MCQ 1011 Mark
The law $a + b = b + a$ is called:
AnswerThe law $a + b = b + a$ is commutative.
View full question & answer→MCQ 1021 Mark
Let $R$ be a relation on the set $N$ given by $R = \{(a, b): a = b - 2, b > 6\}.$ Then,
- ✓
$(2, 4) \in R$
- B
$(3, 8) \in R$
- C
$(6, 8) \in R$
- D
$(8, 7) \in R$
AnswerCorrect option: A. $(2, 4) \in R$
$a = b - 2$
$ \Rightarrow 6 = 8 - 2$ and $b = 8 > 6$
Hence$, (6, 8) \in R$
View full question & answer→MCQ 1031 Mark
Consider a non$-$empty set consisting of children in a family and a relation $R$ defined as $\text{aRb}$ if a is brother of $b.$ Then$, R$ is:
- A
Symmetric but not transitive.
- B
Transitive but not symmetric.
- C
Neither symmetric nor transitive.
- ✓
Both symmetric and transitive.
AnswerCorrect option: D. Both symmetric and transitive.
We have,
$R = \{(a, b): a$ is brother of $b\}$
Let $(\text{a, b})\in\text{R}.$ Then,
$a$ is brother of $b.$
but $b$ is not necessary brother of $a ($As$, b$ can be sister of $a)$
$\Rightarrow\ (\text{b, a})\notin\text{R}$
So$, R$ is not symmetric.
Also,
Let $(\text{a, b})\in\text{R}$ and $(\text{b, c})\in\text{R}$
$\Rightarrow a$ is brother of $b$ and $b$ is brother of $c$
$\Rightarrow a$ is brother of $c$
$\Rightarrow\ (\text{a, c})\in\text{R}$
So$, R$ is transitive.
View full question & answer→MCQ 1041 Mark
Let $f: R \rightarrow R$ be given by $f(x)=\tan x$. Then $f^{-1}(1)$ is:
AnswerCorrect option: B. $\{\text{n}\pi+\frac{\pi}{4};\text{n }\in\text{ Z}\}$
View full question & answer→MCQ 1051 Mark
Let $R$ be a relation on $N$ defined by $x + 2y = 8.$ The domain of $R$ is:
- A
$\{2, 4, 8\}$
- B
$\{2, 4, 6, 8\}$
- ✓
$\{2, 4, 6\}$
- D
$\{1, 2, 3, 4\}$
AnswerCorrect option: C. $\{2, 4, 6\}$
The relation $R$ is defined as $R = x, y: \text{x, y}\in\text{N}$ and $x + 2y = 8$
$\Rightarrow R = x, y: \text{x, y}\in\text{N}$ and $\text{y}=\frac{8-\text{x}}{2}$
Domain of $R$ is all values of $\text{x}\in\text{N}$ satisfying the relation $R.$
Also, there are only three values of $x$ that result in $y,$ which is a natural number.
These are $\{2, 6, 4\}.$
View full question & answer→MCQ 1061 Mark
Let $f : Z \rightarrow Z$ 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:
AnswerCorrect option: A. Onto but not one$-$one.
Given function is
$\text{f(x)}=\frac{\text{x}}{2}$ if $x$ is even
$= 0$ if $x$ is odd
For $f(3) = 0$ and $f(4) = 0$
$\Rightarrow f(3) = f(4)$
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 1071 Mark
Let $\text{f}:\text{R}-\Big\{-\frac{4}{3}\Big\}\rightarrow\text{R}$ be a function defined as $f(\text{x})=\frac{4\text{x}}{3\text{x}+4}.$ The inverse of $f$ is the map $g:$ Range $\text{f}:\text{R}-\Big\{-\frac{4}{3}\Big\}\rightarrow\text{R}$ given by:
- A
$\text{g}(\text{y})=\frac{3\text{y}}{3-4\text{y}}$
- ✓
$\text{g}(\text{y})=\frac{4\text{y}}{4-3\text{y}}$
- C
$\text{g}(\text{y})=\frac{4\text{y}}{3-4\text{y}}$
- D
$\text{g}(\text{y})=\frac{3\text{y}}{4-3\text{y}}.$
AnswerCorrect option: B. $\text{g}(\text{y})=\frac{4\text{y}}{4-3\text{y}}$
Given: $\text{f}:\text{R}-\Big\{-\frac{4}{3}\Big\}\rightarrow\text{R}$ and $f(\text{x})=\frac{4\text{x}}{3\text{x}+4}$
Now, Range of $\text{f}\rightarrow\text{R}-\Big\{-\frac{4}{3}\Big\}$
Let $\text{y}=f(\text{x})$
$ \therefore\ \text{y}=\frac{4\text{x}}{3\text{x}+4}$
$\Rightarrow\ 3\text{xy}+4\text{y}=4\text{x}$
$\Rightarrow\ \ \ \text{x}(4-3\text{y})=4\text{y}$
$ \Rightarrow\ \text{x}=\frac{4\text{y}}{4-3\text{y}}$
$\therefore\ \ f^{-1}(\text{y})=\text{g(y)}=\frac{4\text{y}}{3-4\text{y}}$
Therefore, option $(B)$ is correct.
View full question & answer→MCQ 1081 Mark
The binary operation $\times$ defined on $N$ by $a \times b = a + b + ab$ for all $a, b \in N$ is:
- A
- B
- ✓
Both commutative and associative.
- D
AnswerCorrect option: C. Both commutative and associative.
View full question & answer→MCQ 1091 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.
View full question & answer→MCQ 1101 Mark
Let $f: R \rightarrow R, g: R \rightarrow R$ be two functions such that $f(x)=2 x-3, g(x)=x^3+5$. The function $($fog $))^{-1}(x)$ is:
- A
$\Big(\frac{\text{x}+7}{2}\Big)^\frac{1}{3}$
- B
$\Big(\text{x}-\frac{7}{2}\Big)^\frac{1}{3}$
- C
$\Big(\frac{\text{x}-2}{7}\Big)^\frac{1}{3}$
- ✓
$\Big(\frac{\text{x}-7}{2}\Big)^\frac{1}{3}$
AnswerCorrect option: D. $\Big(\frac{\text{x}-7}{2}\Big)^\frac{1}{3}$
View full question & answer→MCQ 1111 Mark
The relation $'R\ '$ in $N \times N$ such that $(a, b)R(c, d) \Leftrightarrow a + d = b + c$ is:
- A
Reflexive but not symmetric.
- B
Reflexive and transitive but not symmetric.
- ✓
- D
AnswerWe observe the following properties of relation $R.$
Reflexivity: Let $(\text{a, b})\in\text{N}\times\text{N}$
$\Rightarrow\ \text{a, b}\in\text{N}$
$\Rightarrow\ \text{a}+\text{b}=\text{b}+\text{a}$
$\Rightarrow\ (\text{a, b})\in\text{R}$
So$, R$ is reflexive on $N \times N.$
Symmetry: Let $(\text{a, b}),\ (\text{c, d})\in\text{N}\times\text{N}$ such that $(a, b)R(c, d)$
$\Rightarrow\ \text{a}+\text{d}=\text{b}+\text{c}$
$\Rightarrow\ \text{d}+\text{a}=\text{c}+\text{b}$
$\Rightarrow\ (\text{d, c}),\ (\text{b, a})\in\text{R}$
So$, R$ is symmetric on $N \times N.$
Transitivity: Let $(\text{a, b}),\ (\text{c, d}),\ (\text{e, f})\in\text{N}\times\text{N}$ such that $(a, b)R(c, d)$ and $(c, d)R(e, f)$
$\Rightarrow a + d = b + c$ and $c + f = d + e$
$\Rightarrow a + d + c + f = b + c + d + e$
$\Rightarrow a + f = b + e$
$\Rightarrow (a, b)R(e, f)$
So$, R$ is transitive on $N \times N.$
Hence$, R$ is an equivalence relation on $N.$
View full question & answer→MCQ 1121 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. $\frac{5\text{x}+4}{3}$
View full question & answer→MCQ 1131 Mark
Let $f: R \rightarrow R$ be defined as $f(x) = 3x.$ Choose the correct answer.
AnswerCorrect option: A. $f$ is one$-$one onto
$f: R \rightarrow R$ is defined as $f(x) = 3x.$
Let $\text{x},\text{y}\in\text{R}$ such that $f(x) = f(y)$.
$\Rightarrow 3x = 3y$
$ \Rightarrow x = y$
$\therefore f$ is one$-$one.
Also, for any real number $(y)$ in $co-$domain $R,$ there exists $\frac{\text{y}}{3}$ in $R$
such that $f\Big(\frac{\text{y}}{3}\Big)=3\Big(\frac{\text{y}}{3}\Big)=\text{y}$
$\therefore f$ is onto.
Hence, function $f$ is one$-$one and onto.
The correct answer is $A.$
View full question & answer→MCQ 1141 Mark
Let $A = R – \{3\}, B = R – \{1\}.$ Let $f : A \rightarrow B$ be defined by $\text{f(x)}=\frac{\text{x}-2}{\text{x}-3}.$ Then,
AnswerCorrect option: A. $F$ is bijective.
View full question & answer→MCQ 1151 Mark
If the binary operation $\times$ is defind on the set $Q +$ of all positive rational numbers by $\text{a}\times\text{b}=\frac{\text{ab}}{4}.$ Then, $3\times\Big(\frac{1}{5}\times\frac{1}{2}\Big)$ is equal to:
- ✓
$\frac{3}{160}$
- B
$\frac{5}{160}$
- C
$\frac{3}{10}$
- D
$\frac{3}{40}$
AnswerCorrect option: A. $\frac{3}{160}$
View full question & answer→MCQ 1161 Mark
If $(x)=\left(a x^2+b\right)^3$, then the function $g$ such that $f(g(x))=g(f(x))$ 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. $\text{g}(\text{x})=\Big(\frac{\text{x}^\frac{1}{3}-\text{b}}{\text{a}}\Big)^\frac{1}{2}$
View full question & answer→MCQ 1171 Mark
A binary operation $*$ on $Z$ defined by $a^ * b = 3a + b$ for all $a, b \in Z,$ is:
- A
- B
- ✓
- D
Commutative and associative.
AnswerLet $\text{a, b}\in\text{Z}$
$a^ * b = 3a + b$
$b^ * a = 3b + a$
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 1181 Mark
Choose the correct answer from the given four options.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.
Given that$, A = \{1, 2, 3\}$
and $R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)\}$
$\because\ (1,1), (2,2),(3,3)\in\text{R}$
Hence$, R$ is reflexive.
$(1,2)\in\text{R}$ but $(2,1)\notin\text{R}$
Hence$, R$ is not symmetric.
$(1,2)\in\text{R}$ and $(2,3)\in\text{R}$
$\Rightarrow\ (1,3)\in\text{R}$
Hence$, R$ is transitive.
View full question & answer→MCQ 1191 Mark
Which of the following functions from $\text{A}=\{\text{x}:-1\leq\text{x}\leq1\}$ to it self are bijections?
AnswerCorrect option: B. $\text{g(x)}=\sin\big(\frac{\pi\text{x}}{2}\big)$
View full question & answer→MCQ 1201 Mark
The function $f: R \rightarrow R$ defined by $f(x)=2^x+2^{|x|}$ is:
- A
One$-$one and onto.
- B
Many$-$one and onto.
- ✓
One$-$one and into.
- D
Many$-$one and into.
AnswerCorrect option: C. One$-$one and into.
View full question & answer→MCQ 1211 Mark
Choose the correct answer from the given four options.Let us define a relation $R$ in $R$ as $\text{aRb}$ if $a ≥ b.$ Then $R$ is:
- A
- ✓
Reflexive, transitive but not symmetric.
- C
Symmetric, transitive but not reflexive.
- D
Neither transitive nor reflexive but symmetric.
AnswerCorrect option: B. Reflexive, transitive but not symmetric.
We are given that$, \text{aRb}$ if a $\geq b$
$\Rightarrow \text{aRa}$
$ \Rightarrow a \geq$ a which is true.
For relation $\text{aRb}$ to be symmetric, we must have $a \geq b$ and $b \geq a$ which can’t be possible.
Hence$, R$ is not symmetric.
For relation $\text{aRb}$ to be transitive, we must have $\text{aRb}$ and $\text{bRc}.$
$\Rightarrow a \geq b$ and $b \geq c$
$\Rightarrow a \geq c$
Hence$, R$ is transitive.
View full question & answer→MCQ 1221 Mark
Choose the correct answer from the given four options. 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:
View full question & answer→MCQ 1231 Mark
Let $f: R \rightarrow R$ be the functions defined by $f(x)=x^3+5$. Then $f^{-1}(x)$ is:
- A
$(\text{x}+5)^\frac{1}{3}$
- ✓
$(\text{x}-5)^\frac{1}{3}$
- C
$(5-\text{x})^\frac{1}{3}$
- D
$5-\text{x}$
AnswerCorrect option: B. $(\text{x}-5)^\frac{1}{3}$
View full question & answer→MCQ 1241 Mark
If the binary operation $\odot$ is defined on the set $Q^{+}$ of all positive rational numbers by $\text{a}\odot\text{b}=\frac{\text{ab}}4$. Then, $3\odot\Big(\frac{1}5\odot\frac{1}2\Big)$ is equal to:
- ✓
$\frac{3}{160}$
- B
$\frac{5}{160}$
- C
$\frac{3}{10}$
- D
$\frac{3}{40}$
AnswerCorrect option: A. $\frac{3}{160}$
View full question & answer→MCQ 1251 Mark
Let $A = \{1, 2, 3\}$ and $B = \{(1, 2), (2, 3), (1, 3)\}$ be a relation on $A.$ Then, $R$ is:
- A
Neither reflexive nor transitive.
- B
Neither symmetric nor transitive.
- ✓
- D
View full question & answer→MCQ 1261 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 $(a^ * b)^ * c$.
$(a^ * b)^ * c=(1^ * 2)^ * 3=\left(1^3+2^3\right)^ * 3=9^3+3^3=729+27=756$
Apply the given binary operation on $a^ *(b^ * c)$.
$(a^ * b)^ * c=1^ *(2^ * 3)=1^ *\left(2^3+3^3\right)=1^3+35^3=42876$
$(a^ * b)^ * c \neq a^ *(b^ * c)$
So the operation is not associative.
Therefore, the given operation is commutative but not associative.
View full question & answer→MCQ 1271 Mark
For binary operation $\times$ defind on $R – \{1\}$ such that $\text{a}\times\text{b}=\frac{\text{a}}{\text{b}+1}$ is:
- A
- B
- C
- ✓
Both $(a)$ and $(b).$
AnswerCorrect option: D. Both $(a)$ and $(b).$
View full question & answer→MCQ 1281 Mark
$R$ is a relation on the set $Z$ of integers and it is given by $(x, y) \in R \Leftrightarrow | x - y | \leq 1.$ Then$, R$ is:
- A
Reflexive and transitive.
- ✓
- C
Symmetric and transitive.
- D
AnswerReflexivity: Let $\text{x}\in\text{R.}$ Then,
$\text{x}-\text{x}=0<1$
$\Rightarrow\ |\text{x}-\text{x}|\leq1$
$\Rightarrow\ (\text{x, x})\in\text{R}$ for all $\text{x}\in\text{Z}$
So$, R$ is reflexive on $Z.$
Symmetry: Let $\text{x, y}\in\text{R.}$ Then,
$|\text{x}-\text{y}|\leq0$
$\Rightarrow\ |-(\text{y}-\text{x})|\leq1$
$\Rightarrow\ |(\text{y}-\text{x})|\leq1 [$Since $|x - y| = |y - x|]$
$\Rightarrow\ (\text{y, x})\in\text{R}$ for all $\text{x, y}\in\text{Z}$
So$, R$ is symmetric on $Z.$
Transitivity: Let $(\text{x, y})\in\text{R}$ and $(\text{y, z})\in\text{R}.$ Then,
$|\text{x}-\text{y}|\leq1$ and $|\text{y}-\text{z}|\leq1$
$\Rightarrow$ It is not always true that $|\text{x}-\text{y}|\leq1.$
$\Rightarrow\ (\text{x, z})\notin\text{R}$
So$, R$ is not transitive on $Z.$
View full question & answer→MCQ 1291 Mark
The number of binary operation that can be defined on a set of $2$ elements is:
View full question & answer→MCQ 1301 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. $\frac{\text{x}+\sqrt{\text{x}^2-4}}{2}$
View full question & answer→MCQ 1311 Mark
If $f(x) =\frac{3\text{x}+2}{5\text{x}-3}$ then $(fof)(x)$ is:
View full question & answer→MCQ 1321 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:
AnswerCorrect option: D. Neither one$-$one nor onto.
View full question & answer→MCQ 1331 Mark
Subtraction of integers is:
- ✓
Commutative but no associative.
- B
Commutative and associative.
- C
Associative but not commutative.
- D
Neither commutative nor associative.
AnswerCorrect option: A. Commutative but no associative.
View full question & answer→MCQ 1341 Mark
If $A = \{a, b, c\},$ then the relation $R = \{b, c\}$ on $A$ is:
- A
- B
- ✓
- D
Reflexive and transitive only.
View full question & answer→MCQ 1351 Mark
The binary operation $\times$ defind on set $R,$ given by $\text{a}\times\text{b}=\frac{\text{a}+\text{b}}{2}$ for all $a, b \in R$ is:
- ✓
- B
- C
Both $(a)$ and $(b).$
- D
View full question & answer→MCQ 1361 Mark
The maximum number of equivalence relations on the set $A = \{1, 2, 3\}$ are:
View full question & answer→MCQ 1371 Mark
Let $*$ be a binary operation on $N$ defined by $a^ * b = a + b + 10$ for all $a, b \in N.$ The identity element for $*$ in $N$ is:
- A
$−10$
- B
$0$
- C
$10$
- ✓
Non$-$existent.
AnswerCorrect option: D. Non$-$existent.
Given $a^ * b = a + b + 10$
Let the identity element be $e,$ then
$a^ * e = a$
$\Rightarrow a + e + 10 = a$
$\Rightarrow e = -10$
But the operation is defined on the set of natural numbers.
So, the identity element doesn't exist.
View full question & answer→MCQ 1381 Mark
For real numbers $x$ and $y,$ define $\text{xRy}$ if $\text{x}-\text{y}+\sqrt{2}$ is an irrational number. Then the relation $R$ is:
AnswerWe 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 1391 Mark
Let $[x]$ denote the greatest integer less than or equal to $x$. If $f(x)=\sin ^{-1} x, g(x)=\left[x^2\right]$ and $h(x)=2 x, \frac{1}{2} \leq x \leq \frac{1}{\sqrt{2}}$, then
- A
$\text{fogoh(x)}=\frac{\pi}{2}$
- B
$\text{fogoh(x)}=\pi$
- ✓
$\text{hofog}=\text{hogof}$
- D
$\text{hofog}\neq\text{hogof}$
AnswerCorrect option: C. $\text{hofog}=\text{hogof}$
$\operatorname{hogof}(x)=h(f(g(x)))$$=h(f([x]))$
$=h\left(\sin ^{-1}[x]\right)$
$=2 \sin ^{-1}[x]$
$=2 \times 0=0$
$f(x)=\sin ^{-1} x$
$\operatorname{hogof}(x)=\operatorname{hogo}(x)=0$
View full question & answer→MCQ 1401 Mark
The binary operation $*$ is defined by $a^ * b=a^2+b^2+a b+1$, then $(2^ * 3)^ * 2$ is equal to:
AnswerGiven: $a{ }^* b=a^2+b^2+a b+1$
$2^ * 3=2^2+3^2+2 \times 3+1$
$=4+9+6+1$
$=20$
$(2^ * 3)^ * 2=20^ * 2$
$=20^2+2^2+20 \times 2+1$
$=400+4+40+1$
$=445$
View full question & answer→MCQ 1411 Mark
Let $\times$ be a binary operation on $Q,$ defined by $\text{a}\times\text{b}=\frac{3\text{ab}}{5}$ is:
- A
- B
- ✓
Both $(a)$ and $(b).$
- D
AnswerCorrect option: C. Both $(a)$ and $(b).$
View full question & answer→MCQ 1421 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 $b$ for all $a, b \in T.$ Then$, R$ is:
- A
Reflexive but not symmetric.
- B
Transitive but not symmetric.
- ✓
- D
AnswerGiven that $R$ is $T$ be the set of all triangle in the Euclidean plane, and a relation $R$ on $T$ be defined as $\text{aRb}$ if $a$ is congruent to $b$ for all $a, b \in T.$
Here, congruency of triangles follows reflexive, symmetric and transitive property.
Hence, it is an equivalence relation.
View full question & answer→MCQ 1431 Mark
The function $f : A \rightarrow B$ defined by $f(x) = 4x + 7, x \in R$ is:
AnswerCorrect option: A. One$-$one
View full question & answer→MCQ 1441 Mark
$f : R \rightarrow R$ given by $\text{f(x)}=\text{x}+\sqrt{\text{x}^2}$ is:
Answer$\text{f(x)}=\text{x}+\sqrt{\text{x}^2}=\text{x}\pm\text{x}=0 $ or $2\text{x}$
$\Rightarrow$ Each element of the domain has $2$ images.
$f$ is not a function.
View full question & answer→MCQ 1451 Mark
Let $g(x)=x^2-4 x-5$, then:
AnswerCorrect option: B. $G$ is not one$-$one on $R.$
View full question & answer→MCQ 1461 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 $\text{aRb},$ if a is congruent to $\text{b}\ \forall\ \text{a},\ \text{b}\in\text{T}.$ Then$, R$ is:
- A
Reflexive but not transitive.
- B
Transitive but not symmetric.
- ✓
- D
AnswerConsider that $\text{aRb},$ 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 $\text{aRb}$ and $\text{bRc}$
$\Rightarrow\ \text{a}\cong\text{b}$ 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 1471 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 that
$a^ * e = a = e^ * a$, $\forall\text{ a}\in\text{Q}^+$
$a^ * e = a$ and $e^ * a = a, \forall\text{ a}\in\text{Q}^+$
$\frac{\text{ae}}{100}=\text{a}$ 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 1481 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.
View full question & answer→MCQ 1491 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 $(g^ \circ f)^{-1}$ is:
- ✓
$f^{-1^ \circ} g^{-1}$
- B
$fog$
- C
$g^{-1^ \circ} f^{-1}$
- D
$gof$
AnswerCorrect option: A. $f^{-1^ \circ} g^{-1}$
Given that, $f : A \rightarrow B$ and $g : B \rightarrow C$ be the bijective functions.
$(\text{f}^{-1}\text{o}\text{g}^{-1})\text{o}(\text{gof})=\text{f}^{-1}\text{o}(\text{g}^{-1}\text{ogof})$
$=\text{f}^{-1}\text{o}(\text{g}^{-1}\text{og}) of ($As composition of functions is associative$)$
$=(\text{f}^{-1}\text{o}\text{I}_{\text{B}}\text{of}) ($where Iis 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 1501 Mark
If $f: R \rightarrow R$ and $g: R \rightarrow R$ defined by $f(x)=2 x+3$ 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. $\pm2$
View full question & answer→