MCQ 1011 Mark
The function $f : R \rightarrow R$ defined by $f(x) = 2^x + 2^{|x|}$ is:
AnswerThe function $f : R \rightarrow R$ defined by $f(x) = 2^x + 2^{|x|}$
Here, for each value of x we will get different values of $f(x).$
Hence, it is one-one function.
Also, each element of codomain is mapped to at most one element of the domain.
Function is one-one and into.
View full question & answer→MCQ 1021 Mark
Let $f : R \rightarrow R$ be a function defined by $\text{f(x)}=\frac{\text{x}^2-8}{\text{x}^2+2}.$ Then, $f$ is:
- ✓
One $-$ one but not onto.
- B
One $-$ one and onto.
- C
Onto but not one $-$ one.
- D
Neither one $-$ one nor onto.
AnswerCorrect option: A. One $-$ one but not onto.
Injectivity: Let $x$ and $y$ be two elements in the domain $(R),$ such that
$f(x) = f(y)$
$\frac{\text{x}^2-8}{\text{x}^2+2}=\frac{\text{y}^2-8}{\text{y}^2+2}$
$\Rightarrow (x^2 - 8)(y^2 + 2) = (y^2 - 8)(x^2 + 2)$
$\Rightarrow x^2y^2 + 2x^2 - 8y^2 - 16 = x^2y^2 + 2y^2 - 8x^2 - 16$
$\Rightarrow 10x^2 = 10y^2$
$\Rightarrow x^2 = y^2$
$\Rightarrow\ \text{x}=\pm\text{y}$
So $,f$ is not one $-$ one.
Surjectivity: $\text{f}(-1)=\frac{(-1)^2-8}{(-1)^2+2}=\frac{1-8}{1+2}=\frac{-7}{3}$
and $\text{f(1)}=\frac{(1)^2-8}{(1)^2+2}=\frac{1-8}{1+2}=\frac{-7}{3}$
$\Rightarrow\ \text{f}(-1)=\text{f}(1)=\frac{-7}{3}$
$\Rightarrow f$ is not onto.
View full question & answer→Question 1031 Mark
If A = {a, b, c}, then the relation R = {(b, c)} on A is:
- Reflexive only.
- Symmetric only.
- Transitive only.
- Reflexive and transitive only.
Answer
- Transitive only.
Solution:
The relation R = {(b, c)} is neither reflexive nor symmetric because every element of A is not related to itself. Also, the ordered pair of R obtained by interchanging its elements is not contained in R.
We observe that R is transitive on A because there is only one pair.
View full question & answer→Question 1041 Mark
The binary operation × defind on set R, given by $\text{a}\times\text{b}=\frac{\text{a}+\text{b}}{2}$ for all a, b $\epsilon$ R is:
- Commutative.
- Associative.
- Both (a) and (b).
- None of these.
View full question & answer→Question 1051 Mark
The maximum number of equivalence relations on the set A = {1, 2, 3} are:
- 1
- 2
- 3
- 5
Question 1061 Mark
Let * be a binary operation on N defined by a * b = a + b + 10 for all a, b ∈ N. The identity element for * in N is:
- −10
- 0
- 10
- Non-existent.
Answer
- Non-existent.
Solution:
Given a * b = a + b + 10
Let the identity element be e, then
a * e = a
⇒ a + e + 10 = a
⇒ e = -10
But the operation is defined on the set of natural numbers.
So, the identity element doesn't exist.
View full question & answer→Question 1071 Mark
For real numbers x and y, define xRy if $\text{x}-\text{y}+\sqrt{2}$ is an irrational number. Then the relation R is:
- Reflexive.
- Symmetric.
- Transitive.
- None of these.
Answer
- Reflexive.
Solution:
We have,
$\text{R} = \big\{(\text{x, y}):\text{x}-\text{y}+\sqrt{2}$ $$ is an irrational number, $\text{x, y}\in\text{R}\big\}$
As, $\text{x}-\text{x}+\sqrt{2}=\sqrt{2},$ which is an irrational number
$\Rightarrow\ (\text{x, x})\in\text{R}$
So, R is reflexive relation.
Since, $\Big(\sqrt{2},2\Big)\in\text{R}$
i.e. $\sqrt{2}-2+\sqrt{2}=2\sqrt{2}-2,$ which is an irrational number
but $2-\sqrt{2}+\sqrt{2}=2,$ which is a rational number
$\Rightarrow\ \Big(2,\sqrt{2}\Big)\notin\text{R}$
So, R is not symmetric relation.
Also, $\Big(\sqrt{2},2\Big)\in\text{R}$ and $\Big(2,2\sqrt{2}\Big)\in\text{R}$
$\Rightarrow\ \Big(\sqrt{2},2\sqrt{2}\Big)\notin\text{R}$
So, R is not transitive relation. View full question & answer→MCQ 1081 Mark
The range of the function ${f(x)}=^{7-{x}}\text{P}_{{x}-3}$ is:
- A
$\{1, 2, 3, 4, 5\}$
- B
$\{1, 2, 3, 4, 5, 6\}$
- C
$\{1, 2, 3, 4\}$
- ✓
$\{1, 2, 3\}$
AnswerCorrect option: D. $\{1, 2, 3\}$
We know that
$7-{x}>0;\ {x}-3\geq0$ and $7-{x}\geq {x}-3$
$\Rightarrow\ {x}<7;\ {x}\geq3$ and $2 {x}\leq10$
$\Rightarrow\ {x}<7;\ {x}\geq3$ and $ {x}\leq5$
Therefore $, x = 3, 4, 5$
Range of $\text{f}=\Big\{^{(7-3)}\text{P}_{(3-3)},\ ^{(7-4)}\text{P}_{(4-3)},\ ^{(5-3)}\text{P}_{(7-5)}\Big\}$
$= \{4P_0, 3P_1, 2P_2\}$
$= \{1, 3, 2\}$
$= \{1, 2, 3\}$
View full question & answer→MCQ 1091 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 $\text{a, b, c}\in$ set ${R}$
Hence $,S$ is an equivalence relation on $R$.
View full question & answer→Question 1101 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,
- $\text{f}^{-1}(\text{x})=\text{f}(\text{x})$
- $\text{f}^{-1}(\text{x})=-\text{f}(\text{x})$
- $(\text{fof})\text{x}=-\text{x}$
- $\text{f}^{-1}\text{x}=\frac{1}{19}\text{f}(\text{x})$
Answer
- $\text{f}^{-1}(\text{x})=\text{f}(\text{x})$
Solution:
We have, $\text{f}(\text{x})=\frac{3\text{x}+2}{5\text{x}-3}=\text{y}\ (\text{let})$
$\Rightarrow\ 3\text{x}+2=5\text{xy}-3\text{y}$
$\Rightarrow\ \text{x}(3-5\text{y})=-3\text{y}-2$
$\Rightarrow\ \text{x}=\frac{3\text{y}+2}{5\text{y}-3}$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{3\text{x}+2}{5\text{x}-3}$
$\therefore\ \text{ f}^{-1}\text{x}=\text{f}(\text{x})$ View full question & answer→Question 1111 Mark
The relation R = {(1, 1), (2, 2), (3, 3)} on the set {1, 2, 3} is:
- Symmetric only.
- Reflexive only.
- An equivalence relation.
- Transitive only.
Answer
- An equivalence relation.
Solution:
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→Question 1121 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:
- $1$
- $\frac{\text{a}-1}{\text{a}}$
- $\frac{\text{a}}{\text{a}-1}$
- $0$
Answer
- $0$
Solution:
Let 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→Question 1131 Mark
The number of commutative binary operation that can be defined on a set of 2 elements is:
- 8
- 6
- 4
- 2
Answer
- 2
Solution:
The 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 1141 Mark
Choose the correct answer from the given four options. The maximum number of equivalence relations on the set $A = \{1, 2, 3\}$ are:
AnswerGiven that, $A = \{1, 2, 3\}$
Now, number of equivalence relations as follows
$R_1 = \{(1, 1), (2, 2), (3, 3)\}$
$R_2 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$
$R_3 = \{(1, 1), (2, 2), (3, 3), (1, 3), (3, 1)\}$
$R_4 = \{(1, 1), (2, 2), (3, 3), (2, 3), (3, 2)\}$
$R_5 = \{(1, 2, 3) \leftrightarrow A \times A = A^2\}$
$\therefore$ Maximum number of equivalence relation is $'5'$.
View full question & answer→MCQ 1151 Mark
Choose the correct answer from the given four options. Let $f : R \rightarrow R$ be given by $f(x) = \tan x$. Then $f^{-1}(1)$ is:
AnswerCorrect option: A. $\frac{\pi}{4}$
Given that$, f(x) = \tan x$
Let $y = \tan x$
$\Rightarrow x = \tan^{-1}y$
$\Rightarrow f^{-1}(x) = \tan^{-1}x$
$\Rightarrow f^{-1}(1) = \tan^{-1}1$
$\Rightarrow\ \tan^{-1}\tan\frac{\pi}{4}=\frac{\pi}{4}\ \Big[\because\ \tan\frac{\pi}{4}=1\Big]$
View full question & answer→MCQ 1161 Mark
If $f : R \rightarrow R, g : R \rightarrow R$ and $h : R \rightarrow R$ are such that $f(x) = x^2, \text{g(x)}=\tan\text{x}$ and $\text{h(x)}=\log\text{x},$ then the value of $(go(foh)) (x),$ if $x = 1$ will be:
View full question & answer→Question 1171 Mark
Let f : R → R be defind by $\text{f(x)}=\frac{1}{\text{x}}\forall\times\epsilon\text{ R}.$ Then f is:
- One-one.
- Onto.
- Bijective.
- F is not defined.
View full question & answer→Question 1181 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:
- 1
- 0
- 2
- none of these.
Answer
- 2
Solution:
Given 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→Question 1191 Mark
The identity element for the binary operation × defined on Q - {0} as $\text{a}\times\text{b}=\frac{\text{ab}}{2}\ \forall$ a, b $\epsilon$ Q - {0} is:
- 1
- 0
- 2
- None of these.
View full question & answer→MCQ 1201 Mark
If $g(x) = x^2 + x - 2$ and $\frac{1}{2}\text{gof(x)}=2{x}^2-5{x}+2,$ then $f(x)$ is equal to:
- ✓
$2x^2−5x+2$
- B
$2x + 3$
- C
$2x^2 + 3x + 1$
- D
$2x^2 - 3x - 1$
AnswerCorrect option: A. $2x^2−5x+2$
We will solve this problem by the trial $-$ and $-$ error method.
Let us check option $(a)$ first.
If $f(x) = 2x - 3$
$\frac{1}{2}(\text{gof})(x)=\text{g(f(x))}$
$=\frac{1}{2}\text{g}(2{x}-3)$
$=\frac{1}{2}\big[(2{x}-3)^2+(2{x}-3)-2\big]$
$=\frac{1}{2}[4{x}^2+9-12{x}+2{x}-3-2]$
$=\frac{1}{2}[4{x}^2-10{x}+4]$
$=2{x}^2-5 {x}+2$
The given condition is satisfied by $(a)$.
View full question & answer→MCQ 1211 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→Question 1221 Mark
The function $\text{f}:[0,\infty)\rightarrow\ \text{R}$ given by $\text{f(x)}=\frac{\text{x}}{\text{x}+1}$ is:
- One-one and onto.
- One-one but not onto.
- Onto but not one-one.
- Onto but not one-one.
Answer
- One-one but not onto.
Solution:
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)
$\Rightarrow\ \frac{\text{x}}{\text{x}+1}=\frac{\text{y}}{\text{y}+1}$
⇒ xy + x = xy + y
⇒ x = y
Hence, f is one-one.
If y = f(x)
$\text{y}=\frac{\text{x}}{\text{x}+1}$
⇒ xy + y = x
⇒ xy - x = -y
x(y - 1) = -y
$\text{x}=\frac{-\text{y}}{\text{y}-1}\neq\text{f(x)}$
It is not onto. View full question & answer→MCQ 1231 Mark
Which of the following functions form $Z$ to itself are bijections?
- A
$f(x) = x^3$
- B
$f(x) = x + 2$
- C
$f(x) = 2x + 1$
- D
$f(x) = x^2 + x$
Answer
- f is not because for ${y}=3\in$ Co $-$ domain ${(Z)},$ there is no value of ${x}\in$ Domain ${(Z)}$
${x}^3=3$
$\Rightarrow\ {x}=\sqrt[3]{3}\notin\text{Z}$
$\Rightarrow f$ is not onto.
So $,f $ is not a bijection.
- Injectivity: Let $x$ and $y$ be two elements of the domain $(Z),$ such that
$x + 2 = y + 2$
$\Rightarrow x = y$
So $,f$ is one $-$ one.
Surjectivity: Let y be an element in the co $-$ domain $(Z),$ such that
$y = f(x)$
$\Rightarrow y = x + 2$
$\Rightarrow\ {x}={y}-2\in\text{Z} \ ($Domain$)$
$\Rightarrow f$ is onto.
So $,f$ is a bijection.
- $f(x) = 2x + 1$ is not onto because if we take $4\in\text{Z} \ ($co domain$),$ then $4 = f(x)$
$\Rightarrow4 = 2{x} + 1$
$\Rightarrow 2{x} = 3$
$\Rightarrow\ {x}=\frac{3}{2}\notin\text{Z}$
So $,f$ is not a bijection.
- $f(0) = 0^2 + 0 = 0$
$\Rightarrow$ and $f(-1) = (-1)^2 + (-1) = 1 - 1 = 0$
$\Rightarrow 0$ and $-1$ have the same image.
$\Rightarrow f$ is not one $-$ one.
So $,f$ is not a bijection. View full question & answer→Question 1241 Mark
Let × be a binary operation on set of integers I, defined by a × b = a + b - 3, then find the value of 3 × 4.
- 2
- 4
- 7
- 6
Question 1251 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:
- 1
- 2
- 3
- 4
Answer
- 1
Solution:
Relation 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→Question 1261 Mark
Total number of equivalence relations defined in the set S = {a, b, c} is:
- 5
- 3!
- 23
- 33
Question 1271 Mark
Range of $\text{f(x)}=\sqrt{(1-\cos\text{x})\sqrt{(1-\cos\text{x})\sqrt{(1-\cos\text{x}).....\infty}}}$
- [0, 1]
- (0, 1)
- [0, 2]
- (0, 2)
View full question & answer→MCQ 1281 Mark
Let $f : \text{R}-\{\frac{3}{5}\}\rightarrow\text{R}$ be defined by $\text{f(x)}=\frac{3\text{x}+2}{5\text{x}-2}.$ Then:
AnswerCorrect option: A. $f^{-1}(x) = f(x)$
$f^{-1}(x) = f(x)$
View full question & answer→MCQ 1291 Mark
Let $\text{A}=\{\text{x}\in\text{R}:-1\leq\text{x}\leq1\}=\text{B}$ and $\text{C}=\{\text{x}\in\text{R}:\text{x}\geq0\}$ and let $\text{S}=\{(\text{x, y})\in\text{A}\times\text{B}:\text{x}^2+\text{y}^2=1\}$ and $\text{S}_0=\{(\text{x, y})\in\text{A}\times\text{C}:\text{x}^2+\text{y}^2=1\}.$ Then,
- ✓
$S$ defines a function from $A$ to $B$.
- B
$S_0$ defines a function from $A$ to $C$.
- C
$S_0$ defines a function from $A$ to $B$.
- D
$S$ defines a function from $A$ to $C$.
AnswerCorrect option: A. $S$ defines a function from $A$ to $B$.
Given that $\text{A}=\{\text{x}\in\text{R}:-1\leq\text{x}\leq1\}=\text{B}$ and $\text{C}=\{\text{x}\in\text{R}:\text{x}\geq0\}$ and $\text{S}=\{(\text{x, y})\in\text{A}\times\text{B}:\text{x}^2+\text{y}^2=1\}$ and $\text{S}_0=\{(\text{x, y})\in\text{A}\times\text{C}:\text{x}^2+\text{y}^2=1\}$
$\text{x}^2+\text{y}^2=1$
$\Rightarrow\ \text{y}^2=1-\text{x}^2$
$\Rightarrow\ \text{y}=\sqrt{1-\text{x}^2}$
$\text{y}\in\text{B}$
Hence $, S$ defines a function from $A$ to $B$.
View full question & answer→Question 1301 Mark
Let A = {2, 3, 4, 5, ..., 17, 18}. Let $'\simeq'$ be the equivalence relation on A × A, cartesian product of A with itself, defined by $(\text{a, b})\simeq(\text{c, d)}$ if ad = bc. Then, the number of ordered pairs of the equivalence class of (3, 2) is:
- 4
- 5
- 6
- 7
Answer
- 6
Solution:
The ordered pairs of the equivalence class of (3, 2) are {(3, 2), (6, 4), (9, 6), (12, 8), (15, 10), (18, 12)}.
We observe that these are 6 pairs.
View full question & answer→MCQ 1311 Mark
If $f$ is an invertible function defined as $\text{f(x)}=\frac{3\text{x}-4}{5},$ then $f^{-1}(x)$ is:
- A
$5x + 3$
- B
$5x + 3$
- ✓
$\frac{5\text{x}+4}{3}$
- D
$\frac{3\text{x}+2}{3}$
AnswerCorrect option: C. $\frac{5\text{x}+4}{3}$
$\frac{5\text{x}+4}{3}$
View full question & answer→MCQ 1321 Mark
Let $[x]$ denote the greatest integer less than or equal to $x$. If $f(x) = \sin^{-1}x, g(x) = [x^2]$ and $\text{h(x)}=2\text{x},\frac{1}{2}\leq\text{x}\leq\frac{1}{\sqrt{2}},$ then
- A
$\text{fogoh(x)}=\frac{\pi}{2}$
- B
$\text{fogoh(x)}=\pi$
- C
$\text{hofog}=\text{hogof}$
- ✓
$\text{hofog}\neq\text{hogof}$
AnswerCorrect option: D. $\text{hofog}\neq\text{hogof}$
$\text{hofog}=\text{hogof}$
$\text{hogof}(x) = h(f(g(x)))$
$= h(f([x]))$
$= h(\sin^{-1}[x])$
$= 2\sin^{-1}[x]$
$= 2 \times 0 = 0$
$f(x) = \sin^{-1}x$
$\text{hogof}(x) = \text{hogo}(x) = 0$
View full question & answer→Question 1331 Mark
f : R → R given by $\text{f(x)}=\text{x}+\sqrt{\text{x}^2}$ is:
- Injective.
- Surjective.
- Bijective.
- None of these.
Answer
- None of these.
Solution:
$\text{f(x)}=\text{x}+\sqrt{\text{x}^2}=\text{x}\pm\text{x}=0\text{ or }2\text{x}$
⇒ Each element of the domain has 2 images.
f is not a function. View full question & answer→MCQ 1341 Mark
Let $g(x) = x^2 - 4x - 5,$ then:
AnswerCorrect option: B. $G$ is not one$-$one on $R.$
$G$ is not one$-$one on $R.$
View full question & answer→Question 1351 Mark
Choose the correct answer out of the given four options.Let T be the set of all triangles in the Euclidean plane and let a relation R on T be defined as aRb, if a is congruent to $\text{b}\ \forall\ \text{a},\ \text{b}\in\text{T}.$ Then, R is:
- Reflexive but not transitive.
- Transitive but not symmetric.
- Equivalence.
- None of these.
Answer
- Equivalence.
Solution:
Consider that aRb, if a is congruent to b, $\forall\ \text{a, b}\in\text{T}$
Then, $\text{aRa}\Rightarrow\ \text{a}\cong\text{a},$
Which is true for all $\text{a}\in\text{T}$
So, R is reflexive, ....(i)
Let $\text{aRb}\Rightarrow\ \text{a}\cong\text{b}$
$\Rightarrow\ \text{b}\cong\text{a}\Rightarrow\ \text{b}\cong\text{a}$
$\Rightarrow\ \text{bRa}$
So, R is symmetric. ...(ii)
Let aRb and bRc
$\Rightarrow\ \text{a}\cong\text{b}\text{ and }\text{b}\cong\text{c}$
$\Rightarrow\ \text{a}\cong\text{c}\Rightarrow\ \text{aRc}$
So, R is transitive. .....(iii)
Hence, R is equivalence relation. View full question & answer→MCQ 1361 Mark
Let $*$ be a binary operation on $Q^+$ defined by $\text{a}*\text{b}=\frac{\text{ab}}{100}\forall\text{ a, b}\in\text{Q}^+$. The inverse of $0.1$ is:
AnswerCorrect option: A. $10^5$
Let e be the identity element in $Q^+$ with respect to $*$ such thata $* e = a = e$ * $a, \forall\text{ a}\in\text{Q}^+$
$a * e = a$ and $e * a = a, \forall\text{ a}\in\text{Q}^+$
$\frac{\text{ae}}{100}=\text{a}$ and $\frac{\text{ea}}{100}=\text{a},\forall\text{ a}\in\text{Q}^+$
$\text{e}=100,\forall\text{ a}\in\text{Q}^+$
Thus, $100$ is the identity element in $Q^+$ with repect to $*$.
$0.1 * b = e = b * 0.1$
$0.1 * b = e$ and $b * 0.1 = e$
$\frac{(0.1)\text{b}}{100}=100\text{ and }\frac{\text{b}(0.1)}{100}=100$
$\text{b}=\frac{100\times100}{0.1}$
$=10^5\in\text{Q}^+$
Thus, $10^5$ is the inverse of $0.1.$
View full question & answer→MCQ 1371 Mark
Let $\text{A}=\{\text{x}:-1\leq\text{x}\leq1\}$ and $f : A \rightarrow A$ such that $\text{f(x)}=\text{x}|\text{x}|,$ then $f$ is:
- ✓
- B
Injective but not surjective.
- C
Surjective but not injective.
- D
Neither injective nor surjective.
AnswerGiven function is $\text{A}=\{\text{x}:-1\leq\text{x}\leq1\}$ and $f : A \rightarrow A$ such that $\text{f(x)}=\text{x}|\text{x}|$
For the mod function we have to check three cases as $x < 0, x = 0, x > 0.$
For example,
$x < 0$
$f(x) = x|x| < 0$
$|x| = -x$
$y = -x^2$
$\text{x}=-\sqrt{-\text{y}}$ which is not possible for $x > 0$
Hence $, f$ is onto.
$\Rightarrow f$ is bijection.
View full question & answer→Question 1381 Mark
For binary operation × defind on R – {1} such that $\text{a}\times\text{b}=\frac{\text{a}}{\text{b}+1}$ is:
- Not associative.
- Not commutative.
- Commutative.
- Both (a) and (b).
View full question & answer→MCQ 1391 Mark
Choose the correct answer from the given four options.Let $f : A \rightarrow B$ and $g : B \rightarrow C$ be the bijective functions. Then $\text{(gof)}^{-1}$ is:
- ✓
$f^{-1}\text{og}^{-1}$
- B
$\text{fog}$
- C
$g^{-1} \text{of}^{-1}$
- D
$\text{gof}$
AnswerCorrect option: A. $f^{-1}\text{og}^{-1}$
$f^{-1}\text{og}^{-1}$
Given that, $f : A \rightarrow B$ and $g : B \rightarrow C$ be the bijective functions.
$(\text{f}^{-1}\text{o}\text{g}^{-1})\text{o}(\text{gof})=\text{f}^{-1}\text{o}(\text{g}^{-1}\text{ogof})$
$=\text{f}^{-1}\text{o}(\text{g}^{-1}\text{og})\text{of} \ ($As composition of functions is associative$)$
$=(\text{f}^{-1}\text{o}\text{I}_{\text{B}}\text{of}) \ ($where $I_B$ is identity function on $B)$
$=(\text{f}^{-1}\text{o}\text{I}_{\text{B}})\text{of}$
$=\text{f}^{-1}\text{of}$
$=\text{I}_{\text{A}}$
Thus $(\text{gof})^{-1}=\text{f}^{-1}\text{og}^{-1}$
View full question & answer→MCQ 1401 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 1411 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 1421 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 1431 Mark
If $f\text{(x)} = \text{(ax^2 + b)}^3,$ then the function $g$ such that $f(g\text{(x)}) = \text{g(f(x))}$ is given by:
- A
$\text{g}(\text{x})=\Big(\frac{\text{b}-\text{x}^\frac{1}{3}}{\text{a}}\Big)$
- B
$\text{g}(\text{x})=\frac{1}{(\text{ax}^2+\text{b})^3}$
- C
$\text{g}(\text{x})=(\text{ax}^2+\text{b})^\frac{1}{3}$
- ✓
$\text{g}(\text{x})=\Big(\frac{\text{x}^\frac{1}{3}-\text{b}}{\text{a}}\Big)^\frac{1}{2}$
AnswerCorrect option: D. $\text{g}(\text{x})=\Big(\frac{\text{x}^\frac{1}{3}-\text{b}}{\text{a}}\Big)^\frac{1}{2}$
$\text{g}(\text{x})=\Big(\frac{\text{x}^\frac{1}{3}-\text{b}}{\text{a}}\Big)^\frac{1}{2}$
View full question & answer→MCQ 1441 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→Question 1451 Mark
If a * b denote the bigger among a and b and if ab = (a * b) + 3, then 4.7 =
- 14
- 31
- 10
- 8
Answer
- 10
Solution:
4.7 = (4 * 7) + 3
= 7 + 3
= 10
View full question & answer→Question 1461 Mark
Which of the following functions from $\text{A}=\{\text{x}\in\text{R}:-1\leq\text{x}\leq1\}$ to itself are bijections?
- $\text{f(x)}=|\text{x}|$
- $\text{f(x)}=\sin\frac{\pi\text{x}}{2}$
- $\text{f(x)}=\sin\frac{\pi\text{x}}{4}$
- $\text{None of these}$
Answer
- $\text{f(x)}=\sin\frac{\pi\text{x}}{2}$
Solution:
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→Question 1471 Mark
Let R = {(a, a), (b, b), (c, c), (a, b)} be a relation on set A = a, b, c. Then, R is:
- Identify relation.
- Reflexive.
- Symmetric.
- Antisymmetric.
Answer
- Reflexive.
Solution:
Reflexivity: 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.
⇒ R is not antisymmetric on A.
Also, R is not an identity relation on A. View full question & answer→MCQ 1481 Mark
If $f(x) = \sin^2x$ and the composite function $\text{g(f(x))} = |\sin\text{x}|,$ then $g(x)$ is equal to:
- A
$\sqrt{{x}-1}$
- ✓
$\sqrt{{x}}$
- C
$\sqrt{{x}+1}$
- D
$-\sqrt{{x}}$
AnswerCorrect option: B. $\sqrt{{x}}$
Given that $\text{f(x)}=\sin^2\text{x}$ and the composite function $\text{g(f(x))}=|\sin {x}|$
We will do it using trial and error method.
If we take $\text{g(x)}=-\sqrt{{x}}$ and $\text{f(x)}=\sin^2{x}$
$\text{g(f(x))}=\text{g}(\sin^2{x})$
$=-\sin {x}$
Which contradicts to the $\text{g(f(x))}=|\sin {x}|$
Hence, we take $\text{g(x)}=\sqrt{{x}}$
$\text{g(f(x))}=\text{g}(\sin^2 {x})$
$=\sqrt{\sin^2{x}}=|\sin {x}|$
View full question & answer→Question 1491 Mark
Let $\text{f(x)}=\frac{\text{x}-1}{\text{x}+1},$ then f(f(x)) is:
- $\frac{1}{\text{x}}$
- $-\frac{1}{\text{x}}$
- $\frac{1}{\text{x}+1}$
- $\frac{1}{\text{x}-1}$
View full question & answer→MCQ 1501 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→