Question 513 Marks
Find fog and gof if:f(x) = x + 1, g(x) = sinx
Answerf(x) = x + 1, g(x) = sinxRange of f = R $\subset$ Domain of g = R ⇒ gof exists
Range of g = [-1, 1] $\subset$ Domain of f = R ⇒ fog exists
Now,
fog(x) = f(g(x)) = f(sinx) = sinx + 1
And
gof(x) = g(f(x)) = g(x + 1) = sin(x + 1)
View full question & answer→Question 523 Marks
Find fog and gof if: $f(x) = x^2, g(x) = \cos x$
Answer$f(x) = x^2, g(x) = \cos \ x$
Domain of $f$ and Domain of $g = R$
Range of $\text{f}=(0,\infty)$
Range of $g = (-1, 1)$
$\therefore$ Range of $f \subset$ domain of $g \Rightarrow \text{gof}$ exist
Range of $g \subset$ domain of $f \Rightarrow \text{fog}$ exist
Now,
$\text{gof}(x) = g(f(x)) = g(x^2) = \cos \ x^2$
And
$\text{fog}(x) = f(f(x)) = f(cosx) = \cos^2x$
View full question & answer→Question 533 Marks
Let f : $R \rightarrow R$ be defined as $\text{f(x)}=\frac{2\text{x}-3}{4}.$ Write $fof^{-1}(1).$
AnswerLet $f : R \rightarrow R$, defined by $\text{f(x)}=\frac{2\text{x}-3}{4}$
$\Rightarrow\ \text{f}^{-1}\frac{(2\text{x}-3)}{4}=\text{x}$
$\Rightarrow\ \text{f}^{-1}(2\text{x})=4\text{x}+3$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{4\text{x}+3}{2}$
Now, $fof^{-1}(x) = f(f^{-1}(x))$
$=\text{f}\Big(\frac{4\text{x}+3}{2}\Big)$
$=\frac{2\big(\frac{4\text{x}+3}{2}\big)-3}{4}$
$\Rightarrow fof^{-1}(x) = x$
$\therefore fof^{-1}(1) = 1$
View full question & answer→Question 543 Marks
Find fog and gof if:$f(x) = \sin^{-1}x, g(x) = x^2$
Answer$f(x) = \sin^{-1}x, g(x) = x^2 \text{f}:[-1,1]\rightarrow\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big];\ \text{g}:\text{R}\rightarrow[0,\infty)$
Computing fog: Clearly, the range of $g$ is not a subset of the domain of $f.$
Domain $(fog) = x: \text{x}\in$ domain of $g$ and $\text{g(x)}\in$ domain of $f$
Domain $(fog) = x: \text{x}\in\text{R}$ and $\text{x}^2\in[-1,1]$
Domain $(fog) = x: \text{x}\in\text{R}$ and $\text{x}\in[-1,1]$
Domain of $(fog) = [-1, 1]$
$fog : [-1, 1] \rightarrow R$
$(fog)(x) = f(g(x))$
$= f(x^2)$
$= \sin^{-1}(x^2)$
Computing gof: Clearly, the range of f is a subset of the domain of g.
fog : $[-1, 1] \rightarrow R$
$(gof)(x) = g(f(x))$
$= g(\sin^{-1}x)$
$= (\sin^{-1}x)^2$
View full question & answer→Question 553 Marks
Suppose $f_1$ and $f_2$ are non$-$zero one-one functions from $R$ to $R$. Is $\frac{\text{f}_1}{\text{f}_2}$ necessarily one$-$one? Justify your answer. Here, $\frac{\text{f}_1}{\text{f}_2}:\text{R}\rightarrow\ \text{R}$ is given by $\Big(\frac{\text{f}_1}{\text{f}_2}\Big)(\text{x})=\frac{\text{f}_1(\text{x})}{\text{f}_2(\text{x})}$ for all $\text{x}\in\text{R}.$
AnswerWe know that $f_1 : R \rightarrow R,$ given by $f_1(x) = x^3$ and $f_2(x) = x$ are one$-$one.
Injectivity of $f_1$ : Let $x$ and $y$ be two elements in the domain $R,$ such that
$f_1(x) = f_2(y)$
$\Rightarrow x^3 = y$
$\Rightarrow\ \text{x}=\sqrt[3]{\text{y}}\in\text{R}$
So$, f_{1}$ is one$-$one.
Injectivity of $f_2$ :
Let $x$ and $y$ be two elements in the domain $R,$ such that
$f_2(x) = f_2(y)$
$\Rightarrow x = y$
$\text{x} \in \text{R}$
So$, f_2$ is one$-$one.
Proving $\frac{\text{f}_1}{\text{f}_2}$ is not one-one.
Given that $\frac{\text{f}_1}{\text{f}_2}(\text{x})=\frac{\text{f}_1(\text{x})}{\text{f}_2(\text{x})}=\frac{\text{x}^3}{\text{x}}=\text{x}^2$
Let $x$ and $y$ be two elements in the domain $R,$ such that
$\frac{\text{f}_1}{\text{f}_2}(\text{x})=\frac{\text{f}_1}{\text{f}_2}(\text{y})$
$\Rightarrow\ \text{x}^2=\text{y}^2$
$\Rightarrow\ \text{x}=\pm\text{y}$
So, $\frac{\text{f}_1}{\text{f}_2}$ is not one$-$one.
View full question & answer→Question 563 Marks
If A = {1, 2, 3, 4} define relations on A which have properties of being:
Reflexive, transitive but not symmetric.
AnswerThe relation on A having properties of being reflexive, transitive, but not symmetric is, R = {(1, 1), (2, 2), (3, 3), (4, 4), (2, 1)} Relation R satisfies reflexivity and transitivity. $ \Rightarrow(1, 1), (2, 2), (3, 3) \in\text{R}$$$and $(1, 1), (2, 1) \in \text{R}\Rightarrow(1, 1)\in \text{R}$
However, $(2,1)\in\text{R},$ but $(1,2)\notin\text{R}$
View full question & answer→Question 573 Marks
Show that f : R → R, given by f(x) = x - [x], is neither one-one nor onto.
Answerf : R → R, given by f(x) = x - [x]Injectivity: f(x) = 0 for all $\text{x}\in\text{Z}$
Therefore, f is not one-one.Surjectivity: Range of f = (0, 1) ≠ R.
Co-domain of f = R Both are not same. Therefore, f is not onto.
View full question & answer→Question 583 Marks
Let $A = \{1, 2, 3\},$ and let $R_1 = \{(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)\}.$ Find whether or not the relations $R_{1 }$ on $A$ is:
- Reflexive.
- Symmetric.
- Transitive.
AnswerWe have $A = \{1, 2, 3\},$ and $R_1 = \{(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)\}$
$\therefore (1, 1), (2, 2)$ and $(3, 3) \in\text{R}_1$
$\therefore R_1$ is not reflexive.
Now,
$\therefore\ (2,1)\in\text{R}_1$ but $(1,2)\notin\text{R}_1$
$\therefore R_1$ is not symmetric.
Again,
$\therefore\ (2,1)\in\text{R}_1$ and $(1,3)\in\text{R}_1$ but $(2,3)\notin\text{R}_1$
$\therefore R_1$ is not transitive.
View full question & answer→Question 593 Marks
Construct the composition table for $\times _5$ on $Z_5 = \{0, 1, 2, 3, 4\}.$
AnswerHere,
$1\times _51 =$ Remainder obtained by dividing $1 \times 1$ by $5 = 1$
$3\times _54 =$ Remainder obtained by dividing $3 \times 4$ by $5 = 2$
$4\times _54 =$ Remainder obtained by dividing $4 \times 4$ by $5 = 1$
Therefore,
The composition table is as follows:
| $\times _5$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
| $0$ |
$0$ |
$0$ |
$0$ |
$0$ |
$0$ |
| $1$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
| $2$ |
$0$ |
$2$ |
$4$ |
$1$ |
$3$ |
| $3$ |
$0$ |
$3$ |
$1$ |
$4$ |
$2$ |
| $4$ |
$0$ |
$4$ |
$3$ |
$2$ |
$1$ |
View full question & answer→Question 603 Marks
For the binary operation $\times _7$ on the set $S = \{1, 2, 3, 4, 5, 6\},$ compute $3^{−1} \times _7 4.$
AnswerFinding identity element:
Here,
$1\times _71 =$ Remainder obtained by dividing $1 \times 1$ by $7 = 1$
$3\times _74 =$ Remainder obtained by dividing $3 \times 4$ by $7 = 5$
$4\times _75 =$ Remainder obtained by dividing $4 \times 5$ by $7 = 6$
So, the composition table is as follows:
| $\times _7$ |
$1$ |
$2$ |
$3$ |
$4$ |
$5$ |
$6$ |
| $1$ |
$1$ |
$2$ |
$3$ |
$4$ |
$5$ |
$6$ |
| $2$ |
$2$ |
$4$ |
$6$ |
$1$ |
$3$ |
$5$ |
| $3$ |
$3$ |
$6$ |
$2$ |
$5$ |
$1$ |
$4$ |
| $4$ |
$4$ |
$1$ |
$5$ |
$2$ |
$6$ |
$3$ |
| $5$ |
$5$ |
$3$ |
$1$ |
$6$ |
$4$ |
$2$ |
| $6$ |
$6$ |
$5$ |
$4$ |
$3$ |
$2$ |
$1$ |
We observe that all the elements of the first row of the composition table are same as the top$-$most row.
So, the identity element is $1.$
Also, $3\times _{7 }5 = 1$
So, $3^{-1} = 5$
Now,
$3^{-1}\times _7 4 $
$= 5\times _7 4 $
$= 6$ View full question & answer→Question 613 Marks
The binary operation * is defined by $\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{7}$ on the set Q of all rational numbers. Show that * is associative.
AnswerThe binary operator * is defined as,
$\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{7}$ for all $\text{a, b}\in\text{Q}$
Now,
Associativity: Let $\text{a, b, c}\in\text{Q},$ then
$(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}=\frac{\text{ab}}{7}\ ^*\ \text{c}=\frac{\text{abc}}{49}\ ....(\text{i})$
and $\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=\text{a}\ ^*\ \frac{\text{bc}}{7}=\frac{\text{abc}}{49}\ .....(\text{ii})$
From (i) and (ii)
(a * b) * c = a * (b * c)
⇒ '*' is associative on Q.
View full question & answer→Question 623 Marks
For the binary operation multiplication modulo $10 (\times _{10})$ defined on the set $S = \{1, 3, 7, 9\},$ write the inverse of $3.$
Answer$1 \times _{10}1 =$ Remainder obtained by dividing $1 \times 1$ by $10 = 1$
$3 \times _{10}1 =$ Remainder obtained by dividing $3 \times 1$ by $10 = 3$
$7 \times _{10}3 =$ Remainder obtained by dividing $7 \times 3$ by $10 = 1$
$3 \times _{10}3 =$ Remainder obtained by dividing $3 \times 3$ by $10 = 9$
So, the composition table is as follows:
| $\times _{10}$ |
$1$ |
$3$ |
$7$ |
$9$ |
| $1$ |
$1$ |
$3$ |
$7$ |
$9$ |
| $3$ |
$3$ |
$9$ |
$1$ |
$7$ |
| $7$ |
$7$ |
$1$ |
$9$ |
$3$ |
| $9$ |
$9$ |
$7$ |
$3$ |
$1$ |
We observe that the first row of the composition table coincides with the top$-$most row and the first column coincides with the left$-$most column.
These two intersect at $1.$
$\Rightarrow a ^* 1 = 1 ^* a = a, \forall\text{ a}\in\text{S}$
So, the identity element is $1.$
Also,
$3 \times _{10}7 = 1$
$3^{-1} = 7$ View full question & answer→Question 633 Marks
Let $*$ be a binary operation on $Q_0 $(set of non $-$ zero rational numbers$)$ defined by $\text{a}\ ^* \ \text{b}=\frac{\text{ab}}{5}$ for all $\text{a, b}\in\text{Q}_0.$ Show that $*$ is commutative as well as associative. Also, find its identity element if it exists.
AnswerCommutativity: Let $\text{a, b}\in\text{Q}_0$
$\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{5}$
$=\frac{\text{ba}}{5}$
$=\text{b}\ ^*\ \text{a}$
Therefore, $\text{a}\ ^*\ \text{b}=\text{b}\ ^*\ \text{a},\forall\ \text{a, b}\in\text{Q}_0$
Thus $, *$ is commutative on $Q_0.$
Associativity: Let $\text{a, b, c}\in\text{Q}_0$
$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=\text{a}\ ^*\ \Big(\frac{\text{bc}}{5}\Big)$
$=\frac{\text{a}\big(\frac{\text{bc}}{5}\big)}{5}$
$=\frac{\text{abc}}{25}$
$(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}=\Big(\frac{\text{ab}}{5}\Big)\ ^*\ \text{c}$
$=\frac{\big(\frac{\text{ab}}{5}\big)\text{c}}{5}$
$=\frac{\text{abc}}{25}$
Therefore,
$a * (b * c) = (a * b) * c, $ for all $ \text{a, b, c}\in\text{Q}_0$
Thus $,*$ is associative on $Q_0$.
Finding identity element:
Let e be the identity element in $Z$ with respect to $*$ such that,
$a * e = a = e * a,$ for all $\text{a}\in\text{Q}_0$
$a * e = a$ and $e * a = a,$ for all $\text{a}\in\text{Q}_0$
Implies that $\frac{\text{ae}}{5}=\text{a}$ and $\frac{\text{ea}}{5}=\text{a},\forall\ \text{a}\in\text{Q}_0$
Implies that $\text{e}=5,\forall\ \text{a}\in\text{Q}_0\ [\because\ \text{a}\neq0]$
Thus $, 5$ is the identity element in with respect to $*$.
View full question & answer→Question 643 Marks
Given a non $-$ empty set $X,$ let $*: P(X) \times P(X) \rightarrow P(X)$ be defined as $\text{A}*\text{B} =(\text{A – B})\cup(\text{B – A}),\forall\text{A},\text{B}\in\text{P(X)}. $Show that the empty set $\phi$ is the identity for the operation $*$ and all the elements $A$ of $P(X)$ are invertible with $A^{–1} = A.$
$($Hint: $(\text{A}-\phi)\cup(\phi-\text{A})=\text{A} $ and $(\text{A}-\text{A})\cup(\text{A}-\text{A})=\text{A}*\text{A}=\phi ).$
AnswerIt is given that $*: P(X) \times P(X) \rightarrow P(X)$ is defined as $\text{A}*\text{B} =(\text{A – B})\cup(\text{B – A}),\forall\text{A},\text{B}\in\text{P(X)}.$
Let $\text{A}\in\text{P(X)}.$
Then, we have : $\text{A}*\phi=(\text{A}-\phi)\cup(\phi-\text{A})=\text{A}\cup\phi=\text{A}$
$\phi*\text{A}=(\phi-\text{A})\cup(\text{A}-\phi)=\phi\cup\text{A}=\text{A}$
$\therefore\text{A}*\phi=\text{A}=\phi*\text{A}.\forall\text{A}\in\text{P(X)}$ Thus, $\phi$ is the identity element for the given operation $*$.
Now, an element $\text{A}\in\text{P(X)}$ will be invertible if there exists $\text{B}\in\text{P(X)}$ such that $\text{A}*\text{B}=\phi=\text{B}*\text{A}.\ $
As $(\phi)$ is the identity element
Now, we observed that $\text{A}*\text{A}=(\text{A}-\text{A})\cup(\text{A}-\text{A})=\phi\cup\phi=\phi\ \forall\text{A}\in\text{P(X)}$
Hence, all the elements $A$ of $P(X)$ are invertible with $A^{-1} = A.$
View full question & answer→Question 653 Marks
Find the identity element in the set of all rational numbers except -1 with respect to * defined by a * b = a+ b + ab.
AnswerLet R - {-1} be the set and * be a binary operator, given by a * b = a + b + ab for all $\text{a, b}\in\text{R}-\{-1\}$ Now, Let $\text{a}\in\text{R}-\{-1\}$ and $\text{e}\in\text{R}-\{-1\}$ be the identity element with respect to *. by identity property, we have, a * e = e * a = a ⇒ a + e + ae = a ⇒ e(1 + a) = 0 ⇒ e = 0 $[\because\ 1+\text{a}\neq0\text{ as a }\neq-1]$$\therefore$ The required identity element is 0.
View full question & answer→Question 663 Marks
Let A = [-1, 1]. Then, discuss whether the following functions from A to itself are one-one, onto or bijective:
g(x) = |x|
Answerg(x) = |x|Injection test: Let x and y be any two elements in the domain (A), such that f(x) = f(y).
f(x) = f(y) |x| = |y| $\text{x}=\pm\text{y}$ So, f is not one-one. Surjection test: For y = -1, there is no value of x in A. So, f is not onto. So, f is not bijective.
View full question & answer→Question 673 Marks
Prove that the Greatest Integer Function f: R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
Answerf: R → R is given by,f(x) = [x]
It is seen that f(1.2) = [1.2] = 1, f(1.9) = [1.9] = 1. $\therefore$ f(1.2) = f(1.9), but $1.2\neq1.9$ $\therefore$ f is not one-one. Now, consider $0.7\in\text{R}.$
It is known that f(x) = [x] is always an integer. Thus, there does not exist any element $\text{x}\in\text{R}$ such that f(x) = 0.7.
$\therefore$ f is not onto. Hence, the greatest integer function is neither one-one nor onto.
View full question & answer→Question 683 Marks
Show that the function $f: R \rightarrow R$ given by $f(x) = x^3$ is injective.
Answer$f: R \rightarrow R$ is given as $f(x) = x^3$
Suppose $f(x) = f(y),$ where $\text{x},\text{y}\in\text{R}.$
$\Rightarrow x^3 = y^{3 }...(1)$
Now, we need to show that $x = y.$
Suppose $\text{x}\neq\text{y},$ their cubes will also not be equal.
$\Rightarrow\text{x}^3\neq\text{y}^3$
However, this will be a contradiction to $(1).$
$\therefore x = y$
Hence$, f $is injective.
View full question & answer→Question 693 Marks
Show that the Signum Function f: R → R, given by $\text{f(x)}=\begin{cases}1,&\text{if }\text{x }>0\\0,&\text{if }\text{x }=0\\-1,&\text{if }\text{x }<0\end{cases}$ is neither one-one nor onto.
Answerf: R → R is given by,
$\text{f(x)}=\begin{cases}1,&\text{if }\text{x }>0\\0,&\text{if }\text{x }=0\\-1,&\text{if }\text{x }<0\end{cases}$
It is seen that f(1) = f(2) = 1, but $1\neq2.$
$\therefore$ f(-1) = f(1), but $-1\neq1.$
$\therefore$ f is not one-one.
Now, as f(x) takes only 3 values (1, 0, or -1) for the element -2 in co-domain R, there does not exist any x in domain R such that f(x) = -2.
$\therefore$ f is not onto.
Hence, the signum function is neither one-one nor onto.
View full question & answer→Question 703 Marks
On the set Z of integers, if the binary operation * is defined by a * b = a + b + 2, then find the identity element.
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 + 2 = a and e + a + 2 = a, $\forall\ \text{a}\in\text{Z}$e = -2, $\forall\ \text{a}\in\text{Z}$
Thus, -2 is the identity element in Z with respect to *.
View full question & answer→Question 713 Marks
Check the commutativity and associativity of the following binary operations:
$'⊙\ '$ on $Q$ defined by $a ⊙ b = a^2 + b^2$ for all $a, b \in Q.$
AnswerCommutativity: Let $\text{a, b}\in\text{Q}$ then, $a ⊙ b = a^2 + b^2 = b^2 + a^2 = b ⊙ a$
Therefore $, a ⊙ b = b ⊙ a$, for all $\text{a, b}\in\text{Q}$ Thus, $⊙$ is commuatative on $Q$. Associativity : Let $\text{a, b, c}\in\text{Q.} a ⊙ (b ⊙ c) = a ⊙ (b^2 + c^2) = ab2 + (b^2 + c^2)^2 = ab^2 + b^4 + c^4 + 2b^2c^2 (a ⊙ b) ⊙ c$
$= (a^2 + b^2) ⊙ c = (a^2 + b^2)^2 + c^2 = a^4 + b^4 + 2a^2b^2 + c^2$
Therefore,$\text{a}\odot\text{b}\odot\text{c}\neq\text{a}\odot\text{b}\odot\text{c}$
Thus $, ⊙$ is not associative on $Q$.
View full question & answer→Question 723 Marks
Classify the following functions as injection, surjection or bijection:
$f : R \rightarrow R,$ defined by $f(x) = \sin x$
Answer$f : R \rightarrow R,$ given by $f(x) = \sin x$
Injective:
Let $\text{x, y}\in\text{R}$ such that $f(x) = f(y)$
$\Rightarrow \sin x = \sin y$
$\Rightarrow\ \text{x}=\text{n}\pi+(-1)^{\text{n}}\text{y}$
$\Rightarrow\ \text{x}\neq\text{y}$
$\therefore f$ is not one$-$one.
Surjective:
Let $\text{y}\in\text{R}$ be arbitrary such that
$f(x) = y$
$\Rightarrow \sin x = y$
$\Rightarrow x = \sin^{-1}y$
Now, for $\text{y}>1\times\notin\text{R} \ ($domain$)$
$\therefore f$ is not onto.
View full question & answer→Question 733 Marks
Let $A = {1, 2, 3},$ and let $R_3 = {(1, 3), (3, 3)}$. Find whether or not the relations $R_{3 }$ on $A$ is:
- Reflexive.
- Symmetric.
- Transitive.
Answer$R_3 = {(1, 3), (3, 3)}$
$\therefore\ (1,1)\notin\text{R}_3$
$\Rightarrow R_3$ is not reflexive.
Now, $(1,3)\in\text{R}_3$ but $(3,1)\in\text{R}_3$
$\therefore R_3$ is not symmetric.
Again, it is clear that $R_3$ is transitive.
View full question & answer→Question 743 Marks
Show that the relation $''\geq''$ on the set R of all real numbers is reflexive and transitive but not symmetric.
AnswerWe have,
relation $\text{R}=\ ''\geq''$ on the set R of all real numbers
Reflexivity: Let $\text{a}\in\text{R}$
$\Rightarrow\ \text{a}\geq\text{a}$
$\Rightarrow\ ''\geq''$ is reflexive.
Symmetric: Let $\text{a, b}\in\text{R}$
Such that $\text{a}\geq\text{b}\Rightarrow\ \text{b}\geq\text{a}$
$\therefore\ ''\geq''$ not symmetric.
Transitivity: Let $\text{a, b, c}\in\text{R}$
and $\text{a}\geq\text{b}\ \&\ \text{b}\geq\text{c}$
$\Rightarrow\ \text{a}\geq\text{c}$
$\Rightarrow\ ''\geq''$ is transitive.
View full question & answer→Question 753 Marks
Write the multiplication table for the set of integers modulo $5$.
Answer$Z_5 = {0, 1, 2, 3, 4}$
$a\times _5 b$ is the remainder when the product of $ab$ is divided by $5$.
The composition table for $\times _5$ on $Z_5 = {0, 1, 2, 3, 4}$
| $\times _5$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
| $0$ |
$0$ |
$0$ |
$0$ |
$0$ |
$0$ |
| $1$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
| $2$ |
$0$ |
$2$ |
$4$ |
$1$ |
$3$ |
| $3$ |
$0$ |
$3$ |
$1$ |
$4$ |
$2$ |
| $4$ |
$0$ |
$4$ |
$3$ |
$2$ |
$1$ |
View full question & answer→Question 763 Marks
On Q, the set of all rational numbers, * is defined by $\text{a}\ ^*\ \text{b}=\frac{\text{a}-\text{b}}{2},$ shown that * is no associative.
AnswerThe binary operator * defined as, $\text{a}\ ^*\ \text{b}=\frac{\text{a}-\text{b}}{2},$ for all $\text{a, b}\in\text{Q}.$ Now, Associativity: Let $\text{a, b, c}\in\text{Q}.$ Then, $(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}=\frac{\text{a}-\text{b}}{2}\ ^*\ \text{c}=\frac{\frac{\text{a}-\text{b}}{2}-\text{c}}{2}$ $=\frac{\text{a}-\text{b}-2\text{c}}{4}\ ....(\text{i})$ and, $\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=\text{a}\ ^*\ \frac{\text{b}-\text{c}}{2}=\frac{\text{a}-\frac{\text{b}-\text{c}}{2}}{2}$ $=\frac{2\text{a}-\text{b}+\text{c}}{4}\ .....(\text{ii})$ From (i) and (ii),$(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}\neq\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})$
Hence, '*' is not associative on Q.
View full question & answer→Question 773 Marks
Consider f : N → N, g : N → N and h : N → R defined as f(x) = 2x, g(y) = 3y + 4 and $\text{h(z)}=\sin\text{z}$ for all $\text{x, y, z}\in\text{N.}$ Show that ho(gof) = (hog)of.
AnswerGiven, f : N → N, g : N → N and h : N → R
⇒ gof : N → N and hog : N → R
⇒ ho(gof) : N → R and (hog) of : N → R
So, both have the same domains.
(gof)(x) = g(f(x)) = g(2x) = 3(2x) + 4 = 6x + 4 ....(1)
(hog)(x) = h(g(x)) = h(3x + 4) = sin(3x + 4) ......(2)
Now,
(ho(gof))(x) = h((gof)(x)) = h(6x + 4) = sin(6x + 4) [from (1)]
((hog)of)(x) = (hog)(f(x)) = (hog)(2x) = sin(6x + 4) [from (2)]
So, (ho(gof))(x) = ((hog)of)(x), $\forall\text{ x}\in\text{N}$
Hence, ho(gof) = (hog)of
View full question & answer→Question 783 Marks
Find $f^{-1}$ if it exists: $f : A \rightarrow B,$ where, $A = \{1, 3, 5, 7, 9\}; B = \{0, 1, 9, 25, 49, 81\}$ and $f(x) = x^2.$
Answer$A = \{1, 3, 5, 7, 9\}; B = \{0, 1, 9, 25, 49, 81\}$
$f : A \rightarrow B$ be $a$ function defined by $f(x) = x^2$
Since different elements of $A$ have different images in $B.$
$\therefore f $is one$-$one.
Again, $0\in\text{B}$ does not have a preim$-$age in $A.$
$\therefore f $is not onto.
Hence$, f^{-1}$ does not exist.
View full question & answer→Question 793 Marks
Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective:
{(x, y): x is a person, y is the mother of x}
Answerf = {(x, y): x is a person, y is the mother of x}
As, for each element x in domain set, there is a unique related element y in co-domain set.
Therefore, f is the function.
Injection test: As, y can be mother of two or more persons.
Therefore,
f is not injective.
Surjection test: For every mother y defined by (x, y), there exists a person x for whom y is mother. Therefore, f is surjective.
View full question & answer→Question 803 Marks
Let $A$ and $B$ be sets. Show that $f: A \times B \rightarrow B \times A$ such that $f (a, b) = (b, a)$ is bijective function.
Answer$f: A \times B \rightarrow B \times A$ is defined as $f(a, b) = (b, a).$
Let $(\text{a}_1,\text{b}_1),(\text{a}_2,\text{b}_2)\in \text{A}\times\text{B}$ such that $f(a_1, b_1) = f(a_2, b_2)$
$\Rightarrow (b_1, a_1) = (b_2, a_2)$
$\Rightarrow b_1 = b_2$ and $a_1 = a_2$
$\Rightarrow (a_1, b_1) = (a_2, b_2)$
$\therefore f$ is one$-$one.
Now, let $(\text{b},\text{a})\in\text{B}\times\text{A}$ be any element.
Then, there exists $(\text{a},\text{b})\in\text{A}\times\text{B}$ such that $f(a, b) = (b, a). [$By definition of $f]$
$\therefore f$ is onto.
Hence, $f$ is bijective.
View full question & answer→Question 813 Marks
Let A = {1, 2, 3}. Write all one-one from A to itself.
AnswerWe have,
ho(gof)(x) = h(gof(x)) = h(g(f(x)))
= h(g(2x)) = h(3(2x) + 4)
= h(6x + 4) = sin(6x + 4) $\forall\ \text{x}\in\text{N}$
((hog)of)(x) = (hog)(f(x)) = (hog)(2x)
= h(g(2x)) = h(3(2x) + 4)
= h(6x + 4) = sin(6x + 4) $\forall\ \text{x}\in\text{N}$
This shows, ho(gof) = (hog)of
View full question & answer→Question 823 Marks
Let f, g, h be real functions given by f(x) = sinx, g(x) = 2x and h(x) = cosx. Prove that fog = go(fh).
Answerf, g and h are real fuctions given by f(x) = sinx, g(x) = 2x and h(x) = cosx
To prove: fog = go(fh)
L.H.S. fog(x) = f(g(x))
= f(2x) = sin2x
⇒ fog(x) = 2sinx.cosx .....(A)
R.H.S. go(fh)(x) = go(f(x).h(x))
= g(sinx.cosx)
⇒ go(fh)(x) = 2sinx.cosx ......(B)
from (A) & (B)
fog(x) = go(fh)(x)
View full question & answer→Question 833 Marks
Construct the composition table for $\times 6$ on set $S = {0, 1, 2, 3, 4, 5}$.
AnswerHere,
$1\times _61 =$ Remainder obtained by dividing $1 \times $1 by $6 = 1$
$3\times _64 =$ Remainder obtained by dividing $3 \times 4$ by $6 = 0$
$4\times _65 =$ Remainder obtained by dividing $4 \times 5$ by $6 = 2$
So, the composition table is as follows:
| $\times _6$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
$5$ |
| $0$ |
$0$ |
$0$ |
$0$ |
$0$ |
$0$ |
$0$ |
| $1$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
$5$ |
| $2$ |
$0$ |
$2$ |
$4$ |
$0$ |
$2$ |
$4$ |
| $3$ |
$0$ |
$3$ |
$0$ |
$3$ |
$0$ |
$3$ |
| $4$ |
$0$ |
$4$ |
$2$ |
$0$ |
$4$ |
$2$ |
| $5$ |
$0$ |
$5$ |
$4$ |
$3$ |
$2$ |
$1$ |
View full question & answer→Question 843 Marks
The following relation are defined on the set of real numbers.
aRb if $|\text{a}|\leq\text{b}$
Find whether these relation are reflexive, symmetric or transitive.
AnswerWe have aRb if $|\text{a}|\leq\text{b}$ Reflexive: Let $\text{a}\in\text{R}$$\Rightarrow\ |\text{a}|\nleq\text{a}$ $[\therefore|-2|=2>-2|]$
⇒ R is not reflexive. Symmetric: Let aRb $\Rightarrow\ |\text{a}|\leq\text{b}$ $\Rightarrow\ |\text{b}|\leq\text{a}$ $\begin{bmatrix}\therefore\ \ \ \text{Let a}=4, \text{b}=6 \\\ \ \ \ \ \ \ \ \ |4|\leq 8 \text{ but } |8|>4\end{bmatrix}$ ⇒ R is not symmetric. Transitive: Let aRb and bRc$\Rightarrow\ |\text{a}|\leq\text{b}$ and $|\text{b}|\leq\text{c}$
$\Rightarrow\ |\text{a}|\leq|\text{b}|\leq\text{c}$
$\Rightarrow\ |\text{a}|\leq\text{c}$
$\Rightarrow\ \text{aRc}$
⇒ R is transitive.
View full question & answer→Question 853 Marks
Find the inverse of $5$ under multiplication modulo $11$ on $Z_{11}$.
Answer$Z_{11} = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}$
Multiplication modulo $11$ is defined as follows:
For $\text{a},\text{b}\in\text{Z}_{11}$,
$a\times _{11}b$ is the remainder when $a \times b$ is divided by $11$.
Here,
$1\times _{11}1 =$ Remainder obtained by dividing $1 \times 1$ by $11 = 1$
$3\times _{11}4 =$ Remainder obtained by dividing $3 \times 4$ by $11 = 1$
$4\times _{11}5 =$ Remainder obtained by dividing $4 \times 5$ by $11 = 9$
| $\times _{11}$ |
$1$ |
$2$ |
$3$ |
$4$ |
$5$ |
$6$ |
$7$ |
$8$ |
$9$ |
$10$ |
| $1$ |
$1$ |
$2$ |
$3$ |
$4$ |
$5$ |
$6$ |
$7$ |
$8$ |
$9$ |
$10$ |
| $2$ |
$2$ |
$4$ |
$6$ |
$8$ |
$10$ |
$1$ |
$3$ |
$5$ |
$7$ |
$9$ |
| $3$ |
$3$ |
$6$ |
$9$ |
$1$ |
$4$ |
$7$ |
$10$ |
$2$ |
$5$ |
$8$ |
| $4$ |
$4$ |
$8$ |
$1$ |
$5$ |
$9$ |
$2$ |
$6$ |
$10$ |
$3$ |
$7$ |
| $5$ |
$5$ |
$10$ |
$4$ |
$9$ |
$3$ |
$8$ |
$2$ |
$7$ |
$1$ |
$6$ |
| $6$ |
$6$ |
$1$ |
$7$ |
$2$ |
$8$ |
$3$ |
$9$ |
$4$ |
$10$ |
$5$ |
| $7$ |
$7$ |
$3$ |
$10$ |
$6$ |
$2$ |
$9$ |
$5$ |
$1$ |
$8$ |
$4$ |
| $8$ |
$8$ |
$5$ |
$2$ |
$10$ |
$7$ |
$4$ |
$1$ |
$9$ |
$6$ |
$3$ |
| $9$ |
$9$ |
$7$ |
$5$ |
$3$ |
$1$ |
$10$ |
$8$ |
$6$ |
$4$ |
$2$ |
| $10$ |
$10$ |
$9$ |
$8$ |
$7$ |
$6$ |
$5$ |
$4$ |
$3$ |
$2$ |
$1$ |
We observe that the first row of the composition table is same as the top $-$ most row.
Therefore,
The identity element is $1$.
Also,
$5\times _{11}9 = 1$
Hence $, 5 - 1 = 9$ View full question & answer→Question 863 Marks
State with reasons whether the following functions have inverse:
g : {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}
Answerg : {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}
g(5) = g(7) = 4
⇒ f is not one-one.
⇒ f is not a bijection.
So, f does not have an inverse.
View full question & answer→Question 873 Marks
Let f: R → R be the Signum Function defined as $f(\text{x})=\begin{cases}1,&\text{x}>0\\0,&\text{x}=0\\-1,&\text{x}<0\end{cases}$ and g: R → R be the Greatest Integer Function given by g(x) = [x], where [x] is greatest integer less than or equal to x. Then, does fog and gof coincide in (0, 1]?
AnswerIt is given that,
f: R → R is defined as $f(\text{x})=\begin{cases}1,&\text{x}>0\\0,&\text{x}=0\\-1,&\text{x}<0\end{cases}$
Also, g: R → R is defined as g(x) = [x], where [x] is the greatest integer less than or equal to x.
Now, let $\text{x}\in(0,\ 1].$
Then, we have:
[x] = 1 if x = 1 and [x] = 0 if 0 < x < 1.
$\therefore fo\text{g(x)}=f(\text{g(x)})=f([\text{x}])=\begin{cases}f(1),&\text{if x}=1\\f(0),&\text{if x}\in(0,\ 1)\end{cases}=\begin{cases}1,&\text{if x}=1\\0,&\text{if x}\in(0,\ 1)\end{cases}$
gof(x) = g(f(x))
= g(1) [x > 0]
=[1] = 1
Thus, when $\text{x}\in(0,\ 1),$ we have fog(x) = 0 and gof(x) = 1.
Hence, fog and gof do not coincide in (0, 1].
View full question & answer→Question 883 Marks
Let $f : R \rightarrow R$ be the function defined by $f(x) = 2x – 3$ for all $x \in R$. write $f^{-1}$.
AnswerGiven $f(x) = 2x - 3$ for all $x \in R$ Now, Let $a, b \in R$ such that $f(a) = f(b)$
$\Rightarrow 2a - 3 = 2b - 3$
$\Rightarrow a = b$
$\Rightarrow f(x)$ is One $-$ One.
Also, If $x, y \in R$ such that $f(x) = y$
$\Rightarrow 2x - 3 = y $
$\Rightarrow\ \text{x}=\frac{\text{y}+3}{2}=\text{g}(\text{y})\ \forall\ \text{y}\in\text{R}$
$\Rightarrow f(x)$ is Onto and therefore is bijective implies $f(x)$ has an inverse Let $f^{-1}$ denote the inverse of $f(x)$
then $,\text{f}^{-1}\text{x}=\text{g}(\text{x})=\frac{\text{x}+3}{2}$ for all $\text{x}\in\text{R}$
View full question & answer→Question 893 Marks
If A = {1, 2, 3}, show that a onto function f : A → A must be one-one.
AnswerA = {1, 2, 3}
Possible onto function from A to A can be the following:
- {(1, 1), (2, 2), (3, 3)}
- {(1, 1), (2, 3), (3, 2)}
- {(1, 2), (2, 2), (3, 3)}
- {(1, 2), (2, 1), (3, 3)}
- {(1, 3), (2, 2), (3, 1)}
- {(1, 3), (2, 1), (3, 2)}
Here, in each function, different elements of the domain have different images.
Therefore, all the function are one-one.
View full question & answer→Question 903 Marks
On the set Z of integers a binary operation * is defined by a * b = ab + 1 for all a, b ∈ Z. Prove that * is not associative on Z.
AnswerLet $\text{a, b, c}\in\text{Z}$
a * (b * c) = a * (bc + 1)
= a(bc + 1) + 1
= abc + a + 1
(a * b) * c = (ab + 1) * c
= (ab + 1)c + 1
= abc + c + 1
Thus, $\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})\neq(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}$
Thus, * is not associative on Z.
View full question & answer→Question 913 Marks
Let $f: X \rightarrow Y$ be an invertible function. Show that the inverse of $f ^{–1}$ is $f,$ i.e., $(f ^{–1})^{–1} = f$.
AnswerLet $f: X \rightarrow Y$ be an invertible function.
Then $f$ is one $-$ one and onto
$\Rightarrow g: y \rightarrow X$ where g is also one $-$ one and onto such that
$\text{gof}(x) = I_x$ and $\text{fog}(y) = I_y$
$\Rightarrow g = f^{-1}$
Now $, f^{-1}o(f^{-1})^{-1} = I$ and $\text{fof} \ [f^{-1}o(f^{-1})^{-1}] = \text{fof}$
$\Rightarrow [\text{fof}^{-1}]0(f^{-1})^{-1} = f $
$\Rightarrow Io(f^{-1})^{-1} = f \Rightarrow (f^{-1})^{-1} = f$
View full question & answer→Question 923 Marks
Let f be a function from R to R, such that f(x) = cos(x + 2). Is f invertible? Justify your answer.
AnswerGiven: A and B are two sets with finite elements.
f : A → B and g : B → A are injective map.
To prove: f is bijective.
Proof: Since, f : A → B is injective we need to show f in surjective only.
Now,
g : B → A is injective.
⇒ Each element of B has image in A.
View full question & answer→Question 933 Marks
Give an example of a relation which is,
Reflexive and symmetric but not transitive.
AnswerLet A = {4, 6, 8} Define a relation R on A as: A = {(4, 4), (6, 6), (8, 8), (4, 6), (6, 4), (6, 8), (8, 6)}Relation R is reflexive since for every $\text{a}\in\text{A},\ (\text{a, a})\in\text{R}$ i.e., (4, 4), (6, 6), (8, 8) $\in\text{R}$
Relation R is symmetric since $(\text{a, b})\in\text{R}\Rightarrow\ (\text{b, a})\in\text{R}$ for all $\text{a, b}\in\text{R.}$
Relation R is not transitive since (4, 6), (6, 8) $\in\text{R,}$ but $(4,8)\notin\text{R.}$
Hence, relation R is reflexive and symmetric but not transitive.
View full question & answer→Question 943 Marks
Let $f : R \rightarrow R$ and $g : R \rightarrow R$ be defined by $f(x) = x + 1$ and $g(x) = x - 1$. Show that $\text{fog = gof} = I_R$.
AnswerGiven $, f : R \rightarrow R$ and $g : R \rightarrow R\ \text{fog} : R \rightarrow R$ and $\text{gof} : R \rightarrow R \ ($Also, we know that $I_R : R \rightarrow R)$
Therefore, the domains of all $\text{fog},$ and $I_R$ are the same.
$\text{(fog)}(x) = f(g(x)) = f(x - 1) = x - 1 + 1 = x = I_R(x) ..... (1)$
$\text{(gof)}(x) = g(f(x)) = g(x + 1) = x + 1 -1 = x = I_R(x) ..... (2)$
From $(1)$ and $(2), \text{(fog)}(x) = \text{(gof)}(x) = I_R(x), $ for all $\text{ x}\in\text{R}$
Hence, $\text{fog = gof} = I_R$
View full question & answer→Question 953 Marks
Let $A = [-1, 1].$ Then, discuss whether the following functions from $A$ to itself are one-one, onto or bijective:$h(x) = x^2$
Answer$h(x) = x^2$
Injection test: Let $x$ and $y$ be any two elements in the domain $(A),$
such that$ f(x) = f(y). f(x) = f(y) x^2 = y^2\text{x}=\pm\text{y}$
So, $f$ is not one$-$one.
Surjection test: For $y = -1,$ there is no value of $x$ in $A.$
So, $f $ is not onto.
So, $f$ is not bijective.
View full question & answer→Question 963 Marks
State with reasons whether the following functions have inverse:
f : {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}
Answerf : {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}
We have,
f(1) = f(2) = f(3) = f(4) = 10
⇒ f is not one-one.
⇒ f is not a bijection.
So, f does not have an inverse.
View full question & answer→Question 973 Marks
Find which of the binary operations are commutative and which are associative.
Let A = N × N and * be the binary operation on A defined by:
(a, b) * (c, d) = (a + c, b + d)
AnswerA = N × N and * is a binary operation defined on A.
(a, b) * (c, d) = (a + b, c + d) = (c + a, d + b) = (c, d) * (a, b)
$\therefore$ The operation is commutative
Again, [(a, b) * (c, d)] * (e, f) = (a + c, b + d) * (e, f) = (a + c + e, b + d + f)
And (a, b)[(c, d) * (e, f)] = (a, b) * (c + e, e + f) = (a + c + e, b + d + f)
Here, [(a, b) * (c, d)] * (e, f) = (a, b)[(c, d) * (e, f)]
$\therefore$ The operation is associative.
Let the identity function be (e, f), then (a, b) * (e, f) = (a + e, b + f)
For identity function a = a + e $\Rightarrow\ \ \text{e}=0$
And for b + f = b $\Rightarrow\ \ \text{f}=0$
As $0\neq\text{N},$ therefore identity element does not exist.
View full question & answer→Question 983 Marks
Let $A = [-1, 1].$ Then, discuss whether the following functions defined on $A$ are one$-$one, onto or bijective: $\text{f}(\text{x})=\frac{\text{x}}{2}$
AnswerLet $f(x_1) = f(x_2)$
$\Rightarrow\ \frac{\text{x}_1}{2}=\frac{\text{x}_2}{2}$
$\Rightarrow\ \text{x}_1=\text{x}_2$
So, $f(x)$ is one$-$one.
Now, let $\text{y}=\frac{\text{x}}{2}$
$\Rightarrow\ \text{x}=2\text{y}\notin\text{A},\ \forall\ \text{y}\in\text{A}$
As for $\text{y}=1\in\text{A},\ \text{x}=2\notin\text{A}$
So, $f(x)$ is not onto.Also, $f(x)$ is not bijective as it is not onto.
View full question & answer→Question 993 Marks
Let f be an invertible real function. Write $(f^{-1}of)(1) + (f^{-1}of)(2) + ..... + (f^{-1}of)(100)$.
AnswerGiven that $f$ is an invertible real function. $f^{-1}of = I$, where $I$ is an identity function.
Therefore $,(f^{-1}of)(1) + (f^{-1}of)(2) + ....... + (f^{-1}of)(100) = I(1) + I(2) + .... + I(100) = 1 + 2 + .... + 100$
As $({Ix}=\text{x},\ \forall\ \text{x}\in\text{R})$
$=\frac{1001(1001+1)}{2}$ $\Big[$ Sum of first $n$ natural numbers $=\frac{\text{n}(\text{n}+1)}{2}\Big]= 5050$
View full question & answer→Question 1003 Marks
If R and S are transitive relations on a set A, then prove that $\text{R}\cup\text{S}$ may not be a transitive relation on A.
AnswerLet A = {a, b, c} and R and S be two relations on a, given by R = {(a, a), (a, b), (b, a), (b, b)}
And S = {(b, b), (b, c), (c, b), (c, c)}
Here, the relations R and S are transitive on A.
$\text{a, b}\in\text{R}\cup\text{S}$ and $\text{b, c}\in\text{R}\cup\text{S}$ But $\text{a, c}\notin\text{R}\cup\text{S}$
Hence, $\text{R}\cup\text{S}$ is not a transitive relation on A.
View full question & answer→