on the set $N$ of natural numbers defined by $\text{nRm}$ 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 $m; m, n \in N\}$
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 $n$
$\Rightarrow p$ divides $m$
$\Rightarrow\ (\text{m, p})\in\text{R}$
So$, R$ is transitive.
View full question & answer→MCQ 2001 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$: a ^* b = 3a - b$
$2 ^* 3 = 3 (2) - 3$
$= 6 - 3$
$= 3$
$(2 ^* 3) ^* 4 = 3 ^* 4$
$= 3(3) - 4$
$= 9 - 4$
$= 5$
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