Solution

Here,

$ R=\lbrace (a, a),(b, c),(a, b) \rbrace $

for reflexivity we have to add; $(b, b),(c, c)$ and for transitivity we have to add; $(a, c)$

Hence, the required 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 \quad -x^{2} \ge-25 $

$\Rightarrow \quad x^{2} \le 25 $

$\Rightarrow \quad -5 \le x \le 5$

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

Solution

Here, $f(x)=2 x+1$ and $g(x)=x^{2}-2$

$ \begin{aligned} \therefore \quad gof & =g[f(x)] =[2 x+1]^{2}-2 \\ \\
& =4 x^{2}+4 x+1-2 \\ \\ & =4 x^{2}+4 x-1 \end{aligned} $

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

Solution

$\text{Here,} \quad f(x)=2 x-3$

$\text{Let,} \quad f(x)=y=2 x-3 $

$ \begin{aligned} & \Rightarrow \quad \quad y+3=2 x \\ \\ & \Rightarrow \quad x=\dfrac{y+3}{2} \\ \\ & \therefore \quad f^{-1}(y)=\dfrac{y+3}{2} \text{ or } f^{-1}(x)=\dfrac{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 & =\lbrace (a, b),(b, d),(c, a),(d, c) \rbrace \\ \\ \text{then}\quad f^{-1} & =\lbrace (b, a),(d, b),(a, c),(c, d) \rbrace \end{aligned} $

Solution

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

$\begin{aligned} \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 \end{aligned} $

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

Solution

Yes, $g=\lbrace (1,1),(2,3),(3,5),(4,7) \rbrace $ is a function.

Here, $g(x)=\alpha x+\beta$

For $(1,1)$,

$g(1)=\alpha .1+\beta$

$1=\alpha+\beta$

For $(2,3), \quad g(2)=\alpha .2+\beta \quad \text{…(1)}$

$\hspace{2.2cm} 3=2 \alpha+\beta \quad \text{…(2)}$

Solving eq. (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 unique 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=\lbrace (2,5),(5,2),(1,5) \rbrace $

Solution

Here,

$ \begin{aligned} f(1) & =|1|=1 \\ \\ f(-1) & =|-1|=1 \\ \\ f(1) & =f(-1) \end{aligned} $

$ \begin{aligned} \text{ Here, } 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 $\quad \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2} \right ] \in f(x)$

$ \begin{aligned} & f \left (-\dfrac{\pi}{2} \right )=\cos \left (-\dfrac{\pi}{2} \right )=\cos \dfrac{\pi}{2}=0 \\ \\ & \cos \left (\dfrac{\pi}{2} \right )=\cos \dfrac{\pi}{2}=0 \\ \\ & f \left (-\dfrac{\pi}{2} \right )=f \left (\dfrac{\pi}{2} \right )=0 \end{aligned} $

But, $ -\dfrac{\pi}{2} \ne \dfrac{\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=\lbrace 1,2,3 \rbrace , Y=\lbrace 4,5 \rbrace $

$\therefore X \times Y=\lbrace (1,4),(1,5),(2,4),(2,5),(3,4),(3,5) \rbrace $

(i) $f=\lbrace (1,4),(1,5),(2,4),(3,5) \rbrace $

$f$ is not a function because there is no unique image of each element of domain under $f$.

(ii) $g=\lbrace (1,4),(2,4),(3,4) \rbrace $

Yes, $g$ is a function because each element of its domain has a unique image.

(iii) $h=\lbrace (1,4),(2,5),(3,5) \rbrace $

Yes, it is a function because each element of its domain has a unique image.

(iv) $k=\lbrace (1,4),(2,5) \rbrace $

Clearly $k$ is also a function.

Solution

Let $x_1, x_2 \in$ gof

$ gof\lbrace f(x_1) \rbrace =gof\lbrace f(x_2) \rbrace $

$ \Rightarrow \quad g(x_1)=g(x_2) \qquad[\because g \circ f=I_A] $

$ \therefore \qquad x_1=x_2$

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 $f(x)=\dfrac{1}{2-\cos x},\ \forall\ x \in R$.

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

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

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

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

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

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

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

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

$\Rightarrow \quad \quad-3 \le-\dfrac{1}{y} \le-1 $

$\Rightarrow \quad 3 \ge \dfrac{1}{y} \ge 1 $

$\Rightarrow \quad \dfrac{1}{3} \le y \le 1$

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

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 equivalence relation if it is reflexive, symmetric and transitive.

(i) Reflexive:

$a R a \Rightarrow \quad (a-a)=0$ 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)

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

$\Rightarrow \quad b-a$ is divisible by $n$

$\Rightarrow \quad b R a$

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 $n$

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

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

Hence, $R$ is transitive.

So, $R$ is an equivalence relation.

Solution

Given that $A=\lbrace 1,2,3,4 \rbrace $

$\therefore \quad ARA=\lbrace (1,1),(2,2),(3,3),(4,4),(1,2),(1,3),(1,4),(2,3)$,

$ (2,4),(3,4),(2,1),(3,1),(4,1),(3,2),(4,2),(4,3) \rbrace $

(a) Let $R_1=\lbrace (1,1),(2,2),(1,2),(2,3),(1,3) \rbrace $

So, $R_1$ is reflexive and transitive but not symmetric.

(b) Let $R_2=\lbrace (2,3),(3,2) \rbrace $

So, $R_2$ is only symmetric.

(c) Let $R_3=\lbrace (1,1),(1,2),(2,1),(2,4),(1,4) \rbrace $

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=\lbrace 1,2,3,4,5, \ldots, 20 \rbrace $

and $\quad $ Range $=\lbrace 39,37,35,33,31, \ldots, 1 \rbrace $

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=\lbrace 2,3,4 \rbrace $ and $B=\lbrace 2,5,6,7 \rbrace $

(i) Let $f: A \to B$ be the mapping from $A$ to $B$

$f=\lbrace (x, y): y=x+3 \rbrace $

$\therefore\quad f=\lbrace (2,5),(3,6),(4,7) \rbrace $ which is an injective mapping.

(ii) Let $g: A \to B$ be the mapping from $A \to B$ such that $g=\lbrace (2,5),(3,5),(4,2) \rbrace $ which is not an injective mapping.

(iii) Let $h: B \to A$ be the mapping from $B$ to $A$

$h=\lbrace (y, x): x=y-2 \rbrace $

$h=\lbrace (5,3),(6,4),(7,3) \rbrace $ 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 \quad x_1^{2}=x_2^{2} $

$\Rightarrow \quad x_1^{2}-x_2^{2}=0$

$ \Rightarrow \quad (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 \quad x_1=x_2 \\ \\ \therefore & f(x_1)=f(x_2) & \Rightarrow \quad x_1=x_2 \end{matrix} $

So, $f(x)$ is one to one function.

Now, Let $f(x)=5 \in N$

then $\quad x^{2}=5 \Rightarrow \quad x= \pm \sqrt{5} \notin N$

So, $f$ is not onto.

Hence, $f(x)=x^{2}$ is one-one but not onto.

(ii) Let the function $ f: N \rightarrow \quad N $, given by $ f(1)=f(2)=1 $

Here, $ f(x)=f(1)=1 $ and

$ \Rightarrow \quad f(x)=f(2)=1 $

Since, different elements 1,2 have same image 1 ,

$ \therefore f $ is not one-one.

Let $ f(x)=y $, such that $ y \in N $

Here, $ y $ is a natural number and for every $ y $, there is a value of $ x $ which is natural number.

Hence $ f $ is onto.

So, the function $ f: N \rightarrow \quad N $, given by $ f(1)=f(2)=1 $ 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 \quad x^{2}=-2 $

$\therefore \quad x= \pm \sqrt{-2} \notin R$

Which is not possible, so $f$ is not onto.

Hence, $f$ is neither one-one nor onto.

Solution

Here, $A \in R-\lbrace 3 \rbrace , B=R-\lbrace 1 \rbrace $

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

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

$ \begin{aligned} & \therefore \quad f(x_1)=f(x_2) \\ \\ & \Rightarrow \quad \quad \dfrac{x_1-2}{x_1-3}=\dfrac{x_2-2}{x_2-3} \\ \\ & \Rightarrow \quad \quad (x_1-2)(x_2-3)=(x_2-2)(x_1-3) \\ \\ & \Rightarrow \quad \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 \quad -x_1=-x_2 \Rightarrow \quad x_1=x_2 \end{aligned} $

So, it is injective function.

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

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

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

$\Rightarrow \quad x(y-1)=3 y-2 $

$\Rightarrow \quad x=\dfrac{3 y-2}{y-1}$

$ f(x)=\dfrac{x-2}{x-3}=\dfrac{\dfrac{3 y-2}{y-1}-2}{\dfrac{3 y-2}{y-1}-3} $

$\Rightarrow \quad \dfrac{3 y-2-2 y+2}{3 y-2-3 y+3} \Rightarrow \quad y$

$\Rightarrow \quad \quad f(x)=y \in B$

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.

$ f(x_1)=\dfrac{1}{x_1}\quad \text{ and }\quad f(x_2)=\dfrac{1}{x_2} $

$f(x_1)=f(x_2) $

$\Rightarrow \ \dfrac{1}{x_1}=\dfrac{1}{x_2} $

$ \Rightarrow \quad x_1=x_2$

Let, $ \quad f(x)=y=\dfrac{x}{2} $

$\Rightarrow \quad \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, $ \ g(x)=|x|$

$g(x_1)=g(x_2) $

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

$\Rightarrow \quad x_1= \pm x_2$

So, $g(x)$ is not one-one function.

Let, $\quad g(x)=y=|x|$

$ \Rightarrow \quad 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, $ \ h(x) =x|x| $

$h(x_1) =h f(x_2)$

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

$\Rightarrow \quad x_1=x_2 $

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

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

$\Rightarrow \quad \quad x= \pm \sqrt{-y} \notin A\ \forall\ y \in A$

$\therefore h(x)$ 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 \quad x_1= \pm x_2 $

So, $k(x)$ is not one-one function.

Now,

let, $ \ k(x)=y=x^{2} $

$\Rightarrow \quad x= \pm \sqrt{y}$

If $y=-1 \ \Rightarrow \quad x= \pm \sqrt{-1} \notin A\ \forall\ y \in A$

$\therefore \quad k(x)$ 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 $y \ngtr x\ \forall\ x, y \in N$

$\Rightarrow \quad x R y$ but $y$ Z $x$

So, it is not symmetric relation.

For transitivity, $\quad x R y, y R z $

$\Rightarrow \quad x R z\ \forall\ x, y, z \in N$

$ \Rightarrow \quad x>y, y>z $

$\Rightarrow \quad x>z $

So, it is transitive relation.

(ii) Here, $\quad R=\lbrace (x, y): x+y=10\ \forall\ x, y \in N \rbrace $

$R=\lbrace (1,9),(2,8),(3,7),(4,6),(5,5),(6,4),(7,3),(8,2),(9,1) \rbrace $

For reflexive: $5+5=10,5 R 5 $

$\Rightarrow \quad (x, x) \in R$

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, $R=\lbrace (x, y): x y$ is a square of an integer, $x, y \in N \rbrace $

For reflexive: $x R x=x . x=x^{2}$ is an integer

So, $R$ is reflexive.

$[\because\quad$ Square 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 $y R z \quad \Rightarrow \quad x R z$

Let,

$ \begin{aligned} & x y=k^{2}\quad \text{ and }\quad y z=m^{2} \\ \\ & x=\dfrac{k^{2}}{y} \quad \text{ and } \quad z=\dfrac{m^{2}}{y} \end{aligned} $

$\therefore \quad x z=\dfrac{k^{2} m^{2}}{y^{2}}$ which is again a square of an integer.

(iv) Here,

$ \begin{aligned} & R=\lbrace (x, y): x+4 y=10, x, y \in N \rbrace \\ \\ & R=\lbrace (2,2),(6,1) \rbrace \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: $(x, y) \in R \quad \text{ but } \quad (y, z) \notin R \quad \text{ and } \quad (x, z) \in R$

So, $R$ is not transitive.

Solution

Here,

$ A=\lbrace 1,2,3, \ldots, 9 \rbrace $

and $R \to A \times A$ defined by $(a, b) R(c, d)$

$\quad\Rightarrow \quad a+d=b+c$

$\forall\ (a, b),(c, d) \in A \times A$

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. $\quad c+b=d+a$

L.H.S. $=$ R.H.S. So, $R$ is symmetric.

For transitive: $(a, b) R(c, d)$ and $(c, d) R(e, f) \Leftrightarrow(a, b) R(e, f)$

$\Rightarrow \quad \quad a+d=b+c$ and $c+f=d+e$

$\Rightarrow \quad \quad a+d=b+c$ and $d+e=c+f$

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

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

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

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

So, $R$ is transitive.

Hence, $R$ is an equivalence relation.

Equivalent class of $\lbrace (2,5) \rbrace $ is $\lbrace (1,4),(2,5),(3,6),(4,7),(5,8),(6,9) \rbrace $

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) $\quad fog \quad \Rightarrow \quad f[g(x)]=[g(x)]^{2}+3[g(x)]+1$

(ii) $\quad gof \quad \Rightarrow \quad g[f(x)]=2[x^{2}+3 x+1]-3$

$\hspace{3.5cm} \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) $\quad f \circ f \Rightarrow \quad f[f(x)]=[f(x)]^{2}+3[f(x)]+1$

$\hspace{3.5cm} \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) $gog \quad \Rightarrow \quad g[g(x)]=2[g(x)]-3=2(2 x-3)-3=4 x-6-3=4 x-9$

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

So, $*$ is commutative.

Solution

(i): Given that

$a * b=1+a b \quad \forall\ a, b \in R$

$b * a=1+b a \quad \forall\ a, b \in R$

$a * b=b * a=1+a b$

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 $a R a \quad \Rightarrow \quad a \cong a$ 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 $b R c$.

$\Rightarrow \quad a \cong b$ and $b \cong a $

$\Rightarrow \quad a \cong c\ \forall\ a, b, c \in T$

So, $R$ is transitive.

Hence, $R$ is equivalence 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.

Solution

Here,

$a R b \quad \Rightarrow \quad a$ is a brother of $b$.

$a R a \quad \Rightarrow \quad a$ is a brother of $a$ which is not true.

So, $R$ is not reflexive.

$a R b \quad \Rightarrow \quad a$ is a brother of $b$.

$b R a \quad \Rightarrow \quad $ which is not true because $b$ may be sister of $a$.

$\qquad\quad \Rightarrow \quad a R b \ne b R a$

So, $R$ is not symmetric.

Now,

$a R b, b R c \quad \Rightarrow \quad a R c$

$\Rightarrow \quad a$ is the brother of $b$ and $b$ is the brother of $c$.

$\therefore\quad a$ 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.

Solution

Here,

$ A=\lbrace 1,2,3 \rbrace $

The number of equivalence relations are as follows:

$R_1=\lbrace (1,1),(1,2),(2,1),(2,3),(1,3) \rbrace $

$R_2=\lbrace (2,2),(1,3),(3,1),(3,2),(1,2) \rbrace $

$R_3=\lbrace (3,3),(1,2),(2,3),(1,3),(3,2) \rbrace $

Hence, correct answer is $(d)$

• Option (a) is incorrect because there is more than one equivalence relation possible on the set $ A = \lbrace 1, 2, 3 \rbrace $. 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 = \lbrace 1, 2, 3 \rbrace $. 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 = \lbrace 1, 2, 3 \rbrace $. The correct number of equivalence relations is determined by the Bell number for a set of size 3, which is 5.

Solution

Given that: $R=\lbrace (1,2) \rbrace $

$a$ $\not R $ $a$, so it is not reflexive.

$a \mathbb{R} b$ but $b \mathbb{R} a$, so it is not symmetric.

$a R b$ and $b R c \quad \Rightarrow \quad a R c$ 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 $\lbrace 1, 2, 3 \rbrace $, it must include the pairs $(1,1)$, $(2,2)$, and $(3,3)$. Since $ R = \lbrace (1,2) \rbrace $ 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.

Solution

Here, $a R b$ if $a \ge b$

$ a R a \quad \Rightarrow \quad a \ge a$ which is true, so it is reflexive.

Let, $a R b \quad \Rightarrow \quad a \ge b$, but $b \ngeq a$, so $b\not Ra$

$R$ is not symmetric.

Now, $a \ge b, b \ge c \quad \Rightarrow \quad a \ge c$ 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.

Solution

Given that: $R=\lbrace (1,1),(2,2),(3,3),(1,2),(2,3),(1,3) \rbrace $

Here, $1R1, 2R2$ and $3R3$, 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 $1 R 2 \quad \Rightarrow \quad 1 R 3$, 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 $.

Solution

Given that: $a * b=\dfrac{a b}{2}\ \forall\ a, b \in -\lbrace 0 \rbrace $

Let $e$ be the identity element

$ \therefore \quad a * e=\dfrac{a e}{2}=a $

$\Rightarrow \quad 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 = \dfrac{a \cdot 1}{2} = \dfrac{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 = \dfrac{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 = \dfrac{a \cdot 2}{2} = a $ for all $ a $. Therefore, this option is incorrect.

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

$ \begin{cases} n ! , \quad if\ m=n \ 0, \quad if\ m \ne n \ \end{cases} $

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

As, the number of surjections from $ A $ to $ B $ is equal to the number of functions from $ A $ to $ B $ minus the number of functions from $ A $ to $ B $ whose images are proper subsets of $ B $.

And, the number of functions from a set with $ n $ number of elements into a set with $ m $ number of elements $ =m^{n} $.

So, the number of surjections from $ A $ into $ B $ where,

$ A=\lbrace 1,2, \ldots, n \rbrace $ and $ B=\lbrace a, b \rbrace $ is $ 2^{n}-2\ $ (As, two functions can be many-one into functions )

Hence, Correct option is (b).

Solution

Given that $f(x)=\dfrac{1}{x}$

Put $x=0 \quad \therefore \quad f(x)=\dfrac{1}{0}=\infty$

So, $f(x)$ is not defined.

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

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

• (b) onto: The function $ f(x) = \dfrac{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 one-one and onto. Since $ f(x) = \dfrac{1}{x} $ is not defined at $ x = 0 $, it cannot be considered bijective over the entire domain $ \mathbb{R} $.

Solution

Here, $f(x)=3 x^{2}-5$ and $g(x)=\dfrac{x}{x^{2}+1}$

$ \begin{aligned} \therefore \quad g \circ f & =gof(x)=g[3 x^{2}-5] \\ \\ & =\dfrac{3 x^{2}-5}{(3 x^{2}-5)^{2}+1}=\dfrac{3 x^{2}-5}{9 x^{4}+25-30 x^{2}+1} \\ \\ \therefore \quad gof & =\dfrac{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 $9x^{4} - 30x^{2} + 26$.

Solution

Given that $f: Z \to Z$

Let $x_1, x_2 \in Z $

$\Rightarrow \quad f(x_1)=x_1+2, f(x_2)=x_2+2$

$\Rightarrow \quad f(x_1)=f(x_2) $

$\Rightarrow \quad x_1+2=x_2+2 $

$\Rightarrow \quad x_1=x_2$

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

Now, let $y=x+2 $

$\therefore \quad x=y-2 \in Z \quad \forall y \in Z$

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

$\therefore \quad f(x)$ 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 one-to-one 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 one-to-one and onto.

• (d) $ f(x) = x^2 + 1 $:

  • Reason: This function is not a bijection. It is not one-to-one 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.

Solution

Given that

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

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

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

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

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

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) $(5-x)^{1 / 3}$ 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) $5-x$ 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 $.

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.

Solution

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

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

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(\dfrac{3x + 2}{5x - 3}\right) = \dfrac{3\left(\dfrac{3x + 2}{5x - 3}\right) + 2}{5\left(\dfrac{3x + 2}{5x - 3}\right) - 3} $ Simplifying this expression does not yield $-x$. Therefore, $(f \circ f)(x) \neq -x$.

• Option (d): The statement $f^{-1}(x) = \dfrac{1}{19} f(x)$ is incorrect because we have already established that $f^{-1}(x) = f(x)$. Therefore, $f^{-1}(x)$ cannot be equal to $\dfrac{1}{19} f(x)$.

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.

Solution

Given that $f(x)=x^{2}-4 x+5$

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

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

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.

Solution

Here,

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

$ \therefore \quad \begin{aligned} g \circ f(x) & =g[f(x)] \\ \\ & =f(x)+2 \\ \\ & =\dfrac{2 x-1}{2}+2=\dfrac{2 x+3}{2} \\ \\ gof(\dfrac{3}{2}) & =\dfrac{2 \times \dfrac{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( \dfrac{3}{2} \right) $, the result is 3, not 1. The calculation involves substituting $ \dfrac{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( \dfrac{3}{2} \right) $ results in 3, not -1. The calculation does not produce a negative value.

• Option (c) $ \dfrac{7}{2} $: This is incorrect because the correct value of $ g \circ f \left( \dfrac{3}{2} \right) $ is 3, not $ \dfrac{7}{2} $. The calculation of the composed function does not yield $ \dfrac{7}{2} $.

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.

Solution

Given that $f(x)=\tan x$

Let $\quad f(x)=y=\tan x \Rightarrow \quad x=\tan ^{-1} y$

$\Rightarrow \quad \quad f^{-1}(x)=\tan ^{-1}(x)$

$\Rightarrow \quad \quad f^{-1}(1)=\tan ^{-1}(1)$

$\Rightarrow \quad \quad f^{-1}(1)=\tan ^{-1}\left[\tan \left(\dfrac{\pi}{4}\right)\right]=\dfrac{\pi}{4}$

Hence, the correct answer is (a).

• Option (b) $\lbrace n \pi+\dfrac{\pi}{4}: n \in Z \rbrace $ 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 $\dfrac{\pi}{4}$ 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 $\dfrac{\pi}{4}$.

• Option (d) “None of these” is incorrect because the correct answer is indeed provided in option (a), which is $\dfrac{\pi}{4}$.

Solution

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

$ \begin{matrix} &\Rightarrow \quad 3 b=30-2 a \\ \\ &\Rightarrow \quad b=\dfrac{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, $ R=\lbrace (3,8),(6,6),(9,4),(12,2) \rbrace $

Solution

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

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

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

Solution

Here, $f=\lbrace (1,2),(3,5),(4,1) \rbrace $ and $g=\lbrace (2,3),(5,1),(1,3) \rbrace $

$ \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 & =\lbrace (1,3),(3,1),(4,3) \rbrace \\ \\ 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 & =\lbrace (2,5),(5,2),(1,5) \rbrace \end{aligned} $

Solution

Here, $f(x)=\dfrac{x}{\sqrt{1+x^{2}}} \forall x \in R$

$ \begin{aligned} & fofof(x)=fof[f(x)]=f[f\lbrace f(x) \rbrace ] \\ \\ & =f\left[f(\dfrac{x}{\sqrt{1+x^{2}}})\right]=f\left[\dfrac{\dfrac{x}{\sqrt{1+x^{2}}}}{\sqrt{1+\dfrac{x^{2}}{1+x^{2}}}}\right] \\ \\ & =f\left[\dfrac{\dfrac{x}{\sqrt{1+x^{2}}}}{\dfrac{\sqrt{1+x^{2}+x^{2}}}{\sqrt{1+x^{2}}}}\right]=f\left[\dfrac{x}{\sqrt{1+2 x^{2}}}\right] \\ \\ & =\left[\dfrac{\dfrac{x}{\sqrt{1+2 x^{2}}}}{\sqrt{1+\dfrac{x^{2}}{1+2 x^{2}}}}\right]=\left[\dfrac{\dfrac{x}{\sqrt{1+2 x^{2}}}}{\dfrac{\sqrt{1+2 x^{2}+x^{2}}}{\sqrt{1+2 x^{2}}}}\right]=\dfrac{x}{\sqrt{1+3 x^{2}}} \end{aligned} $

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

Solution

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

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

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

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

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

Solution

Here, $\quad R=\lbrace (3,1),(1,3),(3,3) \rbrace $

$(3,3) \in R$, 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=\lbrace 1,2,3 \rbrace $

$ R=\lbrace (1,2),(2,1),(2,3),(1,3) \rbrace $

Here, $(1,2) \in R$ and $(2,1) \in R$, so $R$ is symmetric.

$(1,2) \in R,(2,3) \in R \Rightarrow \quad (1,3) \in R$, so $R$ is transitive.

But $(1,1) \notin R,(2,2) \notin R$ and $(3,3) \notin R$.

Hence, the statement is ‘False’.

Solution

$ R= \lbrace \text{(m,n)}$: m is an integral multiple of n \rbrace

Here, $\quad m=k n$

If $k=1 \quad m=n$, so $R$ is reflexive.

(where $k$ is an integer)

Clearly $R$ is not symmetric but $R$ 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\lbrace (1,1),(1,2),(2,1),(3,3) \rbrace $

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\lbrace goh(x) \rbrace =fo\lbrace g(2 x+3) \rbrace \\ \\ & =f(2 x+3-1)=f(2 x+2)=2(2 x+2)=4 x+4 \\ \\ \text{and} \quad & (fog)oh(x)=(fog)\lbrace h(x) \rbrace \\ \\ & =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’.