Question 13 Marks
If the function $f: R\rightarrow R$ be given by $f(x) = x^2 + 2$ and $g: R \rightarrow R$ be given by $g(x) = \frac{\text{x}}{\text{x} - 1 },\text{x}\neq1 ,$ find fog and gof and hence find fog $(2)$ and gof $(–3).$
Answergetting fog $(x) = \text{f}\bigg(\frac{\text{x}}{\text{x} - 1 }\bigg) = \bigg(\frac{\text{x}}{\text{x} - 1}\bigg)^{2} + 2 $
$fog(2) = 6$
getting g of $(x)=\text{g} (\text{x}^{2} + 2 ) = \frac{\text{x}^{2} + 2 }{\text{x}^{2} + 1 }$
$\text{g of }(-3) = \frac{11}{10}.$
View full question & answer→Question 23 Marks
Consider $\text{f}:\text{R}_{+}\rightarrow[4,\infty)$ given by $f (x) = x^2 + 4$. Show that f is invertible with the inverse $f^{–1 }$ of $f$ given by $f^{–1} (y) =\sqrt{\text{y} - 4 },$ where $R_+$ is the set of all non $-$ negative real numbers.
AnswerFor one $-$ one
Let ${x}_{1},{x}_{2}\in\text{R} ($Domain$)$
$\text{f}({x}_{1}) =\text{f}({x}_{2})$
$\Rightarrow {x}_{1}^{2} + 4 = {x}_{2}^{2} + 4 $
$\Rightarrow\text{x}_{1}^{2} ={x}_{2}^{2}$
$\Rightarrow {x}_{1} ={x}_{2} \ [\because {x}_{1}, {x}_{2}$ are $+ve$ real number$]$
$\therefore f$ is one $-$ one function.
For onto
Let $\text{y}\in[4,\infty]\text{s.t.}$
$\text{y} = \text{f}(\text{x) }\forall\text{ x}\in\text{R}_{+}$ set of non $-$ negative reals
$\Rightarrow\text{y} = \text{x}^{2} + 4 $
$\Rightarrow\text{x} = \sqrt{\text{y} - 4 } \ [\because x$ is $+ ve$ real number$]$
Obviously, $\forall \text{y}\in [4,\infty], x$ is real number $ÎR \ ($domain$)$
i.e., all elements of codomain have pre image in domain.
$\Rightarrow f$ is onto.
Hence $f$ is invertible being one $-$ one onto.
For inverse function: If $f^{-1}$ is inverse of $f,$ then
$fof^{-1} = I \ ($Identity function$)$
$\Rightarrow\text{fof}^{-1}(\text{y}) = \text{y }\forall\text{ y}\in[4,\infty)$
$\Rightarrow\text{f}(\text{f}^{-1}(\text{y})) =\text{y}$
$\Rightarrow(\text{f}^{-1}(\text{y}))^{2} + 4 = \text{y }[\because\text{f}(\text{x}) =\text{x}^{2} + 4 ]$
$\Rightarrow\text{f}^{-1}(\text{y}) = \sqrt{\text{y} - 4 }$
Therefore, required inverse function is $\text{f}^{-1}:[4,\infty]\rightarrow R$ defined by
$\text{f}^{-1}(\text{y}) = \sqrt{\text{y} - 4}\forall \text{ y}\in[4,\infty).$
View full question & answer→Question 33 Marks
Find the value of k, for which $\text{f}(\text{x}) = $ $ \begin{matrix} \frac{\sqrt{1 + \text{kx}} - \sqrt{1 - \text{kx}}}{\text{x}} , \text{if} - 1\leq\text{x} < 0\\ \frac{2\text{x} + 1}{\text{x} - 1} , \text{ if}0\leq\text{x}< 1 \end{matrix} $ is continuous at x = 0.
Answer$\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow0^{-}}\text{f}(\text{x}) = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\text{f}(0 - \text{h}) [ \text{ Let x} = 0 - \text{h},\text{x}\rightarrow0^{-}\Rightarrow\text{h}\rightarrow0]$
$ = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\text{f}(-\text{h}) = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\frac{\sqrt{1 + \text{k}(-\text{h})} - \sqrt{1 - \text{k}(-\text{h})}}{-\text{h}}$
$ = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\frac{\sqrt{1 - \text{kh}} - \sqrt{1 + \text{kh}}}{-\text{h}}\times\frac{\sqrt{1 - \text{kh}} + \sqrt{1 + \text{kh}}}{\sqrt{1 - \text{kh}} + \sqrt{1 + \text{kh}}}$
$ = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\frac{(1 - \text{kh)} - (1 + \text{kh)}}{-\text{h}\left\{\sqrt{1 - \text{kh}} + \sqrt{1 + \text{kh}}\right\}} = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\frac{2\text{k}}{\left\{\sqrt{1 - \text{kh}} + \sqrt{1 + \text{kh}}\right\}}$
$ =\frac{2\text{k}}{2}$
$ \Rightarrow \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow0^{-}}\text{f}(\text{x}) = \text{k}$ - - - - - - - (i)
Again $ \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow0^{+}}\text{f}(\text{x}) = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\text{f}(0 + \text{h}) [ \text{Let x } = 0 \text{ h},\text{x}\rightarrow0^{+}\Rightarrow\text{h}\rightarrow0]$
$ \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\text{f}( - \text{h}) = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0} \frac{2\text{h} + 1 }{\text{h} - 1 } = \frac{1}{-1}$
$ \Rightarrow\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow0^{+}}\text{f}(\text{x}) = - 1 $ - - - - - - - - - (ii)
Also $\text{f}(0) = \frac{2\times0 + 1 }{0 - 1 } = - 1 $
$\because\text{ f is continuous at x } = 0$
$ \therefore\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow0^{-}}\text{f}(\text{x}) = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow0^{+}}\text{f}(\text{x}) = \text{f}(0)\Rightarrow\text{k} = - 1.$
View full question & answer→Question 43 Marks
Let $f : R \rightarrow R$ be defined as $f(x) = 10x + 7.$ Find the function $g : R \rightarrow R$ such that go $f = f\ o\ g = I_R.$
AnswerLet $y = 10x + 7$
$\therefore\text{ x}=\frac{1}{10}\text{(y - 7)}$
Let $g(y) =\frac{1}{10}\text{(y - 7)}$
$\therefore$ go $f(x) = g(10x + 7)$
$=\frac{1}{10} (10x + 7 - 7) = x $
$\Rightarrow I_{R }= go\ f$ and $\therefore$ fog $(y)$
$= f \Bigg(\frac{1}{10}\text{(y - 7)}\Bigg)=10\Bigg(\frac{1}{10}\text{(y - 7)}\Bigg)+7=\text{y}$
$\Rightarrow\text{I}_{\text{R}}=\text{fog}$
Hence $g (y) =\frac{1}{10}\text{(y - 7)}$.
View full question & answer→Question 53 Marks
Show that the relation S in the set A = {x $\in $ Z : 0 < x < 12} given by S = {(a, b): a, b $\in $ Z, | a – b | is divisible by 4} is an equivalence relation. Find the set of all elements related to 1.
Answer
- For all a $\in $ A,(a, a)$\in $ S ($\because$ a - a = 0 is divisible by 4)
$\therefore$S is reflexive in A
- For all a, b $\in $ A, if (a, b) $\in $ S then |a-b| is divisible by 4.
Hence |b-a| is also divisible by 4 $\Rightarrow$ S is symmetric in A
- $\forall$ a, b, c $\in $ A, Let (a, b) $\in $ S and (b, c) S
i.e.|a-b| is divisible by 4 and |b - c| is divisible by
$\Rightarrow$(a–b) = + 4p, (b–c) = + 4q, adding to get a – c = 4m$\Rightarrow$ (a, c) $\in $S
$\Rightarrow$S is transitive in A
Hence S is an equivalence relation
Elements related to 1 are {1, 5, 9} View full question & answer→Question 63 Marks
Show that the relation R defined by (a, b) R (c, d) $\Rightarrow$ a + d = b + c on the set N × N is an equivalence relation.
Answer
- (a, b) R (c,d) ⇒ a+d = b+c
where (a,b), (c,d) ∈ N x N
(a, b) R (a,b) ⇒ a+b = b+a ⇒ True
R is Reflexive.
- (a, b) R (c,d) ⇒ a+d = b+c ⇒ b+c = a+d
= 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)
⇒ a+d = b+c and c+f = d+e
Adding we get
a+d + c+f = b+c +d+e
⇒ a+f = b+e ⇒ (a,b) R (e,f)
∴ R is transitive
∴ R is an equivalence relation.
View full question & answer→Question 73 Marks
Let A = {1, 2, 3,....., 9} and R be the relation in A x A defined by (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) in A x A. Prove that R is an equivalence relation. Also obtain the equivalence class [(2,5)]
Answer$\forall(\text{a},\text{b})\in\text{A}\times\text{A}$
a + b = b + a $\therefore$(a, b) R (a, b) $\therefore$R is reflexive
For (a, b), (c, d) $\in\text{A}\times\text{A}$
If (a, b) R (c, d) i.e. a + d = b + c $\Rightarrow$c + b = d + a
then (c, d) R (a, b) $\therefore$R is symmetric
For (a, b), (c, d), (e, f) $\in\text{A}\times\text{A}$
If (a, b) R (c, d) & (c, d) R (e, f) i.e. a + d = b + c & c + f = d + e
Adding, a + d + c + f = b + c + d + e $\Rightarrow$a + f = b + e
then (a, b) R (e, f) $\therefore$R is transitive
$\therefore$Ris reflexive, symmetric and transitive henceRis an equivalance relation
[(2, 5)] = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}.
View full question & answer→Question 83 Marks
If $\text{x} = \text{a}\sin\text{t} \text{ and } \text{y = a} \bigg(\cos\text{t} + \log\tan\frac{\text{t}}{2}\bigg),\text{ find } \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}.$
AnswerHere, x = a sin t, y = a $\bigg[\cos\text{t} + \log\bigg(\tan\frac{\text{t}}{2}\bigg)\bigg]$
$\because$ x = a sin t
Differentiating both sides w.r.t. t, we get
$\frac{\text{dx}}{\text{dt}}= \text{a}\cos\text{t}$ - - - - - (i)
Again, $\because\text{y = a }\bigg[\cos\text{t} + \log\bigg(\tan\frac{\text{t}}{2}\bigg)\bigg]$
Differentiating both sides w.r.t. t we get
$\frac{\text{dy}}{\text{dt}} = \text{a}\bigg[-\sin\text{t} + \frac{1}{\tan\frac{\text{t}}{2}}.\sec^{2}\frac{\text{t}}{2}.\frac{\text{t}}{2}\bigg]$
$\Rightarrow\frac{\text{dy}}{\text{dt}} = \text{a}\bigg[ -\sin\text{t} + \frac{1}{\sin\text{t}}\bigg]\Rightarrow\frac{\text{dy}}{\text{dt}} = \frac{\text{a}(1 - \sin^{2}\text{t})}{\sin\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dt}} = \frac{\text{a}\cos^{2}\text{t}}{\sin\text{t}}$ - - - - - -(ii)
$\because\frac{\text{dy}}{\text{dx}} =\frac{\text{dy}/\text{dt}}{\text{dx}/\text{dt}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}} =\frac{\text{a}\cos^{2}\text{t}}{\sin\text{t}}\times\frac{1}{\text{a}\cos\text{t}}$ [From (i) and (ii)]
$\Rightarrow\frac{\text{dy}}{\text{dx}} = \cot\text{t}$
Differentiating again w.r.t. x we get
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} = 0-\text{cosec}^{2}\text{t}.\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} = -\text{cosec}^{2}\text{t}.\frac{1}{\text{a}\cos\text{t}} = \frac{-\text{cosec}^{2}\text{t}}{\text{a}\cos\text{t}}.$
View full question & answer→Question 93 Marks
Show that the function $f$ in $\text{A} = |\text{R} -\left\{\frac{2}{3}\right\}$ defined as $\text{f}(\text{x}) =\frac{4\text{x} + 3}{6\text{x} - 4 }$is one$-$one and onto. Hence find $f^{-1}$.
AnswerLet $x_1, x_2 \in A$
Now $\text{f}(\text{x}_{1}) = \text{f}(\text{x}_{2}) = \frac{4\text{x}_{1} + 3}{6\text{x}_{1} - 4 } = \frac{4\text{x}_{2}+ 3}{6\text{x}_{2} - 4 }$
$\Rightarrow24\text{x}_{1}\text{x}_{2} + 18 \text{x}_{2} - 16\text{x}_{1} - 12 = 24 \text{x}_{1}\text{x}_{2} + 18 \text{x}_{1} - 16 \text{x}_{2} - 12 $
$\Rightarrow - 34 \text{x}_{1} = - 34\text{x}_{2}$
$\Rightarrow\text{x}_{1} =\text{x}_{2}$
Hence $f$ is one$-$one function
For onto
Let $\text{y} =\frac{4\text{x} + 3}{6\text{x} - 4}$
$\Rightarrow6\text{xy} - 4\text{y} = 4\text{x} + 3 $
$\Rightarrow6\text{xy} - 4 \text{x} = 4\text{y} + 3 $
$\Rightarrow\text{x}(6\text{y} - 4 ) = 4\text{y} + 3 $
$\Rightarrow\text{x} = \frac{4\text{y} + 3 }{6\text{y} - 4}$
$\Rightarrow\forall\text{y}\in\text{ codomain} \exists\text{x}\in\text{ Domain }\bigg[\therefore\text{x}\neq\frac{2}{3}\bigg]$
$\Rightarrow f$ in onto function.
Thus $f$ is one$-$one onto function.
Also$, \text{f}^{-1}\text{(x)} =\frac{4\text{x} + 3 }{6\text{x} - 4}.$
View full question & answer→Question 103 Marks
Show that the function $\text{f}(\text{x}) = |\text{x} - 3 |,\text{x}\in|\text{R},$is continuous but not differentiable at x = 3.
AnswerHere, f(x) =|x - 3|
$\text{f}(\text{x}) - (\text{x} - 3 ) ,\text{x} < 3 $
$0 ,\text{x} = 3 $
$(\text{x} - 3 ),\text{x} > 3 $
Now, $\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow3^{+}}\text{f}(\text{x}) = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\text{f}(3 + \text{h})$
[Let x = 3 + h and $\text{x}\rightarrow3^{+}\Rightarrow\text{h}\rightarrow0]$
$\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}(3 + \text{h} - 3 ) = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\text{h} = 0 $
$\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow3^{+}}\text{f}(\text{x}) = 0 $ - - - - - (i)
$\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow3^{-}}\text{f}(\text{x}) = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\text{f}(3 - \text{h})$
[Let x = 3 - h and x$\text{x}\rightarrow3^{-}\Rightarrow\text{h}\rightarrow0]$
$ = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0} - (3 - \text{h} - 3) = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}=\text{h} = 0$
$ = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow{3}^{+}} \text{f}(\text{x})$ - - - -- (ii)
Also, f(3) = 0 - - - --- (iii)
From equation (i), (ii) and (iii)
$ = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow{3}^{+}} \text{f}(\text{x}) = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow{3}^{-}} \text{f}(\text{x}) =\text{f}(3)$
Hence, f(x) is continuous at x = 3
At x = 3
RHD $ = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0} \frac{\text{f}(3 + \text{h}) - \text{f}(3)}{\text{h}} = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\frac{(3 + \text{h} - 3)- 0}{\text{h}}$
$ = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0} \frac{\text{h}}{\text{h}}$ [$\because$|h|= h,|0|= 0]
$ = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}1$
RHD = 1 - - - - - - - (iv)
LHD $ = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\frac{\text{f}(3 - \text{h})- \text{f}(3)}{-\text{h}} = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0} \frac{-(3-\text{h}-3)-0}{-\text{h}}$
$ = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\frac{\text{h}}{-\text{h}}$ [$\because$|h|= h]
$ = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}(-1)$
LHD = – 1 - - - - - - -(v)
Equation (iv) and (v) $\Rightarrow$RHD $\neq$LHD at x = 3.
Hence f(x) is not differentiable at x = 3
Therefore, f(x) =|x - 3|, x$\in$R is continuous but not differentiable at x = 3.
View full question & answer→Question 113 Marks
Let $A = IR – \{3\}$ and $B = IR – \{1\}$. Consider the function $f: A \rightarrow B$ defined by $\text{f(x)}=\Bigg(\frac{\text{x - 2}}{\text{x - 3}}\Bigg)$. Show that fis one$-$one and onto and hence find $f^{–1}$.
AnswerLet $x_1, x_2 \in A$ and $f(x_1) = f(x_2)$
$\Rightarrow\frac{\text{x}_{1}- 2}{\text{x}_{1}-3}$
$=\frac{\text{x}_{2}-2}{\text{x}_{2}-3}$
$\therefore x_1 x_2 – 2x_2 – 3x_1 $
$= x_1 x_2 – 2x_1 – 3x_2$
$\Rightarrow x_1= x_2$
Hence $f$ is $1 – 1$
Let $y \in B,$
$\therefore y = f(x)$
$\Rightarrow \text{y}=\frac{\text{x - 2}}{\text{x - 3}}$
$\Rightarrow xy – 3y = x – 2$
Or $\text{x}=\frac{\text{3y - 2}}{\text{y - 1}}$
since $y \neq 1$ and $\frac{\text{3y - 2}}{\text{y - 1}}\neq3$
$\therefore\text{x}\in\text{A}$
Hence $f$ is $\text{ONTO}$
and $f ^{–1}(y) = \frac{\text{3y - 2}}{\text{y - 1}}.$
View full question & answer→Question 123 Marks
Consider the binary operation * on the set {1, 2, 3, 4, 5} defined by a * b =min. {a, b}. Write the operation table of the operation *.
Question 133 Marks
Let Z be the set of all integers and R be the relation on Z defined as R = {(a, b): a, b ∈ Z, and (a – b) is divisible by 5.} Prove that R is an equivalence relation.
AnswerR = {(a, b): a, b ∈ Z and (a - b) is divisible by 5}
- a - a = 0 which is divisible by 5
$\therefore$ R is reflexive.
- a – b is divisible by 5 and so is b – a
$\therefore$ R is symmetric.
- a – c = (a – b) + (b – c)
let a – b = 5m and b – c = 5 n
$\therefore$ R is transitive
$\therefore$ R is an equivalence - relation. View full question & answer→Question 143 Marks
Prove that the relation R in the set $\text{A } = (1, 2, 3, 4, 5)$ given by $\text{R} = (\text{a, b)} : |\text{a-b|} \text{is even},$ is an equivalence relation.
Answer$\text{(i) for all a} \in \text{A, (a ,a)} \in \text{R} \therefore | \text{a - a}| = \text{o is even} $$\therefore \text{R is reflexive in A}$
$\text{(ii) for all a, b} \in \text{A, (a, b)} \in \text{R} \Rightarrow \text{b, a)} \in \text{R} \because \text{if | a -b| is even then|b- a| is also even } \Rightarrow \text{R is symmetric in A} $
$\text{(iii) for all a, b, c} \in \text{A}$
$\text{(a, b)} \in \text{R and (b,c)} \in \text{R then (a, c) } \in \text{R}$
$\because \text{|a - b| is even, | b - c| is even, then |a - c| will also be even}$
$\text{Hence, R is an equivalence relation in A}$
View full question & answer→Question 153 Marks
- Is the binary operation $\ast,$ defined on set N, given by a$a^{\ast}b = \frac{a + b}{2}$ for all $a, b\varepsilon N,$ commutative?
- Is the above binary operation $\ast$ associative?
Answer$(i) \text{Given N be the set}$$a^{\ast}b = \frac{a + b}{2}\forall a, b\varepsilon N$
To find $\ast$ is commutative of or not.
Now, $a^{\ast}b \frac{a +b}{2} =\frac{b + a}{2} \therefore \text{(addition is commutative on N)}$
$= b^{\ast}a$
$\text{So a}^{\ast}b = b^{\ast}a$
$\therefore \ast \text{is commutative}$
To find $a^{\ast}(b^{\ast}c) = (a^{\ast}b)^\ast c $ or Not
Now $a^{\ast}(b^{\ast}c) = a^{\ast} = \frac{b + c}{2} =\frac{a+\bigg(\frac{b + c}{2}\bigg)}{2}= \frac{2a + b + c}{4}\dots\dots\text{(i)}$
$(a^{\ast}b)^{\ast}c = \bigg(\frac{a + b}{2}\bigg)^{\ast} c = \frac{\frac{a + b}{2} + c}{2}$
$= \frac{a + b + 2c}{4} \dots\dots\text{(ii)}$
From (i) and (ii)
$(a^{\ast}b)^{\ast} c \neq a^{\ast}(b^{\ast}c)$
Hence the operation is not associative.
View full question & answer→Question 163 Marks
Show that the relation R on defined as $\text{R}=\big\{(\text{a},\text{b}):\text{a}\leq\text{b}\big\},$ is reflexive, and transitive but not symmetric.
Answer$\text{R}=\big\{(\text{a},\text{b}):\text{a}\leq\text{b}\big\}$
Clearly $(\text{a},\text{a})\in\text{R}$ as $\text{a}=\text{a}.$
$\therefore\ $R is reflexive.
Now,
$(2,4)\in\text{R}$ as $(2<4)$
But, $(4,2)\notin\text{R}$ R as 4 is greater than 2.
$\therefore\ $R is not symmetric.
Now, let $(\text{a},\text{b}),(\text{b},\text{c})\in\text{R}.$
Then,
$\text{a}\leq\text{b}$ and $\text{b}\leq\text{c}$
$\Rightarrow\text{a}\leq\text{c}$
$\Rightarrow(\text{a},\text{c})\in\text{R}$
$\therefore\ $R is transitive.
Hence, R is reflexive and transitive but not symmetric.
View full question & answer→Question 173 Marks
Let N be the set of natural numbers and R be the relation on N × N defined by (a, b) R (c, d) iff ad = bc for all a, b, c, d $\in$ N. Show that R is an equivalence relation.
AnswerLet R be defined on N × N as (a, b) R (c, d) ⇔ ad(b + c) = bc(a + d) ....(1)Reflexivity:
We can write ab(b + a) = ba(a + b) for all a, b $\in$ N Since, sum and product of natural numbers obeys commutative property Hence, by def (1), we can write (a, b) R (a, b) for all (a, b) $\in$ N × N Hence, R is reflexive. Symmmetry: Let (a, b) R (c, d) ⇒ ad(b + c) = bc(a + d) ⇒ da(c + b) = cb(d + a) (Since, sum and product of natural numbers obeys commutative property) or cb(d + a) = da(c + b) ⇒ (c, d) R (a, b) Hence, R is symmetric.Transitivity:
Let (a, b), (c, d), (e, f) $\in$ N × N Let (a, b) R (c, d) and (c, d) R (e, f) ad(b + c) = bc(a + d) and cf(d + e) = de(c + f) $\Rightarrow\frac{\text{ab}}{\text{a}-\text{b}}=\frac{\text{cd}}{\text{c}-\text{d}}$ and $\frac{\text{cd}}{\text{c}-\text{d}}=\frac{\text{ef}}{\text{e}-\text{f}}$ $\Rightarrow\frac{\text{ab}}{\text{a}-\text{b}}=\frac{\text{ef}}{\text{e}-\text{f}}$ $\Rightarrow\text{af}(\text{b}+\text{e})=\text{be}(\text{a}+\text{f})$ $\Rightarrow(\text{a},\text{b})\text{ R }(\text{e},\text{f})$ Hence, R is transitive $\therefore$ R is Equivalence Relation.
View full question & answer→Question 183 Marks
Prove that the function $f : N \rightarrow N,$ defined by $f(x) = x^2 + x + 1$ is one$-$one but not onto. Find inverse of $f : N \rightarrow S,$ where $S$ is range of $f.$
AnswerThe given function is
$\text{f} : \text{N} \rightarrow \text{N}$
$\text{f(x)} = \text{x}^2 + \text{x} + 1$
Let $\text{ x}_1, \text{x}_2 6\in\text{N}$
So $\text{ let} \text{ f(x}_1) = \text{f(x}_2)$
$\text{x}_1^2+\text{x}_1+1=\text{x}_2^2 +\text{x}_2+1$
$\text{x}_1^2-\text{x}_2^2+\text{x}_1-\text{x}_2=0$
$(\text{x}_1-\text{x}_2)(\text{x}_1-\text{x}_2+1)=0$
$\because\text{x}_2=\text{x}_1$
or $\text{x}_2=-\text{x}_1-1$
$\because\text{x}_1\in\text{N}$
$\therefore-\text{x}_1-1\in\text{N}$
So $\text{x}_2\neq-\text{x}_1-1$
$\because\text{f}(\text{x}_2)=\text{f}(\text{x}_1)$ only for $\text{x}_1=\text{x}_2$
So f(x) is one$-$one function.
$\because\text{f}(\text{x})=\text{x}^2+\text{x}+1$
which is an increasing function.
$\text{f}(1)=3$
$\because\ $ Range of $f(x)$ will be $\{3, 7, .....\}$
Which is subset of $N.$
so, it is an subset of $N.$
i.e.,$ f(x)$ is not an onto function.
Let $\text{y}=\text{x}^2+\text{x}+1$
$\text{x}^2+\text{x}+1-\text{y}=0$
$\text{x}=\frac{-1\pm\sqrt{1-4(1-\text{y})}}{2}$
$\text{x}=\frac{-1\pm\sqrt{4\text{y}-3}}{2}$
So, two possibilities are their for $\text{f}^{-1}(\text{x})$
$\text{f}^{-1}(\text{x})=\frac{-1+\sqrt{4\text{x}-3}}{2}$
and we know $\text{f}^{-1}(3)=1$ because $\text{f}(1)=3$
So, $\text{f}^{-1}(\text{x})=\frac{-1+\sqrt{4\text{x}-3}}{2}$
View full question & answer→Question 193 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 203 Marks
Check the commutativity and associativity of the following binary operations:
'*' on Q defined by $\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{4}$ for all a, b ∈ Q.
AnswerCommutativity: Let $\text{a, b}\in\text{Q}.$ Then,$\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{4}$
$=\frac{\text{ba}}{4}$
$=\text{b}\ ^*\ \text{a}$
Therefore,
$\text{a}\ ^*\ \text{b}=\text{b}\ ^*\ \text{a},\ \forall\ \text{a, b}\in\text{Q}$
Thus '*' is commutative on Z.
Associativity: Let $\text{a, b, c}\in\text{Q}.$ Then,
$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=\text{a}\ ^*\ \Big(\frac{\text{bc}}{4}\Big)$
$=\frac{\text{a}\big(\frac{\text{bc}}{4}\big)}{4}$
$=\frac{\text{abc}}{16}$
$(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}=\frac{\text{ab}}{4}\ ^*\ \text{c}$
$=\frac{\big(\frac{\text{ab}}{4}\big)\text{c}}{4}$
$=\frac{\text{abc}}{16}$
Therefore,
$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=(\text{a}\ ^*\ \text{b})\ ^*\ \text{c},\ \forall\ \text{a, b, c}\in\text{Q}$
Thus, '*' is associative on Q.
View full question & answer→MCQ 213 Marks
Let $*$ be the binary operation defined on $Q$. Find which of the following binary operations are commutative:
- A
$a * b = a – b \forall a, b \in Q$
- B
$a * b = a^2 + b^2 \forall a, b \in Q$
- C
$a * b = a + ab \forall a, b \in Q$
- D
$a * b = (a – b)^2 \forall a, b \in Q$
AnswerGiven that $*$ be the binary operation defined on $Q$.
- $a * b = a – b \forall a, b \in Q$
$= -b + a$
$= -(b - a)$
$= -b * a$
$\therefore\ \text{a}\ ^*\ \text{b}\neq\text{b}\ ^*\ \text{a}$
Hence, $*$ is not commutative.
- $a * b = a^2 + b^2$
$= b^2 + a^2$
$= b * a$
Hence $, *$ is commutative.
- We have $a * b = a + ab$ and $b * a = b + ab$
Clearly, $\text{a}+\text{ab}\neq\text{b}+\text{ab}$
So, $*$ is not communicative.
- We have $a * b = (a – b)^2 \forall a, b \in Q$
$= (-b + a)^2$
$= {-(b - a)}^2$
$= (b - a)^2$
$= b * a$
Hence $, *$ is communicative. View full question & answer→Question 223 Marks
Show that each of the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by
- R = {(a, b) : |a – b| is a multiple of 4}
- R = {(a, b) : a = b}
is an equivalence relation. Find the set of all elements related to 1 in each case.
Answer
- It is given that the relation R in the set $\text{A}=\{\text{x}\in\text{Z}:0\leq\text{x}\leq12\},$ given by
R = {(a, b) : |a - b| is a multiple of 4}
For any element $\text{a}\in\text{A},$ we have $(\text{a},\text{a})\in\text{R}$ as |a - a| = 0 is a multiple of 4.
Therefore, R is reflexive.
Now, Let $(\text{a},\text{a})\in\text{R}$
⇒ |a - b| is a multiple of 4
⇒ |b - a| = |a - b| is a multiple of 4
$\Rightarrow(\text{b},\text{a})\in\text{R}$
Therefore, R is symmetric.
Now, Let $(\text{a},\text{b}),(\text{b},\text{c})\in\text{R}$
⇒ |a - b| is a multiple of 4 and |b - c| is a multiple of 4
⇒ |a - c| = |(a - b) + (b - c)| is a multiple of 4
$\Rightarrow(\text{a},\text{c})\in\text{R}$
Therefore, R is transitive.
Therefore, R is an equivalence relation.
The set of elements related to 1 is {1, 5, 9}
|1 - 1| = 0 is multiple of 4
|5 - 1| = 4 is multiple of 4
|9 - 1| = 8 is multiple of 4.
- It is given that the relation R in the set $\text{A}=\{\text{x}\in\text{Z}:0\leq\text{x}\leq12\},$ given by
R = {(a, b) : a = b}
For any element $\text{a}\in\text{A},$ we have $(\text{a},\text{a})\in\text{R}$ as a = a
Therefore, R is reflexive.
Now, Let $(\text{a},\text{a})\in\text{R}$
⇒ a = b
⇒ b = a
$\Rightarrow(\text{b},\text{a})\in\text{R}$
Therefore, R is symmetric.
Now, Let $(\text{a},\text{b}),(\text{b},\text{c})\in\text{R}$
⇒ a = b and b = c
⇒ a = c
$\Rightarrow(\text{a},\text{c})\in\text{R}$
Therefore, R is transitive.
Therefore, R is an equivalence relation.
The set of elements related to 1 will be those elements from set Awhich are equal to 1.
Therefore, the set of elements related to 1 is {1}. View full question & answer→Question 233 Marks
Check the commutativity and associativity of the following binary operations : $'*'$ on $Q$ defined by $a * b = (a - b)^2$ for all $a, b \in Q$.
AnswerCommutativity:
Let $\text{a, b}\in\text{Q}.$ Then$, a * b = (a - b)^2$
$= (b - a)^2$
$= b * a$
Therefore,
$a * b = b * a, \forall\ \text{a, b}\in\text{Q}$
Thus$, *$ is commutative on $Q$.
Associativity:
Let $\text{a, b, c}\in\text{Q}.$ Then,
$a * (b * c) = a * (b - c)^2$
$= a * (b^2 + c^2 - 2bc)$
$= (a - b^2 - c^2 + 2bc)^2$
$(a * b) * c = (a - b)^2 * c$
$= (a^2 + b^2 - 2ab) * c$
$= (a^2 + b^2 - 2ab - c)^2$
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 243 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 that $a * e = a = e * a, \forall\text{ a}\in\text{R}$
$a * e = a$ 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 253 Marks
Classify the following functions as injection, surjection or bijection:$f : R \rightarrow R,$ defined by $f(x) = 1 + x^2$
Answer$f : R \rightarrow R,$ 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 263 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 273 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 283 Marks
Give an example of a map :
- Which is one$-$one but not onto.
- Which is not one$-$one but onto.
- Which is neither one$-$one nor onto.
Answer
- Let $f : N \rightarrow N,$ be a mapping defined by $f(x) = 2x$
Which is one$-$one
For $f(x_1) = f(x_2)$
$\Rightarrow 2x_1 = 2x_2$
$x_1 = x_2$
Further $f$ is not onto, as for $1\in\text{N},$ there does not exist any $x$ in $N$ such that $f(x) = 2x + 1.$
- Let $f : N \rightarrow N,$ given by $f(1) = f(2) = 1$ and $f(x) = x - 1$ for every $x > 2$ is onto but not one$-$one. $f$ is not one$-$one as $f(1) = f(2) = 1$. But $f$ is onto.
- The mapping $f : R \rightarrow R$ defined as $f(x) = x^2,$ is neither one$-$one not onto.
View full question & answer→Question 293 Marks
Construct the composition table for $+_5$ on set $S = \{0, 1, 2, 3, 4\}.$
Answer$a +_5 b =$ the remainder when $a + b$ is divided by $5.$eg. $2 + 4 = 6$
$ \Rightarrow 2 +_5 4 = 1$ $\because[$we get $1$ as remainder when $6$ is divided by $5]$
$2 + 4 = 7$
$ \Rightarrow 3 +_5 4 = 2 \because[$we get $2$ as remainder when $7$ is divided by $5]$
The composition table for $+_5$ on set $S = {0, 1, 2, 3, 4}.$
| $+_5$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
| $0$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
| $1$ |
$1$ |
$2$ |
$3$ |
$4$ |
$0$ |
| $2$ |
$2$ |
$3$ |
$4$ |
$0$ |
$1$ |
| $3$ |
$3$ |
$4$ |
$0$ |
$1$ |
$2$ |
| $4$ |
$4$ |
$0$ |
$1$ |
$2$ |
$3$ |
View full question & answer→Question 303 Marks
Consider the binary operation $*$ and $o$ defined by the following tables on set $S = \{a, b, c, d\}.$
| $o$ |
$a$ |
$b$ |
$c$ |
$d$ |
| $a$ |
$a$ |
$a$ |
$a$ |
$a$ |
| $b$ |
$a$ |
$b$ |
$c$ |
$d$ |
| $c$ |
$a$ |
$c$ |
$d$ |
$b$ |
| $d$ |
$a$ |
$d$ |
$b$ |
$c$ |
AnswerCommutativity:The table is symmetrical about the leading element.
It means that $o$ is commutative on $S$.
$a o (b o c) = a o c$
$= a$
$(a o b) o c = a o c$
$= a$
thus,
$a o (b o c) = (a o b) o$
$\text{c }\forall\text{ a, b, c}\in\text{S}$
Associativity:
Therefore$, o$ is associative on $S.$
Finding identity element:
We observe that the second row of the composition table coincides with the top$-$most row and the first column coincides with the left$-$most column.
These two intersect at $b.$
Implies that $x o b = b o x$
$=\text{x, }\forall\text{ x}\in\text{S}$
Therefore,
$b$ is the identity element.
Finding inverse elements:
In the first row, we don't have $b,$
i.e. there does not exist an element $x$ such that $a o x = x o a = b.$
Therefore,
$a^{-1}$ does not exists.
$b o b = b$
Implies that $b^{-1 }= b$
$c o d = b$
Implies that $c^{-1} = d$
$d o c = b$
Implies that $d^{-1} = c$
View full question & answer→Question 313 Marks
Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective.
- {(x, y): x is a person, y is the mother of x}.
- {(a, b): a is a person, b is an ancestor of a}.
Answer
- We have {(x, y): x is a person, y is the mother of x}.
Clearly each person 'x' has only one biological mother.
So above set of ordered pair is a function.
Now more than one person may have same mother. So function is many-one and surjective.
- We have {(a, b): a is a person, b is an ancestor of a}
Clearly any person 'a' has more than one ancestors.
So, it does not represent function.
View full question & answer→Question 323 Marks
Three relation $R_1$ is defined in set $A = \{a, b, c\}$ as follows:
$R_1 = \{(a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c)\}$
Find whether or not the relation $R_{1 }$ on $A$ is:
- Reflexive.
- Symmetric.
- Transitive.
AnswerConsider $R,$ Reflexive: Clearly, $(a, a), (b, b)$ and $(c, c) \in\text{R}_1$
Therefore, $R_1$ is reflexive.
Symmetric: We see that the ordered pairs obtained by interchanging the components of $R_1$ are also in $R_1$
Therefore, $R_1$ is symmetric.
Transitive: Here, $\text{a, b}\in\text{R}_1,\ \text{b, c}\in\text{R}_1$ and also $\text{a, c}\in\text{R}_1$
Therefore, $R_1$ is transitive.
View full question & answer→Question 333 Marks
Each of the following defines a relation on N:
$\text{x}+\text{y}=10,\ \text{x},\ \text{y}\in\text{N}$
Determine which of the above relations are reflexive, symmetric and transitive.
Answer$\therefore\ \text{R}=\{(\text{x},\text{y});\ \text{x}+\text{y}=10,\ \text{x},\text{y}\in\text{N}\}$$\therefore$ R = {(1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)}
Clearly $(1,1)\notin\text{R}$
So, R is not reflexive.
$(\text{x},\text{y})\in\text{R}\Rightarrow\ (\text{y},\text{x})\in\text{R}$
Thus, R is symmetric.
Now $(1,9)\in\text{R},\ (9,1)\in\text{R},$ but $(1,1)\notin\text{R}$
Hence, R is not transitive.
View full question & answer→Question 343 Marks
Let S be the set of all rational numbers of the for $\frac{\text{m}}{\text{n}},$ where $\text{m}\in\text{Z}$ and n = 1, 2, 3. Prove that * on sdefined by a * b = ab is not a binary operation.
Answer$\text{S}=\Big\{\text{a}=\frac{\text{m}}{\text{n}}:\text{m}\in\text{Z},\text{ n}\in\{1, 2, 3\}\Big\}$Let $\text{a}=\frac{1}{3},\ \text{b}=\frac{5}{3}\in\text{S}$
$\text{a}\ ^*\ \text{b}=\text{ab}$
$=\frac{1}{3}\times\frac{5}{3}$
$=\frac{5}{9}\notin\text{S}\ \big[\because\ 9\notin\{1,2,3\}\big] $
Therefore, $\exists\text{ a, b}\in\text{S},$ such that $\text{a}\ ^*\ \text{b}\notin\text{S}$
Thus, * is not a binary operation.
View full question & answer→Question 353 Marks
Let '*' be a binary operation on N defined by a * b = 1.c.m. (a, b) for all $\text{a, b}\in\text{N}.$
Check the commutativity and associativity of '*' on N.
AnswerCommutativity: Let $\text{a, b}\in\text{N}$a * b = 1.c.m. a, b
= 1.c.m. b, a
= b * a
Therefore,
$\text{a}\ ^*\ \text{b}=\text{b}\ ^*\ \text{a}\ \forall\ \text{a, b}\in\text{N}$
Thus, * is Commutative on N.
Associativity: Let $\text{a, b, c}\in\text{N}$
a * b * c = a * 1.c.m. b, c
= 1.c.m. a, b, c
a * b * c = 1.c.m. a, b * c
= 1.c.m. a, b, c
Therefore,
$\text{a}\ ^*\ \text{b}\ ^*\ \text{c}=\text{a}\ ^*\ \text{b}\ ^*\ \text{c}\ \forall\ \text{a, b, c}\in\text{N}$
Thus, * is associative on N.
View full question & answer→Question 363 Marks
Given A = {2, 3, 4}, B = {2, 5, 6, 7}. Construct an example of each of the following:
- An injective mapping from A to B.
- A mapping from A to B which is not injective.
- A mapping from B to A.
AnswerGiven that, A = {2, 3, 4}, B = {2, 5, 6, 7}
- Let f : A → B denote a mapping
$\text{f}\equiv\{(\text{x},\text{y}):\text{y}=\text{x}+3\}$
or $\text{f}\equiv\{(2,5),(3,6),(4,7)\},$ which is an injective mapping.
- Let g : A → B denote a mapping such that g = {(2, 2), (3, 2), (4, 5)}, which is not an injective mapping.
- Let h : B → A denote a mapping such that h = {(2, 2), (5, 3), (6, 4), (7, 4)}, which is one of the mappings from B to A.
View full question & answer→Question 373 Marks
Check the commutativity and associativity of the following binary operations:
$'*'$ on $Q$ defined by $a * b = ab^2$ for all $a, b \in Q$.
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 383 Marks
Let $R_0$ denote the set of all non $-$ zero real numbers and let $A = R_0 \times R_0$. If $'*'$ is a binary operation on adefined by,
$(a, b) * (c, d) = (ac, bd)$ for all $(a, b), (c, d) \in A$
Show that '*' is both commutative and associative on $A$.
AnswerCommutativity : Let $\text{a}, \text{b}\ \&\ \text{c}, \text{d}\in\text{A}\forall\text{ a, b, c, d}\in\text{R}_0$. Then,
$(a, b) * (c, d) = (ac, bd)$
$= (ca, db)$
$= (c, d) * (a, b)$
$\therefore (a, b) * (c, d) = (c, d) * (a, b)$
Thus $, *$ is commutative on $A$.
Associativity :
Let $(a, b), (c, d) (e, f) \in\text{A}$ for all $\text{ a, b, c, d, e, f}\in\text{R}_0$. Then,
$(a, b) * ((c, d) * (e, f)) = (a, b) * (ce, df)$
$= \text{(ace, bdf)}$
$((a, b) * (c, d)) * (e, f) = (ac, bd) * (e, f)$
$= \text{(ace, bdf)}$
$\therefore (a, b) * ((c, d) * (e, f)) = ((a, b) * (c, d) * (e, f))$
Thus $, *$ is associative on $A$.
View full question & answer→Question 393 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 ${x}=(\text{a},\text{b})$ 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,$ for all $ \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 $\text{Z}=(\text{e},\text{f}),\forall\ \text{a},\text{c},\text{e}\in\text{R}_0\ \ \text{ b},\text{d},\text{f}\in\text{R}$
$x ⊙ Y ⊙ Z = (a, b) ⊙ (ce, de + f)$
$= \text{(ace, bce + de + f)}$
$x ⊙ Y ⊙ Z = (ac, bc + d) ⊙ (e, f)$
$= \text{(ace, (bc + d)e + f)}$
$= \text{(ace, bce + de + f)}$
$\therefore x ⊙ Y ⊙ Z = x ⊙ Y ⊙ Z, $ for all $\text{X},\text{Y},\text{Z}\in\text{A}$
Thus $, ⊙$ is accosiative on $A$.
View full question & answer→Question 403 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 413 Marks
Test whether the following relations $R_{1 }$ are:
- Reflexive.
- Symmetric.
- Transitive.
$R_1$ on $Q_0$ defined by $(\text{a, b})\in\text{R}_1\Leftrightarrow\ \text{a}=\frac{1}{\text{b}}.$ Answer$\text{R}_1=\Big\{(\text{x, y}),\ \text{x, y}\in\text{Q}_0, \text{x}=\frac{1}{\text{y}}\Big\}$ Reflexivity: Let, $\text{x}\in\text{Q}_0$
$\Rightarrow\ \text{x}\neq\frac{1}{\text{x}}$
$\Rightarrow\ (\text{x, x})\in\text{R}_1$
$\therefore R_1$ is not reflexive.
Symmetric: Let, $(\text{x, y})\in\text{R}_1$
$\Rightarrow\ \text{x}=\frac{1}{\text{y}}$
$\Rightarrow\ \text{y}=\frac{1}{\text{x}}$
$\Rightarrow\ (\text{y, x})\in\text{R}_1$
$\therefore R_1$ is Symmetric.
Transitive: Let, $(\text{x, y})\in\text{R}_1$ and $(\text{y, z})\in\text{R}_1$
$\Rightarrow\ \text{x}=\frac{1}{\text{y}}$ and $\text{y}=\frac{1}{\text{z}}$
$\Rightarrow\ \text{x}=\text{z}$
$\Rightarrow\ (\text{x, z})\notin\text{R}_1$
$\therefore R_1$ is not Transitive.
View full question & answer→Question 423 Marks
Check the commutativity and associativity of the following binary operations:
'*' on N defined by a * b = gcd(a, b) for all a, b ∈ N.
AnswerCommutativity: Let $\text{a, b}\in\text{N}.$ Then,a * b = gcd(a, b)
= gcd(b, a)
= b * a
Therefore,
a * b = b * a, $\forall\ \text{a, b}\in\text{N}$
Thus '*' is commutative on N.
Associativity: Let $\text{a, b, c}\in\text{N}.$ Then,
a * (b * c) = a * [gcd(a, b)]
= gcd(a, b, c)
(a * b) * c = [gcd(a, b)] * c
= gcd(a, b, c)
Therefore,
a * (b * c) = (a * b) * c, $\forall\ \text{a, b, c}\in\text{N}$
Thus, '*' is not associative on N.
View full question & answer→Question 433 Marks
Classify the following functions as injection, surjection or bijection:f : R → R, defined by f(x) = 3 - 4x
Answerf : R → R, defined by f(x) = 3 - 4xInjection test: Let x and y be any two elements in the domain (R), such that f(x) = f(y).
f(x) = f(y)
3 - 4x = 3 - 4y
-4x = -4y
x = y
Therefore, f is 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
3 - 4x = y
4x = 3 - y
$\text{x}=\frac{3-\text{y}}{4}\in\text{R}$
Therefore, f is a surjection and f is a bijection.
View full question & answer→Question 443 Marks
If $f : R \rightarrow (0, 2)$ defined by $\text{f(x)}=\frac{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+1$ is invertible, find $f^{-1}$.
Answer$\text{A}=\{\text{x}\in\text{R}:-1\leq\text{x}\leq1\}$ and $f : A \rightarrow A, g : A \rightarrow A$ are two functions defined by $f(x) = x^2$ and $\text{g(x)}=\sin\Big(\frac{\pi\text{x}}{2}\Big)$
Here$, f : A \rightarrow A$ is defined by
$f(x) = x^2$
Clearly $f$ in not injective,
$\because\ \text{f}(1)=\text{f}(-1)=1$
So$, f$ is not bijective and hence not invertible.
Hence$, f^{-1}$ does not exist.
View full question & answer→Question 453 Marks
Find fog and gof if:f(x) = x + 1, g(x) = 2x + 3
Answerf(x) = x + 1, g(x) = 2x + 3f : R → R; g : R → R
Computing fog: Clearly, the range of g is a subset of the domain of f.
⇒ fog : R → R
(fog)(x) = f(g(x))
= f(2x + 3)
= 2x + 3 + 1
= 2x + 4
Computing gof: Clearly, the range of f is a subset of the domain of g.
⇒ fog : R → R
(gof)(x) = g(f(x))
= g(x + 1)
= 2(x + 1) + 3
= 2x + 5
View full question & answer→Question 463 Marks
Find fog and gof if:$\text{f}(\text{x})=\text{c},\text{c}\in \text{R},\text{g(x)}=\sin \text{x}^2$
Answer$\text{f} \ \text{x}=\text{c} = \sin \text{x} \ 2\ \text{f}:\text{R}\ \rightarrow{\ } \ \text{c};\text{g}:\text{R}\ \rightarrow{\ } \ 0,1$
Computing fog: Clearly, the range of g is a subset of the domain of f.
$.\text{fog}:\text{R}\ \rightarrow{\ }\ \text{x}=\text{f}\ \text{g}\text{ x }=\text{f} \ \sin \text{x}^2=\text{c}$
Computing gof: Clearly, the range of f is a subset of the domain of g.
$\Rightarrow \text{fog}: \text{R}\ \rightarrow{\ }\text{x}=\text{g}\ \text{f}\ \text{x}=\text{g}\ \text{c}=\sin \text{c}^2$
View full question & answer→Question 473 Marks
Find fog and gof if : $f(x) = x^2 + 2, \text{g(x)}=1-\frac{1}{1-\text{x}}$
Answer$f(x) = x^2 + 2$ and $\text{g(x)}=1-\frac{1}{1-\text{x}}$
Range of $\text{f}=(2,\infty)\ \subset$ Domain of $g = R$
$\Rightarrow$ gof exist
Range of $g = R - [1] \subset$ Domain of $f = R$
$\Rightarrow$ fog exist
Now,
$fog(x) = f(g(x))$
$=\text{f}\Big(\frac{-\text{x}}{1-\text{x}}\Big)=\frac{\text{x}^2}{(1-\text{x})^2}+2$
And,
$gof(x) = g(f(x))$
$=\text{g}(\text{x}^2+2)=\frac{-(\text{x}^2+2)}{1-(\text{x}^2+2)}$
$\Rightarrow\ \text{gof(x)}=\frac{\text{x}^2+2}{\text{x}^2+1}$
View full question & answer→Question 483 Marks
Show that the logarithmic function $\text{f}:\text{R}0^+\rightarrow \text{R}$ given by $f(x) = log_a x, a > 0$ is a bijection.
AnswerWe have, $f : A \rightarrow B$ and $g : B \rightarrow C$ are one $-$ one functions.
Now we have to prove: gof : $A \rightarrow C$ in one $-$ one.
Let $\text{x, y}\in\text{A}$ such that
$\text{gof(x) = gof(y)}$
$\Rightarrow g(f(x)) = g(f(y))$
$\Rightarrow f(x) = f(y) \ [\because g$ in one $-$ one$]$
$\Rightarrow x = y \ [\because f $ in one $-$ one$]$
$\therefore \text{gof}$ is one $-$ one function.
View full question & answer→Question 493 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 503 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→