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 $(a, c)$

Solution

Here, $f(x)=\sqrt{25-x^{2}}$

For real value of $f(x), 25-x^{2} \ge 0$

$\Rightarrow-x^{2} \ge-25 \Rightarrow x^{2} \le 25 \Rightarrow-5 \le x \le 5$

Hence, $D \in-5 \le x \le 5$ or $[-5,5]$

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+1-2=4 x^{2}+4 x-1 $

Hence, $g o f=4 x^{2}+4 x-1$

Solution

Here,

$ f(x)=2 x-3 $

Let

$ f(x)=y=2 x-3 $

$ \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} $

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} $

Solution

Here, $f(x)=x^{2}-3 x+2$

$\therefore \quad f[f(x)]=[f(x)]^{2}-3 f(x)+2$

$=(x^{2}-3 x+2)^{2}-3(x^{2}-3 x+2)+2$

$=x^{4}+9 x^{2}+4-6 x^{3}+4 x^{2}-12 x-3 x^{2}+9 x-6+2$

$=x^{4}-6 x^{3}+10 x^{2}-3 x$

Hence, $f[f(x)]=x^{4}-6 x^{3}+10 x^{2}-3 x$

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, $\alpha=2, \beta=-1$

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.

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, $\quad f o g={(2,5),(5,2),(1,5)}$

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 & \ne-1 \end{aligned} $

Therefore, it is not one-one.

Now, let $f(z)=y=|z|$. Here, there is no pre-image of negative numbers. Hence, it is not onto.

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 one-one. Also it is not onto function as no pre-image of any real number belongs to the range of $\cos x$ i.e., $[-1,1]$.

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.

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 one-one. But $g$ is not onto as there is no pre-image of $A$ in $B$ under $g$.

Solution

Given function is $f(x)=\frac{1}{2-\cos x}, \forall x \in R$.

Range of $\cos x$ is $[-1,1]$

Let $\quad f(x)=y=\frac{1}{2-\cos x}$

$\Rightarrow \quad 2 y-y \cos x=1 \quad \Rightarrow \quad y \cos x=2 y-1$

$\Rightarrow \quad \cos x=\frac{2 y-1}{y}=2-\frac{1}{y}$

Now $-1 \le \cos x \le 1$

$\Rightarrow \quad-1 \le 2-\frac{1}{y} \le 1 \Rightarrow-1-2 \le-\frac{1}{y} \le 1-2$

$\Rightarrow \quad-3 \le-\frac{1}{y} \le-1 \Rightarrow 3 \ge \frac{1}{y} \ge 1 \Rightarrow \frac{1}{3} \le y \le 1$

Hence, the range of $f=[\frac{1}{3}, \mathbf{1}]$.

Solution

Here, $\forall a, b \in Z$ and $a R b$ if and only if $a-b$ is divisible by $n$. The given relation is an euivalence relation if it is reflexive, symmetric and transitive.

(i) Reflexive:

$a R a \Rightarrow(a-a)=0$ divisible by $n$

So, $R$ is reflexive.

(ii) Symmetric:

$a R b=b R a \quad \forall a, b \in Z$

$a-b$ is divisible by $n$ (Given)

$\Rightarrow-(b-a)$ is divisible by $n$ $\Rightarrow b-a$ is divisible by $n$

$\Rightarrow b R a$

Hence, $R$ is symmetric.

(iii) Transitive:

$a R b$ and $b R c \quad \Leftrightarrow \quad a R c \quad \forall a, b, c \in Z$

$a-b$ is divisible by $n$

$b-c$ is also divisible by $n$

$\Rightarrow(a-b)+(b-c)$ is divisible by $n$

$\Rightarrow(a-c)$ is divisible by $n$

Hence, $R$ is transitive.

So, $R$ is an euivalence relation.

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.

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.

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 $A$

$h={(y, x): x=y-2}$

$h={(5,3),(6,4),(7,3)}$ which is the mapping from B to A.

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_1-x_2)=0 $

Since $x_1, x_2 \in N$, so $x_1+x_2=0$ is not possible.

$ \begin{matrix} \therefore & x_1-x_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 one-one 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 one-one.

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 y-1 \qquad \text{ if } y \text{ is even } \\ n=2 y \qquad \text{ if } y \text{ is odd or even } \\ \Rightarrow n=\begin{cases} 2 y-1 \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 pre-image

$ n=\begin{cases} 2 y-1 \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 one-one 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 \ne-2 \end{aligned} $

So, it is not one-one 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 one-one nor onto.

Solution

Here, $A \in R-{3}, B=R-{1}$

Given that $f: A \to B$ defined by $f(x)=\frac{x-2}{x-3} \forall x \in A$.

Let $x_1, x_2 \in f(x)$

$ \begin{aligned} & \therefore \quad f(x_1)=f(x_2) \\ & \Rightarrow \quad \frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3} \\ & \Rightarrow \quad (x_1-2)(x_2-3)=(x_2-2)(x_1-3) \\ & \Rightarrow \quad {\not x_1 \not x_2}-3 x_1-2 x_2+\not 6=\not x_1 \not x_2-3 x_2-2 x_1+\not 6 \\ & \Rightarrow \quad -x_1=-x_2 \Rightarrow x_1=x_2 \end{aligned} $

So, it is injective function.

Now, Let $\quad y=\frac{x-2}{x-3}$

$\Rightarrow x y-3 y=x-2 \Rightarrow x y-x=3 y-2$

$\Rightarrow x(y-1)=3 y-2 \Rightarrow x=\frac{3 y-2}{y-1}$

$ f(x)=\frac{x-2}{x-3}=\frac{\frac{3 y-2}{y-1}-2}{\frac{3 y-2}{y-1}-3} \Rightarrow \frac{3 y-2-2 y+2}{3 y-2-3 y+3} \Rightarrow y \\ \Rightarrow \quad f(x)=y \in B$

So, $f(x)$ is surjective function.

Hence, $f(x)$ is a bijective function.

Solution

(i) Given that $-1 \le x \le 1$

Let $x_1, x_2 \in f(x)$

So, $f(x)$ is one-one 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 \Rightarrow|x_1|=|x_2| \Rightarrow x_1= \pm x_2 \end{aligned} $

So, $g(x)$ is not one-one 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) & =x|x| \\ h(x_1) & =h f(x_2) \end{aligned} $

$ \Rightarrow \quad x_1|x_1|=x_2|x_2| \Rightarrow x_1=x_2 $

So, $h(x)$ is one-one function.

Now, let $h(x)=y=x|x|=x^{2}$ 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 one-one 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.

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.

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 $d+e=c+f$

$\Rightarrow(a+d)-(d+e)=(b+c)-(c+f)$

$\Rightarrow \quad a-e=b-f$

$\Rightarrow \quad a+f=b+e$

$\Rightarrow \quad (a, b) R(e, f)$

So, $R$ is transitive.

Hence, $R$ is an euivalence relation.

Euivalent class of $\{(2,5)\}$ is $\{(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)\}$

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.

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 x-3)^{2}+3(2 x-3)+1 \\ & =4 x^{2}+9-12 x+6 x-9+1=4 x^{2}-6 x+1 \\ & =2[x^{2}+3 x+1]-3 \\ & =2 x^{2}+6 x+2-3=2 x^{2}+6 x-1 \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) $gog \Rightarrow g[g(x)]=2[g(x)]-3=2(2 x-3)-3=4 x-6-3=4 x-9$

Solution

(i)

$ a * b=a-b \in \quad \forall a, b \in . $

So, $*$ is binary operation.

$a * b=a-b$ and $b * a=b-a \quad \forall a, b \in $

$ a-b \ne b-a $

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=(a-b)^{2} \in $, so $*$ is binary operation.

$a * b=(a-b)^{2}$ and $b * a=(b-a)^{2}$

$a * b=b * a \Rightarrow(a-b)^{2}=(b-a)^{2} \quad \forall a, b \in $.

So, $*$ is commutative.

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.

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).

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 $(b)$.

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 $(d)$

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 $(b)$.

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 $(b)$.

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 $(a)$.

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).

Solution

If $A$ and $B$ sets have $m$ and $n$ elements respectively, then the number of one-one 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).

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 !(n-2) !} \times 2 !=\frac{n(n-1)(n-2) !}{2 !(n-2) !} \times 2=n(n-1)=n^{2}-n $

Hence, the correct answer is $(d)$.

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 $(d)$.

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}+25-30 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 $(a)$.

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 one-one function.

Now, let $y=x+2 \therefore x=y-2 \in Z \quad \forall y \in Z$

So, $f(x)$ is onto function.

$\therefore f(x)$ is bijective function.

Hence, the correct answer is (b).

Solution

Given that

$ f(x)=x^{3}+5 $

Let $\quad y=x^{3}+5 \Rightarrow x^{3}=y-5$

$\therefore \quad x=(y-5)^{1 / 3} \Rightarrow f^{-1}(x)=(x-5)^{1 / 3}$

Hence, the correct answer is $(b)$.

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).

Solution

Given that $f(x)=\frac{3 x+2}{5 x-3} \forall x \ne \frac{3}{5}$

$ \begin{matrix} \text{ Let } & & y & =\frac{3 x+2}{5 x-3} \\ \Rightarrow & & y(5 x-3) & =3 x+2 \\ \Rightarrow & 5 x y-3 y & =3 x+2 \\ \Rightarrow & 5 x y-3 x & =3 y+2 \\ \Rightarrow & & x(5 y-3) & =3 y+2 \\ \Rightarrow & & x & =\frac{3 y+2}{5 y-3} \\ \Rightarrow & & f^{-1}(x) & =\frac{3 x+2}{5 x-3} \\ \Rightarrow & & f^{-1}(x) & =f(x) \end{matrix} $

Hence, the correct answer is (a).

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).

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+5-y=0 \\ & \Rightarrow \quad x=\frac{-(-4) \pm \sqrt{(-4)^{2}-4 \times 1 \times(5-y)}}{2 \times 1} \\ & =\frac{4 \pm \sqrt{16-20+4 y}}{2} \\ & =\frac{4 \pm \sqrt{4 y-4}}{2}=\frac{4 \pm 2 \sqrt{y-1}}{2}=2 \pm \sqrt{y-1} \end{aligned} $

$\therefore$ For real value of $x, y-1 \ge 0 \Rightarrow y \ge 1$.

So, the range is $[1, \infty)$.

Hence, the correct answer is $(b)$.

Solution

Here,

$ f(x)=\frac{2 x-1}{2} \text{ and } g(x)=x+2 $

$ \therefore \quad \begin{aligned} g \circ f(x) & =g[(f(x)] \\ & =f(x)+2 \\ & =\frac{2 x-1}{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 $(d)$.

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 $(a)$.

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).

Solution

Given that $a R b: 2 a+3 b=30$

$\begin{matrix} \Rightarrow & \quad 3 b=30-2 a \\ \Rightarrow \quad & \quad b=\frac{30-2 a}{3} \\ \text{ for } & a=3, b=8 \\ & a=6, b=6 \\ & a=9, b=4 \\ & a=12, b=2\end{matrix} $

Hence, $\quad R=\{(3,8),(6,6),(9,4),(12,2)\}$

Solution

Given that $A={1,2,3,4,5}$ and $R={(a, b):|a^{2}-b^{2}|<8}$

So, clearly, $\quad R=\{(1,1),(1,2),(2,1),(2,2),(2,3),(3,2),(4,3)$

$(3,4),(4,4),(5,5)\}$

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} $

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, $fofof(x)=\frac{x}{\sqrt{3 x^{2}+1}}$

Solution

Given that, $f(x)=[4-(x-7)^{3}]$

Let $ y=[4-(x-7)^{3}] $

$\Rightarrow$ $ (x-7)^{3}=4-y $

$\Rightarrow$ $ x-7=(4-y)^{1 / 3} \Rightarrow x=7+(4-y)^{1 / 3} $

Hence, $\quad f^{-1}(x)=7+(4-x)^{1 / 3}$

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’.

Solution

Given that: $f(x)=\sin (3 x+2) \forall x \in R$,

$f(x)$ is not one-one.

Hence, the statement is ‘False’.

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’.

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’.

Solution

Given that $A=[0,1]$

$f(2 n-1)=0$ and $f(2 n)=1 \quad \forall n \in N$

So, $f: N \to A$ is a onto function.

Hence, the statement is ‘True’.

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’.

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’.

Solution

Let $f(x)=2 x, g(x)=x-1$ and $h(x)=2 x+3$

$ \begin{aligned} & fo\{goh(x)\}=fo\{g(2 x+3)\} \\ & =f(2 x+3-1)=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+3-1)=f(2 x+2)=2(2 x+2)=4 x+4 \end{aligned} $

Hence, the statement is ‘True’.

Solution

Only bijective functions are invertible.

Hence, the statement is ‘False’.

Solution

’ + ’ is a binary operation on the set $N$ but it has no identity element.

Hence, the statement is ‘False’.