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 $\text{x}_{1},\text{x}_{2}\in\text{R}\text{(Domain)}$
$\text{f}(\text{x}_{1}) =\text{f}(\text{x}_{2})\Rightarrow\text{x}_{1}^{2} + 4 = \text{x}_{2}^{2} + 4 $
$\Rightarrow\text{x}_{1}^{2} =\text{x}_{2}^{2}$
$\Rightarrow\text{x}_{1} =\text{x}_{2}[\because\text{x}_{1},\text{x}_{2}\text{ are +ve real number}]$
$\therefore$ f is one-one function.
For onto
Let $\text{y}\in[4,\infty)\text{s.t.}$
$\text{y} = \text{f}(\text{x) }\forall\text{ x}\in\text{R}_{+}(\text{ set of non-negative reals})$
$\Rightarrow\text{y} = \text{x}^{2} + 4 $
$\Rightarrow\text{x} = \sqrt{\text{y} - 4 } [ \because\text{ x is + ve real number }]$
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
fo $f^{-1}$ = 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 $gof = fog = I_R.$
AnswerLet $y = 10x + 7 \therefore\text{ x}=\frac{1}{10}\text{(y - 7)}$
Let $g(y) =\frac{1}{10}\text{(y - 7)}$
$\therefore gof(x) = g(10x + 7) = \frac{1}{10} (10x + 7 - 7) = x \Rightarrow I_R= gof$
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 $ONTO$
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 $f : N → N$, defined by $f(x) = x^2 + x + 1$ is one-one but not onto. Find inverse of $f : N → S$, where S is range of f.
AnswerThe given function is
$\text{f} : \text{N} \rightarrow \text{N}$
$\text{f(x)} = \text{x}^2 + \text{x} + 1$
$\text{Let} \text{ x}_1, \text{x}_2 6\in\text{N}$
$\text{So} \text{ let} \text{ f(x}_1) = \text{f(x}_2)$
$\text{x}_1^2+\text{x}_1+1=\text{x}_2^2 +\text{x}_2+1$
$\text{x}_1^2-\text{x}_2^2+\text{x}_1-\text{x}_2=0$
$(\text{x}_1-\text{x}_2)(\text{x}_1-\text{x}_2+1)=0$
$\because\text{x}_2=\text{x}_1$
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
If A = {1, 2, 3, 4} define relations on A which have properties of being:
Reflexive, transitive but not symmetric.
AnswerThe relation on A having properties of being reflexive, transitive, but not symmetric is, R = {(1, 1), (2, 2), (3, 3), (4, 4), (2, 1)} Relation R satisfies reflexivity and transitivity. $ \Rightarrow(1, 1), (2, 2), (3, 3) \in\text{R}$$$and $(1, 1), (2, 1) \in \text{R}\Rightarrow(1, 1)\in \text{R}$
However, $(2,1)\in\text{R},$ but $(1,2)\notin\text{R}$
View full question & answer→Question 203 Marks
Test whether the following relations $R_2$ are:
- Reflexive.
- Symmetric.
- Transitive.
$R_2$ on Z defined by $(\text{a, b})\in\text{R}_2\Leftrightarrow\ |\text{a}-\text{b}|\leq5$ AnswerReflexivity: Let a be an arbitrary element of $R_2$. Then,$\text{a}\in\text{R}_2$
$\Rightarrow\ |\text{a}-\text{a}|=0\leq5$
So, $R_2$ is reflexive.
Symmetry: Let $(\text{a, b})\in\text{R}_2$
$\Rightarrow\ |\text{a}-\text{b}|\leq5$
$\Rightarrow\ |\text{b}-\text{a}|\leq5$ [Since, |a - b| = |b - a|]
$\Rightarrow\ (\text{b, a})\in\text{R}_2$
So, $R_2$ is symmetric.
Transitivity: Let $(1, 3)\in\text{R}_2$ and $(3,7)\in\text{R}_2$
$\Rightarrow\ |1-3|\leq5$ and $|3-7|\leq5$
But $|1-7|\nleq5$
$\Rightarrow\ (1,7)\notin\text{R}_2$
So, $R_2$ is transitive.
View full question & answer→Question 213 Marks
Show that f : R → R, given by f(x) = x - [x], is neither one-one nor onto.
Answerf : R → R, given by f(x) = x - [x]Injectivity: f(x) = 0 for all $\text{x}\in\text{Z}$
Therefore, f is not one-one.Surjectivity: Range of f = (0, 1) ≠ R.
Co-domain of f = R Both are not same. Therefore, f is not onto.
View full question & answer→Question 223 Marks
Let $A = \{1, 2, 3\},$ and let $R_1 = \{(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)\}$. Find whether or not the relations $R_1$ on A is:
- Reflexive.
- Symmetric.
- Transitive.
AnswerWe have A = {1, 2, 3}, and $R_1$ = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}
$\therefore$ (1, 1), (2, 2) and (3, 3) $\in\text{R}_1$
$\therefore R_1$ is not reflexive.
Now,
$\therefore\ (2,1)\in\text{R}_1$ but $(1,2)\notin\text{R}_1$
$\therefore R_1$ is not symmetric.
Again,
$\therefore\ (2,1)\in\text{R}_1$ and $(1,3)\in\text{R}_1$ but $(2,3)\notin\text{R}_1$
$\therefore R_1$ is not transitive.
View full question & answer→Question 233 Marks
Construct the composition table for $\times _5$ on $Z_5 = \{0, 1, 2, 3, 4\}.$
AnswerHere,
$1\times _51 =$ Remainder obtained by dividing $1 \times 1$ by $5 = 1$
$3\times _54 =$ Remainder obtained by dividing $3 \times 4$ by $5 = 2$
$4\times _54 =$ Remainder obtained by dividing $4 \times 4$ by $5 = 1$
Therefore,
The composition table is as follows:
| $\times _5$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
| $0$ |
$0$ |
$0$ |
$0$ |
$0$ |
$0$ |
| $1$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
| $2$ |
$0$ |
$2$ |
$4$ |
$1$ |
$3$ |
| $3$ |
$0$ |
$3$ |
$1$ |
$4$ |
$2$ |
| $4$ |
$0$ |
$4$ |
$3$ |
$2$ |
$1$ |
View full question & answer→Question 243 Marks
For the binary operation $\times _7$ on the set $S =\{1, 2, 3, 4, 5, 6\},$ compute $3^{−1} \times _7 4.$
AnswerFinding identity element:
Here,
$1 \times _7 1 =$ Remainder obtained by dividing $1 \times 1$ by $7 = 1$
$3 \times _7 4 =$ Remainder obtained by dividing $3 \times 4$ by $7 = 5$
$4 \times _7 5 =$ Remainder obtained by dividing $4 \times 5$ by $7 = 6$
So, the composition table is as follows:
| $\times _7$ |
$1$ |
$2$ |
$3$ |
$4$ |
$5$ |
$6$ |
| $1$ |
$1$ |
$2$ |
$3$ |
$4$ |
$5$ |
$6$ |
| $2$ |
$2$ |
$4$ |
$6$ |
$1$ |
$3$ |
$5$ |
| $3$ |
$3$ |
$6$ |
$2$ |
$5$ |
$1$ |
$4$ |
| $4$ |
$4$ |
$1$ |
$5$ |
$2$ |
$6$ |
$3$ |
| $5$ |
$5$ |
$3$ |
$1$ |
$6$ |
$4$ |
$2$ |
| $6$ |
$6$ |
$5$ |
$4$ |
$3$ |
$2$ |
$1$ |
We observe that all the elements of the first row of the composition table are same as the top-most row.
So, the identity element is $1$.
Also, $3 \times _7 5 = 1$
So, $3 ^{- 1 } = 5$
Now,
$ 3 ^{- 1 }\times _7 4 = 5 \times _7 4 = 6$ View full question & answer→Question 253 Marks
The binary operation * is defined by $\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{7}$ on the set Q of all rational numbers. Show that * is associative.
AnswerThe binary operator * is defined as,
$\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{7}$ for all $\text{a, b}\in\text{Q}$
Now,
Associativity: Let $\text{a, b, c}\in\text{Q},$ then
$(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}=\frac{\text{ab}}{7}\ ^*\ \text{c}=\frac{\text{abc}}{49}\ ....(\text{i})$
and $\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=\text{a}\ ^*\ \frac{\text{bc}}{7}=\frac{\text{abc}}{49}\ .....(\text{ii})$
From (i) and (ii)
(a * b) * c = a * (b * c)
⇒ '*' is associative on Q.
View full question & answer→Question 263 Marks
For the binary operation multiplication modulo $10 (\times _{10})$ defined on the set $S = \{1, 3, 7, 9\},$ write the inverse of $3.$
Answer$1 \times _{10}1 =$ Remainder obtained by dividing $1 \times 1$ by $10 = 1$
$3 \times _{10}1 =$ Remainder obtained by dividing $3 \times 1$ by $10 = 3$
$7 \times _{10}3 = $ Remainder obtained by dividing $7 \times 3$ by $10 = 1$
$3 \times _{10}3 =$ Remainder obtained by dividing $3 \times 3$ by $10 = 9$
So, the composition table is as follows:
| $\times _{10}$ |
$1$ |
$3$ |
$7$ |
$9$ |
| $1$ |
$1$ |
$3$ |
$7$ |
$9$ |
| $3$ |
$3$ |
$9$ |
$1$ |
$7$ |
| $7$ |
$7$ |
$1$ |
$9$ |
$3$ |
| $9$ |
$9$ |
$7$ |
$3$ |
$1$ |
We observe that the first row of the composition table coincides with the top-most row and the first column coincides with the left-most column.
These two intersect at $1.$
$⇒ a * 1 = 1 * a = a, \forall\text{ a}\in\text{S}$
So, the identity element is $1.$
Also,
$3 \times _{10}7 = 1$
$3^{-1} = 7$ View full question & answer→Question 273 Marks
Let $*$ be a binary operation on $Q_0 ($set of non-zero rational numbers$)$ defined by $\text{a}\ ^* \ \text{b}=\frac{\text{ab}}{5}$ for all $\text{a, b}\in\text{Q}_0.$ Show that $*$ is commutative as well as associative. Also, find its identity element if it exists.
AnswerCommutativity: Let $\text{a, b}\in\text{Q}_0$
$\text{a}\ ^*\ \text{b}=\frac{\text{ab}}{5}$
$=\frac{\text{ba}}{5}$
$=\text{b}\ ^*\ \text{a}$
Therefore, $\text{a}\ ^*\ \text{b}=\text{b}\ ^*\ \text{a},\forall\ \text{a, b}\in\text{Q}_0$
Thus, $*$ is commutative on $Q_0.$
Associativity: Let $\text{a, b, c}\in\text{Q}_0$
$\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=\text{a}\ ^*\ \Big(\frac{\text{bc}}{5}\Big)$
$=\frac{\text{a}\big(\frac{\text{bc}}{5}\big)}{5}$
$=\frac{\text{abc}}{25}$
$(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}=\Big(\frac{\text{ab}}{5}\Big)\ ^*\ \text{c}$
$=\frac{\big(\frac{\text{ab}}{5}\big)\text{c}}{5}$
$=\frac{\text{abc}}{25}$
Therefore,
$a * (b * c) = (a * b) * c, \forall\ \text{a, b, c}\in\text{Q}_0$
Thus, * is associative on $Q_0.$
Finding identity element:
Let e be the identity element in $Z$ with respect to $*$ such that,
$a * e = a = e * a, \forall\text{a}\in\text{Q}_0$
$a * e = a$ and $e * a = a, \forall\text{a}\in\text{Q}_0$
Implies that $\frac{\text{ae}}{5}=\text{a}$ and $\frac{\text{ea}}{5}=\text{a},\forall\ \text{a}\in\text{Q}_0$
Implies that $\text{e}=5,\forall\ \text{a}\in\text{Q}_0\ [\because\ \text{a}\neq0]$
Thus, 5 is the identity element in with respect to $*$.
View full question & answer→Question 283 Marks
Find which of the binary operations are commutative and which are associative.
Let A = N × N and * be the binary operation on A defined by:
(a, b) * (c, d) = (a + c, b + d)
AnswerA = N × N and * is a binary operation defined on A.
(a, b) * (c, d) = (a + b, c + d) = (c + a, d + b) = (c, d) * (a, b)
$\therefore$ The operation is commutative
Again, [(a, b) * (c, d)] * (e, f) = (a + c, b + d) * (e, f) = (a + c + e, b + d + f)
And (a, b)[(c, d) * (e, f)] = (a, b) * (c + e, e + f) = (a + c + e, b + d + f)
Here, [(a, b) * (c, d)] * (e, f) = (a, b)[(c, d) * (e, f)]
$\therefore$ The operation is associative.
Let the identity function be (e, f), then (a, b) * (e, f) = (a + e, b + f)
For identity function a = a + e $\Rightarrow\ \ \text{e}=0$
And for b + f = b $\Rightarrow\ \ \text{f}=0$
As $0\neq\text{N},$ therefore identity element does not exist.
View full question & answer→Question 293 Marks
State with reasons whether the following functions have inverse:
g : {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}
Answerg : {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}
g(5) = g(7) = 4
⇒ f is not one-one.
⇒ f is not a bijection.
So, f does not have an inverse.
View full question & answer→Question 303 Marks
Given a non-empty set $X,$ let $*: P(X) \times P(X) \rightarrow P(X)$ be defined as $\text{A}*\text{B} =(\text{A – B})\cup(\text{B – A}),\forall\text{A},\text{B}\in\text{P(X)}.$Show that the empty set $\phi$ is the identity for the operation $*$ and all the elements $A$ of $P(X)$ are invertible with $A^{–1} = A. (\text{Hint: }(\text{A}-\phi)\cup(\phi-\text{A})=\text{A}\ \text{and }(\text{A}-\text{A})\cup(\text{A}-\text{A})=\text{A}*\text{A}=\phi).$
AnswerIt is given that $*: P(X) \times P(X) \rightarrow P(X)$ is defined as $\text{A}*\text{B} =(\text{A – B})\cup(\text{B – A}),\forall\text{A},\text{B}\in\text{P(X)}.$
Let $\text{A}\in\text{P(X)}.$
Then, we have: $\text{A}*\phi=(\text{A}-\phi)\cup(\phi-\text{A})=\text{A}\cup\phi=\text{A}$
$\phi*\text{A}=(\phi-\text{A})\cup(\text{A}-\phi)=\phi\cup\text{A}=\text{A}$
$\therefore\text{A}*\phi=\text{A}=\phi*\text{A}.\forall\text{A}\in\text{P(X)}$ Thus,
$\phi$ is the identity element for the given operation $*$.
Now, an element $\text{A}\in\text{P(X)}$ will be invertible if there exists $\text{B}\in\text{P(X)}$ such that $\text{A}*\text{B}=\phi=\text{B}*\text{A}.\ (\text{As }\phi\text{ is the identity element})$
Now, we observed that $\text{A}*\text{A}=(\text{A}-\text{A})\cup(\text{A}-\text{A})=\phi\cup\phi=\phi\ \forall\text{A}\in\text{P(X)}$
Hence, all the elements $A$ of $P(X)$ are invertible with $A^{-1} = A.$
View full question & answer→Question 313 Marks
Find the identity element in the set of all rational numbers except -1 with respect to * defined by a * b = a+ b + ab.
Question 323 Marks
Let A = [-1, 1]. Then, discuss whether the following functions from A to itself are one-one, onto or bijective:
g(x) = |x|
Answerg(x) = |x|Injection test: Let x and y be any two elements in the domain (A), such that f(x) = f(y).
f(x) = f(y) |x| = |y| $\text{x}=\pm\text{y}$ So, f is not one-one. Surjection test: For y = -1, there is no value of x in A. So, f is not onto. So, f is not bijective.
View full question & answer→Question 333 Marks
Prove that the Greatest Integer Function f: R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
Answerf: R → R is given by,f(x) = [x]
It is seen that f(1.2) = [1.2] = 1, f(1.9) = [1.9] = 1. $\therefore$ f(1.2) = f(1.9), but $1.2\neq1.9$ $\therefore$ f is not one-one. Now, consider $0.7\in\text{R}.$
It is known that f(x) = [x] is always an integer. Thus, there does not exist any element $\text{x}\in\text{R}$ such that f(x) = 0.7.
$\therefore$ f is not onto. Hence, the greatest integer function is neither one-one nor onto.
View full question & answer→Question 343 Marks
Let the function f : R → R be defined by f(x) = cosx, ∀ x ∈ R. Show that f is neither one-one nor onto.
AnswerWe are given, f(x) = cosx, ∀ x ∈ RFor $\frac{\pi}{2},$ we have
$\text{f}\Big(\frac{\pi}{2}\Big)=\cos\frac{\pi}{2}=0$
For $-\frac{\pi}{2},$ we have
$\text{f}\Big(\frac{-\pi}{2}\Big)=\cos\frac{\pi}{2}=0$
$\therefore\ \text{f}\Big(\frac{\pi}{2}\Big)=\text{f}\Big(\frac{-\pi}{2}\Big)$
But, $\frac{\pi}{2}\neq\frac{-\pi}{2}$
Hence, f(x) is not one-one.
We also know that, there is not any pre-image for any real number which does not belong to the range of cosine function i.e., [-1, 1].
View full question & answer→Question 353 Marks
Show that the function $f: R \rightarrow R$ given by $f(x) = x^3$ is injective.
Answer$f: R → R$ is given as $f(x) = x^3$
Suppose $f(x) = f(y),$ where $\text{x},\text{y}\in\text{R}.$
$\Rightarrow x^3 = y^3...(1)$
Now, we need to show that $x = y.$
Suppose $\text{x}\neq\text{y},$ their cubes will also not be equal.
$\Rightarrow\text{x}^3\neq\text{y}^3$
However, this will be a contradiction to $(1).$
$\therefore x = y$
Hence, $f$ is injective.
View full question & answer→Question 363 Marks
Show that the Signum Function f: R → R, given by $\text{f(x)}=\begin{cases}1,&\text{if }\text{x }>0\\0,&\text{if }\text{x }=0\\-1,&\text{if }\text{x }<0\end{cases}$ is neither one-one nor onto.
Answerf: R → R is given by,
$\text{f(x)}=\begin{cases}1,&\text{if }\text{x }>0\\0,&\text{if }\text{x }=0\\-1,&\text{if }\text{x }<0\end{cases}$
It is seen that f(1) = f(2) = 1, but $1\neq2.$
$\therefore$ f(-1) = f(1), but $-1\neq1.$
$\therefore$ f is not one-one.
Now, as f(x) takes only 3 values (1, 0, or -1) for the element -2 in co-domain R, there does not exist any x in domain R such that f(x) = -2.
$\therefore$ f is not onto.
Hence, the signum function is neither one-one nor onto.
View full question & answer→Question 373 Marks
On the set Z of integers, if the binary operation * is defined by a * b = a + b + 2, then find the identity element.
AnswerLet e be the identity element in Z with respect to * such that a * e = a = e * a, $\forall\ \text{a}\in\text{Z}$ a * e = a and e * a = a, $\forall\ \text{a}\in\text{Z}$ a + e + 2 = a and e + a + 2 = a, $\forall\ \text{a}\in\text{Z}$e = -2, $\forall\ \text{a}\in\text{Z}$
Thus, -2 is the identity element in Z with respect to *.
View full question & answer→Question 383 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 identity element in $A.$
AnswerLet $E = (x, y)$ be the identity element in A with respect to ⊙,$\forall\ \text{x}\in\text{R}_0\ \&\text{ y}\in\text{R}$ such that
$x ⊙ E = X = E ⊙ X, \forall\text{ x}\in\text{A}$
$\Rightarrow X ⊙ E = X$ and $E ⊙ X = X$
$\Rightarrow (ax, bx + y) = (a, b)$ and $(xa, ya + b) = (a, b)$
Considering $(ax, bx + y) = (a, b)$
$\Rightarrow ax = a$
$\Rightarrow x = 1$
$\&\ bx + y = b$
$\Rightarrow y = 0 [\because\text{ x}=1]$
Considering$ (xa, ya + b) = (a, b)$
$\Rightarrow xa = a$
$\Rightarrow x = 1$
$\&\ ya + b = b$
$\Rightarrow y = 0 [\because\text{ x}=1]$
$[\because\ 1,0]$ is the identity element in A with respect to $⊙$.
View full question & answer→Question 393 Marks
The following relation are defined on the set of real numbers.
$aRb$ if $1 + ab > 0$
Find whether these relation are reflexive, symmetric or transitive.
AnswerWe have aRb if $1 + ab > 0$
Let $R$ be the set of real numbers
Reflexive: Let $\text{a}\in\text{R}$
$\Rightarrow 1 + a^2 > 0$
$\Rightarrow aRa$
$\Rightarrow R$ is reflexive.
Symmetric: Let $aRb$
$\Rightarrow 1 + ab > 0$
$\Rightarrow 1 + ba > 0$
$\Rightarrow bRa$
$\Rightarrow R$ is symmetric
Transitive: Let $aRb$ and $bRc$
$\Rightarrow 1 + ab > 0$ and $1 + bc > 0$
$\Rightarrow 1 + ac > 0$
$\Rightarrow R$ is not transitive.
View full question & answer→Question 403 Marks
Write the composition table for the binary operation multiplication modulo $10 (\times _{10})$ on the set $S = \{2, 4, 6, 8\}.$
Answer$2 \times _{10}4 =$ Remainder obtained by dividing $2 \times 4$ by $10 = 8$
$4 \times _{10}6 =$ Remainder obtained by dividing $4 \times 6$ by $10 = 4$
$2 \times _{10}8 =$ Remainder obtained by dividing $2 \times 8$ by $10 = 6$
$3 \times _{10}4 =$ Remainder obtained by dividing $3 \times 4$ by $10 = 2$
Therefore, the composition table is as follows:
| $\times _{10}$ |
$2$ |
$4$ |
$6$ |
$8$ |
| $2$ |
$4$ |
$8$ |
$2$ |
$6$ |
| $4$ |
$8$ |
$6$ |
$4$ |
$2$ |
| $6$ |
$2$ |
$4$ |
$6$ |
$8$ |
| $8$ |
$6$ |
$2$ |
$8$ |
$4$ |
View full question & answer→Question 413 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 423 Marks
Check the commutativity and associativity of the following binary operations:
$'⊙'$ on $Q$ defined by $a ⊙ b = a^2 + b^2$ for all $a, b \in Q.$
AnswerCommutativity: Let $\text{a, b}\in\text{Q}$
then, $a ⊙ b = a^2 + b^2 = b^2 + a^2 = b ⊙ a$
Therefore, $a ⊙ b = b ⊙ a, \forall\ \text{a, b}\in\text{Q}$ Thus, $⊙$ is commuatative on $Q.$
Associativity: Let $\text{a, b, c}\in\text{Q.}$
$a ⊙ (b ⊙ c)= a ⊙ (b^2 + c^2) = ab2 + (b^2 + c^2)^2= ab^2 + b^4 + c^4 + 2b^2c^2$
$(a ⊙ b) ⊙ c = (a^2 + b^2) ⊙ c = (a^2 + b^2)^2 + c^2 = a^4 + b^4 + 2a^2b^2 + c^2$
Therefore,$\text{a}\odot\text{b}\odot\text{c}\neq\text{a}\odot\text{b}\odot\text{c}$
Thus, $⊙$ is not associative on $Q.$
View full question & answer→Question 433 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 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
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 463 Marks
Classify the following functions as injection, surjection or bijection:
$f : R \rightarrow R,$ defined by $f(x) = \sin x$
Answer$f : R \rightarrow R,$ given by $f(x) = \sin x$
Injective: Let $\text{x, y}\in\text{R}$
such that $f(x) = f(y) \Rightarrow \sin x = \sin y$
$\Rightarrow\ \text{x}=\text{n}\pi+(-1)^{\text{n}}\text{y}$
$\Rightarrow\ \text{x}\neq\text{y}$
$\therefore f$ is not one-one.
Surjective: Let $\text{y}\in\text{R}$ be arbitrary such that
$f(x) = y$
$\Rightarrow sinx = y \Rightarrow x = \sin^{-1}y$
Now, for $\text{y}>1\times\notin\text{R}$ (domain)$\therefore$ f is not onto.
View full question & answer→Question 473 Marks
State with reasons whether the following functions have inverse:
f : {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}
Answerf : {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}
We have,
f(1) = f(2) = f(3) = f(4) = 10
⇒ f is not one-one.
⇒ f is not a bijection.
So, f does not have an inverse.
View full question & answer→Question 483 Marks
Let $A = \{1, 2, 3\},$ and let $R_3 = \{(1, 3), (3, 3)\}$. Find whether or not the relations $R_3$ on $A$ is:
- Reflexive.
- Symmetric.
- Transitive.
Answer$R_3 = \{(1, 3), (3, 3)\}$
$\therefore\ (1,1)\notin\text{R}_3$
$\Rightarrow R_3$ is not reflexive.
Now, $(1,3)\in\text{R}_3$ but $(3,1)\in\text{R}_3$
$\therefore R_3$ is not symmetric.
Again, it is clear that $R_3$ is transitive.
View full question & answer→Question 493 Marks
Show that the relation $''\geq''$ on the set R of all real numbers is reflexive and transitive but not symmetric.
AnswerWe have,
relation $\text{R}=\ ''\geq''$ on the set R of all real numbers
Reflexivity: Let $\text{a}\in\text{R}$
$\Rightarrow\ \text{a}\geq\text{a}$
$\Rightarrow\ ''\geq''$ is reflexive.
Symmetric: Let $\text{a, b}\in\text{R}$
Such that $\text{a}\geq\text{b}\Rightarrow\ \text{b}\geq\text{a}$
$\therefore\ ''\geq''$ not symmetric.
Transitivity: Let $\text{a, b, c}\in\text{R}$
and $\text{a}\geq\text{b}\ \&\ \text{b}\geq\text{c}$
$\Rightarrow\ \text{a}\geq\text{c}$
$\Rightarrow\ ''\geq''$ is transitive.
View full question & answer→Question 503 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.
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