Question 1512 Marks
Write whether f : R → R, given by $\text{f(x)}=\text{x}+\sqrt{\text{x}^2},$ is one-one, many-one, onto or into.
Answerf : R → R given by $\text{f(x)}=\text{x}+\sqrt{\text{x}^2}$
$\because\ \text{f(x)=}\begin{cases}2\text{x};&\text{for x}>0\\0;&\text{for x}<0\end{cases}$
$\therefore$ f is many-one function.
View full question & answer→Question 1522 Marks
Let D be the domain of the real valued function f defined by $\text{f}(\text{x})=\sqrt{25-\text{x}^2}.$ Then, write D.
AnswerConsider the given function, $\text{f}(\text{x})=\sqrt{25-\text{x}^2}$
For f(x) to be real, the term inside the square root can’t be negative
i.e., $25-\text{x}^2\geq0$
$\Rightarrow\ \text{x}^2\leq25$
$\Rightarrow\ 5\leq\text{x}\leq-5$
Therefore, the domain of the function, f(x) is given by D = [-5, 5]
View full question & answer→Question 1532 Marks
For each binary operation * defined below, determine whether * is commutative or associative.
On Z, define a * b = a – b
AnswerFor Commutativity: a * b = a - b and b * a = b - a = -(a - b) $\neq\text{a}*\text{b}$ For associativity: a * (b * c) = a * (b - c) = a - (b - c) = (a - b + c) Also, (a * b) * c = (a - b) * c = (a - b - c) $\therefore\ \ \ \text{a}*(\text{b}*\text{c})\neq(\text{a}*\text{b})*\text{c}$Therefore, the operation * is neither commutative nor associative.
View full question & answer→Question 1542 Marks
For each binary operation * defined below, determine whether * is commutative or associative.
On Q, define a * b = ab + 1
AnswerFor commutativity: a * b = ab + 1 and b * a = ba + 1 = ab + 1 = a * b
For associativity: a * (b * c) = a * (bc + 1) = a(bc + 1) + 1 = abc + a + 1
Also, (a * b) * c = (ab + 1)c + 1 = abc + c + 1
$\therefore\ \ \text{a }*(\text{b }*\text{c})\neq(\text{a }*\text{b })*\text{c}$
Therefore, the operation * is commutative but not associative.
View full question & answer→Question 1552 Marks
Which one the following relations on A = {1, 2, 3} is a function?
f = {(1, 3), (2, 3), (3, 2)}, g = {(1, 2), (1, 3), (3, 1)}
AnswerAs, each element of the domain set has unique image in the relation f = {(1, 3), (2, 3), (3, 2)}
So, f is a function.
Also, the element 1 of the domain set has two images 2 and 3 of the range set in the relation g = {(1, 2), (1, 3), (3, 1)}
So, g is not a function.
View full question & answer→Question 1562 Marks
Is it true that every relation which is symmetric and transitive is also reflexive? Give reasons.
AnswerNo, it is not necessary that a relation which is symmetric and transitive is reflexive as well.
For Example,
Let A = {a, b, c} be a set and
$R_2$ = {(a, a)} is a relation defined on A.
Clearly,
$R_2$ is symmetric and transitive but not reflexive.
View full question & answer→Question 1572 Marks
For the binary operation multiplication modulo $5 (\times _5)$ defined on the set $S = \{1, 2, 3, 4\}.$ Write the value of $(3 \times _5 4^{-1})^{−1}$
AnswerThe composition table for $\times _5$ on the set $S =\{1, 2, 3, 4\}$ is
| $\times _5$ |
$1$ |
$2$ |
$3$ |
$4$ |
| $1$ |
$1$ |
$2$ |
$3$ |
$4$ |
| $2$ |
$2$ |
$4$ |
$1$ |
$3$ |
| $3$ |
$3$ |
$1$ |
$4$ |
$2$ |
| $4$ |
$4$ |
$3$ |
$2$ |
$1$ |
Now,
$(3 \times _5 4^{-1})^{-1} = (3 \times _5 4)^{-1} [\because 4^{-1} = 4]$
$= 2^{-1} [3 \times _5 4 = 2]$
$= 3 [\because 2^{-1} = 3]$ View full question & answer→Question 1582 Marks
Let f be a function from C (set of all complex numbers) to itself given by $f(x) = x^3$. Write $f^{-1}(-1)$.
AnswerLet $f^{-1}(-1) = x .....(1)$
$\Rightarrow f(x) = -1$
$\Rightarrow x^3 = -1$
$\Rightarrow x^3 + 1 = 0$
$\Rightarrow (x + 1)(x^2 - x + 1) = 0$
[Using the identity: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$]
$\Rightarrow\ (\text{x}+1)(\text{x}+\omega)(\text{x}+\omega^2)=0,$ where $\omega=\frac{1\pm\text{i}\sqrt{3}}{2}$
$\Rightarrow\ \text{x}=-1,-\omega,-\omega^2$ $(\text{as x}\in\text{C})$
$\Rightarrow\ \text{f}^{-1}(-1)=\{-1,-\omega,-\omega^2\}$ [from 1]
View full question & answer→Question 1592 Marks
Find gof and fog, if:
f(x) = |x| and g(x) = |5x – 2|
AnswerTo find: gof and fog
f(x) = x and g(x) = |5x − 2|
gof = g[f(x)] = g[|x|] and fog = f[g(x)] = f[(5x - 2)] = |5x - 2| = |5|x|-2|
View full question & answer→Question 1602 Marks
If $f(x) = 4 - (x - 7)^3$, then write $f^{-1}(x)$.
AnswerWe have, $f(x) = 4 - (x - 7)^3$ Let $y = 4 - (x - 7)^3$
$\Rightarrow\ (\text{x} - 7)^3 = 4 - \text{y}$
$\Rightarrow\ \text{x}-7=\sqrt[3]{4-\text{y}}$
$\Rightarrow\ \text{x}=7+\sqrt[3]{4-\text{y}}$
$\Rightarrow\ \text{f}^{-1}(\text{y})=7+\sqrt[3]{4-\text{y}}$
$\therefore\ \text{f}^{-1}(\text{x})=7+\sqrt[3]{4-\text{x}}$
View full question & answer→Question 1612 Marks
Show that the binary operation * on Z defined by a * b = 3a + 7b is not commutative.
AnswerThe binary operator * defined on Z and is given by a * b = 3a + 7b
Commutativity: Let $\text{a, b}\in\text{Z},$ Then,
a * b = 1a + 7b and
b * a = 3b + 7a
$\therefore\ \text{a}\ ^*\ \text{b}\neq\text{b}\ ^*\ \text{a}$
Hence, '*' is not commutative on Z.
View full question & answer→Question 1622 Marks
Which of the following functions from A to B are one-one and onto?
$f_3 = \{(a, x), (b, x), (c, z), (d, z)\}; A = \{a, b, c, d,\}, B = \{x, y, z\}$
Answer$f_3 = \{(a, x), (b, x), (c, z), (d, z)\}$
$A = \{a, b, c, d,\}, B = \{x, y, z\}$
Since, $f_3(a) = x = f_3(b)$ and $f_3(c) = z = f_3(d)$
$\therefore f_3$ in not one-one.
Again, $\text{y}\in\text{B}$ in not the image of any of the element of $A.$
$\therefore f_3$ in not on to.
View full question & answer→Question 1632 Marks
Write the domain of the real function f defined by $\text{f(x)}=\sqrt{25-\text{x}^2}.$
AnswerWe have, $\text{f(x)}=\sqrt{25-\text{x}^2}$ The function is defined only when $25-\text{x}^2\geq0$$\text{x}^2-25\leq0$
$(\text{x}+5)(\text{x}-5)\leq0$
$\text{x}\in[-5,5]$
Therefore, the domain of the given function is [-5, 5].
View full question & answer→Question 1642 Marks
For each binary operation $*$ defined below, determine whether $*$ is commutative or associative.
On $Z^+,$ define $a * b = 2^{ab}$
AnswerFor commutativity: $\mathrm{a}^* \mathrm{~b}=2^{\mathrm{ab}}$ and $b * \mathrm{a}=2^{\mathrm{ba}}=2^{\mathrm{ab}}=\mathrm{a} * \mathrm{~b}$
For associativity: $a^*(b * c)=a * 2^{b c}=(2)$
$ \text { Also, }\left(a^* b\right) * c=\left(2^{a b}\right) * 2=2^{a b} \times c $
$ \therefore a *(b * c) \neq(a * b) * c$
Therefore, the operation * is commutative but not associative.
View full question & answer→Question 1652 Marks
Give an example of a relation which is,
Symmetric and transitive but not reflexive.
AnswerLet R be the relation on A such that,
R = {(1, 2), (2, 1), (1, 3), (3, 1), (2, 3)}
We see that the relation R on A is symmetric and transitive, but not reflexive.
View full question & answer→Question 1662 Marks
Let A = {x ∈ R : -4 ≤ x ≤ 4 and x ≠ 0} and f : A → R be defined by $\text{f(x)}=\frac{|\text{x}|}{\text{x}}.$ Write the range of f.
AnswerWe have, $\text{A}=\{\text{x}\in\text{R}:-4\leq\text{x}\leq4\text{ and x }\neq0\}$
f : A → R defined by $\text{f(x)}=\frac{|\text{x}|}{\text{x}}$
Clearly, $\text{f(x)}=\begin{cases}1;&\text{x}>0\\-1;&\text{x}<0\end{cases}$
$\therefore$ Range of f = {-1, 1}
View full question & answer→Question 1672 Marks
What is the range of the function $\text{f(x)}=\frac{|\text{x}-1|}{\text{x}-1}?$
Answer$\text{f(x)}=\frac{|\text{x}-1|}{\text{x}-1}=\frac{\pm(\text{x}-1)}{\text{x}-1}=\pm1$Range of f = {-1, 1}
View full question & answer→Question 1682 Marks
If $R$ is a symmetric relation on a set $A$, then write a relation between $R$ and $R^{-1}$.
AnswerHere, R is symmetric on the set A.
Let $(\text{a, b})\in\text{R}$
$\Rightarrow\ (\text{b, a})\in\text{R}$ [Since R is symmetric]
$\Rightarrow\ (\text{a, b})\in\text{R}^{-1}$ [By definition of inverse relation]
$\Rightarrow\ \text{R}\subset\text{R}^{-1}$
Let $(\text{x, y})\in\text{R}^{-1}$
$\Rightarrow\ (\text{y, x})\in\text{R}$ [By definition of inverse relation]
$\Rightarrow\ (\text{x, y})\in\text{R}$ [Since R is symmetric]
$\Rightarrow\ \text{R}^{-1}\subset\text{R}$
Thus, $\text{R}=\text{R}^{-1}$
View full question & answer→Question 1692 Marks
Reflexive and symmetric but not transitive.
Answer“is friend of” R = {( x, y) : x is a friend of y}
| It is clear that x is friend of x. |
$\therefore$ |
R is reflexive. |
| Also x is friend of y and y is friend of x. |
$\therefore$ |
R is symmetric. |
| Also if x is friend of y and y is friend of z then |
|
|
| x cannot be friend of z. |
$\therefore$ |
R is not transitive. |
Therefore, R is reflexive and symmetric but not transitive. View full question & answer→Question 1702 Marks
Determine which of the following binary operations are associative and which are commutative:
'*' on N defined by a * b = 1 for all $\text{a, b}\in\text{N}.$
AnswerClearly, by defination a * b = 1 = b * a, $\forall\ \text{a, b}\in\text{N}$
Also, (a * b) * c = (1 * c) = 1
and a * (b * c) = (a * 1) = 1 $\forall\ \text{a, b, c}\in\text{N}$
Hence, N is both associative and commutative.
View full question & answer→Question 1712 Marks
If f : A → A, g : A → A are two bijections, then prove that:
fog is a surjection.
AnswerGiven: A → A, g : A → A are two bijections. Then, fog : A → A Surjectivity of fog: let z be an element in the co-domain of fog (A).Now, $\text{z}\in\text{A}$ (co-domain of f) and f is a surjection.
So, z = f(y), where $\text{y}\in\text{A}$ (domain of f) .....(1)
Now, $\text{y}\in\text{A}$ (co-domain of g) and g is a surjection.
So, y = g(x), where $\text{x}\in\text{A}$ (domain of g) .....(2)
From (1) and (2),
z = f(y) = f(g(x)) = (fog)(x), where $\text{x}\in\text{A}$ (domain of fog)
So, fog is a surjection.
View full question & answer→Question 1722 Marks
Find which of the binary operations are commutative and which are associative.
State whether the following statements are true or false. Justify
If * is a commutative binary operation on N, then a * (b * c) = (c * b) * a
Answer* being a binary operation on N.
$\therefore$ c * b = b * c
$\therefore$ (c * b) * a = (b * c) * a = a * (b * c)
Thus, a * (b * b) = (c * b) * a, therefore, the given statement is true.
View full question & answer→Question 1732 Marks
Let $\text{f}:\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\rightarrow\ \text{R}$ be a function defined by f(x) = cos[x]. write range (f).
Answer$\text{f}:\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\rightarrow\ \text{R}$ given by f(x) = cos[x]$\because\ \cos\text{x}$ in position in $\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)$
$\therefore$ cos[x] will be $\{1, \cos1, \cos2\}$
$\therefore$ Range of $\text{f}=\{1, \cos1, \cos2\}$
View full question & answer→Question 1742 Marks
$f(x) = 8x^3$ and $g(x) =\text{x}^{\frac{1}{3}}.$
Answer$f(x) = 8x^3$ and $g(x) \text{x}^{\frac{1}{3}}$
$gof = g[f(x)] = g[8x^3] (8\text{x}^3)^{\frac{1}{3}}=2\text{x}$
and $fog = f[g(x)] =f\Big[\Big(\text{x}^{\frac{1}{3}}\Big)\Big]=8\Big(\text{x}^{\frac{1}{3}}\Big)^3=8\text{x}$
View full question & answer→Question 1752 Marks
Determine whether the following operations define a binary operation on the given set or not:
'O' on Z defined by $a$ $O$ $b = a^b$ for all $\text{a, b}\in\text{Z.}$
AnswerWe have,
$a\ O\ b = a^b$ for all $\text{a, b}\in\text{Z}$
Let $\text{a}\in\text{Z}$ and $\text{b}\in\text{Z}$
$\Rightarrow\ \text{a}^{\text{b}}\notin\text{Z}\ \Rightarrow\ \text{a O b}\notin\text{Z}$
For example, if $a = 2, b = -2$
$\Rightarrow\ \text{a}^{\text{b}}=2^{-2}=\frac{1}{4}\notin\text{Z}$
$\therefore$ The operation 'O' does not define a binary operation on Z.
View full question & answer→Question 1762 Marks
Determine whether the following operations define a binary operation on the given set or not:
$'\times _6'$ on $S = \{1, 2, 3, 4, 5\}$ defined by, $a \times _6 b =$ Remainder when ab is divided by $6.$
AnswerConsider the composition table,
|
$\times _6$
|
$1$
|
$2$
|
$3$
|
$4$
|
$5$
|
|
$1$
|
$1$
|
$2$
|
$3$
|
$4$
|
$5$
|
|
$2$
|
$2$
|
$4$
|
$0$
|
$2$
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$4$
|
|
$3$
|
$3$
|
$0$
|
$3$
|
$0$
|
$3$
|
|
$4$
|
$4$
|
$2$
|
$0$
|
$4$
|
$2$
|
|
$5$
|
$5$
|
$4$
|
$3$
|
$2$
|
$1$
|
Here all the elements of the table are not in $S.$
For $a =2$ and $b =3$,
$\text{a}\times_6\text{b}= 2 \times_63=$ remainder when $6$ divided by $6=0\neq\text{S}$
Thus, $\times _6$ is not a binary operation on $S.$ View full question & answer→Question 1772 Marks
$F = \{(a, 2), (b, 1), (c, 1)\}.$
Answer$\mathrm{F}: \mathrm{S} \rightarrow \mathrm{T}$ is defined as:
$\mathrm{F}=\{(\mathrm{a}, 2),(\mathrm{b}, 1),(\mathrm{c}, 1)\}$
Since $F(b)=F(c)=1, F$ is not one-one.
Hence, $F$ is not invertible i.e., $\mathrm{F}^{-1}$ does not exist.
View full question & answer→Question 1782 Marks
A = {1, 2, 3, 4, 5, 6, 7, 8} and if R = {(x, y): y is one half of x; x, y ∈ A} is a relation on A, then write R as a set of ordered pairs.
AnswerSince R = {(x, y): y is one half of x; x, y ∈ A}
So, R = {(2, 1), (4, 2), (6, 3), (8, 4)}
View full question & answer→Question 1792 Marks
Let $f, g : R → R$ be defined by $f(x) = 2x + 1$ and $g(x) = x^2 - 2$ for all $x ∈ R,$ respectively. Then, find gof.
AnswerWe have,
$f, g : R → R$ are defined by $f(x) = 2x + 1$ and $g(x) = x^2 - 2$ for all $x ∈ R,$ respectively
Now,
$gof(x) = g(f(x))$
$= g(2x + 1)$
$= (2x + 1)^2 - 2$
$= 4x^2 + 4x + 1 - 2$
$= 4x^2 + 4x - 1$
View full question & answer→Question 1802 Marks
Let *′ be the binary operation on the set {1, 2, 3, 4, 5} defined by a *' b = H.C.F. of a and b. Is the operation *′ same as the operation * defined in Exercise 4 above? Justify your answer.
AnswerLet A = {1, 2, 3, 4, 5} and a *' b = H.C.F. of a and b.
|
*'
|
1
|
2
|
3
|
4
|
5
|
|
1
|
1
|
1
|
1
|
1
|
1
|
|
2
|
1
|
2
|
1
|
2
|
1
|
|
3
|
1
|
1
|
3
|
1
|
1
|
|
4
|
1
|
2
|
1
|
4
|
1
|
|
5
|
1
|
1
|
1
|
1
|
5
|
We observe that the operation *' is the same as the operation * in ex. 4.
View full question & answer→Question 1812 Marks
Determine whether or not the definition of $*$ given below gives a binary operation. In the event that $*$ is not a binary operation give justification of this.
On $Z^+$ define * by $a * b = |a - b|$
Here, $Z^+$ denotes the set of all non-negative integers.
AnswerOn $Z^+, *$ is defined by $a * b = |a - b|.$
It is seen that for each $\text{a, b}\in\text{Z}^{+},$
there is a unique element $|a - b|$ in $Z^+.$
This means that $*$ carries each pair $(a, b)$ to a unique element $a * b = |a - b|$ in $Z^+.$
Therefore, $*$ is a binary operation.
View full question & answer→