Question 1013 Marks
Find fog and gof if:$f(x) = x + 1, g(x) = e^x$
Answer$f(x) = x + 1, g(x) = e^{x }$ Range of $f = R \subset$ Domain of $g = R \Rightarrow gof $ exist
Range of $\text{g}=(0,\infty)\ \subset$ Domain of $f = R \Rightarrow fog$ exist
Now,
$gof(x) = g(f(x)) = g(x + 1) = e^{x+1}$
And,
$fog(x) = f(g(x)) = f(e^x) = e^x + 1$
View full question & answer→Question 1023 Marks
Let $A = R_0 \times R,$ where $R_0 $ denote the set of all non-zero real numbers. A binary operation $'⊙'$ is defined on $A$ as follows:
$(a, b) ⊙ (c, d) = (ac, bc + d)$ for all $(a, b), (c, d) \in R_0 \times R.$
Show that '⊙' is commutative and associative on $A.$
AnswerCommutativity:Let $\text{x}=(\text{a},\text{b})$ and $\text{y}=(\text{c},\text{d})\in\text{A},\forall\text{ a},\text{c}\in\text{R}_0\ \&\text{ b},\text{d}\in\text{R}.$ Then,
$X ⊙ Y = (ac, bc + d )\ \&\ Y ⊙ X = (ca, da + b)$
Therefore,$ x ⊙ Y = Y ⊙ X, \forall\ \text{X},\text{Y}\in\text{A}$
Thus, $⊙$ is commutative on $A.$
Associativity:
Let $\text{X}=(\text{a},\text{b}),\text{Y}=(\text{c},\text{d})$ and $\text{Z}=(\text{e},\text{f}),\forall\ \text{a},\text{c},\text{e}\in\text{R}_0\ \&\text{ b},\text{d},\text{f}\in\text{R}$
$x ⊙ Y ⊙ Z = (a, b) ⊙ (ce, de + f)$
$= (ace, bce + de + f)$
$x ⊙ Y ⊙ Z = (ac, bc + d) ⊙ (e, f)$
$= (ace, (bc + d)e + f)$
$= (ace, bce + de + f)$
$\therefore x ⊙ Y ⊙ Z = x ⊙ Y ⊙ Z, \forall\ \text{X},\text{Y},\text{Z}\in\text{A}$
Thus, ⊙ is accosiative on A.
View full question & answer→Question 1033 Marks
Let $A = [-1, 1].$ Then, discuss whether the following functions defined on $A$ are one-one, onto or bijective:
$k(x) = x^2$
AnswerLet $k(x_1) = k(x_2)$
$\Rightarrow\ \text{x}^2_1=\text{x}^2_2\Rightarrow\ \text{x}_1=\pm\text{x}_2$
Thus, $k(x)$ is not one-one.
Now, let $y = x^2 \Rightarrow\ \text{x}\sqrt{\text{y}}\notin\text{A},\ \forall\ \text{y}\in\text{A}$
As for $\text{y}=-1,\ \text{x}=\sqrt{-1}\notin\text{A}$
Hence, $k(x)$ is neither one-one nor onto.
View full question & answer→Question 1043 Marks
Check whether the relation $R$ on $R$ defined by $R = \{(a, b): a \leq b^3\}$ is reflexive, symmetric or transitive.
AnswerThe relation $R$ on $R$ is defined by $R.$ We observe that $(-2)\leq(-2)^3$ is not true.
Therefore, $R$ is not reflexive. Since $1\leq\Big(3^{\frac{1}{3}}\Big)^3$ but $3^\frac{1}{3}\leq1$
i.e. $\Big(1,3^\frac{1}{3}\Big)\in\text{R.}$
Therefore, $R$ is not symmetric.
Hence, $R$ is not transitive because $(5,2)\in\text{R}$ and $\Big(2,2^\frac{1}{3}\Big)\in\text{R}$ but $\Big(5,2^\frac{1}{3}\Big)\notin\text{R.}$
View full question & answer→Question 1053 Marks
Let * be a binary operation on Z defined by a * b = a + b - 4 for all a, b ∈ Z.
Show that '*' is both commutative and associative.
AnswerCommutativity: Let $\text{a, b}\in\text{Z}.$ Then,
a * b = a + b - 4
= b + a - 4
= b * a
Therefore,
a * b = b * a, $\forall\ \text{a, b}\in\text{Z}$
Thus, * is commutative on Z.
Associativity: Let $\text{a, b, c}\in\text{Z}.$ Then,
a * (b * c) = a * (b + c - 4)
= a + b + c - 4 - 4
= a + b + c - 8
(a * b) * c = (a + b - 4) * c
= a + b - 4 + c - 4
= a + b + c - 8
Therefore, a * (b * c) = (a * b) * c, $\forall\ \text{a, b, c}\in\text{Z}.$
Thus, * is associative on Z.
View full question & answer→Question 1063 Marks
A binary operation $*$ is defined on the set $R$ of all real numbers by the rule $\text{a}\times\text{b}=\sqrt{\text{a}^2+\text{b}^2}\ \forall\text{ a, b}\in\text{R}$.
Write the identity element for $*$ on $R.$
AnswerLet e be the identity element in $R$ with respect to $*$ such thata $* e = a = e * a, \forall\text{ a}\in\text{R}$
$a * e = a$ and $e * a = a, \forall\text{ a}\in\text{R}$
Then,
$\sqrt{\text{a}^2+\text{e}^2}=\text{a}$ and $\sqrt{\text{e}^2+\text{a}^2}=\text{a},\forall\text{ a}\in\text{R}$
Implies that $\sqrt{\text{a}^2+\text{e}}=\text{a}$ and $\sqrt{\text{e}+\text{a}^2}=\text{a},\forall\text{ a}\in\text{R} [ \because e^2 = e]$
Implies that $a^2 + e = a^2$ and $e + a^2 = a^2,$ $\forall\text{ a}\in\text{R}$
Implies that $\text{e}=0\in\text{R},\forall\text{ a}\in\text{R}$
Thus, $0$ is the identity element in $R$ with respect to $*.$
View full question & answer→Question 1073 Marks
If f(x) = sin x and g(x) = 2x be two real functions, then describe gof and fog. Are these equal functions?
AnswerWe know that f : R → -1, 1 and g : R → R gof clearly, the range of f is a subset of the domain of g. gof : R → R, gof(x) = g(f(x)) = g(sinx) = 2sinx fog clearly, the range of g is a subset of the domain of f. fog : R → RSo,
f : R → R, fog(x) = f(g(x)) = f(2x) = sin2x Clearly, $\text{fog}\neq\text{gof}$
View full question & answer→Question 1083 Marks
Classify the following functions as injection, surjection or bijection:$f : R \rightarrow R,$ defined by $f(x) = 1 + x^2$
Answer$f : R → R,$ defined by $f(x) = 1 + x^2$
Injection test: Let $\text{x, y}\in\text{R,}$ such that,
$f(x) = f(y)$
$\Rightarrow 1 + x^2 = 1 + y^2$
$\Rightarrow x^2 - y^2 = 0$
$\Rightarrow (x - y)(x + y) = 0$
either $x = y$ or $x = -y$ or $\text{x}\neq\text{y}$
Therefore, f is not one-one.
Surjection: Let $\text{y}\in\text{R}$ be arbitrary, then
$f(x) = y$
$\Rightarrow 1 + x^2 = y$
$\Rightarrow x^2 + 1 - y = 0$
$\therefore\ \text{x}\pm\sqrt{\text{y}-1}\notin\text{R}$ or $y < 1$
$\therefore f $ is not onto.
View full question & answer→Question 1093 Marks
Find gof and fog when f : R → R and g : R → R are defined by:
f(x) = x and g(x) = |x|
AnswerGiven, f : R → R and g : R → R Therefore, gof : R → R and fog : R → R f(x) = x and g(x) = |x| (gof)(x) = g(f(x))= g(x)
(fog)(x) = f(g(x)) = f|x| = |x|
View full question & answer→Question 1103 Marks
Let $A = R_0 \times R,$ where $R_0$ denote the set of all non-zero real numbers. A binary operation $'⊙'$ is defined on $A$ as follows:
$(a, b) ⊙ (c, d) = (ac, bc + d)$ for all $(a, b), (c, d) \in R_0 \times R.$
Find the invertible elements in $A.$
AnswerLet $F = (m, n)$ be the inverse in $\text{A }\forall\text{ m}\in\text{R}_0\ \&\text{ n}\in\text{R}x ⊙ F = X = E$ and $F ⊙ X = E$
Implies that $(am, bm + n) = (1, 0)$ and $(ma, na + b) = (1,0)$
Considering $(am, bm + n) = (1, 0)$
Implies that $am = 1$
Implies that $\text{m}=\frac{1}{\text{a}}$
&$ bm + n = 0$
Implies that $\text{n}=\frac{-\text{b}}{\text{a}} \Big[\because\ \text{m}=\frac{1}{\text{a}}\Big]$
Considering $(ma, na + b) = (1,0)$
Implies that $ma = 1$
Implies that $\text{m}=\frac{1}{\text{a}}$
& $na + b = 0$
Implies that $\text{n}=\frac{\text{-b}}{\text{a}}$
$\therefore$ The inverse of $(\text{a},\text{b}) \in \text{A}$ with respect to $⊙$ is $\Big(\frac{1}{\text{a}},\frac{-\text{b}}{\text{a}}\Big).$
View full question & answer→Question 1113 Marks
On the set Z of all integers a binary operation * is defined by a * b = a + b + 2 for all a, b ∈ Z. Write the inverse of 4.
AnswerTo find the identity element, let 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, $\forall\text{ a}\in\text{Z}$
Then,
a + e + 2 = a and e + a + 2 = a, $\forall\text{ a}\in\text{Z}$
$\text{e}=-2\in\text{Z},\ \forall\text{ a}\in\text{Z}$
Thus, -2 is the identity element in Z with respect to *.
Now,
Let b ∈ Z be the inverse of 4.
Here,
4 * b = e = b * 4
4 * b = e and b * 4 = e
Then,
4 + b + 2 = -2 and b + 4 + 2 = -2
$\text{b}=-8\in\text{Z}$
Thus, -8 is the inverse of 4.
View full question & answer→Question 1123 Marks
If f : A → B and g : B → C are onto functions, show that gof is a onto function.
AnswerGiven, f : A → B and g : B → C are onto.
Then, gof : A → C
Let us take an element z in the co-domain (C).
Now, z is in C and g : B → C is onto.
So, there exists some element y in B, such that g(y) = z .... (1)
Now, y is in B and f : A → B is onto.
So, there exists some x in A, such that f(x) = y .... (2)
From (1) and (2),
z = g(y) = g(f(x)) = (gof)(x)
So, z = (gof)(x), where x is in A.
Hence, gof is onto.
View full question & answer→Question 1133 Marks
Three relation $R_3$ is defined in set $A = \{a, b, c\}$ as follows:
$R_3 = \{(b, c)\}$
Find whether or not the relation $R_3$ on $A$ is:
- Reflexive.
- Symmetric.
- Transitive.
Answer$R_3$ is Reflexive: Here $\text{b, b}\notin\text{R}_3$ neither $\text{c, c}\notin\text{R}_3$Therefore, $R_3$ is not reflexive.
Symmetric: Here, $\text{b, c}\notin\text{R}_3,$ but $\text{c, c}\notin\text{R}_3$
So, $R_3$ is not symmetric.
Transitive: Here, $R_3$ has only two elements.
Hence, $R_3$ is transitive.
View full question & answer→Question 1143 Marks
Let $f : R \rightarrow R$ be the function defined by $f(x) = 4x - 3$ for all $x \in R.$ Then write $f^{-1}.$
AnswerWe have,
$f : R \rightarrow R$ is the function defined by $f(x) = 4x - 3$ for all $\text{x}\in\text{R}$
Let $f(x) = y.$ Then,
$y = 4x - 3$
$4x = y + 3$
$\text{x}=\frac{\text{y}+3}{4}$
Therefore, $\text{f}^{-1}(\text{y})=\frac{\text{y}+3}{4}$
or, $\text{f}^{-1}(\text{x})=\frac{\text{x}+3}{4}$
View full question & answer→Question 1153 Marks
Classify the following functions as injection, surjection or bijection:
f : Z → Z, defined by f(x) = x - 5
Answerf : Z → Z, defined by f(x) = x - 5Injection test: Let x and y be any two elements in the domain (Z), such that f(x) = f(y).
f(x) = f(y) x - 5 = y - 5 x = y Therefore, f is an injection. Surjection test: Let y be any element in the co domain (Z), such that f(x) = y for some element x in Z (domain). f(x) = y x - 5 = y x = y + 5, which is in Z. Therefore, f is a surjection and f is a bijection.
View full question & answer→Question 1163 Marks
Give an example of a relation which is,Symmetric but neither reflexive nor transitive.
AnswerLet A = {5, 6, 7}.
Define a relation R on A as R = {(5, 6), (6, 5)}.
Relation R is not reflexive as $(5, 5), (6, 6), (7, 7)\notin\text{R.}$ $$
Now, as $(5, 6)\in\text{R}$ and also $(6,5)\in\text{R,}$ R is symmetric.
$\Rightarrow(5, 6), (6, 5)\in\text{R,}\text{but}(5, 5)\notin\text{R}$ $$ $$ $$
Therefore, R is not transitive.
Hence, relation R is symmetric but not reflexive or transitive.
View full question & answer→Question 1173 Marks
Prove that every identity relation on a set is reflexive, but the converse is not necessarily true.
AnswerLet A be a set. Then $\text{I}_\text{A}=\{(\text{a, a});\text{ a}\in\text{A}\}$ is the identity relation on A. Hence, every identity relation on a set is reflexive by defination. Converse: Let A = {(a, b, c)} be a set. Let R = {(a, a), (b, b), (c, c), (a, b)} be a relation defined on A.Clearly R is reflexive on set A, but it is not identity relation on set A as $(\text{a, b})\in\text{R}$
Hence, a reflexive relation need not be identity relation.
View full question & answer→Question 1183 Marks
Find gof and fog when $f : R \rightarrow R$ and $g : R \rightarrow R$ are defined by:
$f(x) = x^2 + 2x - 3$ and $g(x) = 3x - 4$
AnswerGiven, $f : R \rightarrow R$ and $g : R \rightarrow R$
Therefore, $gof : R \rightarrow R$ and $fog : R \rightarrow R$
$f(x) = x^2 + 2x - 3$ and $g(x) = 3x - 4$
Now, $gof(x) = g(f(x)) = g(x^2 + 2x - 3)$
$\therefore gof(x) = 3(x^2 + 2x - 3) - 4$
$\Rightarrow gof(x) = 3x^2 + 6x - 13$
and, $fog(x) = f(g(x)) = f(3x - 4)$
$\therefore fog(x) = (3x - 4)^2 + 2(3x - 4) - 3$
$= 9x^2 + 16 - 24x + 6x - 8 - 3$
$\therefore fog(x) = 9x^2 - 18x + 5$
View full question & answer→Question 1193 Marks
Classify the following functions as injection, surjection or bijection:
$f : R \rightarrow R,$ defined by $f(x) = \sin^2x + \cos^2x$
Answer$f : R \rightarrow R,$ defined by $f(x) = \sin^2x + \cos^2x$
$f(x) = \sin^2x + \cos^2x = 1$
So, $f(x) = 1$ for every $x$ in $R.$
So, for all elements in the domain, the image is $1.$
So, $f$ is not an injection.
Range of $f = \{1\}$
Co-domain of $f = R$
Both are not same.
So, $f$ is not a surjection and f is not a bijection.
View full question & answer→Question 1203 Marks
If f(x) = |x|, prove that fof = f.
AnswerWe have, f(x) = |x|
We assume the domain of f = R
Range of $\text{f}=(0,\infty)$
$\therefore$ Range of f $\subset$ domain of f
$\therefore$ fof exists.
Now,
fof(x) = f(f(x)) = f(|x|) = ||x|| = f(x)
$\therefore$ fof = f
View full question & answer→Question 1213 Marks
Let * be a binary operation on Q - {-1} defined by a * b = a + b + ab for all a, b ∈ Q - {-1}. Then,
Show that '*' is both commutative and associative on Q - {-1}.
AnswerWe have,
a * b = a + b + ab for all a, b ∈ Q - {-1}
Commutativity: Let a, b ∈ Q - {-1}
⇒ a * b = a + b + ab = b + a + ba = b * a
⇒ a * b = b * a
⇒ '*' is commutative on Q - {-1}
Associativity: Let a, b, c ∈ Q - {-1}, then
⇒ (a * b) * c = (a + b + ab) * c
= a + b + ab + c + ac + bc + abc .......(i)
and, a * (b * c) = a * (b + c + bc)
= a + b + c + bc + ab + ac + abc .....(ii)
From (i) and (ii)
(a * b) * c = a * (b * c)
⇒ * is associative on Q - {-1}
View full question & answer→Question 1223 Marks
State with reasons whether the following functions have inverse:
h : {2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}
Answerh : {2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)} Here, different elements of the domain have different images in the co-domain. ⇒ h is one-one. Also, each element in the co-domain has a pre-image in the domain.⇒ h is onto.
⇒ h is bijection. ⇒ h has an inverse and it is given by, h - 1 = {(7, 2), (9, 3), (11, 4), (13, 5)}
View full question & answer→Question 1233 Marks
If the mapping f : {1, 3, 4} → {1, 2, 5} and g : {1, 2, 5} → {1, 3}, given by f = {(1, 2), (3, 5), (4, 1)} and g= {(2, 3), (5, 1), (1, 3)}, then write fog.
AnswerWe have,
f : {1, 3, 4} → {1, 2, 5} and g : {1, 2, 5} → {1, 3}, are given by f = {(1, 2), (3, 5), (4, 1)} and g= {(2, 3), (5, 1), (1, 3)} respectively.
As,
fog(2) = f(g(2)) = f(3) = 5,
fog(5) = f(g(5)) = f(1) = 2,
fog(1) = f(g(1)) = f(3) = 5,
Therefore, fog : {1, 2, 5} → {1, 2, 5} is given by fog = {(2, 5), (5, 2), (1, 5)}
View full question & answer→Question 1243 Marks
Find gof and fog when $f : R \rightarrow R$ and $g : R \rightarrow R$ are defined by:
$f(x) = x^2 + 8$ and $g(x) = 3x^3 + 1$
AnswerGiven, $f : R \rightarrow R$ and $g : R \rightarrow R$
Therefore, $gof : R \rightarrow R$ and $fog : R \rightarrow R$
$f(x) = x^2 + 8$ and $g(x) = 3x^3 + 1$
$(gof)(x) = g(f(x))$
$= g(x^2 + 8)$
$= 3(x^2 + 8)^3 + 1$
$(fog)(x) = f(g(x))$
$= f(3x^3 + 1)$
$= (3x^3 + 1)^2 + 8$
$= 9x^6 + 6x^3 + 1 + 8$
$= 9x^6 + 6x^3 + 9$
View full question & answer→Question 1253 Marks
Construct the composition table for $\times _4$ on set $S = \{0, 1, 2, 3\}.$
AnswerHere,
$1\times _41 =$ Remainder obtained by dividing $1 \times 1$ by $4 = 1$
$0\times _41 = $Remainder obtained by dividing $0 \times 1$ by $4 = 0$
$2\times _43 =$ Remainder obtained by dividing $2 \times 3$ by $4 = 2$
$3\times _43 =$ Remainder obtained by dividing $3 \times 3$ by $4 = 1$
Therefore,
The composition table is as follows:
| $x_4$ |
$0$ |
$1$ |
$2$ |
$3$ |
| $0$ |
$0$ |
$0$ |
$0$ |
$0$ |
| $1$ |
$0$ |
$1$ |
$2$ |
$3$ |
| $2$ |
$0$ |
$2$ |
$0$ |
$2$ |
| $3$ |
$0$ |
$3$ |
$2$ |
$1$ |
View full question & answer→Question 1263 Marks
If $\text{f(x)}=\sqrt{\text{x}+3}$ and $g(x) = x^2 + 1$ be two real functions, then find fog and gof.
Answer$\text{f(x)}=\sqrt{\text{x}+3}$For domain,
$\text{x}+3\geq0$
$\Rightarrow\ \text{x}\geq-3$
Domain of $\text{f}=[-3,\infty)$
Since f is a square root fuction, range of $\text{f}=[0,\infty)$
$\text{f}:[-3,\infty)\rightarrow[0,\infty)$
$g(x) = x^2 + 1$ is a polynomial.
$\Rightarrow g : R \rightarrow R$
Computation of fog: Range of g is not a subset of the doamin of f.
and domain (fog) = {x : x $\in$ domain of g and g(x) $\in$ domain of f(x)}
⇒ Domain (fog) = {x : x $\in\text{R}$ and $x^2 + 1$ $\in[-3,\infty)$}
⇒ Domain (fog) = {x : x $\in\text{R}$ and $x^2 + 1$ $\geq-3$}
⇒ Domain (fog) = {x : x $\in\text{R}$ and $x^2 + 4$ $\geq0$}
⇒ Domain (fog) = {x : x $\in\text{R}$ and $\text{x}\in\text{R}$}
⇒ Domain (fog) = R
fog :$ R \rightarrow R$
(fog)(x) = f(g(x))
$= f(x^2 + 1)$
$=\sqrt{\text{x}^2+1+3}$
$=\sqrt{\text{x}^2+4}$
Computation of gof: Range of f is a subset of the doamin of g.
$\text{gof}:[-3,\infty)\rightarrow\text{R}$
$\Rightarrow\ \text{(gof)(x)}=\text{g(f(x)})$
$=\text{g}(\sqrt{\text{x}+3})$
$=(\sqrt{\text{x}+3})^2+1$
$=\text{x}+3+1$
$=\text{x}+4$
View full question & answer→Question 1273 Marks
On the set Q of all ration numbers if a binary operation * is defined by $\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{5},$ prove that * is associative on Q.
AnswerLet $\text{a, b, c}\in\text{Q}.$ Then,
$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=\text{a} \ ^*\ \Big(\frac{\text{bc}}{5}\Big)$
$=\frac{\text{a}\big(\frac{\text{bc}}{5}\big)}{5}$
$=\frac{\text{abc}}{25}$
$(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}=\Big(\frac{\text{ab}}{5}\Big)\ ^*\ \text{c}$
$=\frac{\big(\frac{\text{ab}}{5}\big)\text{c}}{5}$
$=\frac{\text{abc}}{25}$
Therefore, a * (b * c) = (a * b) * c, $\forall\ \text{a, b, c}\in\text{Q}$
Thus, * is associative on Q.
View full question & answer→Question 1283 Marks
For the binary operation $\times _{10}$ on set $S = \{1, 3, 7, 9\},$ find the inverse of $3.$
Answer$a\times _{10} b =$ the remainder when the product of ab is divided by $10.$The composition table for $\times _{10}$ on set$ S = \{1, 3, 7, 9\}$
| $\times _{10}$ |
$1$ |
$3$ |
$7$ |
$9$ |
| $1$ |
$1$ |
$3$ |
$7$ |
$9$ |
| $3$ |
$3$ |
$9$ |
$1$ |
$7$ |
| $7$ |
$7$ |
$1$ |
$9$ |
$3$ |
| $9$ |
$9$ |
$7$ |
$3$ |
$1$ |
We know that an element $\text{b}\in\text{S}$ will be the inverse of $\text{a}\in\text{S}$
if $a\times _{10} b = 1 [ \because 1$ is the identity element with respect to multiplication.$]$
$\Rightarrow 3\times _{10} b = 1$
From the above table b $= 7$
$\therefore$ Inverse of $3$ is $7.$ View full question & answer→Question 1293 Marks
Let * be a binary operation on Q - {-1} defined by a * b = a + b + ab for all a, b ∈ Q - {-1}. Then,
Show that every element of Q − {−1} is invertible. Also, find the inverse of an arbitrary element.
AnswerWe have,
a * b = a + b + ab for all a, b ∈ Q - {-1}
Let b be the inverse of a ∈ Q - {-1}
Then, a * b = b * a = e [e is the identity element]
⇒ a + b + ab = e
⇒ a + b + ab = 0
⇒ b(1 + a) = -a
$\Rightarrow\text{b}=\frac{-\text{a}}{1+\text{a}}$ $\begin{bmatrix}\because\ \frac{-\text{a}}{1+\text{a}}\neq-1\text{ because if }\frac{-\text{a}}{1+\text{a}}=-1\\\Rightarrow\text{a}=1+\text{a}\Rightarrow1=0\text{ Not possible}\end{bmatrix}$
$\text{b}=\frac{-\text{a}}{1+\text{a}}$ is the inversre of a with respect to *.
View full question & answer→Question 1303 Marks
Check the commutativity and associativity of the following binary operations:
$'*'$ on $Q$ defined by $a * b = ab^2 $ for all $a, b ∈ Q.$
AnswerCommutativity: Let $\text{a, b}\in\text{Q}.$ Then,$a * b = ab^2$
$b * a = ba^2$
Therefore,
$\text{a}\ ^*\ \text{b}\neq\text{b}\ ^*\ \text{a}$
Thus, $*$ is not commutative on $Q.$
Associativity: Let $\text{a, b, c}\in\text{Q}.$ Then,
$a * (b * c) = a * (bc^2)$
$= a(bc^2)^2$
$= ab^2c^4$
$(a * b) * c = (ab^2) * c$
$= ab^2c^2$
Therefore,
$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})\neq(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}$
Thus, $*$ is not associative on $Q.$
View full question & answer→Question 1313 Marks
Three relation $R_4$ is defined in set $A = \{a, b, c\}$ as follows:
$R_4 = \{(a, b), (b, c), (c, a)\}$
Find whether or not the relation $R_4$ on $A$ is:
- Reflexive.
- Symmetric.
- Transitive.
Answer$R_4$ is reflexive: Here, $\text{a}\notin\text{R}_4,\ \text{b},\ \text{b}\notin\text{R}_4,\ \text{c},\ \text{c}\notin\text{R}_4$
Therefore, $R_4$ is not reflexive.
Symmetric: Here, $\text{a, b}\notin\text{R}_4,$ but $\text{b, b}\notin\text{R}_4$
Therefore, $R_4$ is not symmetric.
Transitive: Here, $\text{a, b}\notin\text{R}_4,\ \text{b, c}\in\text{R}_4$
But $\text{a, c}\notin\text{R}_4$
Therefore, $R_4$ is not transitive.
View full question & answer→Question 1323 Marks
The following relation are defined on the set of real numbers.
aRb if a - b > 0
Find whether these relations are reflexive, symmetric or transitive.
AnsweraRb if a - b > 0 Let R be the set of real numbers. Reflexivity: Let $\text{a}\in\text{R}$ $\Rightarrow\ \text{a}-\text{a}=0$ $\Rightarrow\ (\text{a, a})\notin\text{R}$ $\therefore$ R is not reflexive. Symmetric: Let aRb ⇒ a - a > 0 ⇒ b - a < 0$\therefore$ R is not symmetric.
Transitive: Let aRb and bRc
⇒ a - a > and b - c > 0 ⇒ a - c > 0 ⇒ aRc$\therefore$ R is transitive.
View full question & answer→Question 1333 Marks
Let A = [-1, 1]. Then, discuss whether the following functions from A to itself are one-one, onto or bijective:
$\text{f(x)}=\frac{\text{x}}{2}$
Answerf : A → A, given by $\text{f(x)}=\frac{\text{x}}{2}$
Injection test: Let x and y be any two elements in the domain (A), such that f(x) = f(y).
f(x) = f(y)
$\frac{\text{x}}{2}=\frac{\text{y}}{2}$
x = y
So, f is one-one.
Surjection test: Let y be any element in the co-domain (A), such that f(x) = y for some element x in A (domain).
f(x) = y
$\frac{\text{x}}{2}=\text{y}$
x = 2y, which may not be in A.
For example, if y = 1, then
x = 2, which is not in A.
So, f is not onto.
So, f is not bijective.
View full question & answer→Question 1343 Marks
Give examples of two functions f: N → Z and g: Z → Z such that gof is injective but g is not injective.
(Hint: Consider f(x) = x and g(x) = |x|).
AnswerDefine f: N → Z as f(x) = x and g: Z → Z as g(x) = |x|
We first show that g is not injective.
It can be observed that:
g(-1) = |-1| = 1
g(1) = |1| = 1
$\therefore$ g(-1) = g(1), but $-1\neq1.$
$\therefore$ g is not injective.
Now, gof: N → Z is defined as gof(x) = g(f(x)) = g(x) = |x|
Let $\text{x},\text{y}\in\text{N}$ such that gof(x) = gof(y).
⇒ |x| = |y|
Since $\text{x}\ \text{and}\ \text{y}\in\text{N},$ both are positive.
$\therefore$ |x| = |y| ⇒ x = y
Hence, gof is injective.
View full question & answer→Question 1353 Marks
An integer m is said to be related to another integer n if m is a multiple of n.Check if the relation is symmetric, reflexive and transitive.
Answer$\text{R} = \big\{{(\text{m, n})}(1, 1), (2, 1)\Rightarrow(1, 1): \text{m, n} \in\text{Z,}\text{m} \\=\text{kn},\text{where}\text{k}\in\text{N}\big\}$$$$$$$ $$ $$
$$Reflexivity: Let m be an arbitrary element of R. Then, m = km is true for k = 1$\Rightarrow\ (\text{m, m})\in\text{R}$
Thus, R is reflexive. Symmetry: Let $(\text{m, n})\in\text{R}$ ⇒ m = kn for some $\text{k}\in\text{N}$ $\Rightarrow\ \text{n}=\frac{1}{\text{k}}\text{m}$ $\Rightarrow\ (\text{n, m})\notin\text{R}$ Thus, R is not symmetric.Transitivity: Let (m, n) and (n, o) $\in\text{R}$
⇒ m = kn and n = lo for some $\text{k, l}\in\text{N}$ ⇒ m = (kl)o Here, $\text{kl}\in\text{N}$ $\Rightarrow\ (\text{m, o})\in\text{R}$ Thus, R is transitive.
View full question & answer→Question 1363 Marks
Let $+_6 ($addition modulo $6)$ be a binary operation on $S = \{0, 1, 2, 3, 4, 5\}$. Write the value of $2 + _64^{−1}+ _63^{−1}.$
AnswerThe composition table for $+_6 $ on the set $S = \{0, 1, 2, 3, 4, 5\} $ is
| $+_6$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
$5$ |
| $0$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
$5$ |
| $1$ |
$1$ |
$2$ |
$3$ |
$4$ |
$5$ |
$0$ |
| $2$ |
$2$ |
$3$ |
$4$ |
$5$ |
$0$ |
$1$ |
| $3$ |
$3$ |
$4$ |
$5$ |
$0$ |
$1$ |
$2$ |
| $4$ |
$4$ |
$5$ |
$0$ |
$1$ |
$2$ |
$3$ |
| $5$ |
$5$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
$'0'$ is the identity element for $+_6 $ from the table it is clear that
$4^{-1} = 2$ and $3^{-1} = 3$
Now, $2 + _64^{-1} + _63^{-1} = 2 + _62 + _63$
$= 4 + _63$
$= 1$ View full question & answer→Question 1373 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 1383 Marks
Let A be any set containing more than one element. Let '*' be a binary operation on A defined by a * b = b for all a, b ∈ A. Is '*' commutative or associative on A?
AnswerCommutativity: Let $\text{a, b}\in\text{A.}$ Then, $\text{a}\ ^*\ \text{b}=\text{b}$ $\text{b}\ ^*\ \text{a}=\text{a}$ Therefore, $\text{a}\ ^*\ \text{b}\neq\text{b}\ ^*\ \text{a}$ Thus, * is not commutative on A. Associativity: Let $\text{a, b, c}\in\text{A.}$ Then, $\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=\text{a}\ ^*\ \text{c}$ $=\text{c}$ $(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}=\text{b}\ ^*\ \text{c}$ $=\text{c}$ Therefore,$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=(\text{a}\ ^*\ \text{b})\ ^*\ \text{c},\ \forall\ \text{a, b, c}\in\text{A}$
Thus, * is associative on A.
View full question & answer→Question 1393 Marks
Given the relation R = {(1, 2), (2, 3)} on the set A = {1, 2, 3}, add a minimum number of ordered pairs so that the enlarged relation is symmeteric, transitive and reflexive.
AnswerWe have,
R = {(1, 2), (2, 3)}
R can be a transitive only when the elements (1, 3) is added
R can be a reflexive only when the elements (1, 1), (2, 2), (3, 3) are added
R can be a symmetric only when the elements (2, 1), (3, 1) and (3, 2) are added
So, the required enlarged relation, R' = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)} = A × A
View full question & answer→Question 1403 Marks
Check the commutativity and associativity of the following binary operations:
'*' on Q defined by a * b = a + ab for all a, b ∈ Q.
AnswerCommutativity: Let $\text{a, b}\in\text{Q}.$ Then,a * b = a + ab
b * a = b + ba
= b + ab
Therefore,
$\text{a}\ ^*\ \text{b}\neq\text{b}\ ^*\ \text{a}$
Thus, * is not commutative on Q.
Associativity: Let $\text{a, b, c}\in\text{Q}.$ Then,
a * (b * c) = a * (b + bc)
= a + a(b + bc)
= a + ab + abc
(a * b) * c = (a + ab) * c
= (a + ab) + (a + ab)c
= a + ab + ac + abc
Therefore,
$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})\neq(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}$
Thus, * is not associative on Q.
View full question & answer→Question 1413 Marks
Let A be the set of all human beings in a town at a particular time. Determine whether the following relations are reflexive, symmetric and transitive:
R = {(x, y): x is father of and y}
AnswerA be the set of human beings
R = {(x, y): x is father of and y}
Reflexive: Since x can not be father of x
$\therefore\ (\text{x, x})\notin\text{R}$
⇒ R is not reflexive.
Symmetric: Let $(\text{x, y})\in\text{R}$
⇒ x is father of y
⇒ y can not be father of x
$\Rightarrow\ (\text{y, x})\notin\text{R}$
⇒ R is not Symmetric.
Transitive: Let $(\text{x, y})\in\text{R}$ and $(\text{y, z})\in\text{R}$
⇒ x is father of y and y is father of z
⇒ x is grandfather of z
$\Rightarrow\ (\text{x, z})\notin\text{R}$
⇒ R is not transitive.
View full question & answer→Question 1423 Marks
Let $R^+$^ be the set of all non-negative real numbers. If $f : R^+ \rightarrow R^+ $ and $g : R^+ \rightarrow R^+$ are defined as $f(x) = x^2 $ and $\text{g(x)}=+\sqrt{\text{x}},$ find fog and gof. Are they equal functions?
AnswerWe have, $f : R^+ \rightarrow R^+ $ given by
$f(x) = x^2$
$g : R^+ \rightarrow R^+ $ given by
$\text{g}(\text{x})=\sqrt{\text{x}}$
$\therefore fog(x) = f(g(x))$
$=\text{f}(\sqrt{\text{x}})=(\sqrt{\text{x}})^2=\text{x}$
Also, $gof(x) = g(f(x))$
$=\text{g}(\text{x}^2)\sqrt{\text{x}^2}=\text{x}$
Thus, $fog(x) = gof(x)$
View full question & answer→Question 1433 Marks
Determine which of the following binary operations are associative and which are commutative:
* on Q defined by $\text{a}\ ^*\ \text{b}=\frac{\text{a}+\text{b}}{2}$ for all $\text{a, b}\in\text{Q}$
Answer$\text{a}\ ^*\ \text{b}=\frac{\text{a}+\text{b}}{2}=\frac{\text{b}+\text{a}}{2}=\text{b}\ ^*\ \text{a,}$
Which shows * is commutative.
Further, $(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}=\Big(\frac{\text{a}+\text{b}}{2}\Big)\ ^*\ \text{c}$
$=\frac{\big(\frac{\text{a}+\text{b}}{2}\big)+\text{c}}{2}=\frac{\text{a}+\text{b}+2\text{c}}{4}$
Further, $\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=\text{a}\ ^*\ \Big(\frac{\text{b}+\text{c}}{2}\Big)$
$=\frac{\text{a}+\big(\frac{\text{b}+\text{c}}{2}\big)}{2}=\frac{2\text{a}+\text{b}+\text{c}}{2}\neq\frac{\text{a}+\text{b}+2\text{c}}{4}$
Hence, * is not associative.
View full question & answer→Question 1443 Marks
Find gof and fog when $f : R \rightarrow R$ and $g : R \rightarrow R$ are defined by:
$f(x) = 8x^3 $ and $\text{g(x)}=\text{x}^\frac{1}{3}$
AnswerGiven, $f : R \rightarrow R$ and $g : R \rightarrow R$
Therefore, $gof : R \rightarrow R$ and $fog : R \rightarrow R$
$f(x)=8 x^3 \text { and } g(x)=x^{\frac{1}{3}} $
$ (gof)(x)=g(f(x)) $
$ =g\left(8 x^3\right) $
$ =\left(8 x^3\right)^{\frac{1}{3}} $
$ =\left[(2 x)^3\right]^{\frac{1}{3}} $
$ =2 x $
$ (fog)(x)=f(g(x)) $
$ =f\left(x^{\frac{1}{3}}\right) $
$ =8\left(x^{\frac{1}{3}}\right)^3 $
$ =8 x$
View full question & answer→Question 1453 Marks
Show that the relation $R$ defined in the set $A$ of all triangles as $R=\left\{\left(T_1, T_2\right): T_1\right.$ is similar to $\left.T_2\right\}$, is equivalence relation Consider three right angle triangles $T_1$ with sides $3,4,5, T_2$ with sides $5,12,13$ and $T_3$ with sides $6,8,10$. Which triangles among $T_1, T_2$ and $T_3$ are related?
AnswerPart I: $R=\left\{\left(T_1, T_2\right): T_1\right.$ is similar to $\left.T_2\right\}$ and $T_1, T_2$ are triangle.
| We know that each triangle similar to itself and thus $(\text{T}_1,\text{T}_2)\in\text{R}$ |
$\therefore$ |
R is reflexive |
| Also two triangles are similar, then $\text{T}_1\cong\text{T}_2\Rightarrow\text{T}_1\cong\text{T}_2$ |
$\therefore$ |
R is symmetric |
| Again, if then $\text{T}_1\cong\text{T}_2$ and then $\text{T}_2\cong\text{T}_3\Rightarrow\ \text{then}\ \text{T}_1\cong\text{T}_3$ |
$\therefore$ |
R is transitive |
Therefore, $R$ is an equivalent relation.Part II: It is given that $T_1, T_2$ and $T_3$ are right angled triangles. $\Rightarrow T_1$ with sides $3,4,5, T_2$ with sides $5,12,13, T_3$ with sides $6,8,10$ Since, two triangles are similar if corresponding sides are proportional. Therefore, $\frac{3}{6}=\frac{4}{8}=\frac{5}{10}=\frac{1}{2}$ Therefore, $T_1$ and $T_3$ are related. View full question & answer→Question 1463 Marks
Let $f(x) = x^2 + x + 1$ and $g(x) = \sin x.$ Show that $fog \neq gof.$
Answer$ (fog)(x)=f(g(x)) $
$ f(\sin x)=\sin ^2 x+\sin x+1 $
$ \text { and, }(gof)(x)=g(f(x)) $
$ =g\left(x^2+x+1\right) $
$ =\sin x^2+x+1$
$\text { Therefore, fog } \neq \text { gof. }$
View full question & answer→Question 1473 Marks
Check the commutativity and associativity of the following binary operations:
'o' on Q defined by $\text{a o b}=\frac{\text{ab}}{2}$ for all a, b ∈ Q.
AnswerBinary operation 'o' defined on Q, given by $\text{a o b}=\frac{\text{ab}}{2}$ for all a, b ∈ Q.
Commutative: Let $\text{a, b}\in\text{Q},$ Then
$\text{a o b}=\frac{\text{ab}}{2}=\frac{\text{ba}}{2}=\text{b o a}$
$\Rightarrow\ \text{a o b}=\text{b o a}$
$\therefore$ o is commutative on Q.
Associativity: Let $\text{a, b, c}\in\text{Q},$ Then,
$(\text{a o b})\text{ o c}=\Big(\frac{\text{ab}}{2}\Big)\text{ o c}=\frac{\text{abc}}{4}\ .....(\text{i})$
$\text{a o }(\text{b o c})=\text{a o }\Big(\frac{\text{bc}}{2}\Big)=\frac{\text{abc}}{4}\ ....(\text{ii})$
From (i) and (ii) we get
(a o b) o c = a o (b o c)
$\therefore$ 'o' is associative on Q.
View full question & answer→Question 1483 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 $g \Rightarrow gof$ exist
Range of g $\subset$ domain of $f \Rightarrow fog$ exist
Now,
$gof(x) = g(f(x)) = g(x^2) = \cos x^2$
And
$fog(x) = f(f(x)) = f(\cos x) = \cos^2x$
View full question & answer→Question 1493 Marks
If $f{\text{x}}=\frac{(4\text{x}+3)}{(6\text{x}-4)},\ \text{x}\neq\frac{2}{3},$ show that fof(x) = x, for all $\text{x}\neq\frac{2}{3}.$ What is the inverse of f?
AnswerGiven: $f{\text{x}}=\frac{(4\text{x}+3)}{(6\text{x}-4)},\ \text{x}\neq\frac{2}{3},$
L.H.S = fof(x) = f[f(x)] $f\Big[\frac{4\text{x}+3}{6\text{x}-4}\Big]=\frac{4\Big(\frac{4\text{x}+3}{6\text{x}-4}\Big)+3}{6\Big(\frac{4\text{x}+3}{6\text{x}-4}\Big)-4}=\frac{16\text{x}+12+18\text{x}-12}{24\text{x}+18-24\text{x}+16}$
$=\frac{34\text{x}}{34}=\text{x}$ = R.H.S.
| Now, $\text{y}=\frac{4\text{x}+3}{6\text{x}-4}$ |
⇒ |
6xy - 4y = 4x + 3 |
⇒ |
6xy - 4x = 4y + 3 |
| ⇒ x(6y - 4) = 4y + 3 |
⇒ |
$\text{x}=\frac{4\text{y}+3}{6\text{y}-4}$ |
⇒ |
$\text{y}=\frac{4\text{x}+3}{6\text{x}-4}$ |
Hence inverse of f = f. View full question & answer→Question 1503 Marks
Let $A = R - \{3\}$ and $B = R - \{1\}.$ Consider the function $f : A \rightarrow B$ defined by $\text{f(x)}=\frac{\text{x}-2}{\text{x}-3}.$ Show that $f$ is one-one and onto and hence find $f^{-1}.$
AnswerWe have,
$A = R - \{3\}$ and $B = R -\{1\}.$ Consider the function $f : A \rightarrow B$ defined by
$\text{f}(\text{x})=\frac{\text{x}-2}{\text{x}-3},$ Show that f is one-one and onto and hence find $f^{-1}.$
Let $\text{x, y}\in\text{A}$ such that $f(x) = f(y).$ Then,
$\frac{\text{x}-2}{\text{x}-3}=\frac{\text{y}-2}{\text{y}-3}$
$\Rightarrow xy - 3x - 2y + 6 = xy - 2x - 3y + 6$
$\Rightarrow -x = -y$
$\Rightarrow x = y$
$\therefore f $ is one-one.
View full question & answer→