MCQ 11 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, $ {f}({x})= {x}^2-4{x}+5,$
Let $ {y}={x}^2-4 {x}+5$
$\Rightarrow\ {y}={x}^2-4 {x}+4+1$
$=({x}-2)^2+1$
$\Rightarrow\ ({x}-2)^2={y}-1$
$\Rightarrow\ {x}-2=\sqrt{{y}-1}$
$\Rightarrow {x}=2+\sqrt {y}-{1}$
$\therefore {y}-1\geq0,\ {y}\geq1$
Range $=[1,\infty)$
View full question & answer→Question 21 Mark
Let R be the relation “is congruent to” on the set of all triangles in a plane is:
- Reflexive.
- Symmetric.
- Symmetric and reflexive.
- Equivalence.
Question 31 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:
- Bijection.
- Injection but not a surjection.
- Surjection but not an injection.
- Neither an injection nor a surjection.
Answer
- Bijection.
Solution:
$\text{f(x)}=\sin^{-1}(3\text{x}-4\text{x}^3)$
$\Rightarrow\ \text{f(x)}=3\sin^{-1}\text{x}$
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
$\Rightarrow\ 3\sin^{-1}\text{x}=\text{y}$
$\Rightarrow\ \sin^{-1}\text{x}=\frac{\text{y}}{3}$
$\Rightarrow\ \text{x}=\sin\frac{\text{y}}{3}\in\Big[\frac{-1}{2},\frac{1}{2}\Big]$
⇒ f is onto.
⇒ f is a bijection. View full question & answer→MCQ 41 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→Question 51 Mark
If N be the set of all-natural numbers, consider f : N → N such that f(x) = 2x, ∀ x ∈ N, then f is:
- One-one onto.
- One-one into.
- Many-one onto.
- None of these.
MCQ 61 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 71 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→Question 81 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:
- {1, 4, 6, 9}
- {4, 6, 9}
- {1}
- None of these.
Answer
- {1}
Solution:
Here, $\text{R}=\text{x, y}:\text{x}\in\text{A}$ and $\text{y}\in\text{B}:\text{x}>\text{y}$ ⇒ R = 2, 1, 3, 1
Thus, Range of R = {1} View full question & answer→Question 91 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:
- Onto but not one-one.
- One-one but not onto.
- One-one and onto.
- Neither one-one nor onto.
Answer
- Onto but not one-one.
Solution:
Given 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 101 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 111 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→Question 121 Mark
A binary operation * on Z defined by a * b = 3a + b for all a, b ∈ Z, is:
- Commutative.
- Associative.
- Not commutative.
- Commutative and associative.
Answer
- Not commutative.
Solution:
Let $\text{a, b}\in\text{Z}$
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 131 Mark
Which of the following functions from $\text{A}=\{\text{x}:-1\leq\text{x}\leq1\}$ to itself are bijections?
Answer$\text{g(x)}=\sin\big(\frac{\pi\text{x}}{2}\big)$
- 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→Question 141 Mark
Let A = {1, 2, 3} and B = {(1, 2), (2, 3), (1, 3)} be a relation on A. Then, R is:
- Neither reflexive nor transitive.
- Neither symmetric nor transitive.
- Transitive.
- None of these.
Answer
- Transitive.
Solution:
Reflexivity: 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→Question 151 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 $\triangle$
- Commutative and associative without an identity.
- Commutative but not associative with an identity.
- Associative but not commutative without an identity.
- Associative and commutative with an identity.
Answer
- Associative and commutative with an identity.
Solution:
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→Question 161 Mark
The relation S defined on the set R of all real number by the rule aSb iff a ≥ b is:
- An equivalence relation.
- Reflexive, transitive but not symmetric.
- Symmetric, transitive but not reflexive.
- Neither transitive nor reflexive but symmetric.
Answer
- 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 171 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 $\text{A}=\{{x}:-1\leq{x}\leq1\}$ 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 $, x < 0$
$f(x) = x|x| < 0$
$|x| = -x$
$y = -x^2$
${x}=-\sqrt{-{y}}$ which is not possible for $x > 0$
Hence $,f$ is onto.
$\Rightarrow f$ is bijection.
View full question & answer→Question 181 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:
- Reflexive.
- Symmetric.
- Transitive.
- None of these.
Answer
- Symmetric.
Solution:
Given 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→Question 191 Mark
In the set Z of all integers, which of the following relation R is not an equivalence relation?
- xRy : if $\text{x}\leq\text{y}$
- xRy : if x = y
- xRy : if x - y is an even integer
- xRy : if $\text{x}\equiv\text{y}\ (\text{mod 3})$
Answer
- xRy : if $\text{x}\leq\text{y}$
Solution:
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 201 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 $f^{-1}(x) = y$
$f(y) = x$
$\Rightarrow y^3 + 3 = x$
$\Rightarrow y^3 = x - 3$
$\Rightarrow y = (x - 3)^3$
$\Rightarrow\ \text{y}=(\text{x}-3)^\frac{1}{3}$
View full question & answer→Question 211 Mark
Let $\text{f(x)}=\frac{1}{1-\text{x}}.$ Then, {fo(fof)}(x):
- x for all $\text{x}\in\text{R}$
- x for all $\text{x}\in\text{R}-\{1\}$
- x for all $\text{x}\in\text{R}-\{0,1\}$
- None of these.
Answer
- x for all $\text{x}\in\text{R}-\{0,1\}$
Solution:
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→Question 221 Mark
Let A = N × N and × be the binary operation on A defined by (a, b) × (c, d) = (a + c, b + d). Then × is:
- Commutative.
- Associative.
- Both (a) and (b).
- None of these.l
MCQ 231 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 241 Mark
The distinct linear functions that map $[-1, 1]$ onto $[0, 2]$ are:
- ✓
$f(x) = x + 1, g(x) = -x + 1$
- B
$f(x) = x - 1, g(x) = x + 1$
- C
$f(x) = -x - 1, g(x) = x - 1$
- D
AnswerCorrect option: A. $f(x) = x + 1, g(x) = -x + 1$
$f(x) = -x - 1, g(x) = x - 1$
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 $f(x) = y$
$\Rightarrow 4x - x^2 = y$
$\Rightarrow x^2 - 4x = -y$
$\Rightarrow x^2 - 4x + 4 = 4 - y$
$\Rightarrow (x - 2)^2 = 4 - y$
$\Rightarrow {x}-2=\pm4-{y}$
$\Rightarrow {x}=2\pm4-{y}$
This is defined only when $4-{y}\geq0$
$\Rightarrow {y}\leq4,$
$X =$ Range of $f =(-\infty,4]$
View full question & answer→MCQ 251 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→Question 261 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:
- $\begin{bmatrix}-2&3\\-3&-2\end{bmatrix}$
- $\begin{bmatrix}2&3\\-3&2\end{bmatrix}$
- $\begin{bmatrix}\frac{2}{13}&\frac{-3}{13}\\\frac{3}{13}&\frac{2}{13}\end{bmatrix}$
- $\begin{bmatrix}1&0\\0&1\end{bmatrix}$
Answer
- $\begin{bmatrix}\frac{2}{13}&\frac{-3}{13}\\\frac{3}{13}&\frac{2}{13}\end{bmatrix}$
Solution:
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→Question 271 Mark
The function f : R → R defined by f(x) = (x - 1)(x - 2)(x - 3) is:
- One-one but not onto.
- Onto but not one-one.
- Both one and onto.
- Neither one-one nor onto.
Answer
- Onto but not one-one.
Solution:
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)
⇒ 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→Question 281 Mark
Which one of the following function is not invertible?
- $\text{f} : \text{R} \rightarrow \text{R}, \text{f(x)} = 3\text{x} + 1$
- $\text{f} : \text{R} \rightarrow [0,\infty), \text{f(x)} = \text{x}^2$
- $\text{f} : \text{R}^+\rightarrow\text{R}^+, \text{f(x)} =\frac{1}{\text{x}^3}$
- None of these.
View full question & answer→Question 291 Mark
Let us define a relation R in R as aRb if a ≥ b. Then R is:
- An equivalence relation.
- Reflexive, transitive but not symmetric.
- Symmetric, transitive but not reflexive.
- Neither transitive nor reflexive but symmetric.
Answer
- Reflexive, transitive but not symmetric.
View full question & answer→MCQ 301 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:
AnswerCorrect option: B. $[x] - x$
Given 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→Question 311 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:
- 720
- 120
- 0
- None of these.
Answer
- 0
Solution:
As, 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→Question 321 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:
- x
- 1
- f(x)
- g(x)
Answer
- 1
Solution:
When, -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→Question 331 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:
- $\begin{bmatrix}1&1\\1&1\end{bmatrix}$
- $\begin{bmatrix}-\frac{1}2&-\frac{1}2\\-\frac{1}2&-\frac{1}2\end{bmatrix}$
- $\begin{bmatrix}\frac{1}2&\frac{1}2\\\frac{1}2&\frac{1}2\end{bmatrix}$
- $\begin{bmatrix}-1&-1\\-1&-1\end{bmatrix}$
Answer
- $\begin{bmatrix}\frac{1}2&\frac{1}2\\\frac{1}2&\frac{1}2\end{bmatrix}$
Solution:
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→Question 341 Mark
If f(x) $=1-\frac{1}{\text{x}},$ then $\text{f}(\text{f}(\frac{1}{\text{x}}))$
- $\frac{1}{\text{x}}$
- $\frac{1}{1+\text{x}}$
- $\frac{\text{x}}{\text{x}-1}$
- $\frac{1}{\text{x}-1}$
Answer
- $\frac{\text{x}}{\text{x}-1}$
View full question & answer→Question 351 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:
- $-\text{a}$
- $-\frac{\text{a}}{\text{a}-1}$
- $\frac{1}{\text{a}}$
- $\text{a}^2$
Answer
- $-\frac{\text{a}}{\text{a}-1}$
Solution:
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 361 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 371 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→Question 381 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:
- An equivalence relation.
- Reflexive, transitive but not symmetric.
- Symmetric, transitive but not reflexive.
- Neither transitive nor reflexive but symmetric.
Answer
- Reflexive, transitive but not symmetric.
Solution:
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→Question 391 Mark
Subtraction of integers is:
- Commutative but no associative.
- Commutative and associative.
- Associative but not commutative.
- Neither commutative nor associative.
Answer
- Neither commutative nor associative.
Solution:
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→Question 401 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,
- f is one-one but not onto.
- f is onto but not one-one.
- f is both one-one and onto.
- None of these.
Answer
- f is both one-one and onto.
Solution:
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→Question 411 Mark
Let f : N → R : $\text{f}(\text{x})=\frac{(2\text{x}-1)}{2}$ and g : Q → R : g(x) = x + 2 be two functions. Then, (gof) $(\frac{3}{2})$ is:
- 3
- 1
- $\frac{7}{2}$
- None of these.
View full question & answer→MCQ 421 Mark
Let $f : R \rightarrow R$ be a function defined by $f(x) = x^3 + 4,$ then $f$ is$:$
View full question & answer→MCQ 431 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 441 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 451 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 461 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→Question 471 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 ∀ a, b ∈ T. Then R is:
- Reflexive but not transitive.
- Transitive but not symmetric.
- Equivalence.
- None of these.
Question 481 Mark
Number of binary operations on the set {a, b} are:
- 8
- 20
- 10
- 16
Answer
- 16
Solution:
Let 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→Question 491 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:
- 0
- 120
- 720
- None of these.
Answer
- 0
Solution:
Given,
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 501 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→