Question 512 Marks
If the binary operation o is defined by a o b = a + b - ab on the set Q - {-1} of all rational numbers other than 1, shown that o is commutative on Q - [1].
AnswerLet $\text{a, b}\in\text{Q}-1.$ Then,
a o b = a + b - ab
= b + a - ba
= b o a
Therefore,
a o b = b o a, $\forall\ \text{a, b}\in\text{Q}-1$
Thus, o is commutative on Q - {1}.
View full question & answer→Question 522 Marks
Let * be a binary operation on Q - {-1} defined by a * b = a + b + ab for all a, b ∈ Q - {-1}. Then,
Find the identity element in Q − {−1}.
AnswerWe have,
a * b = a + b + ab for all a, b ∈ Q - {-1}
Let e be identity element with respect to *.
By identity property,
a * e = a = e * a for all a ∈ Q - {-1}
⇒ a + e + ae = a
⇒ e(1 + a) = 0 ⇒ e = 0 $[\because\ 1+\text{a}\neq0\text{ as }\text{a}\neq-1]$
e = 0 is the identity element with respect to *.
View full question & answer→Question 532 Marks
Show that the relation R in the set A of all the books in a library of a college, given by R = {(x, y) : x and y have same number of pages} is an equivalence relation.
Answer
| Books x and x have same number of pages $\Rightarrow(\text{x},\text{x})\in\text{R}$ |
$\therefore$ |
R is reflexive. |
| If $(\text{x},\text{y})\in\text{R}\Rightarrow(\text{y},\text{x})\in\text{R},$so (x, y) = (y, x) |
$\therefore$ |
R is symmetric. |
| Now if $(\text{x},\text{y})\in\text{R},(\text{y},\text{z})\in\text{R}\Rightarrow(\text{x},\text{z})\in\text{R}$ |
$\therefore$ |
R is transitive. |
Therefore, R is an equivalence relation. View full question & answer→Question 542 Marks
Determine whether the following operations define a binary operation on the given set or not:
'*' on N defined by a * b = a + b - 2 for all $\text{a, b}\in\text{N.}$
AnswerIf a = 1 and b = 1, a * b = a + b - 2 = 1 + 1 - 2$=0\notin\text{N}$
Thus, there exist a = 1 and b = 1 such that $\text{a}\ ^*\ \text{b}\notin\text{N}$ So, * is not a binary operation on N.
View full question & answer→Question 552 Marks
Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y) : y is divisible by x}
AnswerR = {( x, y) : y is divisible by x} in A = {1, 2, 3, 4, 5, 6}
Clearly R = (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (5, 5), (6, 6)
| Now (x, x) i.e., (1, 1), (2, 2) and $(3,3)\in\text{R},$ |
$\therefore$ |
R is reflexive. |
| Again $(\text{x},\text{y})\ \text{i.e.},((1,2))\in\text{R}\ \text{but}\ (\text{y},\text{x})\notin\text{R}$ |
$\therefore$ |
R is not symmetric. |
| Also $(1,4)\in\text{R}\ \text{and}\ (4,4)\in\text{R}\ \text{but}(1,4)\in\text{R},$ |
$\therefore$ |
R is transitive. |
Therefore, R is reflexive and transitive but not symmetric. View full question & answer→Question 562 Marks
Which of the following functions from $A$ to $B$ are one-one and onto$?$
$f_2 = \{(2, a), (3, b), (4, c)\}; A = \{2, 3, 4\}, B = \{a, b, c\}$
Answer$f_2 = \{(2, a), (3, b), (4, c)\} A = \{2, 3, 4\}, B = \{a, b, c\}$ It in not clear that different elements of $A$ have different images in $B.$
$\therefore f_2$ in not one-one.
Again, each element of $B$ is the image of some element of $A.$
$\therefore f_2$ in not on to.
View full question & answer→Question 572 Marks
Let $R_0$ denote the set of all non-zero real numbers and let $A = R_0 \times R_0.$ If $'*'$ is a binary operation on adefined by,
$(a, b) * (c, d) = (ac, bd)$ for all $(a, b), (c, d) ∈ A$
Find the identity element in $A$
AnswerLet $(x, y)$ be the identity element in $\text{A}\forall\text{ x, y}\in\text{A}$. Then,
$(a, b) * (x, y) = (a, b) = (x, y) * (a, b)$
Implies that $(a, b) * (x, y) = (a, b)$ and $(x, y) * (a, b) = (a, b)$
Implies that $(ax, by) = (a, b)$ and $(xa, yb) = (a, b)$
Implies that $x = 1$ and $y = 1$
Thus, $(1, 1)$ is the identity element of $A.$
View full question & answer→Question 582 Marks
Let * be a binary operation defined by a * b = 3a + 4b − 2. Find 4 * 5.
AnswerGiven: a * b = 3a + 4b - 2
Here,
4 * 5 = 3(4) + 4(5) - 2
= 12 + 20 - 2
= 30
View full question & answer→Question 592 Marks
Which one of the following graphs represents a function?
-
-
AnswerFigure (a) represents a function f : R → R
Whereas fig (b) does not represent a function.
View full question & answer→Question 602 Marks
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer:
$f: R → R$ defined by $f(x) = 3 – 4x$
Answerf: R → R is defined as f(x) = 3 - 4x. Let $\text{x}_1,\text{x}_2\in\text{R}$ such that $f(x_1) = f(x_2) \Rightarrow 3 - 4x_1 = 3 - 4x_2 \Rightarrow -4x_1 = -4x_2 \Rightarrow x_1 = x_2 $
$\therefore$ f is one-one. For any real number (y) in R, there exists $\frac{3-\text{y}}{4}$ in R such that $f\Big(\frac{3-\text{y}}{4}\Big)=3-4\Big(\frac{3-\text{y}}{4}\Big)=\text{y}.$
$\therefore$ f is onto. Hence, f is bijective.
View full question & answer→Question 612 Marks
Define a commutative binary operation on a set.
AnswerCommutativity: Let S be a non-empty set. A function F: S × S → S is said to be binary operation on S.Mathematically: Let * be a binary operation. It is said to be commutative binary operation if it satisfies commutativity with respect to *.
That is, if $\text{a, b}\in\text{S}$, then
a * b = b * a for all $\text{a, b}\in\text{S}$.
View full question & answer→Question 622 Marks
Find the number of all onto functions from the set A = {1, 2, 3, ..., n} to itself.
AnswerWe know that every onto function from A to itself is one-one.
Therefore, the number of one-one functions = number of bijections =n!
View full question & answer→Question 632 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 $R,$ define by $a * b = ab^2.$
Here, $Z^+$ denotes the set of all non-negative integers.
Answer$\text{a, b}\in\text{R}$Implies that $\text{a, b}^2\in\text{R}$
Implies that $\text{ab}^2\in\text{R}$
Implies that $\text{a}\ ^*\ \text{b}\in\text{R}$
Thus, $*$ is a binary operation on $R$.
View full question & answer→Question 642 Marks
Define an equivalence relation.
AnswerA relation R on a set A is said to be equivalence relation on a if R is:
Reflexive, Symmetric and Transitive.
R = {(x, y): x = y} on the set of real numbers is an equivalence relation.
View full question & answer→Question 652 Marks
Determine whether each of the following relations are reflexive, symmetric and transitive:
Relation R in the set A = {1, 2, 3, ..., 13, 14} defined as R = {(x, y): 3x – y = 0}
AnswerR = {( x, y): 3x − y = 0}, in A = {1, 2, 3, 4, 5, 6, ……13, 14}
Clearly R = {(1, 3), (2, 6), (3, 9), (4, 12)}
| Since, $(\text{x},\text{x})\notin\text{R},$ |
$\therefore$ |
R is not reflexive. |
| Again $(\text{x},\text{y})\in\text{R}\ \text{but}\ (\text{y},\text{x})\notin\text{R}$ |
$\therefore$ |
R is not symmetric. |
| Also $(1,3)\in\text{R}\ \text{and}\ (3,9)\in\text{R}\ \text{but}\notin\text{R},$ |
$\therefore$ |
R is not transitive. |
Therefore, R is neither reflexive, nor symmetric and nor transitive. View full question & answer→Question 662 Marks
If f : R → R be defined by $\text{f(x)} = (3 - \text{x}^3)^\frac{1}{3},$ then find fof(x).
Answerf : R → R defined by $\text{f(x)}=(3-\text{x}^3)^\frac{1}{3}$
$\therefore$ fof(x) = f(f(x))
$=\text{f}(3-\text{x}^3)^\frac{1}{3}$
$=\bigg\{3-\Big[(3-\text{x}^3)^\frac{1}{3}\Big]^3\bigg\}^\frac{1}{3}$
$=\{3-3+\text{x}^3\}^\frac{1}{3}$
$\therefore$ fof(x) = x
View full question & answer→Question 672 Marks
The following defines a relation on N:
$\text{x} +\text{y} = 10,\text{x, y}\in\text{N}$ $$
Determine which of the above relations are reflexive, symmetric and transitive.
AnswerA relation R in A is said to be reflexive if aRa for all $\text{a}\in\text{A},$ R is symmetric if aRb ⇒ bRa, for all $\text{a, b}\in\text{A}$ and it is said to be transitive if aRb and bRc ⇒ aRc for all $\text{a, b, c}\in\text{A.}$$\text{x} +\text{y} = 10,\text{x, y}\in\text{N}$
$(\text{x, y})\in\big\{(1, 9), (9, 1), (2, 8), (8, 2), (3, 7), (7, 3),\\ (4, 6), (6, 4), (5, 5)\big\}$ $$
This is not reflexive as (1, 1), (2, 2), .... are absent.
This only follows the condition of symmetric set as $(1,9)\in\text{R}$ also $(9,1)\in\text{R}$ Similarly other cases are also satisfy the condition.
This is not transitive because {(1, 9), (9, 1)} $\in\text{R}$ but (1, 1) is absent.
View full question & answer→Question 682 Marks
Let * be a binary operation on Z defined by a * b = a + b - 4 for all a, b ∈ Z.
Find the invertible elements in Z.
AnswerLet $\text{a}\in\text{Z}$ and $\text{b}\in\text{Z}$ be the inverse of a. Then,
a * b = e = b * a
a * b = e and b * a = e
a + b - 4 = 4 and b + a - 4 = 4
$\text{b}=8-\text{a}\in\text{Z}$
Thus, 8 - a is the inverse of $\text{a}\in\text{Z.}$
View full question & answer→Question 692 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 $Z^+,$ defined $*$ 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 not a binary operation as the image of $(1, 2)$ under $*$ is $1 * 2 = 1 - 2$
$=-1\notin\text{Z}^{+}$
View full question & answer→Question 702 Marks
Let f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof.
Answerf = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}
Now, f(1) = 2, f(3) = 5, f(4) =1 and g(1) = 3, g(2) = 3, g(5) =1
(gof)(n) = g[f(x)] = g[f(1)] = g(2) = 3
g[f(3)] = g(5) = 1 and g[f(4)] = g(1) = 3
Hence, gof = {(1, 3), (3, 1), (4, 3)}
View full question & answer→Question 712 Marks
Let * be a binary operation on the set Q of rational numbers as follows:
$\text{a} * \text{b} = \frac{\text{ab}}{4}$
Answer$\text{a}*\text{b}=\frac{\text{ab}}{4}=\frac{\text{ba}}{4}=\text{b}*\text{a}$
$\therefore$ operation * is commutative.
$(\text{a}*\text{b})*\text{c}=\frac{\text{ab}}{4}*\text{c}=\frac{\frac{\text{ab}}{4}\text{c}}{4}=\frac{\text{abc}}{16}$
And $\text{a}*(\text{b}*\text{c})=\text{a}*\frac{\text{bc}}{4}=\frac{\text{a}\frac{\text{bc}}{4}}{4}=\frac{\text{abc}}{16}$
Here, $(\text{a}*\text{b})*\text{c}=\text{a}*(\text{b}*\text{c})$
$\therefore$ operation * is associative.
View full question & answer→Question 722 Marks
If the binary operation * on the set Z is defined by a * b = a + b - 5, the find the identity element with respect to *.
AnswerLet e be the identity element in Z with respect to * such that,
a * e = a = e * a, $\forall\ \text{a}\in\text{Z}$
a * e = a and e * a = a, $\forall\ \text{a}\in\text{Z}$
a + e - 5 = a and e + a - 5 = a, $\forall\ \text{a}\in\text{Z}$
e = 5, $\forall\ \text{a}\in\text{Z}$
Thus, 5 is the identity element in Z with respect to *.
View full question & answer→Question 732 Marks
If A = {3, 5, 7} and B = {2, 4, 9} and R is a relation given by "is less than", write R as a set ordered pairs.
AnswerSince, R = x, y: x, y $\in\text{N}$ and x < y,
Hence, R = {(3, 4), (3, 9), (5, 9), (7, 9)}
View full question & answer→Question 742 Marks
Give an example of a relation which is,Reflexive and transitive but not symmetric.
AnswerLet R be the relation on A such that
R = {(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 3)}
Clearly, the relation R on A is reflexive and transitive, but not symmetric.
View full question & answer→Question 752 Marks
Let $f : R → R$ and $g : R → R$ be defined by $f(x) = x^2$ and $g(x) = x + 1$. Show that fog $≠$ gof.
AnswerGiven, $f : R → R$ and $g : R → R.$
So, the domains of f and g are the same.
$(fog)(x) = f(g(x)) = f(x + 1) = (x + 1)^2 = x^2 + 1 + 2x$
$(gof)(x) = g(f(x)) = g(x^2) = x^2 + 1$
So, $fog ≠ gof.$
View full question & answer→Question 762 Marks
Let A = {a, b, c} and the relation R be defined on A as follows: R = {(a, a), (b, c), (a, b)}. Then, write minimum number of ordered pairs to be added in R to make it reflexive and transitive.
AnswerWe have,
A = {a, b, c} and R = {(a, a), (b, c), (a, b)}.
R can be a reflexive relation only when elements (b, b) and (c, c) are added to it.
R can be a transitive relation only when the element (a, c) is added to it.
So, the minimum number of ordered pairs to be added in R is 3.
View full question & answer→Question 772 Marks
If f, g : R → R be two functions defined as f(x) = |x| + x and g(x) = |x| – x, $\forall\ \text{x}\in\text{R}.$ Then find fog and gof. Hence find fog(-3), fog(5) and gof(-2).
Answer$\text{fog(x)}=\begin{cases}0,\ \text{x}\geq0\\-4\text{x},\ \text{x}<0\end{cases}$gof(x) = 0, for all x fog(-3) = 12
fog(5) = 0
gof(-2) = 0
View full question & answer→Question 782 Marks
Let $\text{f}:\text{R}-\Big\{-\frac{3}{5}\Big\}\rightarrow\ \text{R}$ be a function defined as $\text{f(x)}=\frac{2\text{x}}{5\text{x}+3}.$ Write $f^{-1}$: Range of $\text{f}\rightarrow\ \text{R}-\Big\{-\frac{3}{5}\Big\}.$
AnswerLet $f^{-1}(x) = y$ ......(1) ⇒ f(y) = x$\Rightarrow\ \frac{2\text{y}}{5\text{y}+3}=3\text{x}$
$⇒ 2y = 5xy + 3x ⇒ 2y - 5xy = 3x ⇒ y(2 - 5x) = 3x \Rightarrow\ \text{y}=\frac{3\text{x}}{2-5\text{x}}$ $\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{3\text{x}}{2-5\text{x}}$ [from 1]
View full question & answer→Question 792 Marks
Define a reflexive relation.
AnswerA relation R on a set A is said to be reflexive if every element of A is related to itself.
Mathematically, reflexive relation is written as R = {(a, a): for all $\text{a}\in\text{A}$}
For example if A = {1, 2, 3}, then a reflexive relation on A will be R = {(1, 1), (2, 2), (3, 3)}
View full question & answer→Question 802 Marks
If A = {1, 2, 3, 4} define relations on A which have properties of being:
Reflexive, symmetric and transitive.
AnswerThe relation on A having properties of being symmetric, reflexive and transitive is,
R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1)}
The relation R is an equivalence relation on A.
View full question & answer→Question 812 Marks
Let $R_0$ denote the set of all non-zero real numbers and let $A = R_0 \times R_0$. If $'*'$ is a binary operation on adefined by,
$(a, b) * (c, d) = (ac, bd)$ for all $(a, b), (c, d) ∈ A$
Find the invertible element in $A.$
AnswerLet $(m, n)$ be the inverse of $(\text{a, b})\forall(\text{a, b})\in\text{A}$. Then,
$(a, b) * (m, n) = (1, 1)$
Implies that $(am, bn) = (1, 1)$
Implies that $am = 1\ \&\ bn = 1$
Implies that $\text{m}=\frac{1}{\text{a}}\text{ and }\text{n}=\frac{1}{\text{b}}$
Thus, $\Big(\frac{1}{\text{a}},\frac{1}{\text{b}}\Big)$ is the inverse of $(\text{a, b})\forall(\text{a, b})\in\text{A}$.
View full question & answer→Question 822 Marks
If $f : C → C$ is defined by $f(x) = (x - 2)^3$, write $f^{-1}(-1)$.
AnswerLet $f^{-1}(1) = x$ ......(1)
$\Rightarrow f(x) = -1$
$\Rightarrow (x - 2)^3 = -1$
$\Rightarrow\ \text{x}-2=-1\ \text{or }-\omega\text{ or }-\omega^2$
as the roots of $(-1)^\frac{1}{3}$ are $-1,-\omega\text{ and }-\omega^2,$ where $\omega=\frac{1\pm\text{i}\sqrt{3}}{2}$
$\Rightarrow\ \text{x}=-1+2\text{ or }2-\omega\text{ or }2-\omega^2=1,2-\omega,2-\omega$
$\Rightarrow\ \text{f}^{-1}(-1)=1,2-\omega,2-\omega^2$ [from 1]
View full question & answer→Question 832 Marks
Let * be the binary operation on N defined by a * b = H.C.F. of a and b. Is * commutative? Is * associative? Does there exist identity for this binary operation on N?
Answera * b = H.C.F. of a and b.
- a * b = H.C.F. of a and b = H.C.F. of b and a = b * a
Therefore, operation * is commutative.
- (a * b) * c = (H.C.F. of a and b) * c = H.C.F. of (H.C.F. of a and b) and c
= H.C.F. of a, b and c = a * (b * c)
Therefore, the operation is associative.
$1*\text{a}=\text{a}*1\neq\text{a}.$ View full question & answer→Question 842 Marks
Let $f: X → Y$ be an invertible function. Show that f has unique inverse. $($Hint: suppose $g_1$ and $g_2$ are two inverses of f. Then for all $y \in Y, fog_1(y) = 1_Y(y) = fog_2(y).$ Use one-one ness of $f).$
AnswerGiven: $f: X → Y$ be an invertible function.
Thus f is $1 – 1$ and onto and therefore $f^{−1}$ exists.
Let $g_1$ and $g_2$ be two inverses of f . Then for all $\text{y}\in\text{Y},$
$fog_1(y) = I_y(y) = fog_2(y) \therefore fog_1(y) = fog_2(y)$
$\Rightarrow f[g_1(y)] = f[g_{2(y)}] \Rightarrow g_1(y) = g_2(y)$
$\therefore$ The inverse is unique and hence f has a unique inverse.
View full question & answer→Question 852 Marks
Determine whether the following operations define a binary operation on the given set or not:$'\odot'$ on N defined by $\text{a}\odot\text{b}=\text{a}^{\text{b}}+\text{b}^{\text{a}}$ for all $\text{a, b}\in\text{N.}$
AnswerLet $\text{a, b}\in\text{N.}$ Then,
$\text{a}^{\text{b}},\text{b}^{\text{a}}\in\text{N}$
$\Rightarrow\ \text{a}^{\text{b}}+\text{b}^{\text{a}}\in\text{N}$ $\big[\because$ Addition is binary operation on N$\big]$
$\Rightarrow\ \text{a}\odot\text{b}\in\text{N}$
Thus, $\odot$ is a binary operation on N.
View full question & answer→Question 862 Marks
Let '*' be a binary operation on N defined by a * b = 1.c.m. (a, b) for all $\text{a, b}\in\text{N}$
Find 2 * 4, 3 * 5, 1 * 6.
Answera * b = 1.c.m. (a, b)
2 * 4 = 1.c.m. (2, 4)
= 4
3 * 5 = 1.c.m. (3, 5)
= 15
1 * 6 = 1.c.m. (1, 6)
= 6
View full question & answer→Question 872 Marks
Write the identity element for the binary operation * on the set $R_0$ of all non-zero real numbers by the rule $\text{a}\times\text{b}=\frac{\text{ab}}{2}$ for all $a, b \in R_0$.
Answer$\because\ \text{a}\times\text{b}=\frac{\text{ab}}{2}$ for all $a, b \in R_0$ Let e be the identity element, then
a * e = a
$\Rightarrow\frac{\text{ae}}{2}=\text{a}\ \Rightarrow\text{e}=2$
Thus, e = 2 is the identity element with respect to *.
View full question & answer→Question 882 Marks
Let C denote the set of all complex numbers. A function $f : C → C$ is defined by $f(x) = x^3$. Write $f^{-1}(1)$.
Answer$f : R → R$ defined by $f(x) = x^3$
$\therefore f^{-1}(x^3) = x$
$\Rightarrow\ \text{f}^{-1}(1)=\{1,\omega,\omega^2\}$ $[\because\ \sqrt[3]{1}=\{1,\omega,\omega^2\}]$
View full question & answer→Question 892 Marks
Find gof and fog when $f : R → R$ and $g : R → R$ are defined by:
$f(x) = 2x + x^2$ and $g(x) = x^3$
AnswerGiven: $f : R → R$ and $g : R → R$
Therefore, $gof : R → R$ and $fog : R → R$
$f(x) = 2x + x^2$ and $g(x) = x^3$
$gof(x) = g(f(x)) = g(2x + x^2)$
$gof(x) = g(2x + x^2)^3$
$fog(x) = f(g(x)) = f(x^3)$
$\therefore fog(x) = 2x^3 + x^6$
View full question & answer→Question 902 Marks
Consider $f: R_+→ [4, \infty )$ given by $f(x) = x^2 + 4$. Show that f is invertible with the inverse $f^{–1}$ of f given by $f^{-1}(\text{y})=\sqrt{\text{y}-4},$ where $R_+$ is the set of all non-negative real numbers.
AnswerConsider $f:\text{R}_{+}\rightarrow[4,\infty]$ and $f(x) = x^2 + 4$.
Let $\text{x}_1,\text{x}_2\in\text{R}\rightarrow[4,\infty],\text{ then }f(\text{x}_1)=\text{x}_{1}^{2}+4\text{ and }f(\text{x}_2)=\text{x}_{2}^{2}+4$
$\Rightarrow\ \text{x}_{1}^{2}+4=\text{x}_{2}^{2}+4\Rightarrow\text{x}_1=\text{x}_2\ \ \ \ \ \therefore f\text{ is one-one.}$
Now $\text{y}=\text{x}^2+4\Rightarrow\text{x}=\sqrt{\text{y}-4}$ as x > 0
$\therefore\ \ f\left(\sqrt{\text{y}-4}\right)=\left(\sqrt{\text{y}-4}\right)^2+4=\text{y}\ \ f(\text{x})=\text{y}\ \ \ \ \therefore\ f\text{ is onto.}$
Therefore, f(x) is invertible and $f^{-1}(\text{y})=\sqrt{\text{y}-4}.$
View full question & answer→Question 912 Marks
If the mappings f and g are given by f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)}, write fog.
AnswerWe are given that, f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)} Now, the domain of g is {2, 5, 1} We know that, fog(x) = f{g(x)}$\therefore$ fog(2) = f{g(2)} = f(3) = 5
$\therefore$ fog(5) = f{g(5)} = f(1) = 2
$\therefore$ fog(1) = f{g(1)} = f(3) = 5
Therefore, fog = {(2, 5), (5, 2), (1, 5)}
View full question & answer→Question 922 Marks
Write the identity element for the binary operation * defined on the set R of all real numbers by the rule $\text{a}\times\text{b}=\frac{3\text{ab}}{7}\ \forall\text{ a, b}\in\text{R}$.
AnswerWe have,
$\text{a}\times\text{b}=\frac{3\text{ab}}{7}$
Let e be the identity element with respect to *. Then
a * e = a
$\Rightarrow\frac{3\text{ae}}{7}=\text{a}\ \Rightarrow\text{e}=\frac{7}{3}$
View full question & answer→Question 932 Marks
Let $f : R - {-1} → R - {1}$ be given by $\text{f(x)}=\frac{\text{x}}{\text{x}+1}.$ Write $f^{-1}(x)$.
Answer$f : R - [-1] → R - [1]$ given by $\text{f(x)}=\frac{\text{x}}{\text{x}+1}$
$\Rightarrow\ \text{f}^{-1}\Big(\frac{\text{x}}{\text{x}+1}\Big)=\text{x}$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{\text{x}}{1-\text{x}}$
$\because$ Let $\frac{\text{x}}{\text{x}+1}=\text{y}$
$\Rightarrow\ \text{x}=\text{xy}+\text{y}$
$\Rightarrow\ \text{x}(1-\text{y})=\text{y}$
$\Rightarrow\ \text{x}=\frac{\text{y}}{1-\text{y}}$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{\text{x}+7}{10}$
View full question & answer→Question 942 Marks
Let * be a binary operation on the set Q of rational numbers as follows:
a * b = a – b
Answera * b = a - b = -(b - a) = -b * a
$\therefore$ operation is not commutative.
(a * b) * c = (a - b) * c = (a - b) - c = a - b - c
And a * b (b * c) = a * (b - c) = a - (b - c) = a - b + c
Here, $(\text{a}*\text{b})*\text{c}\neq\text{a}*(\text{b}*\text{c})$
$\therefore$ operation * is not associative.
View full question & answer→Question 952 Marks
For each binary operation * defined below, determine whether * is commutative or associative.
On Q, define $\text{a} * \text{b} =\frac{\text{ab}}{2}$
AnswerFor commutativity: $\text{a}*\text{b}=\frac{\text{ab}}{2}\ \text{and}\ \text{b}*\text{a}=\frac{\text{ba}}{2}=\frac{\text{ab}}{2}=\text{a}*\text{b}$
For associativity: $\text{a}*(\text{b}*\text{c})=\text{a}*\Big(\frac{\text{bc}}{2}\Big)=\frac{\text{abc/2}}{2}=\frac{\text{abc}}{4}$
Also, $(\text{a}*\text{b})*\text{c}=\Big(\frac{\text{ab}}{2}\Big)*\text{c}=\frac{\text{abc}/2}{2}=\frac{\text{abc}}{4}$
$\therefore$ a * (b * c) = (a * b) * c
Therefore, the operation * is commutative and associative.
View full question & answer→Question 962 Marks
If $f : R → R$ is defined by $f(x) = x^2 - 3x + 2$, write f{f(x)}.
AnswerWe have, $f(x)=x^2-3 x+2$
$\therefore f\{f(x)\}=f\left(x^2-3 x+2\right)$
$=\left(x^2-3 x+2\right)^2-3\left(x^2-3 x+2\right)+2$
$=x^4+9 x^2+4-6 x^3-12 x+4 x^2-3 x^2+9 x-6+2$
$=x^4-6 x^3+10 x^2-3 x$
$\therefore f\{f(x)\}=x^4-6 x^3+10 x^2-3 x$
View full question & answer→Question 972 Marks
Let S = {a, b, c}. Find the total number of binary operations on S.
AnswerNumber of binary operations on a set with n elements is $n^2$.
Here, S = {a, b, c}
Number of elements in S = 3
Number of binary operations on a set with 3 elements is $3^{3^{2}}=3^9$
View full question & answer→Question 982 Marks
Let $f : R → R^+$ be defined by $f(x) = ax, a > 0$ and $\text{a}\neq1.$ Write $f^{-1}(x).$
AnswerLet $f^{-1}(x) = y .......(1)$
$\Rightarrow f(y) = x$
$\Rightarrow a^y = x$
$\Rightarrow\ \text{y}=\log_\text{a}\text{x}$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\log_\text{a}\text{x} [$from $(1)]$
View full question & answer→Question 992 Marks
Find which of the binary operations are commutative and which are associative.
Show that none of the operations given above has identity.
AnswerLet the identity be I.
- $\text{a}*\text{I}=\text{a - I}\neq\text{a}$
- $\text{a}*\text{I}=\text{a}^2-\text{I}^2\neq\text{a}$
- $\text{a}*\text{I}=\text{a + aI}\neq\text{a}$
- $\text{a}*\text{I}=(\text{a - I})^2\neq\text{a}$
- $\text{a}*\text{I}=\frac{\text{aI}}{4}\neq\text{a}$
- $\text{a}*\text{I}=\text{aI}^2\neq\text{a}$
Therefore, none of the operations given above has identity. View full question & answer→Question 1002 Marks
$f: N → N$ given by $f(x) = x^3$
Answer$f: R → R$ is given by,$f(x) = x^3$
It is seen that for $\text{x},\text{y}\in\text{N},$
$f(x) = f(y) \Rightarrow x^3 = y^3 \Rightarrow x = y.$
$\therefore$ f is injective.
Now, $2\in\text{N}.$
But, there does not exist any element $x$ in domain $N$ such that $f(x) = x^3 = 2.$
$\therefore f$ is not surjective.
Hence, function f injective but not surjective.
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