\(\text { Given } f(x)=4 x+3\)
\(\text { Let } f(x)=a \Rightarrow 4 x+3=a \Rightarrow x=\frac{a-3}{4}\)
\(\text { We define } g: R \rightarrow R \text { as } g(x)=\frac{x-3}{4}\)
\(\text { Now. } \ f o g(x)=f(g(x))=f\left(\frac{x-3}{4}\right)=4\left(\frac{x-3}{4}\right)+3=x\)
\(\text { and } \quad g o f(x)=g(f(x))=g(4 x+3)=\frac{4 x+3-3}{4}=x\)
\(\text { Hence, } \quad g(x)=\frac{x-3}{4}\)
or
\(\text { Given, } f(x)=4 x+3, \text { function } f: R \rightarrow R\)
\(\text { For one-one: Let for } x_{1}, x_{2} \in R\)
\(f\left(x_{1}\right) =f\left(x_{2}\right) \Rightarrow 4 x_{1}+3=4 x_{2}+3 \)
\(\Rightarrow 4 x_{1}=4 x_{2} \Rightarrow x_{1}=x_{2} \)
\(\text { As } f\left(x_{1}\right)= f\left(x_{2}\right) \Rightarrow x_{1}=x_{2}\)
Hence, one-one
For onto: Let for y \(\in\) R
There exists x \(\in\) R, such that y = f(x).
\(\Rightarrow y=4 x+3 \Rightarrow y-3=4 x \)
\(\Rightarrow x=\frac{y-3}{4} \in R\)
Hence, onto
\(\text {For inverse, } x=\frac{y-3}{4} \Rightarrow f^{-1}(y)=\frac{y-3}{4}\)
\(\text {So, } f^{-1}(x)=\frac{x-3}{4}\)

\(A=\{x=Z: 0 \leq \leq 12\}\) = {0, 1, 2,3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
R = {(a,b) : |a-b| is a multiple of 4}
For any element a ∈A, we have (a, a) ∈ R as |a - a| = 0 is a multiple of 4.
∴ R is reflexive.
Now, let (a, b) ∈ R ⇒ |a- b| is a multiple of 4.
⇒ |-(a-b)| = |b -a| is a multiple of 4
⇒ (b, a) ∈ R
∴ R is symmetric.
Now, let (a, b), (b, c) ∈ R.
⇒ |a - b| is multiple of 4 and |b - c| is a multiple of 4
⇒ (a-b) is a multiple of 4 and (b-c) is a multiple of 4
⇒ (a-c) = (a - b) + (b - c) is a multiple of 4
⇒ (a,c) ∈R
∴ R is transitive.
Hence, R is an equivalence relation.
The set of elements related to 1 is {1, 5, 9} since
|1 - 1| =  0 is a multiple of 4
|5 - 1| =  4 is a multiple of 4
|9 - 1| =  8 is a multiple of 4

Given, R = {(L_1, L₂): L₁ is parallel to L₂}.
Reflexive R is reflexive as any line L is parallel to itself,
i.e. (L, L) ∈ R.
Symmetric Now,let (L₁, L₂)∈R
⇒ L₁ is parallel to L₂ => L₂ is parallel to L₁
⇒ (L₂, L₁)∈R
So, R is symmetric.
Transitive Now, let (L₁, L₂), (L_2, L₃) ∈R
Then, L_I is parallel to L₂ and L₂ is parallel to L₃.
⇒ L₁ is parallel to L₃.
⇒ (L₁, L₃)∈R.
So, R is transitive.
Hence, R is an equivalence relation.
The set of all lines related to the line y = 2x + 4 is the
set of all lines that are parallel to the line y = 2x + 4.
Slope of line, y = 2x + 4 is m = 2.
It is known that parallel lines have the same slope.
So, the line parallel to the given line will be of the form
y = 2x + c, where c ∈R.
Hence, the set of all lines related to the given line is
given by y = 2x + c, where c ∈R.

We have a relation R in set A = {1, 2, 3, 4, 5} defined as
R = {(a, b): la - bl is divisible by 2}
Clearly, R = {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (5, 1), (5, 3), (5, 5)}
Reflexive For any a ∈ A, we have |a - a| = 0, which is divisible by 2.
⇒ (a, a) ∈ R, ∀ a ∈ A
Thus, R is reflexive.
Symmetric Let a, b∈A,such that(a, b)∈R
\( \Rightarrow |a-b| \) Iis divisible by 2
\( \Rightarrow |a-b| \) = \(2\lambda\) for some \( \lambda \epsilon N\) 
\(\Rightarrow |b-a|=2 \lambda \) for some \( \lambda \epsilon N\) \([\because|a-b|=|b-a|] \) 
Thus, R is symmetric
Transitive Let a, b, c ∈  A, such that (a, b) ∈ R and 
⇒ |a - b| is divisible by 2 and |b - c| is divisible by 2.
⇒ |a - b| = 2λand |b - c| = 2, for some λ μ∈N
⇒ (a-b) = ±2A and (b-c) = ±2μ
Now, |a-c| = |(a-b) + (b-c)| = |±2λ + (±2λ)1
= I± 2λ ± 2μ = 2|± λ ± μ| [some positive number]
⇒ |a-c|is divisible by 2 ⇒ (a, c) ∈ R
Thus, R is transitive.
Hence, R is an equivalence relation.
Now, [a] = {xe A : (a, x)e R}
ஃ Equivalence class [1] = {1, 3, 5}, [2] = {2, 4}, [3] = {1, 3, 5},
[4] = {2, 4} and [5] = {1, 3, 5}
Hence, [1] = [3] = [5] = {1, 3, 5} and [2] = [4] = {2, 4}

Note that the characteristic of sets {1, 4, 7}, {2, 5, 8} and {3, 6, 9} is that difference between any two elements of these sets is a multiple of 3.
Therefore, (x, y) \(\in\) R₁ \(\Rightarrow\) x – y is a multiple of 3 \(\Rightarrow\) {x, y} \(\subset\) {1, 4, 7} or {x, y} \(\subset\) {2, 5, 8} or {x, y} \(\subset\) {3, 6, 9} \(\Rightarrow\) (x, y) \(\in\) R₂.
Hence, R₁ \(\subset\) R₂.
Similarly, {x, y} \(\in\) R₂ Þ {x, y}\(\subset\) {1, 4, 7} or {x, y} \(\subset\) {2, 5, 8} or {x, y} \(\subset\) {3, 6, 9} \(\Rightarrow\) x – y is divisible by 3 \(\Rightarrow\) {x, y} \(\in\) R₁.
This shows that R₂ \(\subset\) R₁.
Hence, R₁ = R₂

It is given that A = {−1, 0, 1, 2}, B = {−4, −2, 0, 2}.
Also, it is given that f, g : A → B are defined by f(x) = x² − x, x ∈ A and \(g(x)=2\left|x-\frac{1}{2}\right|-1, x \in A\)
It is observed that:
\(f(-1)=\left(1^{2}\right)-(-1)=1+1=2 \)
\(g(-1)=2\left|(-1)-\frac{1}{2}\right|-1=2\left(\frac{3}{2}\right)-1=3-1=2 \)
⇒ f(-1) = g(-1)
f(0) = (0)^2 - 0 = 0
\(g(0)=2\left|0-\frac{1}{2}\right|-1=2\left(\frac{1}{2}\right)-1=1-1=0\)
⇒ f(0) = g(0)
f(1) = (1)^2 - 1 = 1 -   1 = 0
\(g(1)=2\left|a-\frac{1}{2}\right|-1=2\left(\frac{1}{2}\right)-1=1-1=0\)
⇒ f(1) = g(1)
f(2) = (2)^2 - 2 = 4 - 2 = 2
\(g(2)=2\left|2-\frac{1}{2}\right|-1=2\left(\frac{3}{2}\right)-1=3-1=2\)
⇒ f(2) = g(2)
f(a) = g(a) ∀ a ∈ A
Hence, the functions f and g are equal.

(fof)(x) = f(f(x)
\(=f\left( \frac { 4x+3 }{ 6x-4 } \right) \)
\(=\frac { 4\left( \frac { 4x+3 }{ 6x-4 } \right) +3 }{ 6\left( \frac { 4x+3 }{ 6x-4 } \right) -4 } =\frac { \left( \frac { 16x+12 }{ 6x-4 } \right) +3 }{ \left( \frac { 24x+18 }{ 6x-4 } \right) -4 } \)
\(=\frac { \left( \frac { 16x+12+3(6x-4) }{ 6x-4 } \right) }{ \left( \frac { 24x+18-4(6x-4) }{ 6x-4 } \right) } \)
\(=\frac { \left( \frac { 16x+12+18x-12 }{ 6x-4 } \right) }{ \left( \frac { 24x+18-24x+16 }{ 6x-4 } \right) } \)
\(=\frac { 34x }{ 34 } z\)
\(x\neq \frac { 2 }{ 3 } \)

Given a * b = min {a, b]
Operation table for * is 

Not reflexive, as (1, 1) \(\notin\) R.
Not symmetric, as (2, 3) \(\in\) R but (3, 2) \(\notin\) R.
Transitive, as relation R in a non empty set containing one element is transitive

We. have \(\left[\begin{array}{r}1 \\ 2 \\ -1\end{array}\right]\) Here,w. see that matrix has threerows and only one column. So, it is a column matrix.

We have:
\(f(x)=x^{ 2 }+2\) and \(g(x)=\frac { x }{ x-1 } \)
\(\therefore \) \(fog(x)=f(g(x))=(g(x))^{ 2 }+2\)
\(=\left( \frac { x }{ x-1 } \right) ^{ 2 }+2=\frac { x^{ 2 }+2(x-1)^{ 2 } }{ (x-1)^{ 2 } } \)
\(=\frac { x^{ 2 }+2(x^{ 2 }-2x+1) }{ (x-1)^{ 2 } } =\frac { 3x^{ 2 }-4x+2 }{ (x-1)^{ 2 } } \)
Hence, fog (2)\(=\frac { 3(4)-4(2)+2 }{ (2-1)^{ 2 } } =\frac { 12-8+2 }{ (1)^{ 2 } } =\frac { 6 }{ 1 } =6\)
And \(gof(x)=g(f(x))=\frac { f(x) }{ f(x)-1 } \)
\(=\frac { x^{ 2 }+2 }{ (x^{ 2 }+2)-1 } =\frac { x^{ 2 }+2 }{ x^{ 2 }+1 } \)
Hence, \(gof(-3)=\frac { (-3)^{ 2 }+2 }{ (-3)^{ 2 }+1 } =\frac { 9+2 }{ 9+1 } =\frac { 11 }{ 10 } \) .

Let \(f:R\rightarrow R\) , where \(X,Y\subseteq R.\).
Let \(y\in Y\) , arbitrarily.
By definition, \(y=10x+7\) for \(x\in X\)
\(\Rightarrow \) \(x=\frac { y-7 }{ 10 } \).
We define \(g:Y\rightarrow X\) by \(g(y)=\frac { y-7 }{ 10 } \) .
Now \((gof)\ (x)\ =g(f(x))=\frac { f(x)-7 }{ 10 } \)
\(=\frac { (10x+7)-7 }{ 10 } =x\)
and  (fog) (y) = f(g(y)) = 10g(y)+7
\(=10\left( \frac { y-7 }{ 10 } \right) +7=y\)
Thus \(gof=fog=I_{ R }\).
Hence, f is invertible and \(g:Y\rightarrow X\) such that
\(g(y)=\frac { y-7 }{ 10 } \)      

We have: f(1) = a, f(2) = b and f(3) = c.
Thus f = {(1,a), (2,b), (3,c)}.
Clearly, f is one-one \(\Rightarrow f\) is invertible
and \(1=f^{ -1 }(a),\ 2=f^{ -1 }(b),\)
\(3=f^{ -1 }(c)\)
\(\Rightarrow \) \(f^{ -1 }:\{ a,b,c\} \rightarrow \{ 1,2,3\} \).
Clearly, \(f^{ -1 }\) os one-one and onto
and \((f^{ -1 })^{ -1 }=\{ (1,a),\ (2,b),\ (3,c)\)
\(\left[ \because \ f^{ -1 }=\{ (a,1),\ (b,2),\ (c,3)\} \right] \)
\(\Rightarrow \) \((f^{ -1 })^{ -1 }=f\).

Let A = {1, 2, 3}.
(i) The relation R = {(2, 3), (3, 2)} is symmetric but neither reflexive nor transitive.
\(\left[ \because \ (1,1)\notin R;(2,3)\in R\ but\ (3,2)\in R\ but\ (2,2)\notin R \right] \)
(ii) The relation R = {(1, 3), (3, 2), (1, 2)} is transitive but neither reflexive nor symmetric
\(\left[ \because \ (1,1)\notin R;(1,3)\in R\ but\ (3,1)\notin R \right] \)
(iii) The relation R={(1,1), (2,2), (3,3), (2,3), (3,2), (1,2), (2,1)} is reflexive and symmetric but not transistive.
\(\left[ \because \ (1,1)\notin R;(2,3)\in R\ but\ (3,2)\in R\ but\ (2,2)\notin R \right] \)
(iv) The relation R = {(1, 1), (2, 2), (3, 3), (1, 2)} is reflexive and transitive but not symetric.
\(\left[ \because \ (1,2)\in R\ but\ (2,1)\notin R \right] \)
(v) The relation R = {(1, 2), (2, 1), (1, 1), (2, 2)} is symmetric and transitive but not reflexive. \(\left[ \because \ (3,3)\notin R \right] \)

Clearly, (x, y) R (x, y), ∀ (x, y) ∈ A, since xy = yx.
This shows that R is reflexive.
Further, (x, y) R (u, v) ⇒ xv = yu ⇒ uy = vx and hence (u, v) R (x, y).
This shows that R is symmetric. Similarly, (x, y) R (u, v) and (u, v) R (a, b) ⇒ xv = yu and 
\(u b=v a \Rightarrow x v \frac{a}{u}=y u \frac{a}{u} \Rightarrow x v \frac{b}{v}=y u \frac{a}{u} \Rightarrow x b=y a\) and hence (x, y) R (a, b).
Thus, R is transitive. Thus, R is an equivalence relation.

\(+:\ R\rightarrow R\) is given by :
\((x,y)\rightarrow x+y\)
 \(-:\ R\rightarrow R\)  is given by :
\((x,y)\rightarrow x-y\)
\(\times :\ R\rightarrow R\) is given by :
\((x,y)\rightarrow xy\)
Since '+' , '-' and '\(\times \)' are functions,
\(\therefore \) these are binary operations on R.
But \(\div :\ R\rightarrow R\)   is given by :
\((x,y)\rightarrow \frac { x }{ y } \) is not a function and consequently it is not a binary operation. \(\left[ \because \ For\ y=0,\frac { x }{ y } \ is\ not\ defined \right] \)
However, \(\div :R*\times R*\rightarrow R*\) given by:
\((x,y)\rightarrow \frac { x }{ y } \)
is a function and hence a binary operation in R*, where R* is the set of non-zero real numbers (i.e. R-{0}).

We have gof (2) = g (f (2)) = g(3) = 7, gof (3) = g(f (3)) = g(4) = 7, gof (4) = g(f (4)) = g(5) = 11 and gof (5) = g(5) = 11.

(a) R = {(x, y): x and y work at the same place}
⇒ (x, x) ∈ R
∴ R is reflexive.
If (x, y) ∈ R, then x and y work at the same place.
⇒ y and x work at the same place.
⇒ (y, x) ∈ R.
∴ R is symmetric.
Now, let (x, y), (y, z) ∈ R
⇒ x and y work at the same place and y and z work at the same place.
⇒ x and z work at the same place.
⇒ (x, z) ∈R
∴ R is transitive.
Hence, R is reflexive, symmetric, and transitive.
(b) R = {(x, y): x and y live in the same locality}
Clearly (x, x) ∈ R as x and x is the same human being.
∴ R is reflexive.
If (x, y) ∈R, then x and y live in the same locality.
⇒ y and x live in the same locality.
⇒ (y, x) ∈ R
∴ R is symmetric.
Now, let (x, y) ∈ R and (y, z) ∈ R.
⇒ x and y live in the same locality and y and z live in the same locality.
⇒ x and z live in the same locality.
⇒ (x, z) ∈ R
∴ R is transitive.
Hence, R is reflexive, symmetric, and transitive.
(c) R = {(x, y): x is exactly 7 cm taller than y}
Now,
(x, x) ∉ R
Since human being x cannot be taller than himself.
∴ R is not reflexive.
Now, let (x, y) ∈R.
⇒ x is exactly 7 cm taller than y.
Then, y is not taller than x.
∴ (y, x) ∉R
Indeed if x is exactly 7 cm taller than y, then y is exactly 7 cm shorter than x.
∴ R is not symmetric.
Now,
Let (x, y), (y, z) ∈ R.
⇒ x is exactly 7 cm taller thany and y is exactly 7 cm taller than z.
⇒ x is exactly 14 cm taller than z .
∴ (x, z) ∉R
∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
(d) R = {(x, y): x is the wife of y}
Now,
(x, x) ∉ R
Since x cannot be the wife of herself.
∴ R is not reflexive.
Now, let (x, y) ∈ R
⇒ x is the wife of y.
Clearly y is not the wife of x.
∴ (y, x) ∉ R
Indeed if x is the wife of y, then y is the husband of x.
∴ R is not symmetric.
Let (x, y), (y, z) ∈ R
⇒ x is the wife of y and y is the wife of z.
This case is not possible. Also, this does not imply that x is the wife of z.
∴ (x, z) ∉ R
∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
(e) R = {(x, y): x is the father of y}
(x, x) ∉ R
As x cannot be the father of himself.
∴ R is not reflexive.
Now, let (x, y) ∈R.
⇒ x is the father of y.
⇒ y cannot be the father of y.
Indeed, y is the son or the daughter of y.
∴(y, x) ∉ R
∴ R is not symmetric.
Now, let (x, y) ∈ R and (y, z) ∈ R.
⇒ x is the father of y and y is the father of z.
⇒ x is not the father of z.
Indeed x is the grandfather of z.
∴ (x, z) ∉ R
∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.

Let y = (x-3)³
Taking cube roots on both sides, we get
\((y)^{1 / 3} =\left((x-3)^{3}\right)^{1 / 3}\) 
\(\Rightarrow \ y^{1 / 3} =x-3 \Rightarrow x=3+y^{1 / 3}\) 
\(\therefore f^{-1}(y) =3+y^{1 / 3}\left[\because y=f(x) \Rightarrow x=f^{-1}(y)\right]\) 
\(\text { or } f^{-1}(x) =3+x^{1 / 3}\)

The given function is f: X → Y and relation on X is R={(a, b): f(a) = f (b)}
Reflexive Since, for every x ∈ X, we have
f'(x) = f(x)
⇒ (xx) ∈ R, ∀ ∈ X Therefore, R is reflexive.
Symmetric Let (x, y) ∈ R
Then, f(x)=f(y)
⇒ f(y)=f(x)
⇒ (y, x) ∈ R
Thus, (x, y) ∈ R ⇒  (y, x)∈ R, ∀x, y∈ X
Therefore, R is symmetric.
Transitive Let x, y, z∈ X such that
(x, y) ∈ R and (y, z) ∈ R
Given a relation 5 in \(N \times N\), defined as
(a, b) S(c, d), if a+d=b+c.
Reflexive Let (a, b) be any arbitrary element of \(N \times N\)
i.e. \((a, b) \in N \times N\), where \(a, b \in N\)
Now, as a+b=b+a
[∴ addition is commutative ]
Therefore \quad(a, b) S(a, b)
So, S is reflexive.
Symmetric \(\operatorname{Let}(a, b),(c, d) \in N \times N\), such that (a, b)
S(c, d). Then, a+d=b+c
\( \Rightarrow  b+c=a+d \Rightarrow c+b=d+a \)
\(\Rightarrow (c, d) S(a, b)\)
So, S is symmetric.
Transitive Let \((a, b),(c, d),(e, f) \in N \times N\) such that (a, b) S(c, d) and (c, d) S(e, f).
Then, a+d=b+c and c+f=d+e
On adding the above equations, we get
a+d+c+f=b+c+d+e 
\( \Rightarrow  a+f=b+e \Rightarrow(a, b) S(e, f)\)
So, S is transitive.
Thus, S is reflexive, symmetric and transitive. Hence, S is an equivalence

We have, R = {(l₁,l₂) :l₁ is parallel to l₂}
(i) If l₁ is parallel to l₂, then l₂ is parallel to l_1.
So, if (l₁, l₂) \(\in R,\) then (l₂, l₁) \(\in R\)
\(\therefore\) R is symmetric.
(ii) If l₁ is parallel to l₂ and l₂ is parallel to l₃, then l₁ is parallel to l₃.
So, if \(\left(l_1, l_2\right) \in R,\left(l_2, l_3\right) \in R\), then (l₁,l₃)\(\in R\)
\(\therefore\) R is transitive.
(iii) Let equation of line parallel to y = 3x + 2 be y = mx + c, where m is the slope of line. Since, y = 3x + 2 and y = mx + c are parallel. Slope of (y =3x + 2) = Slope of(y = mx + c)
\(\Rightarrow\) 3 = m i.e. m = 3
Hence, the required line is
y = 3x + c, where c \(\in R\)
Or
We have, S = {(I₁, I₂) : l₁ is perpendicular to l₂}
For Symmetric If I₁, is perpendicular to I₂, then l₂ is perpendicular to l₁.
So, if (l₁,l₂) \(\in S\), then (l₂, l₁) \(\in S\)
\(\therefore\) S is symmetric.
For Transitive If I₁, is perpendicular to l₂, and l₂, is perpendicular to l₃, then I₁, is not perpendicular to l₃, it is parallel to I₃.
So, if \(\left(l_1, l_2\right) \in S,\left(l_2, l_3\right) \in S \text {, then }\left(l_1, l_3\right) \notin S \text {. }\)
\(\therefore\) S is not transitive.

(i) (a) : For f(x) to be defined \(x-2 \neq 0\) i.e.,\(x \neq 2\) 
\(\therefore\) Domain of f = R - {2}
(ii) (b) : Let y =J(x), then \(y=\frac{x-1}{x-2}\) 
\(\Rightarrow x y-2 y=x-1 \Rightarrow x y-x=2 y-1 \Rightarrow x=\frac{2 y-1}{y-1}\) 
Since, \(x \in R-\{2\}\),therefore \(y \neq 1\)
Hence, range of f = R-{1} 
(iii) (d): We have,g(x) = 2f(x) - 1
\(=2\left(\frac{x-1}{x-2}\right)-1=\frac{2 x-2-x+2}{x-2}=\frac{x}{x-2}\)
(iv) (a) : We have, \(g(x)=\frac{x}{x-2}\) ,
Let  \(g\left(x_{1}\right)=g\left(x_{2}\right) \Rightarrow \frac{x_{1}}{x_{1}-2}=\frac{x_{2}}{x_{2}-2}\) 
 \(\Rightarrow x_{1} x_{2}-2 x_{1}=x_{1} x_{2}-2 x_{2} \Rightarrow 2 x_{1}=2 x_{2} \Rightarrow x_{1}=x_{2}\) 
Thus, \(g\left(x_{1}\right)=g\left(x_{2}\right) \Rightarrow x_{1}=x_{2}\) 
Hence, g(x) is one-one.
(v) (c)