Relations and Functions
Short Answer Type Questions
1. Let $A={a, b, c}$ and the relation $R$ be defined on $A$ as follows:
$ R={(a, a),(b, c),(a, b)} $
Then, write minimum number of ordered pairs to be added in $R$ to make $R$ reflexive and transitive.
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Solution
Here,
$ R={(a, a),(b, c),(a, b)} $
for reflexivity; $(b, b),(c, c)$ and for transitivity; $(a, c)$
Hence, the reuired ordered pairs are $(b, b),(c, c)$ and $(a, c)$
2. Let $D$ be the domain of the real valued function $f$ defined by $f(x)=\sqrt{25x^{2}}$. Then write D.
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Solution
Here, $f(x)=\sqrt{25x^{2}}$
For real value of $f(x), 25x^{2} \ge 0$
$\Rightarrowx^{2} \ge25 \Rightarrow x^{2} \le 25 \Rightarrow5 \le x \le 5$
Hence, $D \in5 \le x \le 5$ or $[5,5]$
3. Let $f, g: R \to R$ be defined by $f(x)=2 x+1$ and $g(x)=x^{2}2 \forall$ $x \in R$, respectively. Then find gof.
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Solution
Here, $f(x)=2 x+1$ and $g(x)=x^{2}2$
$\therefore \quad gof=g[f(x)]$
$ =[2 x+1]^{2}2=4 x^{2}+4 x+12=4 x^{2}+4 x1 $
Hence, $g o f=4 x^{2}+4 x1$
4. Let $f: R \to R$ be the function defined by $f(x)=2 x3 \forall x \in R$. Write $f^{1}$.
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Solution
Here,
$ f(x)=2 x3 $
Let
$ f(x)=y=2 x3 $
$ \begin{aligned} & \Rightarrow \quad y+3=2 x \Rightarrow x=\frac{y+3}{2} \\ & \therefore \quad f^{1}(y)=\frac{y+3}{2} \text{ or } f^{1}(x)=\frac{x+3}{2} \end{aligned} $
5. If $A={a, b, c, d}$ and the function $f={(a, b),(b, d),(c, a),(d, c)}$, write $f^{1}$.
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Solution
$ \begin{aligned} \text{Let}\quad y & =f(x) \quad \therefore x=f^{1}(y) \\ \therefore \quad \text{If} \quad f & ={(a, b),(b, d),(c, a),(d, c)} \\ \text{then}\quad f^{1} & ={(b, a),(d, b),(a, c),(c, d)} \end{aligned} $
6. If $f: R \to R$ is defined by $f(x)=x^{2}3 x+2$, write $f[f(x)]$.
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Solution
Here, $f(x)=x^{2}3 x+2$
$\therefore \quad f[f(x)]=[f(x)]^{2}3 f(x)+2$
$=(x^{2}3 x+2)^{2}3(x^{2}3 x+2)+2$
$=x^{4}+9 x^{2}+46 x^{3}+4 x^{2}12 x3 x^{2}+9 x6+2$
$=x^{4}6 x^{3}+10 x^{2}3 x$
Hence, $f[f(x)]=x^{4}6 x^{3}+10 x^{2}3 x$
7. Is $g={(1,1),(2,3),(3,5),(4,7)}$ a function? If $g$ is described by $g(x)=\alpha x+\beta$, then what value should be assigned to $\alpha$ and $\beta$ ?
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Solution
Yes, $g={(1,1),(2,3),(3,5),(4,7)}$ is a function.
Here, $g(x)=\alpha x+\beta$
For $(1,1)$,
$g(1)=\alpha .1+\beta$
$1=\alpha+\beta$
For $(2,3), \quad g(2)=\alpha .2+\beta \quad \text{…(1)}$
$\qquad 3=2 \alpha+\beta \quad \text{…(2)}$
Solving es. (1) and (2) we get, $\alpha=2, \beta=1$
8. Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective.
(i) $\{(x, y): x$ is a person, $y$ is the mother of $x\}$
(ii) $\{(a, b): a$ is a person, $b$ is an ancestor of $a\}$
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Solution
(i) It represents a function. The image of distinct elements of $x$ under $f$ are not distinct. So, it is not injective but it is surjective.
(ii) It does not represent a function as every domain under mapping does not have a uniue image.
9. If the mapping $f$ and $g$ are given by
$f=\{(1,2),(3,5),(4,1)\} \quad$ and $g=\{(2,3),(5,1),(1,3)\}$ write fog.
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Solution
$ \begin{aligned} f \circ g & =f[g(x)] \\ & =f[g(2)]=f(3)=5 \\ & =f[g(5)]=f(1)=2 \\ & =f[g(1)]=f(3)=5 \end{aligned} $
Hence, $\quad f o g={(2,5),(5,2),(1,5)}$
10. Let $C$ be the set of complex numbers. Prove that the mapping $f: C \to R$ given by $f(z)=z, \forall z \in C$, is neither oneone nor onto.
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Solution
Here,
$ \begin{aligned} f(1) & =1=1 \\ f(1) & =1=1 \\ f(1) & =f(1) \end{aligned} $
$ \text{ Here, } \quad \begin{aligned} f(z) & =z \quad \forall z \in \text{ C } \\ f(1) & =1=1 \\ f(1) & =1=1 \\ f(1) & =f(1) \\ \text{ But } \quad 1 & \ne1 \end{aligned} $
Therefore, it is not oneone.
Now, let $f(z)=y=z$. Here, there is no preimage of negative numbers. Hence, it is not onto.
11. Let the function $f: R \to R$ be defined by $f(x)=\cos x, \forall x \in R$. Show that $f$ is neither oneone nor onto.
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Solution
Here,
$ f(x)=\cos x \forall x \in R $
Let $\quad[\frac{\pi}{2}, \frac{\pi}{2}] \in f(x)$
$ \begin{aligned} & f(\frac{\pi}{2})=\cos (\frac{\pi}{2})=\cos \frac{\pi}{2}=0 \\ & \cos (\frac{\pi}{2})=\cos \frac{\pi}{2}=0 \\ & f(\frac{\pi}{2})=f(\frac{\pi}{2})=0 \end{aligned} $
But
$ \frac{\pi}{2} \ne \frac{\pi}{2} $
Therefore, the given function is not oneone. Also it is not onto function as no preimage of any real number belongs to the range of $\cos x$ i.e., $[1,1]$.
12. Let $X={1,2,3}$ and $Y={4,5}$. Find whether the following subsets of $X \times Y$ are functions from $X$ to $Y$ or not.
(i) $f={(1,4),(1,5),(2,4),(3,5)}$
(ii) $g={(1,4),(2,4),(3,4)}$
(iii) $h={(1,4),(2,5),(3,5)}$
(iv) $k={(1,4),(2,5)}$
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Solution
Here, given that $X={1,2,3}, Y={4,5}$
$\therefore X \times Y={(1,4),(1,5),(2,4),(2,5),(3,4),(3,5)}$
(i) $f={(1,4),(1,5),(2,4),(3,5)}$
$f$ is not a function because there is no uniue image of each element of domain under $f$.
(ii) $g={(1,4),(2,4),(3,4)}$
Yes, $g$ is a function because each element of its domain has a uniue image.
(iii) $h={(1,4),(2,5),(3,5)}$
Yes, it is a function because each element of its domain has a uniue image.
(iv) $k={(1,4),(2,5)}$
Clearly $k$ is also a function.
13. If function $f: A \to B$ and $g: B \to A$ satisfy $g \circ f=I _{A^{\prime}}$, then show that $f$ is oneone and $g$ is onto.
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Solution
Let $x_1, x_2 \in$ gof
$ \begin{aligned} & gof{f(x_1)}=gof{f(x_2)} \\ & \Rightarrow \quad g(x_1)=g(x_2) \quad[\because g \circ f=I_A] \\ & \therefore \quad x_1=x_2 \end{aligned} $
Hence, $f$ is oneone. But $g$ is not onto as there is no preimage of $A$ in $B$ under $g$.
14. Let $f: R \to R$ be the function defined by $f(x)=\frac{1}{2\cos x}$, $\forall x \in R$. Then, find the range of $f$.
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Solution
Given function is $f(x)=\frac{1}{2\cos x}, \forall x \in R$.
Range of $\cos x$ is $[1,1]$
Let $\quad f(x)=y=\frac{1}{2\cos x}$
$\Rightarrow \quad 2 yy \cos x=1 \quad \Rightarrow \quad y \cos x=2 y1$
$\Rightarrow \quad \cos x=\frac{2 y1}{y}=2\frac{1}{y}$
Now $1 \le \cos x \le 1$
$\Rightarrow \quad1 \le 2\frac{1}{y} \le 1 \Rightarrow12 \le\frac{1}{y} \le 12$
$\Rightarrow \quad3 \le\frac{1}{y} \le1 \Rightarrow 3 \ge \frac{1}{y} \ge 1 \Rightarrow \frac{1}{3} \le y \le 1$
Hence, the range of $f=[\frac{1}{3}, \mathbf{1}]$.
15. Let $n$ be a fixed positive integer. Define a relation $R$ in $Z$ as follows $\forall a, b \in Z, a R b$ if and only if $ab$ is divisible by $n$. Show that $R$ is an euivalence relation.
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Solution
Here, $\forall a, b \in Z$ and $a R b$ if and only if $ab$ is divisible by $n$. The given relation is an euivalence relation if it is reflexive, symmetric and transitive.
(i) Reflexive:
$a R a \Rightarrow(aa)=0$ divisible by $n$
So, $R$ is reflexive.
(ii) Symmetric:
$a R b=b R a \quad \forall a, b \in Z$
$ab$ is divisible by $n$ (Given)
$\Rightarrow(ba)$ is divisible by $n$ $\Rightarrow ba$ is divisible by $n$
$\Rightarrow b R a$
Hence, $R$ is symmetric.
(iii) Transitive:
$a R b$ and $b R c \quad \Leftrightarrow \quad a R c \quad \forall a, b, c \in Z$
$ab$ is divisible by $n$
$bc$ is also divisible by $n$
$\Rightarrow(ab)+(bc)$ is divisible by $n$
$\Rightarrow(ac)$ is divisible by $n$
Hence, $R$ is transitive.
So, $R$ is an euivalence relation.
Long Answer Type Questions
16. If $A={1,2,3,4}$, define relations on $A$ which have properties of being.
(a) reflexive, transitive but not symmetric.
(b) symmetric but neither reflexive nor transitive
(c) reflexive, symmetric and transitive.
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Solution
Given that $A={1,2,3,4}$
$\therefore \quad ARA=\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,3),(1,4),(2,3)$,
$ (2,4),(3,4),(2,1),(3,1),(4,1),(3,2),(4,2),(4,3)\} $
(a) Let $R_1=\{(1,1),(2,2),(1,2),(2,3),(1,3)\}$
So, $R_1$ is reflexive and transitive but not symmetric.
(b) Let $R_2=\{(2,3),(3,2)\}$
So, $R_2$ is only symmetric.
(c) Let $R_3=\{(1,1),(1,2),(2,1),(2,4),(1,4)\}$
So, $R_3$ is reflexive, symmetric and transitive.
17. Let $R$ be relation defined on the set of natural number $N$ as follows:
$R={(x, y): x \in N, y \in N, 2 x+y=41}$. Find the domain and range of the relation $R$. Also verify whether $R$ is reflexive, symmetric and transitive.
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Solution
Given that $x \in N, y \in N$ and $2 x+y=41$
$\therefore \quad $ Domain of $R={1,2,3,4,5, \ldots, 20}$
and $\quad $ Range $={39,37,35,33,31, \ldots, 1}$
Here, $\quad (3,3) \notin R$
as $\quad 2 \times 3+3 \ne 41$
So, $R$ is not reflexive.
$R$ is not symmetric as $(2,37) \in R$ but $(37,2) \notin R$
$R$ is not transitive as $(11,19) \in R$ and $(19,3) \in R$
but $(11,3) \notin R$.
Hence, $R$ is neither reflexive, nor symmetric and nor transitive.
18. Given $A={2,3,4}, B={2,5,6,7}$, construct an example of each of the following:
(i) an injective mapping from $A$ to $B$.
(ii) a mapping from $A$ to $B$ which is not injective
(iii) a mapping from $B$ to $A$.
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Solution
Here, $A={2,3,4}$ and $B={2,5,6,7}$
(i) Let $f: A \to B$ be the mapping from $A$ to $B$ $f={(x, y): y=x+3}$
$\therefore f={(2,5),(3,6),(4,7)}$ which is an injective mapping.
(ii) Let $g: A \to B$ be the mapping from $A \to B$ such that $g={(2,5),(3,5),(4,2)}$ which is not an injective mapping.
(iii) Let $h: B \to A$ be the mapping from $B$ to $A$
$h={(y, x): x=y2}$
$h={(5,3),(6,4),(7,3)}$ which is the mapping from B to A.
19. Give an example of a map
(i) which is oneone but not onto.
(ii) which is not oneone but onto.
(iii) which is neither oneone nor onto.
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Solution
(i) Let $f: N \to N$ given by $f(x)=x^{2}$
Let $x_1, x_2 \in N$ then $f(x_1)=x_1^{2}$ and $f(x_2)=x_2^{2}$
Now, $f(x_1)=f(x_2) \Rightarrow x_1^{2}=x_2^{2} \Rightarrow x_1^{2}x_2^{2}=0$
$ \Rightarrow(x_1+x_2)(x_1x_2)=0 $
Since $x_1, x_2 \in N$, so $x_1+x_2=0$ is not possible.
$ \begin{matrix} \therefore & x_1x_2=0 & \Rightarrow x_1=x_2 \\ \therefore & f(x_1)=f(x_2) & \Rightarrow x_1=x_2 \end{matrix} $
So, $f(x)$ is one to one function.
Now, Let $f(x)=5 \in N$
then $\quad x^{2}=5 \Rightarrow x= \pm \sqrt{5} \notin N$
So, $f$ is not onto.
Hence, $f(x)=x^{2}$ is oneone but not onto.
(ii) Let $f: N \times N$, defined by $f(n)= \begin{cases}\frac{n+1}{2} & \text{ if } n \text{ is odd } \\ \frac{n}{2} & \text{ if } n \text{ is even }\end{cases}$
Since $f(1)=f(2)$ but $1 \ne 2$,
So, $f$ is not oneone.
Now, let $y \in N$ be any element.
Then $f(n)=y$
$ \Rightarrow \begin{cases} \frac{n+1}{2} & \text{ if } n \text{ is odd } \\ \frac{n}{2} & \text{ if } n \text{ is even } \end{cases} =y $
$ \Rightarrow n=2 y1 \qquad \text{ if } y \text{ is even } \\ n=2 y \qquad \text{ if } y \text{ is odd or even } \\ \Rightarrow n=\begin{cases} 2 y1 \qquad \text{ if } y \text{ is even } \\ 2 y \qquad \text{ if } y \text{ is odd or even } \end{cases} \in N \forall y \in N. $
$\therefore$ Every $y \in N$ has preimage
$ n=\begin{cases} 2 y1 \text{ if } y \text{ is even } \\ 2 y \text{ if } y \text{ is odd or even } \end{cases} \in N. $
$\therefore f$ is onto.
Hence, $f$ is not oneone but onto.
(iii) Let $f: R \to R$ be defined as $f(x)=x^{2}$
Let $x_1=2$ and $x_2=2$
$ \begin{aligned} & f(x_1)=x_1^{2}=(2)^{2}=4 \\ & f(x_2)=x_2^{2}=(2)^{2}=4 \\ & f(2)=f(2) \quad \text{ but } 2 \ne2 \end{aligned} $
So, it is not oneone function.
Let $f(x)=2 \Rightarrow x^{2}=2 \quad \therefore \quad x= \pm \sqrt{2} \notin R$
Which is not possible, so $f$ is not onto.
Hence, $f$ is neither oneone nor onto.
20. Let $A=R{3}, B=R{1}$. Let $f: A \to B$ be defined by $f(x)=\frac{x2}{x3}, \forall x \in A$. Then, show that $f$ is bijective.
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Solution
Here, $A \in R{3}, B=R{1}$
Given that $f: A \to B$ defined by $f(x)=\frac{x2}{x3} \forall x \in A$.
Let $x_1, x_2 \in f(x)$
$ \begin{aligned} & \therefore \quad f(x_1)=f(x_2) \\ & \Rightarrow \quad \frac{x_12}{x_13}=\frac{x_22}{x_23} \\ & \Rightarrow \quad (x_12)(x_23)=(x_22)(x_13) \\ & \Rightarrow \quad {\not x_1 \not x_2}3 x_12 x_2+\not 6=\not x_1 \not x_23 x_22 x_1+\not 6 \\ & \Rightarrow \quad x_1=x_2 \Rightarrow x_1=x_2 \end{aligned} $
So, it is injective function.
Now, Let $\quad y=\frac{x2}{x3}$
$\Rightarrow x y3 y=x2 \Rightarrow x yx=3 y2$
$\Rightarrow x(y1)=3 y2 \Rightarrow x=\frac{3 y2}{y1}$
$ f(x)=\frac{x2}{x3}=\frac{\frac{3 y2}{y1}2}{\frac{3 y2}{y1}3} \Rightarrow \frac{3 y22 y+2}{3 y23 y+3} \Rightarrow y \\ \Rightarrow \quad f(x)=y \in B$
So, $f(x)$ is surjective function.
Hence, $f(x)$ is a bijective function.
21. Let $A=[1,1]$, then discuss whether the following functions defined on $A$ are oneone, onto or bijective.
(i) $f(x)=\frac{x}{2}$
(ii) $g(x)=x$
(iii) $h(x)=xx$ (iv) $k(x)=x^{2}$
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Solution
(i) Given that $1 \le x \le 1$
Let $x_1, x_2 \in f(x)$
So, $f(x)$ is oneone function.
$ \begin{aligned} & f(x_1)=\frac{1}{x_1} \text{ and } f(x_2)=\frac{1}{x_2} \\ & f(x_1)=f(x_2) \Rightarrow \frac{1}{x_1}=\frac{1}{x_2} \Rightarrow x_1=x_2 \end{aligned} $
Let
$ f(x)=y=\frac{x}{2} \quad \Rightarrow \quad x=2 y $
For $y=1, x=2 \notin[1,1]$
So, $f(x)$ is not onto. Hence, $f(x)$ is not bijective function.
(ii) Here,
$ \begin{aligned} & g(x)=x \\ & g(x_1)=g(x_2) \quad \Rightarrowx_1=x_2 \Rightarrow x_1= \pm x_2 \end{aligned} $
So, $g(x)$ is not oneone function.
Let $g(x)=y=x \Rightarrow x= \pm y \notin A \forall y \in A$
So, $g(x)$ is not onto function.
Hence, $g(x)$ is not bijective function.
(iii) Here,
$ \begin{aligned} h(x) & =xx \\ h(x_1) & =h f(x_2) \end{aligned} $
$ \Rightarrow \quad x_1x_1=x_2x_2 \Rightarrow x_1=x_2 $
So, $h(x)$ is oneone function.
Now, let $h(x)=y=xx=x^{2}$ or $x^{2}$
$\Rightarrow \quad x= \pm \sqrt{y} \notin A \forall y \in A$
$\therefore h(x)$ is not onto function.
Hence, $h(x)$ is not bijective function.
(iv) Here,
$ \begin{aligned} k(x) & =x^{2} \\ k(x_1) & =k(x_2) \end{aligned} $
$ \Rightarrow \quad x_1^{2}=x_2^{2} \Rightarrow x_1= \pm x_2 $
So, $k(x)$ is not oneone function.
Now, let $k(x)=y=x^{2} \Rightarrow x= \pm \sqrt{y}$
If $y=1 \Rightarrow x= \pm \sqrt{1} \notin A \forall y \in A$
$\therefore k(x)$ is not onto function.
Hence, $k(x)$ is not a bijective function.
22. Each of the following defines a relation of $N$
(i) $x$ is greater than $y, x, y \in N$
(ii) $x+y=10, x, y \in N$
(iii) $x y$ is suare of an integer $x, y \in N$
(iv) $x+4 y=10, x, y \in N$.
Determine which of the above relations are reflexive, symmetric and transitive.
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Solution
(i) $x$ is greater than $y, \quad x, y \in N$
For reflexivity $x>x \forall x \in N$ which is not true
So, it is not reflexive relation.
Now, $x>y$ but $y \ngtr x \forall x, y \in N$
$\Rightarrow x R y$ but $y$ Z $x$
So, it is not symmetric relation.
For transitivity, $\quad x R y, y R z \Rightarrow x R z \forall x, y, z \in N$
$ \Rightarrow x>y, y>z \Rightarrow x>z $
So, it is transitive relation.
(ii) Here, $\quad R={(x, y): x+y=10 \forall x, y \in N}$
$R=\{(1,9),(2,8),(3,7),(4,6),(5,5),(6,4),(7,3),(8,2),(9,1)\}$
For reflexive: $5+5=10,5 R 5 \Rightarrow(x, x) \in R$
So, $R$ is reflexive.
For symmetric: $(1,9) \in R$ and $(9,1) \in R$
So, $R$ is symmetric.
For transitive: $(3,7) \in R,(7,3) \in R$ but $(3,3) \notin R$
So, $R$ is not transitive.
(iii) Here, $R=\{(x, y): x y$ is a suare of an integer, $x, y \in N\}$
For reflexive: $x R x=x . x=x^{2}$ is an integer
So, $R$ is reflexive.
$[\because$ Suare of an integer is also an integer $]$
For symmetric: $x R y=y R x \forall x, y \in \mathbf{N}$
$\therefore \quad x y=y x \quad $ (integer)
So, it is symmetric.
For transitive: $x R y$ and $y R z \Rightarrow x R z$
Let
$ \begin{aligned} & x y=k^{2} \text{ and } y z=m^{2} \\ & x=\frac{k^{2}}{y} \quad \text{ and } \quad z=\frac{m^{2}}{y} \end{aligned} $
$\therefore \quad x z=\frac{k^{2} m^{2}}{y^{2}}$ which is again a suare of an integer. (iv) Here,
$ \begin{aligned} & R={(x, y): x+4 y=10, x, y \in N} \\ & R={(2,2),(6,1)} \end{aligned} $
For reflexivity: $(2,2) \in R$
So, $R$ is reflexive.
For symmetric: $\quad (x, y) \in R$ but $(y, x) \notin R$
$ (6,1) \in R \quad \text{ but } \quad (1,6) \notin R $
So, $R$ is not symmetric.
For transitive: $(x, y) \in R$ but $(y, z) \notin R$ and $(x, z) \in R$
So, $R$ is not transitive.
23. Let $A={1,2,3, \ldots, 9}$ and $R$ be the relation in $A \times A$ defined by $(a, b) R(c, d)$ if $a+d=b+c$ for $(a, b),(c, d)$ in $A \times A$. Prove that $R$ is an euivalence relation and also obtain euivalent class $[(2,5)]$.
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Solution
Here,
$ A={1,2,3, \ldots, 9} $
and $R \to A \times A$ defined by $(a, b) R(c, d) \Rightarrow a+d=b+c$
$\forall(a, b),(c, d) \in A \times A$
For reflexive: $(a, b) R(a, b)=a+b=b+a \quad \forall a, b \in A$ which is true. So, $R$ is reflexive.
For symmetric: $(a, b) R(c, d)=(c, d) R(a, b)$
L.H.S. $\quad a+d=b+c$
R.H.S. $c+b=d+a$
L.H.S. $=$ R.H.S. So, $R$ is symmetric.
For transitive: $(a, b) R(c, d)$ and $(c, d) R(e, f) \Leftrightarrow(a, b) R(e, f)$
$\Rightarrow \quad a+d=b+c$ and $c+f=d+e$
$\Rightarrow \quad a+d=b+c$ and $d+e=c+f$
$\Rightarrow(a+d)(d+e)=(b+c)(c+f)$
$\Rightarrow \quad ae=bf$
$\Rightarrow \quad a+f=b+e$
$\Rightarrow \quad (a, b) R(e, f)$
So, $R$ is transitive.
Hence, $R$ is an euivalence relation.
Euivalent class of $\{(2,5)\}$ is $\{(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)\}$
24. Using the definition, prove that the function $f: A \to B$ is invertible if and only if $f$ is both oneone and onto.
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Solution
A function $f: X \to Y$ is said to be invertible if there exists a function $g: Y \to X$ such that $gof=I_X$ and $fog=I_Y$ and then the inverse of $f$ is denoted by $f^{1}$.
A function $f: X \to Y$ is said to be invertible iff $f$ is a bijective function.
25. Function $f, g: R \to R$ are defined, respectively, by $f(x)=x^{2}+3 x+1$, $g(x)=2 x3$, find
(i) $fog$
(ii) $gof$
(iii) fof
(iv) $gog$
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Solution
(i) $\quad fog \Rightarrow f[g(x)]=[g(x)]^{2}+3[g(x)]+1$
(ii) $\quad gof \Rightarrow g[f(x)]=2[x^{2}+3 x+1]3$
$ \begin{aligned} & =(2 x3)^{2}+3(2 x3)+1 \\ & =4 x^{2}+912 x+6 x9+1=4 x^{2}6 x+1 \\ & =2[x^{2}+3 x+1]3 \\ & =2 x^{2}+6 x+23=2 x^{2}+6 x1 \end{aligned} $
(iii) $\quad f \circ f \Rightarrow f[f(x)]=[f(x)]^{2}+3[f(x)]+1$
$ \begin{aligned} & =(x^{2}+3 x+1)^{2}+3(x^{2}+3 x+1)+1 \\ & =x^{4}+9 x^{2}+1+6 x^{3}+6 x+2 x^{2}+3 x^{2}+9 x+3+1 \\ & =x^{4}+6 x^{3}+14 x^{2}+15 x+5 \end{aligned} $
(iv) $gog \Rightarrow g[g(x)]=2[g(x)]3=2(2 x3)3=4 x63=4 x9$
26. Let $*$ be the binary operation defined on . Find which of the following binary operations are commutative.
(i) $a * b=ab \forall a, b \in $
(ii) $a * b=a^{2}+b^{2} \forall a, b \in $
(iii) $a * b=a+a b \forall a, b \in $
(iv) $a * b=(ab)^{2} \forall a, b \in $
Show Answer
Solution
(i)
$ a * b=ab \in \quad \forall a, b \in . $
So, $*$ is binary operation.
$a * b=ab$ and $b * a=ba \quad \forall a, b \in $
$ ab \ne ba $
So, $*$ is not commutative.
(ii) $a * b=a^{2}+b^{2} \in $, so $*$ is a binary operation.
$ a * b=b * a $
$ \Rightarrow \quad a^{2}+b^{2}=b^{2}+a^{2} \quad \forall a, b \in $
Which is true. So, $*$ is commutative.
(iii) $a * b=a+a b \in $, so $*$ is a binary operation.
$ a * b=a+a b \text{ and } b * a=b+b a $
$a+a b \ne b+b a \Rightarrow a * b \ne b * a \quad \forall a, b \in $.
So, $*$ is not commutative.
(iv) $a * b=(ab)^{2} \in $, so $*$ is binary operation.
$a * b=(ab)^{2}$ and $b * a=(ba)^{2}$
$a * b=b * a \Rightarrow(ab)^{2}=(ba)^{2} \quad \forall a, b \in $.
So, $*$ is commutative.
27. If * be binary operation defined on $R$ by $a * b=1+a b \forall a, b \in R$.
Then, the operation $*$ is
(i) commutative but not associative
(ii) associative but not commutative
(iii) neither commutative nor associative
(iv) both commutative and associative
Show Answer
Solution
(i): Given that
and $\quad $ $a * b=1+a b \quad \forall a, b \in R$   :—   $b * a=1+b a \quad \forall a, b \in R$   $a * b=b * a=1+a b$ 
So, $*$ is commutative.
Now $a *(b * c)=(a * b) * c \quad \forall a, b, c \in R$
L.H.S. $a *(b * c)=a *(1+b c)=1+a(1+b c)=1+a+a b c$
R.H.S. $(a * b) * c=(1+a b) * c=1+(1+a b) . c=1+c+a b c$
L.H.S. $\ne$ R.H.S.
So, $*$ is not associative.
Hence, $*$ is commutative but not associative.
Objective Type Questions
28. Let $T$ be the set of all triangles in the Euclidean plane and let a relation $R$ on $T$ be defined as $a R b$, if $a$ is congruent to $b, \forall a$, $b \in T$. Then $R$ is
(a) Reflexive but not transitive
(b) Transitive but not symmetric
(c) Euivalence
(d) None of these
Show Answer
Solution
If $a \cong b \forall a, b \in T$
then $a R a \Rightarrow a \cong a$ which is true for all $a \in T$
So, $R$ is reflexive.
Now, $a R b$ and $b R a$.
i.e., $a \cong b$ and $b \cong a$ which is true for all $a, b \in T$
So, $R$ is symmetric.
Let $a R b$ and $b R c$.
$\Rightarrow a \cong b$ and $b \cong a \Rightarrow a \cong c \forall a, b, c \in T$
So, $R$ is transitive.
Hence, $R$ is euivalence relation.
So, the correct answer is (c).

Option (a) Reflexive but not transitive:
 This option is incorrect because the relation $R$ is indeed transitive. If $a R b$ and $b R c$, then $a \cong b$ and $b \cong c$ imply $a \cong c$, satisfying the transitivity property.

Option (b) Transitive but not symmetric:
 This option is incorrect because the relation $R$ is symmetric. If $a R b$, then $a \cong b$ implies $b \cong a$, satisfying the symmetry property.

Option (d) None of these:
 This option is incorrect because the relation $R$ is an equivalence relation, which means it is reflexive, symmetric, and transitive. Therefore, option (c) is the correct answer.
29. Consider the nonempty set consisting of children in a family and a relation $R$ defined as $a R b$, if $a$ is brother of $b$. Then $R$ is
(a) symmetric but not transitive
(b) transitive but not symmetric
(c) neither symmetric nor transitive
(d) both symmetric and transitive
Show Answer
Solution
Here, $a R b \Rightarrow a$ is a brother of $b$.
$a R a \Rightarrow a$ is a brother of $a$ which is not true.
So, $R$ is not reflexive.
$a R b \Rightarrow a$ is a brother of $b$.
$b R a \Rightarrow$ which is not true because $b$ may be sister of $a$.
$\Rightarrow a R b \ne b R a$
So, $R$ is not symmetric.
Now, $a R b, b R c \Rightarrow a R c$
$\Rightarrow a$ is the brother of $b$ and $b$ is the brother of $c$.
$\therefore a$ is also the brother of $c$.
So, $R$ is transitive.
Hence, correct answer is $(b)$.

Option (a) symmetric but not transitive:
 This option is incorrect because the relation $R$ is not symmetric. If $a$ is a brother of $b$, it does not necessarily mean that $b$ is a brother of $a$ (since $b$ could be a sister). Therefore, $R$ is not symmetric.

Option (c) neither symmetric nor transitive:
 This option is incorrect because the relation $R$ is transitive. If $a$ is a brother of $b$ and $b$ is a brother of $c$, then $a$ is also a brother of $c$. Therefore, $R$ is transitive.

Option (d) both symmetric and transitive:
 This option is incorrect because the relation $R$ is not symmetric. As mentioned earlier, if $a$ is a brother of $b$, it does not necessarily mean that $b$ is a brother of $a$ (since $b$ could be a sister). Therefore, $R$ is not symmetric.
30. The maximum number of euivalence relations on the set $A={1,2,3}$ are
(a) 1
(b) 2
(c) 3
(d) 5
Show Answer
Solution
Here,
$ A={1,2,3} $
The number of euivalence relations are as follows:
$R_1={(1,1),(1,2),(2,1),(2,3),(1,3)}$
$R_2={(2,2),(1,3),(3,1),(3,2),(1,2)}$
$R_3={(3,3),(1,2),(2,3),(1,3),(3,2)}$
Hence, correct answer is $(d)$

Option (a) is incorrect because there is more than one equivalence relation possible on the set ( A = {1, 2, 3} ). Specifically, there are multiple ways to partition the set into equivalence classes.

Option (b) is incorrect because there are more than two equivalence relations possible on the set ( A = {1, 2, 3} ). The number of equivalence relations corresponds to the number of ways to partition the set, which is more than two.

Option (c) is incorrect because there are more than three equivalence relations possible on the set ( A = {1, 2, 3} ). The correct number of equivalence relations is determined by the Bell number for a set of size 3, which is 5.
31. If a relation $R$ on the set ${1,2,3}$ be defined by $R={(1,2)}$, then $R$ is
(a) reflexive
(b) transitive
(c) symmetric
(d) None of these
Show Answer
Solution
Given that: $R={(1,2)}$
$a$ Z $a$, so it is not reflexive.
$a R b$ but $b \mathbb{R} a$, so it is not symmetric.
$a R b$ and $b R c \Rightarrow a R c$ which is true.
So, $R$ is transitive.
Hence, correct answer is $(b)$.

Reflexive: A relation ( R ) on a set is reflexive if every element is related to itself. In this case, for ( R ) to be reflexive on the set ({1, 2, 3}), it must include the pairs ((1,1)), ((2,2)), and ((3,3)). Since ( R = {(1,2)}) does not include these pairs, it is not reflexive.

Symmetric: A relation ( R ) on a set is symmetric if for every pair ((a, b) \in R), the pair ((b, a)) is also in ( R ). Here, ( (1,2) \in R ) but ( (2,1) \notin R ). Therefore, ( R ) is not symmetric.

None of these: This option is incorrect because the relation ( R ) is transitive, as shown in the given answer.
32. Let us define a relation $R$ in $R$ as $a R b$ if $a \ge b$. Then $R$ is
(a) an euivalence relation
(b) reflexive, transitive but not symmetric
(c) symmetric, transitive but not reflexive
(d) neither transitive nor reflexive but symmetric.
Show Answer
Solution
Here, $a R b$ if $a \ge b$
$\Rightarrow a R a \Rightarrow a \ge a$ which is true, so it is reflexive.
Let $a R b \Rightarrow a \ge b$, but $b \ngeq a$, so $b \not R a$
$R$ is not symmetric.
Now, $a \ge b, b \ge c \Rightarrow a \ge c$ which is true.
So, $R$ is transitive.
Hence, correct answer is $(b)$.

Option (a): an equivalence relation
An equivalence relation must be reflexive, symmetric, and transitive. While the relation ( R ) is reflexive and transitive, it is not symmetric. Therefore, ( R ) cannot be an equivalence relation.

Option (c): symmetric, transitive but not reflexive
The relation ( R ) is reflexive, as ( a \ge a ) is always true. However, ( R ) is not symmetric because if ( a \ge b ), it does not necessarily mean that ( b \ge a ). Therefore, this option is incorrect.

Option (d): neither transitive nor reflexive but symmetric
The relation ( R ) is reflexive, as ( a \ge a ) is always true, and it is transitive because if ( a \ge b ) and ( b \ge c ), then ( a \ge c ). However, ( R ) is not symmetric. Therefore, this option is incorrect.
33. Let $A={1,2,3}$ and consider the relation
$R={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)}$, then $R$ is
(a) reflexive but not symmetric
(b) reflexive but not transitive
(c) symmetric and transitive
(d) neither symmetric nor transitive.
Show Answer
Solution
Given that: $R={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)}$
Here, 1 R 1, 2 R 2 and 3 R 3, so R is reflexive.
$1 R 2$ but $2 R^{\prime} 1$ or $2 R 3$ but $3 R^{\prime} 2$, so, $R$ is not symmetric.
$1 R 1$ and $1 R 2 \Rightarrow 1 R 3$, so, $R$ is transitive.
Hence, the correct answer is $(a)$.

Option (b) is incorrect because the relation ( R ) is transitive. For example, ( 1 R 2 ) and ( 2 R 3 ) imply ( 1 R 3 ), which is present in ( R ).

Option (c) is incorrect because the relation ( R ) is not symmetric. For example, ( 1 R 2 ) is in ( R ), but ( 2 R 1 ) is not in ( R ).

Option (d) is incorrect because the relation ( R ) is reflexive. For example, ( (1,1) ), ( (2,2) ), and ( (3,3) ) are all in ( R ).
34. The identity element for the binary operation $*$ defined on $ \sim{0}$ as $a * b=\frac{a b}{2} \forall a, b \in \sim{0}$ is
(a) 1
(b) 0
(c) 2
(d) None of these
Show Answer
Solution
Given that: $a * b=\frac{a b}{2} \forall a, b \in {0}$
Let $e$ be the identity element
$ \therefore \quad a * e=\frac{a e}{2}=a \Rightarrow e=2 $
Hence, the correct answer is (c).

Option (a) 1: If 1 were the identity element, then for any ( a ), we would have ( a * 1 = a ). However, ( a * 1 = \frac{a \cdot 1}{2} = \frac{a}{2} ), which is not equal to ( a ) for all ( a ). Therefore, 1 cannot be the identity element.

Option (b) 0: If 0 were the identity element, then for any ( a ), we would have ( a * 0 = a ). However, ( a * 0 = \frac{a \cdot 0}{2} = 0 ), which is not equal to ( a ) for all ( a ). Therefore, 0 cannot be the identity element.

Option (d) None of these: This option suggests that there is no identity element for the given operation. However, we have already shown that 2 is indeed the identity element because ( a * 2 = \frac{a \cdot 2}{2} = a ) for all ( a ). Therefore, this option is incorrect.
35. If the set A contains 5 elements and set B contains 6 elements, then the number of oneone and onto mapping from $A$ to $B$ is
(a) 720
(b) 120
(c) 0
(d) None of these
Show Answer
Solution
If $A$ and $B$ sets have $m$ and $n$ elements respectively, then the number of oneone and onto mapping from $A$ to $B$ is
$n$ ! if $m=n$
and 0 if $m \ne n$
Here,
$ \begin{aligned} m & =5 \text{ and } n=6 \\ 5 & \ne 6 \end{aligned} $
So, number of mapping $=0$
Hence, the correct answer is (c).

Option (a) 720: This option is incorrect because it assumes that there are 720 oneone and onto mappings from set A to set B. However, since set A has 5 elements and set B has 6 elements, it is impossible to have a oneone and onto mapping when the number of elements in the two sets is different. Therefore, the number of such mappings is 0, not 720.

Option (b) 120: This option is incorrect because it assumes that there are 120 oneone and onto mappings from set A to set B. Similar to option (a), since set A has 5 elements and set B has 6 elements, it is impossible to have a oneone and onto mapping when the number of elements in the two sets is different. Therefore, the number of such mappings is 0, not 120.

Option (d) None of these: This option is incorrect because it implies that none of the given options, including 0, is correct. However, as explained, the correct number of oneone and onto mappings from set A to set B is indeed 0 when the number of elements in the two sets is different. Therefore, option (c) is the correct answer, making option (d) incorrect.
36. Let $A={1,2,3, \ldots, n}$ and $B={a, b}$. Then the number of surjections from $A$ to $B$ is
(a) ${ }^{n} P_2$
(b) $2^{n}2$
(c) $2^{n}1$
(d) None of these
Show Answer
Solution
Here, $A={1,2,3, \ldots, n}$ and $B={a, b}$
Let $m$ be the number of elements of set $A$ and $n$ be the number of elements of set $B$
$\therefore$ Number of surjections from $A$ to $B$ is
${ }^{n} C_m \times m$ ! as $n \ge m$
Here, $m=2$ (given)
$\therefore$ Number of surjections from $A$ to $B={ }^{n} C_2 \times 2$ !
$ =\frac{n !}{2 !(n2) !} \times 2 !=\frac{n(n1)(n2) !}{2 !(n2) !} \times 2=n(n1)=n^{2}n $
Hence, the correct answer is $(d)$.

Option (a) ${ }^{n} P_2$: This represents the number of permutations of $n$ items taken 2 at a time, which is not related to the number of surjections from $A$ to $B$. Surjections require every element in $B$ to be mapped to by at least one element in $A$, which is not guaranteed by permutations.

Option (b) $2^{n}2$: This expression represents the total number of functions from $A$ to $B$ minus the two nonsurjective functions (all elements of $A$ mapping to $a$ or all elements of $A$ mapping to $b$). However, this does not correctly account for the requirement that each element in $B$ must be mapped to by at least one element in $A$.

Option (c) $2^{n}1$: This expression represents the total number of functions from $A$ to $B$ minus one function (all elements of $A$ mapping to either $a$ or $b$). This also does not correctly account for the requirement that each element in $B$ must be mapped to by at least one element in $A$.
37. Let $f: R \to R$ be defined by $f(x)=\frac{1}{x}, \forall x \in R$ then $f$ is
(a) oneone
(b) onto
(c) bijective
(d) $f$ is not defined
Show Answer
Solution
Given that $f(x)=\frac{1}{x}$
Put $x=0 \quad \therefore \quad f(x)=\frac{1}{0}=\infty$
So, $f(x)$ is not defined.
Hence, the correct answer is $(d)$.

(a) oneone: The function ( f(x) = \frac{1}{x} ) is indeed oneone (injective) for all ( x \neq 0 ). However, since ( f(x) ) is not defined at ( x = 0 ), the function cannot be considered oneone over the entire domain ( \mathbb{R} ).

(b) onto: The function ( f(x) = \frac{1}{x} ) is onto (surjective) for all ( x \neq 0 ). However, since ( f(x) ) is not defined at ( x = 0 ), the function cannot be considered onto over the entire domain ( \mathbb{R} ).

(c) bijective: A function is bijective if it is both oneone and onto. Since ( f(x) = \frac{1}{x} ) is not defined at ( x = 0 ), it cannot be considered bijective over the entire domain ( \mathbb{R} ).
38. Let $f: R \to R$ be defined by $f(x)=3 x^{2}5$ and $g: R \to R$ by $g(x)=\frac{x}{x^{2}+1}$, then $g \circ f$ is
(a) $\frac{3 x^{2}5}{9 x^{4}30 x^{2}+26}$
(b) $\frac{3 x^{2}5}{9 x^{4}6 x^{2}+26}$
(c) $\frac{3 x^{2}}{x^{4}+2 x^{2}4}$
(d) $\frac{3 x^{2}}{9 x^{4}+30 x^{2}2}$
Show Answer
Solution
Here, $f(x)=3 x^{2}5$ and $g(x)=\frac{x}{x^{2}+1}$
$ \begin{aligned} \therefore \quad g \circ f & =gof(x)=g[3 x^{2}5] \\ & =\frac{3 x^{2}5}{(3 x^{2}5)^{2}+1}=\frac{3 x^{2}5}{9 x^{4}+2530 x^{2}+1} \\ \therefore \quad gof & =\frac{3 x^{2}5}{9 x^{4}30 x^{2}+26} \end{aligned} $
Hence, the correct answer is $(a)$.

Option (b) is incorrect because the denominator is given as $9x^{4}  6x^{2} + 26$, but the correct denominator should be $9x^{4}  30x^{2} + 26$.

Option (c) is incorrect because the numerator is given as $3x^{2}$, but the correct numerator should be $3x^{2}  5$. Additionally, the denominator is given as $x^{4} + 2x^{2}  4$, which is not the correct expression derived from $(3x^{2}  5)^{2} + 1$.

Option (d) is incorrect because the denominator is given as $9x^{4} + 30x^{2}  2$, but the correct denominator should be $9x^{4}  30x^{2} + 26$.
39. Which of the following functions from $Z$ to $Z$ are bijections?
(a) $f(x)=x^{3}$
(b) $f(x)=x+2$
(c) $f(x)=2 x+1$
(d) $f(x)=x^{2}+1$
Show Answer
Solution
Given that $f: Z \to Z$
Let $x_1, x_2 \in f(x) \Rightarrow f(x_1)=x_1+2, f(x_2)=x_2+2$
$f(x_1)=f(x_2) \Rightarrow x_1+2=x_2+2 \Rightarrow x_1=x_2$
So, $f(x)$ is oneone function.
Now, let $y=x+2 \therefore x=y2 \in Z \quad \forall y \in Z$
So, $f(x)$ is onto function.
$\therefore f(x)$ is bijective function.
Hence, the correct answer is (b).

(a) ( f(x) = x^3 ):
 Reason: This function is indeed a bijection. For any integer ( y ), there exists a unique integer ( x ) such that ( x^3 = y ). Therefore, this function is both onetoone and onto.

(c) ( f(x) = 2x + 1 ):
 Reason: This function is also a bijection. For any integer ( y ), there exists a unique integer ( x ) such that ( 2x + 1 = y ). Therefore, this function is both onetoone and onto.

(d) ( f(x) = x^2 + 1 ):
 Reason: This function is not a bijection. It is not onetoone because ( f(x) = f(x) ) for any ( x \in Z ). For example, ( f(2) = 5 ) and ( f(2) = 5 ). It is also not onto because there is no integer ( x ) such that ( x^2 + 1 ) equals any negative integer.
40. Let $f: R \to R$ be the functions defined by $f(x)=x^{3}+5$. Then $f^{1}(x)$ is
(a) $(x+5)^{1 / 3}$
(b) $(x5)^{1 / 3}$
(c) $(5x)^{1 / 3}$
(d) $5x$
Show Answer
Solution
Given that
$ f(x)=x^{3}+5 $
Let $\quad y=x^{3}+5 \Rightarrow x^{3}=y5$
$\therefore \quad x=(y5)^{1 / 3} \Rightarrow f^{1}(x)=(x5)^{1 / 3}$
Hence, the correct answer is $(b)$.

Option (a) $(x+5)^{1 / 3}$ is incorrect because it suggests that the inverse function is obtained by adding 5 to ( x ) and then taking the cube root. However, the correct process involves subtracting 5 from ( x ) before taking the cube root.

Option (c) $(5x)^{1 / 3}$ is incorrect because it suggests that the inverse function is obtained by subtracting ( x ) from 5 and then taking the cube root. This does not correctly reverse the original function ( f(x) = x^3 + 5 ).

Option (d) $5x$ is incorrect because it suggests that the inverse function is a linear function, which does not correctly reverse the cubic nature of the original function ( f(x) = x^3 + 5 ).
41. Let $f: A \to B$ and $g: B \to C$ be the bijective functions. Then $(gof)^{1}$ is
(a) $f^{1} og^{1}$
(b) $fog$
(c) $g^{1}$ of ${ }^{1}$
(d) $gof$
Show Answer
Solution
Here, $f: A \to B$ and $g: B \to C$
$\therefore \quad (g o f)^{1}=f^{1} o g^{1}$
Hence, the correct answer is (a).

Option (b) $fog$ is incorrect because it represents the composition of $f$ and $g$, not the inverse of the composition $(gof)$. The inverse of a composition is not simply the composition itself.

Option (c) $g^{1}$ of ${ }^{1}$ is incorrect because it is not a valid mathematical notation. The correct notation for the inverse of the composition $(gof)$ is $(gof)^{1}$, which is equal to $f^{1} o g^{1}$.

Option (d) $gof$ is incorrect because it represents the composition of $g$ and $f$, not the inverse of the composition $(gof)$. The inverse of a composition is not the composition itself.
42. Let $f: R{\frac{3}{5}} \to R$ be defined by $f(x)=\frac{3 x+2}{5 x3}$, then
(a) $f^{1}(x)=f(x)$
(b) $f^{1}(x)=f(x)$
(c) $(f o f) x=x$
(d) $f^{1}(x)=\frac{1}{19} f(x)$
Show Answer
Solution
Given that $f(x)=\frac{3 x+2}{5 x3} \forall x \ne \frac{3}{5}$
$ \begin{matrix} \text{ Let } & & y & =\frac{3 x+2}{5 x3} \\ \Rightarrow & & y(5 x3) & =3 x+2 \\ \Rightarrow & 5 x y3 y & =3 x+2 \\ \Rightarrow & 5 x y3 x & =3 y+2 \\ \Rightarrow & & x(5 y3) & =3 y+2 \\ \Rightarrow & & x & =\frac{3 y+2}{5 y3} \\ \Rightarrow & & f^{1}(x) & =\frac{3 x+2}{5 x3} \\ \Rightarrow & & f^{1}(x) & =f(x) \end{matrix} $
Hence, the correct answer is (a).

Option (b): The statement $f^{1}(x) = f(x)$ is incorrect because we have already established that $f^{1}(x) = f(x)$. Therefore, $f^{1}(x)$ cannot be equal to $f(x)$.

Option (c): The statement $(f \circ f)(x) = x$ is incorrect. To verify this, we need to compute $(f \circ f)(x)$: [ f(f(x)) = f\left(\frac{3x + 2}{5x  3}\right) = \frac{3\left(\frac{3x + 2}{5x  3}\right) + 2}{5\left(\frac{3x + 2}{5x  3}\right)  3} ] Simplifying this expression does not yield $x$. Therefore, $(f \circ f)(x) \neq x$.

Option (d): The statement $f^{1}(x) = \frac{1}{19} f(x)$ is incorrect because we have already established that $f^{1}(x) = f(x)$. Therefore, $f^{1}(x)$ cannot be equal to $\frac{1}{19} f(x)$.
43. Let $f:[0,1] \to[0,1]$ be defined by $f(x)=\begin{cases} x, & \text{ if } x \text{ is rational } \\ 1x, & \text{ if } x \text{ is irrational } \end{cases} .$
Then $(fof) x$ is
(a) constant
(b) $1+x$
(c) $x$
(d) None of these
Show Answer
Solution
Given that $f:[0,1] \to[0,1]$
$\therefore \qquad f=f^{1}$
$So, \qquad (fof)x=x \qquad \text{(identity element)}$
Hence, correct answer is (c).

Option (a) constant: This option is incorrect because the function ( (f \circ f)(x) ) is not a constant function. A constant function would mean that ( (f \circ f)(x) ) takes the same value for all ( x ) in ([0,1]). However, ( (f \circ f)(x) = x ), which varies with ( x ).

Option (b) (1 + x): This option is incorrect because ( (f \circ f)(x) ) does not equal ( 1 + x ). For ( x ) in ([0,1]), ( 1 + x ) would exceed the range ([0,1]) for ( x > 0 ), which is not possible since ( f ) maps ([0,1]) to ([0,1]).

Option (d) None of these: This option is incorrect because there is a correct answer among the given options, which is ( (f \circ f)(x) = x ). Therefore, “None of these” is not the correct choice.
44. Let $f:[2, \infty) \to R$ be the function defined by $f(x)=x^{2}4 x+5$, then the range of $f$ is
(a) $R$
(b) $[1, \infty)$
(c) $[4, \infty)$
(d) $[5, \infty)$
Show Answer
Solution
Given that $f(x)=x^{2}4 x+5$
$ \begin{aligned} & \text{ Let } \quad y=x^{2}4 x+5 \\ & \Rightarrow x^{2}4 x+5y=0 \\ & \Rightarrow \quad x=\frac{(4) \pm \sqrt{(4)^{2}4 \times 1 \times(5y)}}{2 \times 1} \\ & =\frac{4 \pm \sqrt{1620+4 y}}{2} \\ & =\frac{4 \pm \sqrt{4 y4}}{2}=\frac{4 \pm 2 \sqrt{y1}}{2}=2 \pm \sqrt{y1} \end{aligned} $
$\therefore$ For real value of $x, y1 \ge 0 \Rightarrow y \ge 1$.
So, the range is $[1, \infty)$.
Hence, the correct answer is $(b)$.

Option (a) $R$: This option is incorrect because the function ( f(x) = x^2  4x + 5 ) is a quadratic function that opens upwards (since the coefficient of ( x^2 ) is positive). The minimum value of this function is 1, which occurs at ( x = 2 ). Therefore, the function cannot take any value less than 1, making the range not equal to all real numbers ( R ).

Option (c) $[4, \infty)$: This option is incorrect because the minimum value of the function ( f(x) = x^2  4x + 5 ) is 1, not 4. The function achieves this minimum value at ( x = 2 ). Therefore, the range starts from 1, not 4.

Option (d) $[5, \infty)$: This option is incorrect because the minimum value of the function ( f(x) = x^2  4x + 5 ) is 1, not 5. The function achieves this minimum value at ( x = 2 ). Therefore, the range starts from 1, not 5.
45. Let $f: N \to R$ be the function defined by $f(x)=\frac{2 x1}{2}$ and $g: \to R$ be another function defined by $g(x)=x+2$ then, gof $(\frac{3}{2})$ is
(a) 1
(b) 1
(c) $\frac{7}{2}$
(d) None of these
Show Answer
Solution
Here,
$ f(x)=\frac{2 x1}{2} \text{ and } g(x)=x+2 $
$ \therefore \quad \begin{aligned} g \circ f(x) & =g[(f(x)] \\ & =f(x)+2 \\ & =\frac{2 x1}{2}+2=\frac{2 x+3}{2} \\ gof(\frac{3}{2}) & =\frac{2 \times \frac{3}{2}+3}{2}=3 \end{aligned} $
Hence, the correct answer is $(d)$.

Option (a) 1: This is incorrect because when calculating ( g \circ f \left( \frac{3}{2} \right) ), the result is 3, not 1. The calculation involves substituting ( \frac{3}{2} ) into the composed function ( g(f(x)) ), which yields 3.

Option (b) 1: This is incorrect because the composed function ( g \circ f \left( \frac{3}{2} \right) ) results in 3, not 1. The calculation does not produce a negative value.

Option (c) ( \frac{7}{2} ): This is incorrect because the correct value of ( g \circ f \left( \frac{3}{2} \right) ) is 3, not ( \frac{7}{2} ). The calculation of the composed function does not yield ( \frac{7}{2} ).
46. Let $f: R \to R$ be defined by $f(x)= \begin{cases}2 x & : x>3 \\ x^{2} & : 1<x \le 3 \\ 3 x & : x \le 1\end{cases}$ then $f(1)+f(2)+f(4)$ is
(a) 9
(b) 14
(c) 5
(d) None of these
Show Answer
Solution
Given that:
$ \begin{aligned} f(x) & = \begin{cases}2 x & : x>3 \\ x^{2} & : 1<x \le 3 \\ 3 x & : x \le 1\end{cases} \\ \therefore f(1)+f(2)+f(4) & =3(1)+(2)^{2}+2(4)=3+4+8=9 \end{aligned} $
Hence, the correct answer is $(a)$.

Option (b) 14 is incorrect because the calculation of ( f(1) + f(2) + f(4) ) does not sum to 14. The correct values are ( f(1) = 3 ), ( f(2) = 4 ), and ( f(4) = 8 ), which sum to 9, not 14.

Option (c) 5 is incorrect because the calculation of ( f(1) + f(2) + f(4) ) does not sum to 5. The correct values are ( f(1) = 3 ), ( f(2) = 4 ), and ( f(4) = 8 ), which sum to 9, not 5.

Option (d) None of these is incorrect because the correct answer is indeed one of the given options, specifically option (a) 9.
47. If $f: R \to R$ be given by $f(x)=\tan x$, then $f^{1}(1)$ is
(a) $\frac{\pi}{4}$
(b) ${n \pi+\frac{\pi}{4}: n \in Z}$
(c) does not exist
(d) None of these
Show Answer
Solution
Given that $f(x)=\tan x$
Let $\quad f(x)=y=\tan x \Rightarrow x=\tan ^{1} y$
$\Rightarrow \quad f^{1}(x)=\tan ^{1}(x)$
$\Rightarrow \quad f^{1}(1)=\tan ^{1}(1)$
$\Rightarrow \quad f^{1}(1)=\tan ^{1}[\tan (\frac{\pi}{4})]=\frac{\pi}{4}$
Hence, the correct answer is (a).

Option (b) ${n \pi+\frac{\pi}{4}: n \in Z}$ is incorrect because it represents the general solution for $\tan x = 1$, which includes all possible values of $x$ where $\tan x = 1$. However, the inverse function $f^{1}(x)$ is defined to return a single value, specifically the principal value, which is $\frac{\pi}{4}$ for $\tan^{1}(1)$.

Option (c) “does not exist” is incorrect because the inverse function $f^{1}(x)$ does exist for $f(x) = \tan x$ and $f^{1}(1)$ specifically exists and is equal to $\frac{\pi}{4}$.

Option (d) “None of these” is incorrect because the correct answer is indeed provided in option (a), which is $\frac{\pi}{4}$.
Fillers
48. Let the relation $R$ be defined in $N$ by $a R b$ if $2 a+3 b=30$. Then $R=$
Show Answer
Solution
Given that $a R b: 2 a+3 b=30$
$\begin{matrix} \Rightarrow & \quad 3 b=302 a \\ \Rightarrow \quad & \quad b=\frac{302 a}{3} \\ \text{ for } & a=3, b=8 \\ & a=6, b=6 \\ & a=9, b=4 \\ & a=12, b=2\end{matrix} $
Hence, $\quad R=\{(3,8),(6,6),(9,4),(12,2)\}$
49. Let the relation $R$ be defined on the set
$A=\{1,2,3,4,5\}$ by $R=\{(a, b):a^{2}b^{2}<8\}$. Then $R$ is given by
Show Answer
Solution
Given that $A={1,2,3,4,5}$ and $R={(a, b):a^{2}b^{2}<8}$
So, clearly, $\quad R=\{(1,1),(1,2),(2,1),(2,2),(2,3),(3,2),(4,3)$
$(3,4),(4,4),(5,5)\}$
50. Let $f=\{(1,2),(3,5),(4,1)\}$ and $g=\{(2,3),(5,1),(1,3)\}$. Then $gof=……$ and $fog=……$
Show Answer
Solution
Here, $f=\{(1,2),(3,5),(4,1)\}$ and $g=\{(2,3),(5,1),(1,3)\}$
$ \begin{aligned} gof(1) & =g[f(1)]=g(2)=3 \\ gof(3) & =g[f(3)]=g(5)=1 \\ gof(4) & =g[f(4)]=g(1)=3 \\ \therefore \quad gof & ={(1,3),(3,1),(4,3)} \\ fog(2) & =f[g(2)]=f(3)=5 \end{aligned} $
$ \begin{aligned} & fog(5)=f[g(5)]=f(1)=2 \\ & fog(1)=f[g(1)]=f(3)=5 \\ & \therefore \quad fog={(2,5),(5,2),(1,5)} \end{aligned} $
51. Let $f: R \to R$ be defined by $f(x)=\frac{x}{\sqrt{1+x^{2}}}$, then
$(fofof)(x)=……$
Show Answer
Solution
Here, $f(x)=\frac{x}{\sqrt{1+x^{2}}} \forall x \in R$
$ \begin{aligned} & fofof(x)=fof[f(x)]=f[f{f(x)}] \\ & =f[f(\frac{x}{\sqrt{1+x^{2}}})]=f[\frac{\frac{x}{\sqrt{1+x^{2}}}}{\sqrt{1+\frac{x^{2}}{1+x^{2}}}}] \\ & =f[\frac{\frac{x}{\sqrt{1+x^{2}}}}{\frac{\sqrt{1+x^{2}+x^{2}}}{\sqrt{1+x^{2}}}}]=f[\frac{x}{\sqrt{1+2 x^{2}}}] \\ & =[\frac{\frac{x}{\sqrt{1+2 x^{2}}}}{\sqrt{1+\frac{x^{2}}{1+2 x^{2}}}}]=[\frac{\frac{x}{\sqrt{1+2 x^{2}}}}{\frac{\sqrt{1+2 x^{2}+x^{2}}}{\sqrt{1+2 x^{2}}}}]=\frac{x}{\sqrt{1+3 x^{2}}} \end{aligned} $
Hence, $fofof(x)=\frac{x}{\sqrt{3 x^{2}+1}}$
52. If $f(x)=[4(x7)^{3}]$, then $f^{1}(x)=……$
Show Answer
Solution
Given that, $f(x)=[4(x7)^{3}]$
Let $ y=[4(x7)^{3}] $
$\Rightarrow$ $ (x7)^{3}=4y $
$\Rightarrow$ $ x7=(4y)^{1 / 3} \Rightarrow x=7+(4y)^{1 / 3} $
Hence, $\quad f^{1}(x)=7+(4x)^{1 / 3}$
True/False
53. Let $R=\{(3,1),(1,3),(3,3)\}$ be a relation defined on the set $A=\{1,2,3\}$. Then $R$ is symmetric, transitive but not reflexive.
Show Answer
Solution
Here, $\quad R=\{(3,1),(1,3),(3,3)\}$
$(3,3) \in R$, so $R$ is reflexive.
$(3,1) \in R$ and $(1,3) \in R$, so $R$ is symmetric.
Now, $(3,1) \in R$ and $(1,3) \in R$ but $(1,1) \notin R$
So, $R$ is not transitive.
Hence, the statement is ‘False’.
54. Let $f: R \to R$ be the function defined by
$f(x)=\sin (3 x+2) \forall x \in R$, then $f$ is invertible.
Show Answer
Solution
Given that: $f(x)=\sin (3 x+2) \forall x \in R$,
$f(x)$ is not oneone.
Hence, the statement is ‘False’.
55. Every relation which is symmetric and transitive is also reflexive.
Show Answer
Solution
Let $R$ be any relation defined on $A={1,2,3}$
$ R=\{(1,2),(2,1),(2,3),(1,3)\} $
Here, $(1,2) \in R$ and $(2,1) \in R$, so $R$ is symmetric.
$(1,2) \in R,(2,3) \in R \Rightarrow(1,3) \in R$, so $R$ is transitive.
But $(1,1) \notin R,(2,2) \notin R$ and $(3,3) \notin R$.
Hence, the statement is ‘False’.
56. An integer $m$ is said to be related to another integer $n$ if $m$ is an integral multiple of $n$. This relation in $Z$ is reflexive, symmetric and transitive.
Show Answer
Solution
Here, $\quad m=k n$
If $k=1 \quad m=n$, so $z$ is reflexive.
(where $k$ is an integer)
Clearly $z$ is not symmetric but $z$ is transitive.
Hence, the statement is ‘False’.
57. Let $A={0,1}$ and $N$ be the set of natural numbers then the mapping $f: N \to$ A defined by $f(2 n1)=0, f(2 n)=1, \forall n \in N$ is onto.
Show Answer
Solution
Given that $A=[0,1]$
$f(2 n1)=0$ and $f(2 n)=1 \quad \forall n \in N$
So, $f: N \to A$ is a onto function.
Hence, the statement is ‘True’.
58. The relation $R$ on the set $A=\{1,2,3\}$ defined as $R=\{(1,1),(1,2),(2,1),(3,3)\}$ is reflexive, symmetric and transitive.
Show Answer
Solution
Here, $\quad R\{(1,1),(1,2),(2,1),(3,3)\}$
Here, $(1,1) \in R$, so $R$ is Reflexive.
$(1,2) \in R$ and $(2,1) \in R$, so $R$ is Symmetric.
$(1,2) \in R$ but $(2,3) \notin R$
So, $R$ is not transitive.
Hence, the statement is ‘False’.
59. The composition of functions is commutative.
Show Answer
Solution
Let $f(x)=x^{2}$ and $g(x)=2 x+3$
$ \begin{aligned} & fog(x)=f[g(x)]=(2 x+3)^{2}=4 x^{2}+9+12 x \\ & gof(x)=g[f(x)]=2 x^{2}+3 \end{aligned} $
So,
$ fog(x) \ne gof(x) $
Hence, the statement is ‘False’.
60. The composition of functions is associative.
Show Answer
Solution
Let $f(x)=2 x, g(x)=x1$ and $h(x)=2 x+3$
$ \begin{aligned} & fo\{goh(x)\}=fo\{g(2 x+3)\} \\ & =f(2 x+31)=f(2 x+2)=2(2 x+2)=4 x+4 \\ \text{and} \quad & (fog)oh(x)=(fog)\{h(x)\} \\ & =fog(2 x+3) \\ & =f(2 x+31)=f(2 x+2)=2(2 x+2)=4 x+4 \end{aligned} $
Hence, the statement is ‘True’.
61. Every function is invertible.
Show Answer
Solution
Only bijective functions are invertible.
Hence, the statement is ‘False’.
62. A binary operation on a set has always the identity element.
Show Answer
Solution
’ + ’ is a binary operation on the set $N$ but it has no identity element.
Hence, the statement is ‘False’.