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→