Question 15 Marks
Show that the binary operation $\ast \text{ on A = R - {-1}}$ defined as a $\text{a} \ast \text{b} = \text{a + b + ab}$ for all $\text{a, b}\in \text{A}$ is communicative and associative on A. Also find the identity element of $\ast$ in A and prove that every element of a is invertible.
AnswerCommutative: For any elements $\text{a, b}\in \text{A}$
$\text{a}\ast\text{b} = \text{a + b + ab = b + a + ba = b}\ast \text{a}.$ Hence $\ast$ is commutative
Associative: For any three elements $\text{a, b, c,}\in \text{A}$
$\text{a}\ast \text{(b}\ast\text{c}) = \text{a}\ast\text{(b + c + bc) = a + b + c + bc + ab + ac + abc}$
$\text{(a} \ast\text{b}) \ast\text{c} = \text{(a + b + ab)}\ast \text{c} = \text{a + b + ab + c + ac + bc + abc}$
$\therefore \text{a}{\ast} \text{(b}{\ast}\text{c}) = \text{(a}{\ast} \text{b}) \ast\text{c}, \text{Hence } \ast \text{ is Associative.}$
Identity element: let e $\in$ A be the identity element them $\text{a}\ast \text{e = e}\ast \text{a = a}$
$\Rightarrow\text{a + e + ae = e + a + ea = a}\Rightarrow\text{e(1 + a) = 0, as a} \neq -1 $
$\text{e = 0}$ is the identity element
Invertible: let $\text{a, b}\in \text{A}$ so that ‘b’ is inverse of a
$\therefore \text{a}\ast \text{b = b}\ast\text{a = e}$
$\Rightarrow\text{a + b + ab = b + a + ba = 0}$
$\text{As a}\neq -1, \text{b} = \frac{\text{-a}}{1 + \text{a}} \in \text{A}.$ Hence every element of A is invertible.
View full question & answer→Question 25 Marks
Determine whether the relation R defined on the set $\Re$ of all real numbers as R =$(\text{a,b) : a, b} \in \Re$ and $\text{a - b} + \sqrt{3} \in \text{S},$where S is the set of all irrational numbers, is reflexive, symmetric and transitive.
AnswerHere R ={ $\text{(a, b) : a, b} \in \Re$ and $\text{a - b + }\sqrt{3} \in \text{S}, $ where S is the set of all irrational numbers.}
$\text{(i)} \forall \text{a}\in \Re, \text{(a, a)} \in \text{R as a - a +} \sqrt{3}$ is irrational
$\therefore \text{R is reflexive}$
$\text{(ii)} \text{Let for a, b} \in \Re, \text{(a, b)} \in \text{R i. e. a - b +}\sqrt{3}$ is irrational
$\text{a - b +}\sqrt{3}$ is irrational $\Rightarrow \text{b - a +}\sqrt{3} \in \text{S} \therefore \text{(b, a)} \in \text{R}$
Hence R is symmetric
$\text{(iii)} \text{Let (a, b)} \in \text{R and (b, c)}\in \text{R for a, b, c}\in \Re$
$\therefore \text{a - b +} \sqrt{3} \in \text{S and b - c +} \sqrt{3} \in \text{S}$
$\text{adding to get a - c +} 2\sqrt{3} \in \text{S Hence (a, c)}\in \text{R}$
$\therefore\text{R is Transitive}$
View full question & answer→Question 35 Marks
Let $\text{A = Q} \times \text{Q}$ and let $*$ be a binary operation on $A$ defined by$\text{(a, b)} * \text{(c, d) = (ac, b + ad)} \text{ for (a, b), (c, d)} \in \text{A}.$ Determine, whether $*$ is commutative and associative. Then, with respect to $*$ on $A$.
- Find the identity element in $A.$
- Find the invertible elements of $A.$
Answer$\text{(a, b)}{*} \text{(c, d) = (ac, b + ad); (a, b), (c, d)} \in \text{A}$
$\text{(c, d)}{*} \text{(a, b) = (ca, d + bc)}$
Since $\text{b + ad} \neq \text{d + bc}$
$\Rightarrow \text{*}$ is $\text{NOT}$ comutative
for associativity, we have,
$\text{(a, b)}{*} \text{[(c, d)}{*} \text{(e, f)] = (a, b)}{*} \text{(ce, d + cf) = (ace, b + ad + acf)}$
$\Rightarrow \text{ }{*}$ is associative
Let $(e, f)$ be the identity element in $A$
Then $\text{(a, b)}{*} \text{(e, f) = (a, b) = (e, f)}{*} \text{(a, b)}$
$\Rightarrow\text{(ae, b + af) = (a, b) = (ae, f + be)}$
$\Rightarrow \text{e = 1, f = 0}$
$\Rightarrow \text{(1, 0)}$ is the identity element
Let $(c, d)$ be the inverse element for $(a, b)$
$\Rightarrow \text{(a, b)}{*} \text{(c, d) = (1, 0) = (c, d)}{*} \text{(a, b)}$
$\Rightarrow \text{(ac, b + ad) = (1, 0) = (ac, d + bc)}$
$\Rightarrow \text{ac} = 1$
$\Rightarrow \text{c} = \frac{1}{\text{a}}$ and $\text{b + ad} = 0$
$\Rightarrow \text{d} = - \frac{\text{b}}{\text{a}}$ and $\text{d + bc =0}$
$\Rightarrow \text{d = -bc = -b} \bigg(\frac{1}{\text{a}}\bigg)$
$\Rightarrow \bigg(\frac{1}{\text{a}}, - \frac{\text{b}}{\text{a}}\bigg), \text{a} \neq 0$ is the inverse of $\text{(a, b)} \in \text{A}$
View full question & answer→Question 45 Marks
Consider $\text{f : R} - \left\{-\frac{4}{3}\right\} \rightarrow \text{R} - \left\{\frac{4}{3}\right\}$ given by $f(x) = \frac{\text{4x + 3}}{\text{3x + 4}}.$ Show that $f$ is bijective. Find the inverse of $f$ and hence find $f ^{–1}(0)$ and $x$ such that $f ^{–1}(x) = 2.$
Answer$\text{Let } \text{x}_{1}, \text{x}_{2} \in \text{R} - \left\{-\frac{4}{3}\right\}$ and $\text{f(x}_{1}) = \text{f(x}_{2})$
$\Rightarrow \frac{\text{4x}_{1} + 3}{\text{3x}_{1} + 4} = \frac{\text{4x}_{2} + 3}{\text{3x}_{2} + 4}$
$\Rightarrow \text{(4x}_{1} + 3) \text{(3x}_{2} + 4) = \text{(3x}_{1} + 4) \text{(4x}_{2} + 3)$
$\Rightarrow \text{12x}_{1} \text{x}_{2} + \text{16x}_{1} + \text{9x}_{2} + 12 = 12_{1} \text{x}_{2} + 16\text{x}_{2} + 9\text{x}_{1} + 12$
$\Rightarrow 16(\text{x}_{1} - \text{x}_{2}) - 9 \text{(x}_{1} - \text{x}_{2}) = 0$
$\Rightarrow \text{x}_{1} - \text{x}_{2} = 0$
$\Rightarrow \text{x}_{1} = \text{x}_{2}$
Hence $f$ is a $1–1$ function
Let $\text{ y} = \frac{\text{4x + 3}}{\text{3x + 4}}, \text{for y} \in \text{R} - \left\{\frac{4}{3}\right\}$
$\text{3xy + 4y = 4x + 3}$
$\Rightarrow \text{4x – 3xy = 4y – 3}$
$\Rightarrow \text{x} = \frac{\text{4y - 3}}{\text{4 - 3y}} \therefore \forall \text{ y} \in \text{R} - \left\{\frac{4}{3}\right\}, \text{x} \in \text{R} - \left\{-\frac{4}{3}\right\}$
Hence $f$ is $\text{ONTO}$ and so bijective
and $\text{f}^{-1} \text{(y)} = \frac{\text{4y - 3}}{\text{4 - 3y}} ; \text{y} \in \text{R} - \left\{\frac{4}{3}\right\}$
$\text{f}^{-1} (0) = - \frac{3}{4}$
and $\text{f}^{-1} \text{(x)} = 2$
$\Rightarrow \frac{\text{4x - 3}}{\text{4 - 3x}} = 2$
$\Rightarrow \text{4x - 3 = 8 - 6x}$
$\Rightarrow \text{10x} = 11$
$ \Rightarrow \text{x} = \frac{11}{10}$
View full question & answer→Question 55 Marks
Show that the binary operation $\ast \text{ on A = R - {-1}}$ defined as a $\text{a} \ast \text{b} = \text{a + b + ab}$ for all $\text{a, b}\in \text{A}$ is communicative and associative on A. Also find the identity element of $\ast$ in A and prove that every element of a is invertible.
AnswerCommutative: For any elements $\text{a, b}\in \text{A}$
$\text{a}\ast\text{b} = \text{a + b + ab = b + a + ba = b}\ast \text{a}.$ Hence $\ast$ is commutative
Associative: For any three elements $\text{a, b, c,}\in \text{A}$
$\text{a}\ast \text{(b}\ast\text{c}) = \text{a}\ast\text{(b + c + bc) = a + b + c + bc + ab + ac + abc}$
$\text{(a} \ast\text{b}) \ast\text{c} = \text{(a + b + ab)}\ast \text{c} = \text{a + b + ab + c + ac + bc + abc}$
$\therefore \text{a}{\ast} \text{(b}{\ast}\text{c}) = \text{(a}{\ast} \text{b}) \ast\text{c}, \text{Hence } \ast \text{ is Associative.}$
Identity element: let e $\in$ A be the identity element them $\text{a}\ast \text{e = e}\ast \text{a = a}$
$\Rightarrow\text{a + e + ae = e + a + ea = a}\Rightarrow\text{e(1 + a) = 0, as a} \neq -1 $
$\text{e = 0}$ is the identity element
Invertible: let $\text{a, b}\in \text{A}$ so that ‘b’ is inverse of a
$\therefore \text{a}\ast \text{b = b}\ast\text{a = e}$
$\Rightarrow\text{a + b + ab = b + a + ba = 0}$
$\text{As a}\neq -1, \text{b} = \frac{\text{-a}}{1 + \text{a}} \in \text{A}.$ Hence every element of A is invertible.
View full question & answer→Question 65 Marks
Let $\text{A} = \Re\times\Re$ and$\ast$ be the binary operation on A defined by $\text(a, b) \ast \text{(c, d)} = \text{(a + c, b + d)}. $ Prove that $\ast$ is commutative and associative. Find the identity element for $\ast$ on A. Also write the inverse element of the element (3, – 5) in A.
Answer$\forall \text{a, b, c, d, e, f}\in \Re$
$\text{((a, b)}\ast \text{(c,d))}\ast(\text{e, f})= \text{a + c, b + d)}\ast\text{(e, f)}$
$= \text{(a + c + e, b + d + f)} \rightarrow\text{(3)}$
$\text{((a, b)}\ast \text{(c,d))}\ast(\text{e, f}) = \text{(a + b)}\ast\text{(c + e, d + f)}$
$ = \text{(a + c = e, b + d + f)} \rightarrow \text{(4)}$
$\therefore \ast \text { is Associative}$
Let (x, y) be on identity element in $\Re\times\Re$
$\Rightarrow \text{(a, b)}\ast\text{(x, y)} = \text{(a, b)} = \text{(x, y)}\ast \text{(a,b)}$
$\Rightarrow \text{a + x = a, b + y = b}$
$\text{x = 0, y = 0}$
$\therefore \text{(0, 0) is identity element} $
Let the inverse element of $\text{(3, -5) be } \text{(x}_{1},\text{y}_{1},) $
$\Rightarrow (3, - 5)\ast \text{(x}_{1},\text{y}_{1},) = (0, 0) = \text{(x}_{1},\text{y}_{1},) \ast(3, -5)$
$\text{3 + x}_{1} = 0,\text{-5 + y}_{1} = 0 $
$\text{x}_{1} = -3, \text{y}_{1} = 5 $
$\Rightarrow (-3, 5) \text{is an inverse of (3, -5)}$
View full question & answer→Question 75 Marks
Let $\text{A} = \text{R} - \left\{3\right\}, \text{B} = \text{R} - \left\{1\right\}. \text{Let f : A} \rightarrow \text{B}$ be defined by $\text{f(x)}\frac{\text{x - 2}}{\text{x - 3}}, \forall \text{ x} \in \text{A}.$ Show that f is bijective. Also, find
- $\text{x, if f}^{-1} \text{(x) = 4}$
- $\text{f}^{-1} (7)$
Answer$\text{x}_{1}, \text{x}_{2} \in \text{A and } \text{f(x}_{1}) = \text{f(x}_{2})$
$\Rightarrow \frac{\text{x}_{1} - 2}{\text{x}_{1} - 3} = \frac{\text{x}_{2} - 2}{\text{x}_{2} - 3} \Rightarrow \text{(x}_{1} - 2) \text{(x}_{2} - 3) = \text{(x}_{1} - 3) \text{(x}_{2} - 2)$
$\Rightarrow \text{x}_{1} \text{x}_{2} - \text{3x}_{1} - 2\text{x}_{2} + 6 = \text{x}_{1} \text{x}_{2} - \text{2x}_{1} - \text{3x}_{2} + 6$
$\Rightarrow \text{x}_{1} = \text{x}_{2}$
Hence f is a one-one function
Let $\text{y} = \frac{\text{x - 2}}{\text{x - 3}} \text{ for y } \in \text{R} - \left\{1\right\}$
$\Rightarrow \text{x} = \frac{\text{3y - 2}}{\text{y - 1}} ; \text{y} \neq 1$
$\therefore \forall \text{ y} \in \text{R} - \left\{1\right\}, \text{x} \in \text{R} - \left\{3\right\}$
i.e. Range of f = co-domain of f.
Hence f is onto and so bijective.
Also, $\text{f}^{-1} \text{(x)} = \frac{\text{3x - 2}}{\text{x - 1}} ; \text{x} \neq 1$
Now, $\text{f}^{-1} \text{(x)} = 4 \Rightarrow \frac{\text{3x - 2}}{\text{x - 1}} = 4 \Rightarrow \text{x = 2}$
and $ \text{f}^{-1}(7) = \frac{19}{6}$
View full question & answer→Question 85 Marks
If f, $\text{f, g : R}\rightarrow \text{R}$ be two functions defined as $\text{f}(x) = |x| + x \text{ and } \text{g} (x) = |x| - x, \forall \text{ }x \in \text{R}.$Then find fog and gof. Hence find fog(–3), fog(5) and gof (–2).
Answer$\text{Given f}(x) = |\text{x}| + \text{x}$
$\text{and g} (x) = |{\text{x}}| - x , \forall x \in \text{R}$
$\text{fog} = \text{f(g(x))} = |\text{x}| + \text{g(x)}$
$ = \text{||x| - x| + (|x| - x)}$
Therefore,
$\text{f(g(x))} = \begin{cases} 0 & \text{x} \geq 0\\ 4x & \text{x} < 0\\ \end{cases}$
$\text{fog} = \begin{cases} 4\text{x} & \text{x} < 0\\ 0 & \text{x} \geq 0\\ \end{cases}$
$\text{gof = g(f(x)) = |f(x)| - f(x)}$
$= ||\text{x| + x| - (|x| + x)}$
Therefore, $\text{g(f(x)) = gof = 0}$
| Now, fog(−3) =(4)(−3) = −12 |
(since, fog = 4x for x < 0) |
| fog(5) = 0 |
(since, fog = 0 for x ≥ 0) |
| gof(−2) = 0 |
(since, gof = 0 for x < 0) |
View full question & answer→Question 95 Marks
Let $\text{A} = \text{R} \times \text{R}$ and let$ _*$ be a binary operation on $A$ defined by $\text{(a, b)} {*} \text{(c, d)} = \text{(ad + bc, bd)}$ for all $\text{(a, b), (c, d)} \in \text{R} \times \text{R}.$
- Show that $_*$ is commutative on $A.$
- Show that $_*$ is associative on $A.$
- Find the identity element of $_*$ in $A.$
Answer
- $\text{(a, b)}{*} \text{(c, d) = (ad + bc, bd)}$
$\text{Now}, \text{(c, d)}{*} \text{(a, b) = (cb + da, db) = (ad + bc, bd) = (a, b)}{*} \text{(c, d)}$
$\Rightarrow \text{ }{*}$ is Communicative
- $\text{[(a, b) }{*} \text{(c, d)]}{*} \text{(e, f) = (ad + bc, bd)}{*} \text{(e, f) = (adf + bcf + bde, bdf)]}$
$\text{(a,b)}{*} \text{[(c, d)}{*} \text{(e, f)] = (a, b)} {*} \text{(cf + de, df) = (adf + bcf + bde, bdf)}$
$\Rightarrow \text{ } {*} $ is associative.
- Let $(e_1, e_2)$ be the identity element of $A.$
$\Rightarrow \text{(a, b}) {*} \text{(e}_{1}, {\text{e}_{2}) = \text{(a, b)} = \text{(e}_{1}, \text{e}_{2}) {*} \text{(a, b)}}$
$\Rightarrow \text{(ae}_{2} + \text{be}_{1} , \text{be}_{2}) = \text{(a, b)} = \text{(e}_{1}\text{b} + \text{e}_{2}\text{a}, \text{e}_{2} \text{b})$
$\Rightarrow \text{a}\text{e}_{2} + \text{b}\text{e}_{1} = \text{a and b}\text{e}_{2} = \text{b} \Rightarrow \text{e}_{1} = 0, \text{e}_{2} = 1$
$\Rightarrow (0, 1)$ is the identity on $A.$ View full question & answer→Question 105 Marks
Let N denote the set of all natural numbers and R be the relation on$\text{N} \times \text{N}$defined by $\text{(a, b) R (c, d)}$ if ad$\text{(b + c) = bc(a + d)}$. Show that R is an equivalence relation.
Answer$\forall \text{a}, \text{b}\in \text{N},\text{(a, b)R (a, b) as ab ( b + a) = ba(a + b)} $
$\therefore \text{R is reflexive} \dots\dots\dots\dots\dots\dots \text{(i)}$
$ \text{Let (a, b) R (c, d) for (a, b), (c, d)} \in \text{N} \times \text{N}$
$\therefore \text{ad (b + c) = bc (a + d)} \dots\dots\dots\dots\dots\dots\text{(ii)}$
$ \text{Also (c, d) R (a, b)} \because\text{cb (d + a) = da (c + b) (using ii)}$
$\therefore \text{R is symmetric} \dots\dots\dots\dots\text{(iii)}$
$\text{Let (a, b) R (c,d) and (c, d) R(e,f), for a, b, c, d, e, f,} \in\text{N}$
$\therefore \text{ad (b + c) = bc (a + d) and cf (d + e) = de (c + f)}$
$\therefore \frac{\text{b + c}}{\text{bc}} = \frac{\text{a + d}}{\text{ad}} \text{and} \frac{\text{d + e}}{\text{de}} = \frac{\text{c + f}}{\text{cf}}$
$\text{i.e} \frac{1}{\text{c}} + \frac{1}{\text{b}} =\frac{1}{\text{d}} +\frac{1}{\text{a}} \text{and} \frac{1}{\text{e}} + \frac{1}{\text{d}} = \frac{1}{\text{f}}+ \frac{1}{\text{c}}$
$\text{adding we get} \frac{1}{\text{c}}+ \frac{1}{\text{b}} + \frac{1}{\text{e}} + \frac{1}{\text{d}} = \frac{1}{\text{d}} + \frac{1}{\text{a}} + \frac{1}{\text{f}} + \frac{1}{\text{c}}$
$\Rightarrow \text{af (b + e) = be (a + f)}$
$\text{Hence (a, b) R (e,f)} \therefore \text{R is transitive}\dots\dots\dots\dots\text{(iv)}$
$\text{From (i),(iii) and (iv) R is an equivalence relation}$
View full question & answer→Question 115 Marks
Discuss the commutativity and associativity of binary operation $^{‘*’}$ defined on A = Q – {1} by the rule $\text{a} ^{*} \text{b = a – b + ab}$ for all a, b $\in$ A. Also find the identity element of $^{*}$ in A and hence find the invertible elements of A.
Answer$\text{a }{*} \text{ b} = \text{a - b + ab } \forall \text{ a}, \text{b} \in \text{A} = \text{Q} - [1]$
$\text{b }{*} \text{ a} = \text{b - a + ba}$
$\text{(a }{*} \text{ b}) \neq \text{b }{*} \text{ a} \Rightarrow$ is not commutative.
$\text{(a }{*} \text{ b) }{*} \text{ c} = \text{a - b + ab) }{*} \text{ c}$
$\text{= a – b – c + ab + ac – bc + abc}$
$\text{a }{*} \text{ (b }{*} \text{ c)} = \text{a }{*} \text{ (b - c + bc)}$
$\text{= a – b + c + ab – ac – bc + abc}$
$\text{(a }{*} \text{ b) }{*} \text{ c} \neq \text{a }{*} \ \text{(b } {*}\text{ c)}$
$\Rightarrow \text{ }{*}$ is not associative.
Existence of identity
| $\text{a }{*} \text{ e = a – e + ae = a}$ |
$\text{e }{*} \text{ a = e – a + ea = a }$ |
| $\Rightarrow \text{e (a - 1) = 0}$ |
$\Rightarrow \text{e (1 + a) = 2a}$ |
| $\Rightarrow \text{e = 0}$ |
$\Rightarrow \text{e} = \frac{\text{2a}}{\text{1 + a}}$ |
$\because$ e is not unique
$\therefore$ No idendity element exists.
$\text{a }{*} \text{ b} = \text{e = b }{*} \text{ a}$
$\therefore$ No identity element exists.
$\Rightarrow$ Inverse element does not exist. View full question & answer→Question 125 Marks
Consider $\text{f} : \text{R}_{+} \rightarrow [ - 5, \infty)$ given by $\text{f}(x) = 9x^{2} + 6x - 5.$ Show that f is invertible with $\text{f}^{-1}\text{(y)} = \bigg(\frac{\sqrt{\text{y} + 6} - 1}{3}\bigg).$
Hence Find:
- $\text{f}^{-1} (10)$
- $\text{y if }\text{f}^{-1} \text{(y)} = \frac{4}{3},$
where $R_{+ }$ is the set of all non$-$negative real numbers. AnswerClearly $\text{f}^{-1} \text{(y) = g (y):} [ -5, \infty) \rightarrow \text{R}_{+} \text{ and},$
$\text{fog (y) = y} \bigg(\frac{\sqrt{\text{y}\text{ + } 6} - 1}{3}\bigg) = 9 \bigg(\frac{\sqrt{\text{y + 6}} - 1}{3} \bigg)^{2} + 6 \bigg(\frac{\sqrt{\text{y + 6}} - 1}{3}\bigg) - 5 = \text{y}$
$\text{and (gof) (x) = g} (\text{9x}^{2} + \text{6x - 5)} = \frac{\sqrt{\text{9x}^{2} + \text{6x + 1}} - 1}{3} = \text{x}$
$\therefore \text{g = f}^{-1}$
- $\text{f}^{-1} (10) = \frac{\sqrt{16} - 1}{3} = 1$
- $\text{f}^{-1} (\text{y)} = \frac{4}{3} \Rightarrow \text{y = 19}$
View full question & answer→Question 135 Marks
$\text{Let f : N}\rightarrow\text{N}$ be a function defined as $\text{f(x)} = 9x^{2} + 6x -5.$ Show that $\text{f : N}\rightarrow\text{S},$ where S is the range of f, is invertible. Find the inverse of f and hence find $\text{f}^{-1}(43) \text{and f}^{-1}(163). $
Answer$\text{Let x}_{1},\text{x}_{2}\in \text{N and f (x}_{1}) =\text{f(x}_{2}) $
$\Rightarrow\text{9x}^{2}_{1} + 6\text{x}_{1} - 5 = \text{9x}^{2}_{2} + 6\text{x}_{2} - 5$
$^2_1 - \text {x}^{2}_{2}) + 6(\text{x}_{1} - \text{x}_{2}) = 0 \Rightarrow\text{(x}_{1} - \text{x}_{2}) \text{(9x}_{1} + \text{9x}_{2} + 6) = 0$
$\Rightarrow\text{x}_{1} - \text{x}_{2} = \text{0 or x}_{1} = \text{x}_{2}\text{ as}\text{(9x}_{1} + \text{9x}_{2} + 6\neq 0,\text{x}_{1}, \text{x}_{2} \in \text{N}$
$\therefore$ f is one one function
$\text{f : N}\rightarrow\text{S is ONTO as co-domain = Range}$
Hence f is invertible
$\text{y} = \text{9x}^{2} + \text{6x} - 5 = (\text{3x + 1)}^{2} - 6 \Rightarrow\text{x} = \frac{\sqrt{\text{y} + 6} -1}{3}$
$\therefore\text{f}^{-1}\text{(y)} = \frac{\sqrt{\text{y} + 6} - 1}{3}, \text{y}\in\text{S}$
$\text{f}^{-1}(43) = \frac{\sqrt{49}-1}{3} = 2$
$\text{f}^{-1}(163) = \frac{\sqrt{169} - 1}{3} = 4$
View full question & answer→Question 145 Marks
Find the values of $x$ for which $f(x) = [x (x - 2)]^2$ is an increasing function. Also, find the points on the curve, where the tangent is parallel to $x-$axis.
Answer$f (x) = [x (x – 2)]^2$
$f′ (x) = 4x (x – 2) (x – 1)$
$f′ (x) = 0$ gives $x = 0, x = 1$ or $x = 2$
$\therefore$ Intervals are $(-\propto,0), (0, 1), (1, 2), (2, \propto)$
Increasing in $[0, 1]$ and $[2, \propto]$
$\Rightarrow\text{OR} 0 < x < 1$ and $x > 2$
The point where tangents are parallel to $x$ axis
are $(0, 0), (1, 1), (2, 0).$
View full question & answer→Question 155 Marks
Show that the function $\text{f} : \text{R}\rightarrow\text{R}$ defined by $\text{f(x})=\frac{\text{x}}{\text{x}^2+1},\forall\text{ x}\in\text{R}$ is neither one-one nor onto. Also, if $\text{g} : \text{R}\rightarrow\text{R}$ is defined as g(x) = 2x – 1, find fog(x).
Answer$\text{f(x})=\frac{\text{x}}{\text{x}^2+1}$
For one-one f(x) = f(y)
$\frac{\text{x}}{\text{x}^2+1}=\frac{\text{y}}{\text{y}^2+1}$
$\text{xy}^2+\text{x}=\text{yx}^2+\text{y}$
$\text{xy}(\text{y}-\text{x})=\text{y}-\text{x}$
$\text{xy}=1$
$\text{x}=\frac{1}{\text{y}}$
$\text{x}\neq\text{y}$
So, not one-one
For onto f(x) = y
$\frac{\text{x}}{\text{x}^2+1}=\text{y}$
$\text{x}=\text{yx}^2+\text{y}$
$\text{x}^2\text{y}+\text{y}-\text{x}=0$
x cannot be express in y so not onto
As g(x) = 2x – 1
$\text{fog(x})=\text{f[g(x})]=\text{f}(2\text{x}-1)=\frac{2\text{x}-1}{(2\text{x}-1)^2+1}$
$=\frac{2\text{x}-1}{4\text{x}^2-4\text{x}+2}$
View full question & answer→Question 165 Marks
Let $\text{A}=\{\text{x}\in\text{Z}:0\leq\text{x}\leq12\}.$ Show that, $\text{R}=\{(\text{a, b}):\text{a, b}\in\text{A},\ |\text{a}-\text{b}|$ is divisible by 4} is an equivalence relation. Find the set of all elements related to 1. Also write the equivalence class [2].
AnswerR = {(a, b) : a, b ∈ A, |a – b| is divisible by 4}
Reflexivity: for any a ∈ A
|a – a| = 0, which is divisible by 4
(a, a) ∈ R.
So, R is reflexive.
Symmetry: Let (a, b) ∈ R
|a – b| is divisible by 4
⇒ |b – a| is also divisible by 4
So, R is symmetry
Transitive: Let (a, b) ∈ R & (b, c) ∈ R
|a – b| is divisible by 4
$|\text{a}-\text{b}|=4\lambda$
$\text{a}-\text{b}=\pm4\lambda\ \ \ .....(1)$
$|\text{b}-\text{c}|$ is divisible by 4
$|\text{b}-\text{c}|=4\mu$
$\text{b}-\text{c}=\pm4\mu\ \ \ .....(2)$
Add (1) & (2)
$\text{a}-\text{b}+\text{b}-\text{c}=\pm4(\lambda+\mu)$
$\text{a}-\text{c}=\pm4(\lambda+\mu)$
$\Rightarrow(\text{a, c})\in\text{R}$
So, Transitive
Hence, R is reflexive, Symmetry & Transitive so, it is an equivalence relation
Let x be an element of A such that (x, 1) ∈ R, then
|x – 1| is divisible by 4
x – 1 = 0, 4, 8, 12
⇒ x = 1, 5, 9
Hence, the set of all element of A which are related to 1 in {1, 5, 9}.
View full question & answer→Question 175 Marks
If R and S are relations on a set A, then prove that:
- R and S are symmetric $\Rightarrow\ \text{R}\cap\text{S}$ and $\text{R}\cup\text{S}$ are symmetric
- R is reflexive and S is any relation $\Rightarrow\ \text{R}\cup\text{S}$ is reflexive.
Answer
- R and S are symmetric relations on the set A.
$\Rightarrow\ \text{R}\subset\text{A}\times\text{A}$ and $\text{S}\subset\text{A}\times\text{A}$
$\Rightarrow\ \text{R}\cap\text{S}\subset\text{A}\times\text{A}$
Thus, $\text{R}\cap\text{S}$ is a relation on A.
Let $\text{a, b}\in\text{A}$ such that $(\text{a, b})\in\text{R}\cap\text{S.}$ Then,
$(\text{a, b})\in\text{R}\cap\text{S}$
$\Rightarrow\ (\text{a, b})\in\text{R}$ and $(\text{a, b})\in\text{S}$
$\Rightarrow\ (\text{b, a})\in\text{R}$ and $(\text{b, a})\in\text{S}$ [Since R and S are symmetric]
$\Rightarrow\ (\text{b, a})\in\text{R}\cap\text{S}$
Thus,
$(\text{a, b})\in\text{R}\cap\text{S}$
$\Rightarrow\ (\text{b, a})\in\text{R}\cap\text{S}$ for all $\text{a, b}\in\text{A}$
So, $\text{R}\cap\text{S}$ is symmetric on A.
Also,
Let $\text{a, b}\in\text{A}$ such that $(\text{a, b})\in\text{R}\cup\text{S}$
$\Rightarrow\ (\text{a, b})\in\text{R}$ or $(\text{a, b})\in\text{S}$
$\Rightarrow\ (\text{b, a})\in\text{R}$ or $(\text{b, a})\in\text{S}$ [Since R and S are symmetric]
$\Rightarrow\ (\text{b, a})\in\text{R}\cup\text{S}$
So, $\text{R}\cup\text{S}$ is symmetric on A.
- R is reflexive and S is any relation.
Suppose $\text{a}\in\text{A.}$ Then,
$(\text{a, a})\in\text{R}$ [Since R is reflexive]
$\Rightarrow\ (\text{a, a})\in\text{R}\cup\text{S}$
$\Rightarrow\ \text{R}\cup\text{S}$ is reflexive on A. View full question & answer→Question 185 Marks
Let A = {1, 2, 3, ... 9} and R be the relation in A × A defined by (a, b)R(c, d) if a + d = b + c for (a, b), (c, d) in A × A. Prove that R is an equivalence relation and also obtain the equivalent class [(2, 5)].
AnswerGiven that, A = {1, 2, 3, ... 9} and (a, b)R(c, d) if a + d = b + c for (a, b) ∈ A × A and (c, d) ∈ A × A.Let (a, b)R(a, b)
⇒ a + b = b + a, ∀ a, b ∈ A which is true for any a, b ∈ A.
Since, the sum of two numbers doesn’t depend on the order.
Hence, R is reflexive.
Let (a, b)R(c, d)
Now, a + d = b + c
Also, c + b = d + a
⇒ (c, d)R(a, b)
Hence, R is symmetric.
Let (a, b)R(c, d) and (c, d)R(e, f)
$\therefore$ a + d = b + c and c + f = d + e
⇒ a + d = b + c and d + e = c + f
⇒ (a + d) - (d + e) = (b + c) - (c + f)
⇒ a - e = b - f
⇒ a + f = b + e
⇒ (a, b)R(e, f)
Hence, R is transitive.
Since, R is symmetric, reflexive and transitive.
Hence, R is an equivalence relation.
The equivalence class containing [(2, 5)] is given by {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}.
View full question & answer→Question 195 Marks
Consider $\text{f}:\text{R}_+\rightarrow[-5,\infty)$ given by $f(x) = 9x^2 + 6x - 5.$ Show that $f$ is invertible with $\text{f}^{-1}(\text{x})=\frac{\sqrt{\text{x}+6}-1}{3}.$
AnswerInjectivity of $f :$ Let $x$ and $y$ be two elements of domain $(R_+),$ such that $f(x) = f(y)$
$\Rightarrow 9x^2 + 6x - 5 = 9y^2 + 6y - 5$
$\Rightarrow 9x^2 + 6x = 9y^2 + 6y$
$\Rightarrow x = y (As, \text{x, y}\in\text{R}_+)$
So, $f$ is one$-$one.
Surjectivity of $f :$ Let $y$ is in the co domain $(Q)$ such that $f(x) = y$
$\Rightarrow 9x^2 + 6x - 5 = y$
$\Rightarrow 9x^2 + 6x = y + 5$
$\Rightarrow 9x^2 + 6x + 1 = y + 6 ($ Adding $1$ on both sides$)$
$\Rightarrow (3x + 1)^2 = y + 6$
$\Rightarrow\ \text{3x}+1=\sqrt{\text{y}+6}$
$\Rightarrow\ 3\text{x}=\sqrt{\text{y}+6}-1$
$\Rightarrow\ \text{x}=\frac{\sqrt{\text{y}+6}-1}{3}\in\text{R}^+ ($domain$)$
$\Rightarrow f$ is onto.
So, $f$ is a bijection and hence, it is invertible.
Finding $f^{ -1}:$ Let $f^{-1}(x) = y .....(1)$
$\Rightarrow x = f(y)$
$\Rightarrow x = 9y^2 + 6y - 5$
$\Rightarrow x + 5 = 9y^2 + 6y$
$\Rightarrow x + 6 = 9y^2 + 6y + 1 ($adding $1$ on both sides$)$
$\Rightarrow x + 6 = (3y + 1)^2$
$\Rightarrow\ 3\text{y}+1=\sqrt{\text{x}+6}$
$\Rightarrow\ 3\text{y}=\sqrt{\text{x}+6}-1$
$\Rightarrow\ \text{y}=\frac{\sqrt{\text{x}+6}-1}{3}$
So, $\text{f}^{-1}(\text{x})=\frac{\sqrt{\text{x}+6}-1}{3} [$from $(1)]$
View full question & answer→Question 205 Marks
If $\text{f(x)}=\sqrt{1-\text{x}}$ and $\text{g(x)}=\log_\text{e}\text{x}$ are two real functions, then describe, functions fog and gof.
Answer$\text{f(x)}={1-\text{x}}$For domain,
$1-\text{x}\geq0$
$\text{x}\leq1$
⇒ domain of g
$\text{f}:(-\infty,1]\rightarrow0,\infty=\log_\text{e}\text{x}$
Clearly, $\text{g}:0,\infty\rightarrow\text{R}$
Computation of fog: Clearly, the range of g is not a subset of the domain of f.
Therefore,
We need to compute the domain of fog.
⇒ Domain fog = x : x $\in$ Domain g and $\text{g}(\text{x})\in$ Domain of f
⇒ Domain fog = x : x : x $\in0,\ \infty$ and $\log_\text{e}\text{x}\in(-\infty,1]$
⇒ Domain fog = x : x : x $\in0,\ \infty$ and $\text{x}\in(0,\text{e}]$
⇒ Domain fog = x : x $\in(0,\text{e}]$
⇒ Domain fog = (0, e]
⇒ fog : 0, e → R
Therefore,
(fog)(x) = f(g(x))
$=\text{f}(\log_\text{e}\text{x})$
$=\sqrt{1-\log_\text{e}\text{x}}$
Computation of gof: Clearly, the range of f is a subset of the domain of g.
$\Rightarrow\ \text{gof}:(-\infty,1]\rightarrow\text{R}$
$(\text{gof})(\text{x})=\text{g(f(x))}$
$=\text{g}(\sqrt{1-\text{x}})$
$\Rightarrow\ \log_\text{e}\sqrt{1-\text{x}}$
$=\log_\text{e}(1-\text{x})^\frac{1}{2}$
$=\frac{1}{2}\log_\text{e}(1-\text{x})$
View full question & answer→Question 215 Marks
Let n be a fixed positive integer. Define a relation R on Z as follows:
$(\text{a, b})\in\text{R}\Leftrightarrow\ \text{a}-\text{b}$ is divisible by n. Show that R is an equivalence relation on Z.
AnswerWe observe the following properties of R.
Reflexivity: Consider $\text{a}\in\text{N}$
Here, a - a = 0 = 0 × n
Implies that a - a is divisible by n
Implies that $\text{a, a}\in\text{R}$
Implies that $\text{a, a}\in\text{R}$ for all $\text{a}\in\text{Z}.$
So, R is reflexive on Z.
Symmetry: Consider $\text{a, b}\in\text{R}$
Here a - b is divisible by n
Implies that a - b = np for some $\text{p}\in\text{Z}$
Implies that b - a = n - p.
Implies that b - a is divisible by n $\big[\text{p}\in\text{Z}$ implies that $-\text{p}\in\text{Z}\big]$
implies that $\text{b, a}\in\text{R}$
So, R is symmetric on Z.
Transitivity: Consider a, b and b, c $\in\text{R}$
Here, a - b is divisible by n and b - c is divisible by n.
implies that a - b = np for some $\text{p}\in\text{Z}$ and b - c = nq for some $\text{q}\in\text{Z}$
Adding the above two
we get a - b + b - c = np + nq
Implies that a - c = n(p + q).
Here, $\text{p}+\text{q}\in\text{Z}$
Implies that $\text{a, c}\in\text{R}$ for all $\text{a, c}\in\text{Z.}$
So, R is transitive on Z.
Hence, R is an equivalence relation on Z.
View full question & answer→Question 225 Marks
Prove that the relation R on Z defined by $(\text{a, b})\in\text{R}\Leftrightarrow\ \text{a}-\text{b}$ is divisible by 5 is an equivalence relation on Z.
AnswerWe observe the following properties of relation R.
Reflexivity: Let a be an arbitrary element of R. Then,
⇒ a - a = 0 = 0 × 5
⇒ a - a is divisible by 5
$\Rightarrow\ (\text{a, a})\in\text{R}$ for all $\text{a}\in\text{Z}$
So, R is reflexive on Z.
Symmetry: Let $(\text{a, b})\in\text{R}$
⇒ a - b is divisible by 5
⇒ a - b = 5p for some $\text{p}\in\text{Z}$
⇒ b - a = 5(-p)
Here, $-\text{p}\in\text{Z}$ [Since $\text{p}\in\text{Z}$]
⇒ b - a is divisible by 5
$\Rightarrow\ (\text{b, a})\in\text{R}$ for all $\text{a, b}\in\text{Z}$
So, R is symmetric on Z.
Transitivity: Let a, b and $\text{b},\text{c}\in\text{R}$
⇒ a - b is divisible by 5
⇒ a - b = 5p for some Z
Also, b - c is divisible by 5
⇒ b - c = 5q for some Z
Adding the above two, we get
a - b + b - c = 5p + 5q
⇒ a - c = 5(p + q)
⇒ a - c is divisible by 5
Here, $\text{p}+\text{q}\in\text{Z}$
$\Rightarrow\ \text{a, c}\in\text{R}$ for all $\text{a, c}\in\text{Z}$
So, R is transitive on Z.
Hence, R is an equivalence relation on Z.
View full question & answer→Question 235 Marks
Let A =R×R and * be a binary operation on A defined by,
(a, b) * (c, d) = (a + c, b + d).
Show that * is commutative and associative. Find the binary element for * on A, if any.
AnswerWe have,
A = R × R and * is a binary operation on A defined by (a, b) * (c, d) = (a + c, b + d).
Now,
(a, b) * (c, d) = (a + c, b + d) = (c + a, d + b)
⇒ (a, b) * (c, d) = (c, d) * (a, b)
So, * is commutative.
Also,
(a, b) * [(c, d) * (e, f)] = (a, b) * (c + e, d + f)
= (a, b) * (c + e, d + f)
= (a + c + e, b + d + f)
= (a + c, b + d) * (e, f)
= [(a, b) * (c, d)] * (e, f)
⇒ (a, b) * [(c, d) * (e, f)] = [(a, b) * (c, d)] * (e, f)
So, * is associative.
Let (x, y) be the binary element for * on A.
(a, b) * (x, y) = (a, b) = (x, y) * (a, b)
⇒ (a + x, b + y) = (a, b)
⇒ a + x = a and b + y = b
⇒ x = 0 and y = 0
Hence, (0, 0) is the binary element for * on A.
View full question & answer→Question 245 Marks
Classify the following functions as injection, surjection or bijection :
$f : R \rightarrow R,$ defined by $f(x) = x^3 - x$
Answer$f : R \rightarrow R,$ defined by $f(x) = x^3 - x$
Injective:
Let $\text{x, y}\in\text{R}$ such that$, f(x) = f(y)$
$\Rightarrow x^3 - x = y^3 - y$
$\Rightarrow x^3 - y^3 - (x - y) = 0$
$\Rightarrow (x - y)(x^2 + xy + y^2 - 1) = 0$
$\because\ \text{x}^2+\text{xy}+\text{y}^2\geq0$
$\Rightarrow\ \text{x}^2+\text{xy}+\text{y}^2-1\geq-1$
$\therefore\ \text{x}^2+\text{xy}+\text{y}^2-1\neq0$
$\Rightarrow \text{x}-\text{y}=0\Rightarrow \text{x}=\text{y}$
$\therefore f$ is one$-$one.
Surjective:
Let $\text{y}\in\text{R},$ then $f(x) = y$
$\Rightarrow x^3 - x - y = 0$
We know that a degree $3$ equation has atleast one real solution.
Let $\text{x}=\alpha$ be that real solution
$\therefore\ \alpha^2-\alpha=\text{y}$
$\Rightarrow\ \text{f}(\alpha)=\text{y}$
$\therefore$ For each $\text{y}\in\text{R,}$ there exist $\text{x}=\alpha\in\text{R}$ such that $\text{f}(\alpha)=\text{y}$
$\therefore f$ is onto.
View full question & answer→Question 255 Marks
Let $f$ be a real function given by $\text{f(x)}=\sqrt{\text{x}-2}.$ Find the following: $(fofof)(38) $ Also, show that $fof \neq f^2.$
AnswerWe have, $\text{f(x)}=\sqrt{\text{x}-2}$
Clearly, Domain $(\text{f})=[2,\infty)$ and Range $(\text{f})=[0,\infty).$
We observe that range $(f)$ is not a subset of domain of $f.$
$\therefore$ Domain of $(fof) = x : x \in$ Domain $(f)$ and $f(x) \in$ Domain $(f) $
$= x : x \in2,\infty$ and $\sqrt{\text{x}-2}\in[2,\infty$
$= x : x \in2,\infty$ and $\sqrt{\text{x}-2}\geq2$
$= x : x \in2,\infty$ and $\text{x}-2\geq4$
$= x : x \in2,\infty$ and $\text{x}\geq6$
$=[6,\infty)$
Clearly, range of $\text{f}=[0,\infty)⊄$ domain of $fof.$
$\therefore$ Domain of $((fof)of) = x : x \in$ domain $(f)$ and $f(x) \in$ Domain $(fof)$
$= x : x \in2,\infty$ and $\sqrt{\text{x}-2}\in[6,\infty$
$= x : x \in2,\infty$ and $\sqrt{\text{x}-2}\geq6$
$= x : x \in2,\infty$ and $\text{x}-2\geq36$
$= x : x \in2,\infty$ and $\text{x}\geq38$
$=[38,\infty)$
Now, $(fof)(x) = f(f(x))$
$=\text{f}(\sqrt{\text{x}-2})$
$=\sqrt{\sqrt{\text{x}-2}-2}$
$(fofof)(x) = (fof)(f(x))$
$=(\text{fof})(\sqrt{\text{x}-2})$
$=\sqrt{\sqrt{\sqrt{\text{x}-2}-2}-2}$
$\therefore\ \text{fofof}:[38,\infty)\rightarrow\text{R}$ defined as
$(\text{fofof})(\text{x})=\sqrt{\sqrt{\sqrt{\text{x}-2}-2}-2}$
$(\text{fofof})(38)=\sqrt{\sqrt{\sqrt{38-2}-2}-2}$
$=\sqrt{\sqrt{\sqrt{36}-2}-2}=\sqrt{\sqrt{6-2}-2}$
$=\sqrt{\sqrt{4}-2}$
$=\sqrt{2-2}$
$=0$
View full question & answer→Question 265 Marks
Let A = {a, b, c}, B = {u, v, w} and let f and g be two functions from A to B and from B to A, respectively, defined as:
f = {(a, v), (b, u), (c, w)}, g = {(u, b), (v, a), (w, c)}.
Show that f and g both are bijections and find fog and gof.
AnswerProving f is a bijection:f = {(a, v), (b, u), (c, w)} and f : A → B
Injectivity of f: No two elements of A have the same image in B.
So, f is one-one.
Surjectivity of f: Co-domain of f = {u v, w}
Range of f = {u v, w}
Both are same.
So, f is onto.
Hence, f is a bijection.
Proving g is a bijection:
g = {(u, b), (v, a), (w, c)} and g : B → A
Injectivity of g: No two elements of B have the same image in A.
So, g is one-one.
Surjectivity of g: Co-domain of g = {a, b, c}
Range of g = {a, b, c}
Both are the same.
So, g is onto.
Hence, g is a bijection.
Finding fog: Co-domain of g is same as the domain of f.
So, fog exists and fog: {u v, w} → {u v, w}
(fog)(u) = f(g(u)) = f(b) = u
(fog)(v) = f(g(v)) = f(a) = v
(fog)(w) = f(g(w)) = f(c) = w
So, fog = {(u, u), (v, v), (w, w)}
Finding gof.
Co-domain of f is same as the domain of g.
So, fog exists and gof: {a, b, c} → {a, b, c}
(gof)(a) = g(f(a)) = g(v) = a
(gof)(b) = g(f(b)) = g(u) = b
(gof)(c) = g(f(c)) = g(w) = c
So, gof = {(a, a), (b, b), (c, c)}
View full question & answer→Question 275 Marks
Let $\text{f}:[-1,\infty)\rightarrow[-1,\infty)$ be given by $f(x) = (x + 1)^2 - 1, \text{x}\geq-1.$ Show that $f$ is invertible. Also, find the set $S = {x : f(x) = f^{-1}(x)}.$
AnswerInjectivity: Let $x$ and $\text{y}\in[-1,\infty),$ such that
$f(x) = f(y)$
$\Rightarrow (x + 1)^2 - 1 = (y + 1)^2 - 1$
$\Rightarrow (x + 1)^2 = (y + 1)^2$
$\Rightarrow (x + 1) = (y + 1)$
$\Rightarrow x = y$
So$, f$ is a injection.
Surjectivity: Let $\text{y}\in[-1,\infty)$
Then$, f(x) = y$
$\Rightarrow (x + 1)^2 - 1 = y$
$\Rightarrow\ \text{x}+1=\sqrt{\text{y}+1}$
$\Rightarrow\ \text{x}=\sqrt{\text{y}+1}-1$
Clearly, $\text{x}=\sqrt{\text{y}+1}-1$ is real for all $\text{y}\geq-1.$
Thus, every element $\text{y}\in[-1,\infty)$ has its pre$-$image $\text{x}\in[-1,\infty)$
given by $\text{x}=\sqrt{\text{y}+1}-1$
$\Rightarrow f$ is a surjection.
So, $f$ is a bijection.
Hence, f is invertible.
Let $f^{-1}(x) = y ......(1)$
$\Rightarrow f(y) = x$
$\Rightarrow (y + 1)^2 - 1 = x$
$\Rightarrow (y + 1)^2 = x + 1$
$\Rightarrow\ \text{y}+1=\sqrt{\text{x}+1}$
$\Rightarrow\ \text{y}=\pm\sqrt{\text{x}+1}-1$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\pm\sqrt{\text{x}+1}-1 [$from $(1)]$
$\text{f}(\text{x})=\text{f}^{-1}(\text{x})$
$\Rightarrow\ (\text{x}+1)^2-1=\pm\sqrt{\text{x}+1}-1$
$\Rightarrow\ (\text{x}+1)^2=\pm\sqrt{\text{x}+1}$
$\Rightarrow (x + 1)^4 = x + 1$
$\Rightarrow (x + 1)[(x + 1)^3 - 1] = 0$
$\Rightarrow x + 1 = 0$ or $(x + 1)^3 - 1 = 0$
$\Rightarrow x = -1$ or $(x + 1)^3 = 1$
$\Rightarrow x = -1$ or $x + 1 = 1$
$\Rightarrow x = -1$ or $x = 0$
$\Rightarrow S = {0, -1}$
View full question & answer→Question 285 Marks
m is said to be related to n if m and n are integers and m - n is divisible by 13. Does this define an equivalence relation?
AnswerWe observe the following properties of relation R. $\text{Let R} = \big\{(\text{m, n}): \text{(m, n)}\in\text{Z}:\text{m}-\text{n}\ \text{is divisible by} 13\big\}$$$$$Relexivity: Let m be an arbitrary element of Z. Then, $\text{m}\in\text{R}$
⇒ m - m = 0 = 0 × 13 ⇒ m - m is divisible by 13 ⇒ (m, m) is reflexive on Z. Symmetry: Let $(\text{m, n})\in\text{R.}$ Then, ⇒ m - n is divisible by 13 ⇒ m - n = 13p Here, $\text{p}\in\text{Z}$ ⇒ n - m = 13(-p) Here, $-\text{p}\in\text{Z}$ ⇒ n - m is divisible by 13$\Rightarrow\ (\text{n, m})\in\text{R}$ for all $\text{m, n}\in\text{Z}$
So, R is symmetric on Z. Transitivity: Let (m, n) and (n, o) $\in\text{R}$ ⇒ m - n and n - o are divisible by 13 ⇒ m - n = 13p and n - o = 13q for some $\text{p, q}\in\text{Z}$ Adding the above two, we get ⇒ m - n + n - o = 13p + 13q ⇒ m - o = 13(p + q) Here, $\text{p}+\text{q}\in\text{Z}$ ⇒ m - o is divisible by 13 $\Rightarrow\ (\text{m, o})\in\text{R}$ for all $\text{m, o}\in\text{Z}$ So, R is transitive on Z. Hence, R is an equivalence relation on Z.
View full question & answer→Question 295 Marks
On R − {1}, a binary operation * is defined by a * b = a + b − ab. Prove that * is commutative and associative. Find the identity element for * on R − {1}. Also, prove that every element of R − {1} is invertible.
AnswerCommutativity:Let, $\text{a},\text{b}\in\text{R}-\{1\}.$ Then,
a * b = a + b - ab
= b + a - ba
= b * a
Therefore,
a * b = b * a, $\forall\ \text{a},\text{b}\in\text{R}-\{1\}$
Thus, * is commutative on R - {1}.
Associativity:
Let, $\text{a},\text{b}\in\text{R}-\{1\}.$ Then,
a * (b * c) = a * (b + c - bc)
= a + b + c - bc - a(b + c - bc)
= a + b + c - bc - ab - ac - abc
(a * b) * c = (a + b - ab) * c
= a + b - ab + c - (a + b - ab)c
= a + b + c - ab - ac - bc + abc
Therefore,
a * (b * c) = (a * b) * c, $\forall\text{ a},\text{b},\text{c}\in\text{R}-\{1\}$
Thus, * is associative on R - {1}.
Finding identity element:
Let e be the element in R - {1} with respect to * such that
a * e = a = e * a, $\forall\text{ a}\in\text{R}-\{1\}$
a * e = a and e * a = a, $\forall\text{ a}\in\text{R}-\{1\}$
⇒ a + e - ae = a and e + a - ea = a, $\forall\text{ a}\in\text{R}-\{1\}$
e(1 - a) = 0, $\forall\text{ a}\in\text{R}-\{1\}$
$\text{e}=0\in\forall\text{ a}\in\text{R}-\{1\},\forall\text{ a}\in\text{R}-\{1\}$ $[\because\ \text{a}\neq1]$
Thus, 0 is the identity element in R - {1} with respect to *.
Finding inverse:
Let $\text{a}\in\text{R}-\{1\}$ and $\text{b}\in\text{R}-\{1\}$ be the inverse of a. Then,
a * b = e = b * a
a * b = e and b * a = e
⇒ a + b - ab = 0 and b + a - ba = 0
⇒ a = ab - b
⇒ a = b(a - 1)
$\Rightarrow\text{b}=\frac{\text{a}}{\text{a}-1}$
Thus, $\frac{\text{a}}{\text{a}-1}$ is inverse of $\text{a}\in\text{R}-\{1\}.$
View full question & answer→Question 305 Marks
If $f : R \rightarrow (-1, 1)$ defined by $\text{f(x)}=\frac{10^{\text{x}}-10^{-\text{x}}}{10^{\text{x}}+10^{-\text{x}}}$ is invertible, find $f^{-1}.$
AnswerInjectivity of $f$: Let $x$ and $y$ be two elements of domain $(R),$ such that $f(x) = f(y)$
$\Rightarrow\ \frac{10^{\text{x}}-10^{-\text{x}}}{10^{\text{x}}-10^{-\text{x}}}=\frac{10^{\text{y}}-10^{-\text{y}}}{10^{\text{y}}-10^{-\text{y}}}$
$\Rightarrow\ \frac{10^{-\text{x}}(10^{2\text{x}}-1)}{10^{-\text{x}}(10^{2\text{x}}+1)}=\frac{10^{-\text{y}}(10^{2\text{y}}-1)}{10^{-\text{y}}(10^{2\text{y}}+1)}$
$\Rightarrow\ \frac{(10^{2\text{x}}-1)}{(10^{2\text{x}}+1)}=\frac{(10^{2\text{y}}-1)}{(10^{2\text{y}}+1)}$
$\Rightarrow (10^{2x} - 1)(10^{2y} + 1) = (10^{2x} + 1)(10^{2y} - 1)$
$\Rightarrow 10^{2x+2y} + 10^{2x} - 10^{2y} - 1 = 10^{2x+2y} - 10^{2x} + 10^{2y} - 1$
$\Rightarrow 2 \times 10^{2x} = 2 \times 10^{2y}$
$\Rightarrow 10^{2x} = 10^{2y}$
$\Rightarrow 2x = 2y$
$\Rightarrow x = y$
So, $f$ is one$-$one.
Surjectivity of $f:$ Let $y$ is in the co domain $(R),$ such that $f(x) = y$
$\Rightarrow\ \frac{10^{\text{x}}-10^{-\text{x}}}{10^{\text{x}}+10^{-\text{x}}}=\text{y}$
$\Rightarrow\ \frac{10^{-\text{x}}(10^{2\text{x}}-1)}{10^{-\text{x}}(10^{2\text{x}}+1)}=\text{y}$
$\Rightarrow\ 10^{2\text{x}}-1=\text{y}\times10^{2\text{x}}+\text{y}$
$\Rightarrow\ 10^{2\text{x}}(1-\text{y})=1+\text{y}$
$\Rightarrow\ 10^{2\text{x}}=\frac{1+\text{y}}{1-\text{y}}$
$\Rightarrow\ 2\text{x}=\log\Big(\frac{1+\text{y}}{1-\text{y}}\Big)$
$\Rightarrow\ \text{x}=\frac{1}{2}\log\Big(\frac{1+\text{y}}{1-\text{y}}\Big)\in\text{R}$ (domain)
$\Rightarrow f$ is onto.
So, f is a bijection and hence, it is invertible.
Finding $f^{-1}:$ Let $f^{-1}(x) = y .......(1)$
$\Rightarrow f(y) = x$
$\Rightarrow\ \frac{10^{\text{y}}-10^{-\text{y}}}{10^{\text{y}}+10^{-\text{y}}}=\text{x}$
$\Rightarrow\ \frac{10^{-\text{y}}(10^{2\text{y}}-1)}{10^{-\text{y}}(10^{2\text{y}}+1)}=\text{x}$
$\Rightarrow\ 10^{2\text{y}}-1=\text{x}\times10^{2\text{y}}+\text{x}$
$\Rightarrow\ 10^{2\text{y}}(1-\text{x})=1+\text{x}$
$\Rightarrow\ 10^{2\text{y}}=\frac{1+\text{x}}{1-\text{x}}$
$\Rightarrow\ 2\text{y}=\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$
$\Rightarrow\ \text{y}=\frac{1}{2}\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$
So, $\text{f}^{-1}(\text{x})=\frac{1}{2}\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big) [$from $(1)]$
View full question & answer→Question 315 Marks
Consider the function $\text{f}:\text{R}^{+}\rightarrow[-9,\infty]$ given by $f(x) = 5x2 + 6x - 9.$ Prove that $f$ is invertible with $\text{f}^{-1}\text{(y)}=\frac{\sqrt{54+5\text{y}}-3}{5}.$
Answer$\text{f}:\text{R}^{+}\rightarrow\ [-9,\infty)$ given by $f(x) = 5x^2 + 6x - 9$ For any $\text{x, y}\in\text{R}^{+}$
$f(x) = f(y)$
$\Rightarrow 5x^2 + 6x - 9 = 5y^2 + 6y - 9$
$\Rightarrow 5(x^2 - y^2) + 6(x - y) = 0$
$\Rightarrow (x - y)[5(x + y) + 6] = 0$
$\Rightarrow (x - y) = 0 [\because5(\text{x}+\text{y})+6\neq0\text{ as x, y}\in\text{R}^{+}]$
$\Rightarrow x = y$
So, $f$ is an injection.
Let $y$ be an arbitrary element of $[-9,\infty).$
$f(x) = y$
$\Rightarrow 5x^2 + 6x - 9 = y$
$\Rightarrow 25x^2 + 30x - 45 = 5y$
$\Rightarrow 25x^2 + 30x + 9 - 54 = 5y$
$\Rightarrow (5x + 3)^2 = 5y + 54$
$\Rightarrow(5\text{x}+3)=\sqrt{5\text{y}+54}$
$\Rightarrow\ \text{x}=\frac{\sqrt{5\text{y}+54}-3}{5}$
Now, $\text{y}\in[-9,\infty)$
$\Rightarrow\ \text{y}\geq-9$
$\Rightarrow\ 5\text{y}+54\geq9$
$\Rightarrow\ \sqrt{5\text{y}+54}\geq3$
$\Rightarrow\ \sqrt{5\text{y}+54}-3\geq0$
$\Rightarrow\ \frac{\sqrt{5\text{y}+54}-3}{5}\geq0$
$\Rightarrow\ \text{x}\geq0\Rightarrow\ \text{x}\in\text{R}^{+}$
Thus, for every $\text{y}\in[-9,\infty)$ there exist $\text{x}=\frac{\sqrt{5\text{y}+54}-3}{5}\in\text{R}^{+}$ such that $f(x) = y.$
$So, \text{f}:\text{R}^{+}\rightarrow[-9,\infty)$ is onto.
Thus, $\text{f}:\text{R}^{+}\rightarrow[-9,\infty)$ is a bijection and hence invertible.
Let $f^{-1}$ denotes the inverse of $f.$
Then,
$(fof^{-1})(y) = y$ for all $\text{y}\in[-9,\infty)$
$f(f^{-1}(y)) = y$ for all $\text{y}\in[-9,\infty)$
$\Rightarrow 5(f^{-1}(y))^2 + 6(f^{-1}(y)) - 9 = y$ for all $\text{y}\in[-9,\infty)$
$\Rightarrow 25(f^{-1}(y))^2 + 30(f^{-1}(y)) - 45 = 5y$ for all $\text{y}\in[-9,\infty)$
$\Rightarrow 25(f^{-1}(y))^2 + 30(f^{-1}(y)) + 9 = 5y + 54$ for all $\text{y}\in[-9,\infty)$
$\Rightarrow {5f^{-1}(y) + 3}^2 = 5y + 54$ for all $\text{y}\in[-9,\infty)$
$\Rightarrow 5f^{-1}(y) + 3 =\sqrt{5\text{y}+54}$ for all $\text{y}\in[-9,\infty)$
$\Rightarrow\ \text{f}^{-1}(\text{y})=\frac{\sqrt{5\text{y}+54}-3}{5}$
View full question & answer→Question 325 Marks
Classify the following functions as injection, surjection or bijection: $f : R \rightarrow R,$ defined by $\text{f(x)}=\frac{{x}}{{x}^2+1}$
Answer$f : R \rightarrow R,$ defined by $\text{f(x)}=\frac{\text{x}}{\text{x}^2+1}$ Injection test:
Let $x$ and $y$ be any two elements in the domain $(R),$ such that $f(x) = f(y)$.
$f(x) = f(y)$
$\frac{\text{x}}{\text{x}^2+1}=\frac{\text{y}}{\text{y}^2+1}$
$xy^2 + x = x^2y + y$
$xy^2 - x^2y + x - y = 0$
$-xy(-y + x) + 1(x - y) = 0$
$(x - y)(1 - xy) = 0$
$x = y$ or ${x}=\frac{1}{\text{y}}$
So,$ f$ is not an injection.
Surjection test: Let $y$ be any element in the co $-$ domain $(R),$ such that $f(x) = y$ for some element $x$ in $R \ ($domain$)$.
$f(x) = y$
$\frac{\text{x}}{\text{x}^2+1}=\text{y}$
$yx^2 - x + y = 0$
${x}=\frac{-(-1)\pm\sqrt{1-4\text{y}^2}}{2\text{y}},$ if $\text{y}\neq0$
$=\frac{1\pm\sqrt{1-4\text{y}^2}}{2\text{y}},$ which may not be in $R$
For example, if $y = 1,$ then
${x}=\frac{1\pm\sqrt{1-4}}{2}=\frac{1\pm\text{i}\sqrt{3}}{2},$ which is not in $R$
So $,f$ is not surjection and $f$ is not bijection.
View full question & answer→Question 335 Marks
$A$ function $f : R \rightarrow R$ is defined as $f(x) = x^3 + 4$. Is it a bijection or not? In case it is a bijection, find $f^{-1}(3)$.
AnswerWe have,
$f : R \rightarrow R$ in a function defined by
$f(x) = x^3 + 4$
Injectivity: Let $f(x_1) = f(x_2)$ for ${x}_1,{x}_2\in\text{R}$
$\Rightarrow\ {x}_1^3+4= {x}_2^3+4$
$\Rightarrow\ {x}_1^3= {x}_2^3$
$\Rightarrow\ {x}_1= {x}_2$
$\Rightarrow f$ is one $-$ one.
Surjectivity: Let $\text{y}\in\text{R}$ be artritrary such that
$f(x) = y$
$\Rightarrow x^3 + 4 = y$
$\Rightarrow x^3 + 4 - y = 0$
We know that an odd degree equation must have a real root.
$\Rightarrow\ \alpha^3+4=\text{y}$
$\Rightarrow\ \text{f}(\alpha)=\text{y}$
$\Rightarrow f $ is onto.
Since f is one-one and onto.
$\Rightarrow f$ is bijective.
finally, $f(x) = y$
$\Rightarrow x^3 + 4 = y$
$\Rightarrow x^3 = y - 4$
$\Rightarrow\ {x}=({y}-4)^\frac{1}{3}$
$\therefore\ \text{f}^{-1}({x})=({x}-4)^\frac{1}{3}$
$\therefore\ \text{f}^{-1}(3)=(3-4)^\frac{1}{3}=-1$
View full question & answer→Question 345 Marks
Let f = {(3, 1), (9, 3), (12, 4)} and g = {(1, 3), (3, 3) (4, 9) (5, 9)}. Show that gof and fog are both defined. Also, find fog and gof.
Answerf = {(3, 1), (9, 3), (12, 4)} and g = {(1, 3), (3, 3) (4, 9) (5, 9)} f : {3, 9, 12} → {1, 3, 4} and g : {1, 3, 4, 5} → {3, 9}Co-domain of f is a subset of the domain of g.
Therefore, exists and (gof) : {3, 9, 12} → {3, 9}
(gof)(3) = g(f(3)) = g(1) = 3 (gof)(9) = g(f(9)) = g(3) = 3 (gof)(12) = g(f(12)) = g(4) = 9 (gof) = {(3, 3), (9, 3), (12, 9)} Co-domain of g is a subset of the domain of f. Therefore, (fog) exists and : {1, 3, 4, 5} → {3, 9, 12} (fog)(1) = f(g(1)) = f(3) = 1 (fog)(3) = f(g(3)) = f(3) = 1 (fog)(4) = f(g(4)) = f(9) = 3 (fog)(5) = f(g(5)) = f(9) = 3 fog = {(1, 1), (3, 1), (4, 3), (5, 3)}
View full question & answer→Question 355 Marks
Give examples of two surjective functions $f_1$ and $f_2$ from $Z$ to $Z$ such that $f_1 + f_2$ is not surjective.
AnswerWe know that $f_1 : R \rightarrow R,$ given by $f_1(x) = x,$ and $f_2(x) = -x$ are surjective functions.
Proving $f_1$ is surjective: Let $y$ be an element in the co $-$ domain $(R),$ such that $f_1(x) = y$.
$f_1(x) = y$
Implies that $x = y,$ which is in $R$.
Therefore, for every element in the co $-$ domain, there exists some pre $-$ image in the domain.
Therefore $, f_1$ is surjective.
Proving $f_2$ is surjective: Let $f_2(x) = y$
$x = y,$ which is in $R$.
Therefore, for every element in the co $-$ domain, there exists some pre $-$ image in the domain.
Therefore $, f_2$ is surjective.
Proving $(f_1 + f_2)$ is not surjective:
Given: $(f_1 + f_2)(x) = f_1(x) + f_2(x) = x + (-x) = 0$
Therefore, for every real number $x, (f_1 + f_2)(x) = 0$
Therefore, the image of every number in the domain is same as $0$.
Implies that Range $= {0}$
Co $-$ domain $= R$
Therefore, both are not same.
Therefore $, f_1 + f_2$ is not surjective.
View full question & answer→Question 365 Marks
Define a binary operation $*$ on the set $\{0, 1, 2, 3, 4, 5\}$ as:
$\text{a}\times\text{b}=\begin{cases}\text{a + b},&\text{if }\text{a + b}<6\\\text{a + b}-6,&\text{if }\text{a + b}\geq6\end{cases}$
Show that $0$ is the identity for this operation and each element $a ≠ 0$ of the set is invertible with $6 − a$ being the inverse of $a$.
AnswerLet $X = \{0, 1, 2, 3, 4, 5\}$.
The operation $*$ on $X$ is defined as:
$\text{a}\times\text{b}=\begin{cases}\text{a + b},&\text{if }\text{a + b}<6\\\text{a + b}-6,&\text{if }\text{a + b}\geq6\end{cases}$
An element $\text{e}\in\text{X}$ is the identity element for the operation $*,$ if
$a * e = a = e * a \forall\text{ a}\in\text{X}$.
For $\text{a}\in\text{X}$, we observed that:
$a * 0 = a + 0 = a [\text{a}\in\text{X}\Rightarrow\text{a}+0<6]$
$0 * a = a = 0 * a [\text{a}\in\text{X}\Rightarrow0+\text{a}<6]$
$\therefore a * 0 = a = 0 * a \forall\text{ a}\in\text{X}$
Thus$, 0$ is the identity element for the given operation $*$.
An element $\text{a}\in\text{X}$ is invertible if there exists be $X$ such that $a * b = 0 = b * a.$
i.e.,$\begin{cases}\text{a + b}=0=\text{b + a},&\text{if }\text{a + b}<6\\\text{a + b}-6=0=\text{b + a}-6,&\text{if }\text{a + b}\geq6\end{cases}$
i.e.,
$a = -b$ or $b = 6 - a$
But$, X = \{0, 1, 2, 3, 4, 5\}$ and $\text{a, b}\in\text{X}$.
Then, $\text{a}\neq-\text{b}$.
Therefore$, b = 6 - a$ is the inverse of $\text{a}\in\text{X}$.
Hence, the inverse of an element $\text{a}\in\text{X},\text{a}\neq0$ is $6 - a i.e., a^{-1} = 6 - a.$
View full question & answer→Question 375 Marks
Show that the function $f : Q \rightarrow Q,$ defined by $f(x) = 3x + 5,$ is invertible. Also, find $f^{-1}.$
AnswerGiven that $f : Q \rightarrow Q$ defined by $f(x) = 3x + 5.$
To prove that $f$ is invertible,
we need to prove that f is one$-$one and onto.
Let $\text{x, y}\in\text{Q}$ be such that, $f(x) = f(y)$
$\Rightarrow 3x + 5 = 3y + 5$
$\Rightarrow x = y$
So, $f$ is an injection.
Let $y$ be an arbitrary element of $Q$ such that $f(x) = y.$
$\Rightarrow 3x + 5 = y$
$\Rightarrow 3x = y - 5$
$\Rightarrow\ \text{x}=\frac{\text{y}-5}{3}$
Thus, for any $\text{y}\in\text{Q}$ there exists $\text{x}=\frac{\text{y}-5}{3}\in\text{Q}$ such that $\text{f(x)}=\text{f}\Big(\frac{\text{y}-5}{3}\Big)=3\frac{\text{y}-5}{3}+5=\text{y}$
Thus, $f : Q \rightarrow Q$ is a bijection and hence invertible.
Let $f^{-1}$ denotes the inverse of $f.$
Thus, fo $f^{-1}(x) = x$ for all $\text{x}\in\text{Q}$
$\Rightarrow f [f^{-1}(x)] = x$ for all $\text{x}\in\text{Q}$
$\Rightarrow 3f^{-1}(x) + 5 = x$ for all $\text{x}\in\text{Q}$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{\text{x}-5}{3}$ for all $\text{x}\in\text{Q}$
View full question & answer→Question 385 Marks
Show that the function $\text{f}:\text{R}_\ast\rightarrow\text{R}_\ast$ defined by $\text{f(x)}=\frac{1}{\text{x}}$ is one-one and onto, where $\text{R}_\ast$ is the set of all non-zero real numbers. Is the result true, if the domain $\text{R}_\ast$ is replaced by N with co-domain being same as $\text{R}_\ast?$
AnswerIt is given that $\text{f}:\text{R}_\ast\rightarrow\text{R}_\ast$ is defined by $\text{f(x)}=\frac{1}{\text{x}}$
One-one:
f(x) = f(y)
$\Rightarrow\frac{1}{\text{x}}=\frac{1}{\text{y}}$
⇒ x = y
$\therefore$ f is one-one.
Onto:
It is clear that for $\text{y}\in\text{R}_\ast,$ there exists$\text{x}=\frac{1}{\text{y}}\in\text{R}_\ast\ (\text{Exists as y}\neq0)$ such that $\text{f(x)}=\frac{1}{\frac{1}{\text{y}}}=\text{y}.$
$\therefore$ f is onto.
Thus, the given function (f) is one-one and onto.
Now, consider function $\text{g}:\text{N}\rightarrow\text{R}_\ast$ defined by
$\text{g(x)}=\frac{1}{\text{x}}.$
We have,
$\text{g}(\text{x}_1)=\text{g}(\text{x}_2)\Rightarrow\frac{1}{\text{x}_1}=\frac{1}{\text{x}_2}\Rightarrow\text{x}_1=\text{x}_2$
$\therefore$ g is one-one.
Further, it is clear that g is not onto as for $1.2\in\text{R}_\ast$ there does not exit any x in N such that $\text{g(x)}=\frac{1}{1.2}.$
Hence, function g is one-one but not onto.
View full question & answer→Question 395 Marks
Relation R in the set A of human beings in a town at a particular time given by
- R = {(x, y) : x and y work at the same place}
- R = {(x, y) : x and y live in the same locality}
- R = {(x, y) : x is exactly 7 cm taller than y}
- R = {(x, y) : x is wife of y}
- R = {(x, y) : x is father of y}
AnswerRelation R in the set A of human being in a town at a particular time.
- R = {( x, y) : x and y work at the same place}
| Since $ (\text{x},\text{x})\in\text{R},$ because x and x work at the same place. |
$\therefore$ R is reflexive. |
| Now, if $ (\text{x},\text{y})\in\text{R}\ \text{and}\ (\text{y},\text{x})\in\text{R},$ since x and y work at |
|
| the same place and y and x work at the same place. |
$\therefore$ R is symmetric. |
| Now, if $ (\text{x},\text{y})\in\text{R}\ \text{and}\ (\text{y},\text{x})\in\text{R}\Rightarrow(\text{x},\text{z})\in\text{R}.$ |
$\therefore$ R is transitive. |
Therefore, R is reflexive, symmetric and transitive.
- R = {( x, y) : x and y live in the same locality}
| Since $ (\text{x},\text{x})\in\text{R},$ because x and x live in the same locality. |
$\therefore$ R is reflexive. |
|
| Also $ (\text{x},\text{y})\in\text{R}\ \Rightarrow (\text{y},\text{x})\in\text{R},$ because x and y live |
|
|
| in same locality and y and x also live in same locality. |
$\therefore$ R is symmetric. |
|
| $\text{Again }(\text{x},\text{y})\in\text{R}\ \ \Rightarrow\ \ (\text{y},\text{x})\in\text{R}\ \Rightarrow\ \ (\text{x},\text{z})\in\text{R}$ |
$\therefore$ R is transitive. |
|
Therefore, R is reflexive, symmetric and transitive.
- R = {( x, y) : x is exactly 7 cm taller than y}
| x is not exactly 7 cm taller than x , so $ (\text{x},\text{x})\notin\text{R},$ |
$\therefore$ |
R is not reflexive. |
| Also x is exactly 7 cm taller than y but y is not |
|
|
| 7 cm taller than x , so $ (\text{x},\text{y})\in\text{R}\ \text{but}\ (\text{y},\text{x})\notin\text{R}$ |
$\therefore$ |
R is not symmetric. |
Now x is exactly 7 cm taller than y and y is
exactly 7 cm taller than z then it does not |
|
|
| imply that x is exactly 7 cm taller than z. |
$\therefore$ |
R is not transitive. |
Therefore, R is neither reflexive, nor symmetric and nor transitive.
- R = {( x, y) : x is wife of y}
| x is not wife of x , so $ (\text{x},\text{x})\notin\text{R},$ |
$\therefore$ |
R is not reflexive. |
| Also x is wife of y but y is not wife of x , |
|
|
| so $ (\text{x},\text{y})\in\text{R}\ \text{but}\ (\text{y},\text{x})\notin\text{R}$ |
$\therefore$ |
R is not symmetric. |
| $\text{Also }(\text{x},\text{y})\in\text{R}\text{ and }(\text{y},\text{z})\in\text{R}\text{ then }(\text{x},\text{z})\notin\text{R}$ |
$\therefore$ |
R is not transitive. |
Therefore, R is neither reflexive, nor symmetric and nor transitive.
- R = {( x, y) : x is father of y}
| x is not father of x , so $ (\text{x},\text{x})\notin\text{R},$ |
$\therefore$ |
R is not reflexive. |
| Also x is father of y but y is not father of x , |
|
|
| so $ (\text{x},\text{y})\in\text{R}\ \text{but}\ (\text{y},\text{x})\notin\text{R}$ |
$\therefore$ |
R is not symmetric. |
| $\text{Also }(\text{x},\text{y})\in\text{R}\text{ and }(\text{y},\text{z})\in\text{R}\text{ then }(\text{x},\text{z})\notin\text{R}$ |
$\therefore$ |
R is not transitive. |
Therefore, R is neither reflexive, nor symmetric and nor transitive. View full question & answer→Question 405 Marks
Let S be the set of all real numbers except -1 and let '*' be an operation defined by a * b = a + b + ab for all a, b ∈ S. Determine whether '*' is a binary operation on S. If yes, check its commutativity and associativity. Also, solve the equation (2 * x) * 3 = 7.
AnswerChecking for binary operation: Let $\text{a, b}\in\text{S.}$ Then, $\text{a, b}\in\text{R}$ and $\text{a}\neq-1,\text{b}\neq-1$ a * b = a + b + ab We need to prove that $\text{a}+\text{b}+\text{ab}\in\text{S.}$ $\big[$For this we have to prove that $\text{a}+\text{b}+\text{ab}\in\text{R}$ and $\text{a}+\text{b}+\text{ab}\neq-1\big]$ Since, $\text{a, b}\in\text{R},\ \text{a}+\text{b}+\text{ab}\in\text{R},$ let us assume that a + b + ab = -1. a + b + ab + 1 = 0 a + ab + b + 1 = 0 a(1 + b) + 1(1 + b) = 0 (a + 1)(b+ 1) = 0 a = -1, b = -1 [which is false] Hence, $\text{a}+\text{b}+\text{ab}\neq-1$ Therefore, $\text{a}+\text{b}+\text{ab}\in\text{S}$ Thus, * is a binary operation on S. Commutativity: Let $\text{a, b}\in\text{S.}$ Then, a * b = a + b + ab = b + a + ba = b * aTherefore,
a * b = b * a, $\forall\ \text{a, b}\in\text{S}$ Thus, * is commutative on N. Associativity: Let $\text{a, b, c}\in\text{S.}$ a * (b * c) = a * (b + c + bc) = a + b + c + bc + a(b + ac + bc) = a + b + c + bc + ab + ac + abc (a * b) * c = (a + b + ab) * c = a + b + ab + c + (a + b + ab)c = a + b + ab + c + ac + bc + abc Therefore, a * (b * c) = (a * b) * c, $\forall\ \text{a, b, c}\in\text{S}$ Thus, * is associative on S. Now, Given: (2 * x) * 3 = 7 Implies that (2 + x + 2x) * 3 = 7 Implies that (2 + 3x) * 3 = 7
View full question & answer→Question 415 Marks
If f : $Q \rightarrow Q, g : Q \rightarrow Q$ are two functions defined by $f(x) = 2x$ and $g(x) = x + 2,$ show that $f$ and $g$ are bijective maps. Verify that $(gof)^{-1} = f^{-1}og^{-1}.$
AnswerInjectivity of $f:$ Let $x$ and $y$ be two elements of domain $(Q),$ such that $f(x) = f(y)$
$\Rightarrow 2x = 2y $
$\Rightarrow x = y$
So, $f$ is one$-$one.
Surjectivity of $f:$ Let $y$ be in the co$-$domain $(Q),$
such that $f(x) = y. $
$\Rightarrow 2x = y$
$\Rightarrow\ \text{x}=\frac{\text{y}}{2}\in\text{Q} ($domain$)$
$\Rightarrow f$ is onto.
So, $f$ is a bijection and hence, it is invertible.
Finding $f^{-1}:$ Let $f^{-1}(x) = y .....(1) $
$\Rightarrow x = f(y) $
$\Rightarrow x = 2y$
$\Rightarrow\ \text{y}=\frac{\text{x}}{2}$
So, $\text{f}^{-1}(\text{x})=\frac{\text{x}}{2} [$from $(1)]$
Injectivity of $g:$ Let $x$ and $y$ be two elements of domain $(Q),$
such that $g(x) = g(y)$
$\Rightarrow x + 2 = y + 2 $
$\Rightarrow x = y$
So, $g$ is one$-$one.
Surjectivity of $g:$ Let y be in the co domain $(Q),$
such that $g(x) = y.$
$\Rightarrow x + 2 = y$
$\Rightarrow\ \text{x}=\text{y}-2\in\text{Q} ($domain$)$
$\Rightarrow g$ is onto.
So, $g$ is a bijection and hence, it is invertible.
Finding $g^{-1}:$ Let $g^{-1}(x) = y .....(2)$
$\Rightarrow x = g(y) $
$\Rightarrow x = y + 2 $
$\Rightarrow y = x - 2$
So, $g^{-1}(x) = x - 2 [$from $(2)]$
Verification of $(gof)^{-1} = f^{-1}og^{-1}:$
$ f(x) = 2x; g(x) = x + 2$ and $\text{f}^{-1}(\text{x})=\frac{\text{x}}{2}; g^{-1}(x) = x - 2$
Now, $(f^{-1}og^{-1})(x) = f^{-1}(g^{-1}(x)) $
$\Rightarrow (f^{-1}og^{-1})(x) = f^{-1}(x - 2) $
$\Rightarrow\ (\text{f}^{-1}\text{og}^{-1})(\text{x})=\frac{\text{x}-2}{2}\ ......(3) $
$(gof)(x) = g(f(x)) = g(2x) = 2x + 2$
Let $(gof)^{-1}(x) = y ......(4)x = (gof)(y)$
$\Rightarrow x = 2y + 2 $
$\Rightarrow 2y = x - 2 $
$\Rightarrow\ \text{y}=\frac{\text{x}-2}{2}$
$\Rightarrow\ (\text{gof})^{-1}(\text{x})=\frac{\text{x}-2}{2}\ .....(5) [$from $(4)]$ From $(3)$ and $(5),$
$(gof)^{-1} = f^{-1}og^{-1}$
View full question & answer→Question 425 Marks
Let S be the set of all rational numbers except 1 and * be defined on S by a * b = a + b - ab, for all a, b ∈ S.
Prove that:
- * is a binary operation on S.
- * is commutative as well as associative.
AnswerWe have, S = R - {1} and * is defined on S as a * b = a + b - ab, for all a, b ∈ S.
- It is seen that for each a, b ∈ S, there is a unique element a + b - ab in S.
This means that * carries each pair (a, b) to a unique element a * b = a + b - ab in S.
Therefore, * is a binary operation on S.
- Commutativity: Let a, b ∈ S. Then,
a * b = a + b - ab
= b + a - ba
= b * a
Therefore, a * b = b * a, $\forall\ \text{a, b}\in\text{S}$
Thus, * is commutative on S.
Associativity: Let a, b, c ∈ S. Then,
a * (b * c) = a * (b + c - bc)
= a + b + c - bc - a(b + c - bc)
= a + b + c - bc - ab - ac + abc
(a * b) * c = (a + b - ab) * c
= a + b - ab + c = (a + b - ab)c
= a + b + c - ab - ac - bc + abc
Therefore,
a * (b * c) = (a * b) * c, $\forall\ \text{a, b, c}\in\text{S}$
Thus, * is associative on S.
Therefore, * is commutative as well as associative. View full question & answer→Question 435 Marks
Show that the function f: R → {x ∈ R: –1 < x < 1} defined by $f(\text{x})=\frac{\text{x}}{1+|\text{x}|},\text{x}\in\text{R}$ is one-one and onto function.
AnswerIt is given that $f:\text{R}\rightarrow\{\text{x}\in\text{R}:-1<\text{x}<1\}$ is defined as $f(\text{x})=\frac{\text{x}}{1+|\text{x}|},\text{x}\in\text{R}.$
Suppose f(x) = f(y), where $\text{x},\text{y}\in\text{R}.$
$\Rightarrow\frac{\text{x}}{1+|\text{x}|}=\frac{\text{y}}{1+|\text{y}|}$
It can be observed that if x is positive and y is negative, then we have: $\frac{\text{x}}{1+\text{x}}=\frac{\text{y}}{1+\text{y}}\Rightarrow2\text{xy}=\text{x}-\text{y}$
Since x is positive and y is negative:
x > y ⇒ x - y > 0
But, 2xy is negative.
Then, $2\text{xy}\neq\text{x}-\text{y}$
Thus, the case of x being positive and y being negative can be ruled out.
Under a similar argument, x being negative and y being positive can also be ruled out
$\therefore$ x and y have to be either positive or negative.
When x and y are both positive, we have:
$f(\text{x})=f(\text{y})\Rightarrow\frac{\text{x}}{1+\text{x}}=\frac{\text{y}}{1+\text{y}}\Rightarrow\text{x}+\text{xy}=\text{y}+\text{xy}\Rightarrow\text{x}=\text{y}$
When x and y both are negative, we have:
$f(\text{x})=f(\text{y})\Rightarrow\frac{\text{x}}{1-\text{x}}=\frac{\text{y}}{1-\text{y}}\Rightarrow\text{x}-\text{xy}=\text{y}-\text{xy}\Rightarrow\text{x}=\text{y}$
$\therefore$ f is one-one.
Now, let $\text{y}\in\text{R}$ such that -1 < y < 1.
If y is negative, then there exists $\text{x}=\frac{\text{y}}{1+\text{y}}\in\text{R}$ such that
$f(\text{x})=f\Big(\frac{\text{y}}{1+\text{y}}\Big) =\frac{\Big(\frac{\text{y}}{1+\text{y}}\Big)}{1+\Big|\frac{\text{y}}{1+\text{y}}\Big|}=\frac{\frac{\text{y}}{1+\text{y}}}{1+\Big(\frac{\text{-y}}{1+\text{y}}\Big)}=\frac{\text{y}}{1+\text{y}-\text{y}}=\text{y}.$
If y is positive, then there exists $\text{x}=\frac{\text{y}}{1+\text{y}}\in\text{R}$ such that
$f(\text{x})=f\Big(\frac{\text{y}}{1-\text{y}}\Big) =\frac{\Big(\frac{\text{y}}{1-\text{y}}\Big)}{1+\Big|\frac{\text{y}}{1-\text{y}}\Big|}=\frac{\frac{\text{y}}{1-\text{y}}}{1+\frac{\text{y}}{1+\text{y}}}=\frac{\text{y}}{1-\text{y}+\text{y}}=\text{y}.$
$\therefore$ f is onto.
Hence, f is one-one and onto.
View full question & answer→Question 445 Marks
Let $f: R \rightarrow R$ be defined as $f(x) = 10x + 7.$ Find the function $g: R \rightarrow R$ such that $gof = fog = 1_R.$
AnswerIt is given that $f: R \rightarrow R$ is defined as $f(x) = 10x + 7.$
One$-$one:
Let $f(x) = f(y),$ where $\text{x},\text{y}\in\text{R}.$
$\Rightarrow 10x + 7 = 10y + 7$
$\Rightarrow x = y$
$\therefore f$ is a one$-$one function.
Onto:
For $\text{y}\in\text{R},$ let $y = 10x + 7.$
$\Rightarrow\text{x}=\frac{\text{y}-7}{10}\in\text{R}$
Therefore, for any $\text{y}\in\text{R},$ there exists $\text{x}=\frac{\text{y}-7}{10}\in\text{R}$ such that
$f(\text{x})=f\Big(\frac{\text{y}-7}{10}\Big)$
$=10\Big(\frac{\text{y}-7}{10}\Big)+7$
$=\text{y}-7+7$
$=\text{y}.$
$\therefore f$ is onto.
Therefore, $f$ is one$-$one and onto.
Thus, $f$ is an invertible function.
Let us define $g: R \rightarrow R$ as $\text{g(y)}=\frac{\text{y}-7}{10}.$
Now, we have:
$gof(x) = g(f(x)) = g(10x + 7) =\frac{(10\text{x}+7)}{10}=\frac{10\text{x}}{10}=\text{x}$
And,
$fo\text{g(y)}=f(\text{g(y)})$
$=f\Big(\frac{\text{y}-7}{10}\Big)$
$=10\Big(\frac{\text{y}-7}{10}\Big)+7$
$=\text{y}-7+7$
$=\text{y}$
$\therefore gof = I_R$ and $fog = I_R$
Hence, the required function $g: R \rightarrow R$ is defined as $\text{g(y)}=\frac{\text{y}-7}{10}.$
View full question & answer→Question 455 Marks
Let Z be the set of integers. Show that the relation R = {(a, b): a, b ∈ Z and a + b is even} is an equivalence relation on Z.
AnswerWe have, Z be set of integers and R = {(a, b): a, b ∈ Z and a + b is even} be a relation on Z. We want to prove that R is an equivalence relation on Z. Now, Reflexivity: Let $\text{a}\in\text{Z}$ ⇒ a + a is even [if a is even ⇒ a + a is even, if a is odd ⇒ a + a is even] $\Rightarrow\ (\text{a, a})\in\text{R}$ ⇒ R is reflexive. Symmetric: Let $\text{a, b}\in\text{Z}$ and $(\text{a, b})\in\text{R}$ ⇒ a + b is even ⇒ b + a is even $\Rightarrow\ (\text{b, a})\in\text{R}$ ⇒ R is symmetric.Transitivity: Let $(\text{a, b})\in\text{R}$ and $(\text{b, c})\in\text{R}$ For some $\text{a, b, c}\in\text{Z}$
⇒ a + b is even and b + c is even [if b is odd, then a and c must be odd ⇒ a + c is even, if b is even, then a and c must be even ⇒ a + c is even] ⇒ a + c is even $\Rightarrow\ (\text{a, c})\in\text{R}$ ⇒ R is transitive. Hence, R is an equivalence relation on Z.
View full question & answer→Question 465 Marks
Consider the binary operations *: R × R → R and o: R × R → R defined as $\text{a}∗\text{b} = |\text{a } – \text{ b}|\text{ and}\text{ a} o \text{b} = \text{a},\forall\text{a},\text{b}\in\text{R}.$ Show that * is commutative but not associative, o is associative but not commutative. Further, show that $\forall\text{a},\text{b},\text{c}\in\text{R},\text{a} *(\text{b}o\text{c}) = (\text{a} *\text{b})o(\text{a}*\text{b}).$ [If it is so, we say that the operation * distributes over the operation o]. Does o distribute over *? Justify your answer.
AnswerIt is given that *: R × R → and o: R × R → R is defined as
a * b = |a - b| and aob = a, $\Box\text{a},\text{b}\in\text{R}.$
For $\text{a},\text{b}\in\text{R},$ we have:
a * b = |a - b|
b * a = |b - a| = |-(a - b)| = |a - b|
$\therefore$ The operation * is commutative.
It can be observed that,
(1 * 2) * 3 = (|1 - 2|) * 3 = 1 * 3 = |1 - 3| = 2
1 * (2 * 3) = 1 * (|2 - 3|) = 1 * 1 = |1 - 1| = 0
$\therefore(1*2) *3\neq1*(2*3)(\text{where }1, 2,3\in\text{R})$
$\therefore$ The operation * is not associative.
Now, consider the operation o:
It can be observed that 1o2 = 1 and 2o1 = 2.
$\therefore1o2\neq2o1\ (\text{where }1,2\in\text{R})$
$\therefore$ The operation o is not commutative.
Let $\text{a},\text{b},\text{c}\in\text{R}.$ Then, we have:
(aob)oc = aoc = a
ao(boc) = aob = a
⇒ (aob)oc = ao(boc)
$\therefore$ The operation o is associative.
Now, let $\text{a},\text{b},\text{c}\in\text{R},$ then we have:
a * (boc) = a * b = |a - b|
(a * b)o(a * c) = (|a - b|)o(|a - c|) = |a - b|
Hence, a * (boc) = (a * b)o(a * c).
Now,
1o(2 * 3) = 1o(|2 - 3|) = 1o1 = 1
(1o2) * (1o3) = 1 * 1 = |1 - 1| = 0
$\therefore1o(2 * 3)\neq(1o2)*(1o3)\ (\text{where }1,2,3\in\text{R})$
$\therefore$ The operation o does not distribute over *.
View full question & answer→Question 475 Marks
Let f = {(1, -1), (4, -2), (9, -3), (16, 4)} and g = {(-1, -2), (-2, -4), (-3, -6), (4, 8)}. Show that gof is defined while fog is not defined. Also, find gof.
AnswerWe have, f = {(1, -1), (4, -2), (9, -3), (16, 4)} and
g = {(-1, -2), (-2, -4), (-3, -6), (4, 8)}
Now,
Domain of f = {1, 4, 9, 16}
Range of f = {-1, -2, -3, 4}
Domain of g = {-1, -2, -3, 4}
Range of g = {-2, -4, -6, 8}
Clearly range of f = domain of g
$\therefore$ gof is defined.
but, range of g $\neq$ domain of f
$\therefore$ fog in not defined.
Now,
gof(1) = g(-1) = -2
gof(4) = g(-2) = -4
gof(9) = g(-3) = -6
gof(16) = g(4) = 8
$\therefore$ gof = {(1, -2), (4, -4), (9, -6), (16, 8)}
View full question & answer→Question 485 Marks
Show that the relation R on the set Z of integers, given by R = {(a, b): 2 divides a - b}, is an equivalence relation.
AnswerWe have, R = {(a, b): a - b is divisible by 2; a, b $\in\text{Z}$} To prove: R is an equivalence relation.Proof:
Reflexivity: Let $\text{a}\in\text{Z}$ ⇒ a - a = 0 ⇒ a - a is divisible by 2 $\Rightarrow\ (\text{a, a})\in\text{R}$ ⇒ R is reflexive. Symmetric: Let $\text{a, b}\in\text{Z}$ and $(\text{a, b})\in\text{R}$ ⇒ a - b is divisible by 2 ⇒ a - b = 2p For some $\text{p}\in\text{Z}$ ⇒ b - a = 2 × (-p)$\Rightarrow\ \text{b}-\text{a}\in\text{R}$
⇒ R is symmetric. Transitive: Let $\text{a, b, c}\in\text{Z}$ and such that $(\text{a, b})\in\text{R}$ and $(\text{b, c})\in\text{R}$ ⇒ a - b = 2p and b - c = q For some $\text{p, q}\in\text{Z}$ ⇒ a - c = 2(p + q) ⇒ a - c is divisible by 2 $\Rightarrow\ (\text{a, c})\in\text{R}$ ⇒ R is transitive.
View full question & answer→Question 495 Marks
Consider $f : R \rightarrow R$ given by $f(x) = 4x + 3.$ Show that $f$ is invertible. Find the inverse of $f.$
AnswerInjectivity of $f:$ Let $x$ and $y$ be two elements of domain $(R),$
such that $f(x) = f(y)$
$\Rightarrow 4x + 3 = 4y + 3$
$\Rightarrow 4x = 4y$
$\Rightarrow x = y$
So, $f$ is one$-$one.
Surjectivity of $f:$ Let $y$ be in the co$-$domain $(R),$
such that $f(x) = y.$
$\Rightarrow 4x + 3 = y$
$\Rightarrow 4x = y - 3$
$\Rightarrow\ \text{x}=\frac{\text{y}-3}{4}\in\text{R} ($Domain$)$
$\Rightarrow f$ is onto.
So, $f$ is a bijection and hence is invertible.
Finding $f^{-1}:$
Let $f^{-1}(x) = y ....(1)$
$\Rightarrow x = f(y)$
$\Rightarrow x = 4y + 3$
$\Rightarrow x - 3 = 4y$
$\Rightarrow\ \text{y}=\frac{\text{x}-3}{4}$
So, $\text{f}^{-1}(\text{x})=\frac{\text{x}-3}{4} [$from $(1)]$
View full question & answer→Question 505 Marks
Let $f : N \rightarrow N$ be a function as $f(x) = 9x^2 + 6x - 5.$ Show that $f : N \rightarrow S,$ where $S$ is the range of $f,$ is invertible. Find the inverse of $f$ and hence find $f^{-1}(43)$ and $f^{-1}(163).$
AnswerWe have, $f : N \rightarrow N$ is a function defined as $f(x) = 9x^2 + 6x - 5.$
Let $y = f(x) = 9x^2 + 6x - 5$
$\Rightarrow y = 9x^2 + 6x - 5 $
$\Rightarrow y = 9x^2 + 6x + 1 - 1 - 5 $
$\Rightarrow y = (9x^2 + 6x + 1) - 6 $
$\Rightarrow y = (3x + 1)^2 - 6$
$\Rightarrow y + 6 = (3x + 1)^2$
$\Rightarrow\ \sqrt{\text{y}+6}=3\text{x}+1\ \ (\because\ \text{y}\in\text{N})$
$\Rightarrow\ \sqrt{\text{y}+6}-1=3\text{x}$
$\Rightarrow\ \text{x}=\frac{\sqrt{\text{y}+6}-1}{3}$
$\Rightarrow\ \text{g(y)}=\frac{\sqrt{\text{y}+6}-1}{3} [$Let $x = g(y)]$ Now, $fog(y) = f[g(y)]$ $=\text{f}\bigg(\frac{\sqrt{\text{y}+6}-1}{3}\bigg)$
$=9\bigg(\frac{\sqrt{\text{y}+6}-1}{3}\bigg)^2+6\bigg(\frac{\sqrt{\text{y}+6}-1}{3}\bigg)-5$
$=9\bigg(\frac{\text{y}+6-2\sqrt{\text{y}+6}+1}{9}\bigg)+2\Big(\sqrt{\text{y}+6}-1\Big)-5$
$=\text{y}+6-2\sqrt{\text{y}+6}+1+2\sqrt{\text{y}+6}-2-5$
$=\text{y}$
$=\text{I}_\text{y} ($Identity function$)\text{gof}(\text{x})=\text{g[f(x)]}$
$=\text{g}(9\text{x}^2+6\text{x}-5)$
$=\frac{\sqrt{(9\text{x}^2+6\text{x}-5)+6}-1}{3}$
$=\frac{(3\text{x}+1)-1}{3}$
$=\frac{3\text{x}}{3}$
$=\text{x}$
$=\text{I}_\text{X} ($Identity function$)$
Since, $fog(y)$ and $gof(x)$ are identity function. Thus, $f$ is invertible.
So, $\text{f}^{-1}(\text{x})=\text{g(x)}=\frac{\sqrt{\text{x}+6}-1}{3}$ Now, $\text{f}^{-1}(43)=\frac{\sqrt{43+6}-1}{3}=\frac{\sqrt{49}-1}{3}$
$=\frac{7-1}{3}=\frac{6}{3}=2$ And, $\text{f}^{-1}(163)=\frac{\sqrt{163+6}-1}{3}=\frac{\sqrt{169}-1}{3}$
$=\frac{13-1}{3}=\frac{12}{3}=4$
View full question & answer→