Question 13 Marks
If the function f: R$\rightarrow$R be given by f(x) = x2 + 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) = x2 + 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
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 → R be defined as f(x) = 10x + 7. Find the function g : R → R such that gof = fog = IR.
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 $\Rightarrow$ IR = 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 x1, x2 $\in$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 x1, x2 $\in$ A and f(x1) = f(x2) $\Rightarrow$$\frac{\text{x}_{1}- 2}{\text{x}_{1}-3}=\frac{\text{x}_{2}-2}{\text{x}_{2}-3}$ $\therefore$
x1 x2 – 2x2 – 3x1 = x1 x2 – 2x1 – 3x2 $\Rightarrow$
x1 = x2 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) = x2 + 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
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 213 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 R1 = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}
$\therefore$ (1, 1), (2, 2) and (3, 3) $\in\text{R}_1$
$\therefore$ R1 is not reflexive.
Now,
$\therefore\ (2,1)\in\text{R}_1$ but $(1,2)\notin\text{R}_1$
$\therefore$ R1 is not symmetric.
Again,
$\therefore\ (2,1)\in\text{R}_1$ and $(1,3)\in\text{R}_1$ but $(2,3)\notin\text{R}_1$
$\therefore$ R1 is not transitive.
View full question & answer→Question 223 Marks
Construct the composition table for ×5 on Z5 = {0, 1, 2, 3, 4}.
AnswerHere,
1×
51 = Remainder obtained by dividing 1 × 1 by 5 = 1
3×
54 = Remainder obtained by dividing 3 × 4 by 5 = 2
4×
54 = Remainder obtained by dividing 4 × 4 by 5 = 1
Therefore,
The composition table is as follows:
| ×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 233 Marks
For the binary operation ×7 on the set S = {1, 2, 3, 4, 5, 6}, compute 3−1 ×7 4.
AnswerFinding identity element:
Here,
1×
71 = Remainder obtained by dividing 1 × 1 by 7 = 1
3×
74 = Remainder obtained by dividing 3 × 4 by 7 = 5
4×
75 = Remainder obtained by dividing 4 × 5 by 7 = 6
So, the composition table is as follows:
| ×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×
7 5 = 1
So, 3
-1 = 5
Now,
3
-1×
7 4 = 5×
7 4 = 6
View full question & answer→Question 243 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 253 Marks
For the binary operation multiplication modulo 10 (×10) defined on the set S = {1, 3, 7, 9}, write the inverse of 3.
Answer1 ×101 = Remainder obtained by dividing 1 × 1 by 10 = 1
3 ×101 = Remainder obtained by dividing 3 × 1 by 10 = 3
7 ×103 = Remainder obtained by dividing 7 × 3 by 10 = 1
3 ×103 = Remainder obtained by dividing 3 × 3 by 10 = 9
So, the composition table is as follows:
| ×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 ×107 = 1
3-1 = 7
View full question & answer→Question 263 Marks
Let * be a binary operation on Q0 (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 Q0.
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 Q0.
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 273 Marks
Given a non-empty set X, let *: P(X) × P(X) → 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) × P(X) → 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 283 Marks
Find the identity element in the set of all rational numbers except -1 with respect to * defined by a * b = a+ b + ab.
AnswerLet R - {-1} be the set and * be a binary operator, given by a * b = a + b + ab for all $\text{a, b}\in\text{R}-\{-1\}$ Now, Let $\text{a}\in\text{R}-\{-1\}$ and $\text{e}\in\text{R}-\{-1\}$ be the identity element with respect to *. by identity property, we have, a * e = e * a = a ⇒ a + e + ae = a ⇒ e(1 + a) = 0 ⇒ e = 0 $[\because\ 1+\text{a}\neq0\text{ as a }\neq-1]$ $\therefore$
The required identity element is 0. View full question & answer→Question 293 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 303 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 313 Marks
Show that the function f: R → R given by f(x) = x3 is injective.
Answerf: R → R is given as f(x) = x3
Suppose f(x) = f(y), where $\text{x},\text{y}\in\text{R}.$
⇒ x3 = y3 ...(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 323 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 333 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 343 Marks
Check the commutativity and associativity of the following binary operations:
'⊙' on Q defined by a ⊙ b = a2 + b2 for all a, b ∈ Q.
AnswerCommutativity: Let $\text{a, b}\in\text{Q}$ then, a ⊙ b = a2 + b2 = b2 + a2 = 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 ⊙ (b2 + c2) = ab2 + (b2 + c2)2 = ab2 + b4 + c4 + 2b2c2 (a ⊙ b) ⊙ c = (a2 + b2) ⊙ c = (a2 + b2)2 + c2 = a4 + b4 + 2a2b2 + c2 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 353 Marks
Classify the following functions as injection, surjection or bijection:
f : R → R, defined by f(x) = sinx
Answerf : R → R, given by f(x) = sinx Injective: Let $\text{x, y}\in\text{R}$ such that f(x) = f(y) ⇒ sinx = siny $\Rightarrow\ \text{x}=\text{n}\pi+(-1)^{\text{n}}\text{y}$
$\Rightarrow\ \text{x}\neq\text{y}$
$\therefore$
f is not one-one. Surjective: Let
$\text{y}\in\text{R}$ be arbitrary such that f(x) = y
⇒ sinx = y ⇒ x = sin-1y Now, for $\text{y}>1\times\notin\text{R}$ (domain) $\therefore$ f is not onto.
View full question & answer→Question 363 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.
AnswerR3 = {(1, 3), (3, 3)}
$\therefore\ (1,1)\notin\text{R}_3$
⇒ R3 is not reflexive.
Now, $(1,3)\in\text{R}_3$ but $(3,1)\in\text{R}_3$
$\therefore$ R3 is not symmetric.
Again, it is clear that R3 is transitive.
View full question & answer→Question 373 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 383 Marks
Write the multiplication table for the set of integers modulo 5.
AnswerZ
5 = {0, 1, 2, 3, 4}
a×
5 b is the remainder when the product of ab is divided by 5.
The composition table for ×
5 on Z
5 = {0, 1, 2, 3, 4}
| ×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 393 Marks
On Q, the set of all rational numbers, * is defined by $\text{a}\ ^*\ \text{b}=\frac{\text{a}-\text{b}}{2},$ shown that * is no associative.
AnswerThe binary operator * defined as, $\text{a}\ ^*\ \text{b}=\frac{\text{a}-\text{b}}{2},$ for all $\text{a, b}\in\text{Q}.$ Now, Associativity: Let $\text{a, b, c}\in\text{Q}.$ Then, $(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}=\frac{\text{a}-\text{b}}{2}\ ^*\ \text{c}=\frac{\frac{\text{a}-\text{b}}{2}-\text{c}}{2}$ $=\frac{\text{a}-\text{b}-2\text{c}}{4}\ ....(\text{i})$ and, $\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})=\text{a}\ ^*\ \frac{\text{b}-\text{c}}{2}=\frac{\text{a}-\frac{\text{b}-\text{c}}{2}}{2}$ $=\frac{2\text{a}-\text{b}+\text{c}}{4}\ .....(\text{ii})$ From (i) and (ii), $(\text{a}\ ^*\ \text{b})\ ^*\ \text{c}\neq\text{a}\ ^*\ (\text{b}\ ^*\ \text{c})$
Hence, '*' is not associative on Q. View full question & answer→Question 403 Marks
Consider f : N → N, g : N → N and h : N → R defined as f(x) = 2x, g(y) = 3y + 4 and $\text{h(z)}=\sin\text{z}$ for all $\text{x, y, z}\in\text{N.}$ Show that ho(gof) = (hog)of.
AnswerGiven, f : N → N, g : N → N and h : N → R
⇒ gof : N → N and hog : N → R
⇒ ho(gof) : N → R and (hog) of : N → R
So, both have the same domains.
(gof)(x) = g(f(x)) = g(2x) = 3(2x) + 4 = 6x + 4 ....(1)
(hog)(x) = h(g(x)) = h(3x + 4) = sin(3x + 4) ......(2)
Now,
(ho(gof))(x) = h((gof)(x)) = h(6x + 4) = sin(6x + 4) [from (1)]
((hog)of)(x) = (hog)(f(x)) = (hog)(2x) = sin(6x + 4) [from (2)]
So, (ho(gof))(x) = ((hog)of)(x), $\forall\text{ x}\in\text{N}$
Hence, ho(gof) = (hog)of
View full question & answer→Question 413 Marks
Find f-1 if it exists: f : A → B, where, A = {1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81} and f(x) = x2.
AnswerA = {1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81}
f : A → B be a function defined by f(x) = x2
Since different elements of A have different images in B.
$\therefore$ f is one-one.
Again, $0\in\text{B}$ does not have a preim-age in A.
$\therefore$ f is not onto.
Hence, f-1 does not exist.
View full question & answer→Question 423 Marks
Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective:
{(x, y): x is a person, y is the mother of x}
Answerf = {(x, y): x is a person, y is the mother of x}
As, for each element x in domain set, there is a unique related element y in co-domain set.
Therefore, f is the function.
Injection test: As, y can be mother of two or more persons.
Therefore,
f is not injective.
Surjection test: For every mother y defined by (x, y), there exists a person x for whom y is mother. Therefore, f is surjective.
View full question & answer→Question 433 Marks
Let A and B be sets. Show that f: A × B → B × A such that f (a, b) = (b, a) is bijective function.
Answerf: A × B → B × A is defined as f(a, b) = (b, a). Let
$(\text{a}_1,\text{b}_1),(\text{a}_2,\text{b}_2)\in \text{A}\times\text{B}$ such that f(a1, b1) = f(a2, b2) ⇒ (b1, a1) = (b2, a2) ⇒ b1 = b2 and a1 = a2 ⇒ (a1, b1) = (a2, b2) $\therefore$ f is one-one. Now, let $(\text{b},\text{a})\in\text{B}\times\text{A}$ be any element. Then, there exists $(\text{a},\text{b})\in\text{A}\times\text{B}$ such that f(a, b) = (b, a). [By definition of f] $\therefore$ f is onto. Hence, f is bijective. View full question & answer→Question 443 Marks
Let A = {1, 2, 3}. Write all one-one from A to itself.
AnswerWe have,
ho(gof)(x) = h(gof(x)) = h(g(f(x)))
= h(g(2x)) = h(3(2x) + 4)
= h(6x + 4) = sin(6x + 4) $\forall\ \text{x}\in\text{N}$
((hog)of)(x) = (hog)(f(x)) = (hog)(2x)
= h(g(2x)) = h(3(2x) + 4)
= h(6x + 4) = sin(6x + 4) $\forall\ \text{x}\in\text{N}$
This shows, ho(gof) = (hog)of
View full question & answer→Question 453 Marks
Let f, g, h be real functions given by f(x) = sinx, g(x) = 2x and h(x) = cosx. Prove that fog = go(fh).
Answerf, g and h are real fuctions given by f(x) = sinx, g(x) = 2x and h(x) = cosx
To prove: fog = go(fh)
L.H.S. fog(x) = f(g(x))
= f(2x) = sin2x
⇒ fog(x) = 2sinx.cosx .....(A)
R.H.S. go(fh)(x) = go(f(x).h(x))
= g(sinx.cosx)
⇒ go(fh)(x) = 2sinx.cosx ......(B)
from (A) & (B)
fog(x) = go(fh)(x)
View full question & answer→Question 463 Marks
Construct the composition table for ×6 on set S = {0, 1, 2, 3, 4, 5}.
AnswerHere,
1×
61 = Remainder obtained by dividing 1 × 1 by 6 = 1
3×
64 = Remainder obtained by dividing 3 × 4 by 6 = 0
4×
65 = Remainder obtained by dividing 4 × 5 by 6 = 2
So, the composition table is as follows:
| ×6 | 0 | 1 | 2 | 3 | 4 | 5 |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 2 | 3 | 4 | 5 |
| 2 | 0 | 2 | 4 | 0 | 2 | 4 |
| 3 | 0 | 3 | 0 | 3 | 0 | 3 |
| 4 | 0 | 4 | 2 | 0 | 4 | 2 |
| 5 | 0 | 5 | 4 | 3 | 2 | 1 |
View full question & answer→Question 473 Marks
The following relation are defined on the set of real numbers.
aRb if $|\text{a}|\leq\text{b}$
Find whether these relation are reflexive, symmetric or transitive.
AnswerWe have aRb if $|\text{a}|\leq\text{b}$ Reflexive: Let $\text{a}\in\text{R}$ $\Rightarrow\ |\text{a}|\nleq\text{a}$ $[\therefore|-2|=2>-2|]$
⇒ R is not reflexive. Symmetric: Let aRb $\Rightarrow\ |\text{a}|\leq\text{b}$ $\Rightarrow\ |\text{b}|\leq\text{a}$ $\begin{bmatrix}\therefore\ \ \ \text{Let a}=4, \text{b}=6 \\\ \ \ \ \ \ \ \ \ |4|\leq 8 \text{ but } |8|>4\end{bmatrix}$ ⇒ R is not symmetric. Transitive: Let aRb and bRc $\Rightarrow\ |\text{a}|\leq\text{b}$
and $|\text{b}|\leq\text{c}$ $\Rightarrow\ |\text{a}|\leq|\text{b}|\leq\text{c}$
$\Rightarrow\ |\text{a}|\leq\text{c}$
$\Rightarrow\ \text{aRc}$
⇒ R is transitive. View full question & answer→Question 483 Marks
Find the inverse of 5 under multiplication modulo 11 on Z11.
AnswerZ11 = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Multiplication modulo 11 is defined as follows:
For $\text{a},\text{b}\in\text{Z}_{11}$,
a×11b is the remainder when a × b is divided by 11.
Here,
1×111 = Remainder obtained by dividing 1 × 1 by 11 = 1
3×114 = Remainder obtained by dividing 3 × 4 by 11 = 1
4×115 = Remainder obtained by dividing 4 × 5 by 11 = 9
| ×11 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 1 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 2 | 2 | 4 | 6 | 8 | 10 | 1 | 3 | 5 | 7 | 9 |
| 3 | 3 | 6 | 9 | 1 | 4 | 7 | 10 | 2 | 5 | 8 |
| 4 | 4 | 8 | 1 | 5 | 9 | 2 | 6 | 10 | 3 | 7 |
| 5 | 5 | 10 | 4 | 9 | 3 | 8 | 2 | 7 | 1 | 6 |
| 6 | 6 | 1 | 7 | 2 | 8 | 3 | 9 | 4 | 10 | 5 |
| 7 | 7 | 3 | 10 | 6 | 2 | 9 | 5 | 1 | 8 | 4 |
| 8 | 8 | 5 | 2 | 10 | 7 | 4 | 1 | 9 | 6 | 3 |
| 9 | 9 | 7 | 5 | 3 | 1 | 10 | 8 | 6 | 4 | 2 |
| 10 | 10 | 9 | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 |
We observe that the first row of the composition table is same as the top-most row.
Therefore,
The identity element is 1.
Also,
5×119 = 1
Hence, 5 - 1 = 9
View full question & answer→Question 493 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 503 Marks
Let f: R → R be the Signum Function defined as $f(\text{x})=\begin{cases}1,&\text{x}>0\\0,&\text{x}=0\\-1,&\text{x}<0\end{cases}$ and g: R → R be the Greatest Integer Function given by g(x) = [x], where [x] is greatest integer less than or equal to x. Then, does fog and gof coincide in (0, 1]?
AnswerIt is given that,
f: R → R is defined as $f(\text{x})=\begin{cases}1,&\text{x}>0\\0,&\text{x}=0\\-1,&\text{x}<0\end{cases}$
Also, g: R → R is defined as g(x) = [x], where [x] is the greatest integer less than or equal to x.
Now, let $\text{x}\in(0,\ 1].$
Then, we have:
[x] = 1 if x = 1 and [x] = 0 if 0 < x < 1.
$\therefore fo\text{g(x)}=f(\text{g(x)})=f([\text{x}])=\begin{cases}f(1),&\text{if x}=1\\f(0),&\text{if x}\in(0,\ 1)\end{cases}=\begin{cases}1,&\text{if x}=1\\0,&\text{if x}\in(0,\ 1)\end{cases}$
gof(x) = g(f(x))
= g(1) [x > 0]
=[1] = 1
Thus, when $\text{x}\in(0,\ 1),$ we have fog(x) = 0 and gof(x) = 1.
Hence, fog and gof do not coincide in (0, 1].
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