The range of the function is {x, y, z}.

This problem can be solved either by property of cardinality or by Venn diagram.
(i) Cardinality: Given that n(A) = 45% of 5000 = 2250
Similarly, n(B) = 1250, n(C) = 500, \(n(A\cap B)=250,n(B\cap C)=200,n(C\cap A)=200\) and \(n(A\cap B \cap C)=150.\)
The number of persons who knows only Language A is
\(n(A\cap B'\cap C')=n\{ A\cap (B\cup C)' \}\)
\(=n(A)-n\{ A\cap(B\cap C)\}\)
\(=n(A)-n(A\cap B)-n(A\cap B)-n(A\cap C)+n(A\cap B\cap C)\)
= 2250 - 250 - 200 + 150 = 1950
The required number of persons = 1950
(ii) Venn Diagram: Draw the Venn Diagrams using percentages.
From the diagram, the percentage of persons who knows only Language A is 39. Therefore, the required number of persons
\(=5000\times{39\over100}=1950\)

Let y = 3x -5.
\(\Rightarrow y+5=3x\Rightarrow \frac { y+5 }{ 3 } =x\)
Let g(y) = \(\frac { y+5 }{ 3 } \)
\(gof(x)=g(f(x))=g(3x-5)=\frac { 3x-5+5 }{ 3 } =\frac { 3x }{ 3 } =y\)
Also f o g(y) = f(g(y)) = \(f\left( \frac { y+5 }{ 3 } \right) =3\left( \frac { y+5 }{ 3 } \right) -5=y+5-5=y\)
Thus g o f = Ix and fog = Iy.
This implies that f and g are bijections and inverses to each other.
Hence f is a bijection and f^-1(y) = \(\frac { y+5 }{ 3 } \)
Replacing y by x we get, f^-1(x) = \(\frac { x+5 }{ 3 } \)

For any two distinct elements x₁, x₂ in \([0,{\pi\over 2}]\)
in x₁ = sin x₂ ⇒ x₁ = x₂
∴ f is one - one...(1)
Also cos x₁ = cos x₂ ⇒ x₁ = x₂
∴ g is also one-one...(2)
Now, (f + g) (x) = f(x) + g(x) = sin x + cos x
⇒ (f+g)(0) = sin0 + cos0 = 0+1=1
and \(f+g\left(\pi\over 2\right)=sin\frac{\pi}{2}+cos{\pi}{2}=1+0=1\)
(f+g)(0) = (f+g)\((\frac{\pi}{2})\) ⇒ 0 ≠ \(\frac{\pi}{2}\)
∴ (f + g) is not one-one...(3)
From (1), (2) and (3), we get f and g are one-one but (f +g) is not one-one.

Let x, y be any two elements of A.Then f(x) = f(y)
\(⇒\ \ {x-1\over x-2}={y-1\over y-2}\)
(x-1)(y-2) = (y-1)(x-2)

-2x - y + 2y + x = 0
-x + y = 0
x = y
∴ f(x) = f(y) ⇒ x = y for all x,y ∈ A
∴ f is one-one.
Let y be an arbitrary element of B.
Then f(x) = \(y⇒{x-1\over x-2}=y\)
⇒ (x - 1) = y (x - 2) ⇒ x-1 = xy - 2y
⇒ x - xy = 1 - 2y
⇒ x (1-y) = 1- 2y
\(⇒\ x={1-2y\over 1-y}\)
Clearly \(x={1-2y\over 1-y}\) is a real number for all y≠1. Also \({1-2y\over 1-y}≠2\)
Thus, every element y in B has is pre-image in A ∴ f is onto.
Hence,f is one-one and onto.

Given relation is 2a + 3b = 30 for all a, b \(\in \) N.
2a + 3b = 30 \(\Rightarrow\) 2a =30 - 3b
\(\Rightarrow a=\frac { 30-3b }{ 2 } \)

a	12	9	6	3
b	2	4	6	8

\(\therefore\) The list of ordered pairs are (12, 2) (9, 4) (6, 6) (3, 8)
Transitivity: Clearly R is not transitive.

Given \(f(x)=\sqrt{x+2}\)and \(g(x)=\sqrt{4-x^2}\) Clearly, f(x) is defined for all x, satisfying x + 2 > 0 ⇒x > -2 ⇒ x ∈[-2, ∞]
∴ Domain of f = [-2, ∞]
Also, g(x) is defined for all x satisfying 4 - x² ≥ 0 ⇒ x² - 4 < 0 ⇒ (x - 2) (x + 2) < 0
⇒ x ∈ [-2, 2]
∴ Domain of g = [-2, 2]
Given \(g(x)=\sqrt{4-x^2},\ x)=0⇒4-x^2=0⇒x=±2\)
So, domain of \(\left(f\over g\right)=[-2, 2] ⇒ -[-2, 2] = (-2, 2)\)
\(∴\left(f\over g\right):(-2, 2)⟶R\) is given by
\(\left(f\over g\right)(x)={f(x)\over g(X)}={\sqrt{x+2}\over\sqrt{4-x^2}}={1\over \sqrt{2-x}}\)

Given \(f(x)=\sqrt{x+2}\)and \(g(x)=\sqrt{4-x^2}\) Clearly,f(x) is defined for all x, satisfying x + 2>0 ⇒x > -2 ⇒ x ∈[-2, ∞)
∴ Domain of f = [-2, ∞)
Also, g(x) is defined for all x satisfying 4 - x² ≥ 0 ⇒ x² - 4 < 0 ⇒ (x - 2) (x + 2) < 0
⇒ x ∈ [-2, 2]
∴ Domain of g = [-2, 2]
(f - g) : [-2, 2] ⟶ R is given by \((f- g)(x) =f(x) - g(x) =\sqrt{x+2}+\sqrt{4-x^2}\)

Given \(f(x)=\sqrt{x+2}\)and \(g(x)=\sqrt{4-x^2}\) Clearly,f(x) is defined for all x, satisfying x + 2 > 0 ⇒x > -2 ⇒ x ∈[-2, ∞)
∴ Domain of f = [-2, ∞)
Also, g(x) is defined for all x satisfying 4 - x² ≥ 0 ⇒ x² - 4 < 0 ⇒ (x - 2) (x + 2) < 0
⇒ x ∈ [-2, 2]
∴ Domain of g = [-2, 2]
(f + g) : [-2, 2] ⟶ R is given by \((f+ g)(x) =f(x) + g(x) =\sqrt{x+2}+\sqrt{4-x^2}\)

The relation R on the set of all polygons is defined as R = {(P₁, P₂): P₁ and P₂ have same number of sides}
Reflexivity : Let P be any polygon in A. Then P and P have same number of sides.
⇒ (P, P) ∈ R
⇒ R is reflexive on A.
Symmetry : Let P₁ and P₂ be two polygons in A such that (P_1, P₂) ∈ R
(P₁, P₂) ∈ R ⇒ P₁ and P₂ have same number of sides.
⇒ P₂ and P₁ have same number of sides.
⇒ (P₂, P₁)∈ R
∴ R is symmetric on A
Transitivity : Let P₁, P₂, P₃ be three polygons in A such that (P₁, P₂) E Rand (P₂, P₃) ∈ R.
⇒ P₁ and P₂ have same number of sides and P2 and P3 have same number of sides.
⇒ P₁ and P₃ have same number of sides.
⇒ (P₁, P₃) ∈ R
∴ R is transitive.
Hence, R is an equivalence relation.

Given R = {(a, b) : |a - b| is a multiple of 4}
Reflexivity: Where a, b ∈ A = {0, 1, 2, ... 12}. For any a ∈ A, we have |a - a| = 0 which is a multiple of 4.
⇒ (a, a) ∈ A for all a ∈ A
∴ R is reflexive
Symmetry: Let (a, b) ∈ R. Then
(a, b) ∈ R
⇒ |a - b| is a multiple of 4.
⇒ |a - b| = 4⋋- for some ⋋∈N.
⇒ Ib - al = 4⋋- for some ⋋∈N.
⇒ (b, a) ∈N
∴ R is symmetricTransitivity: Let (a, b) ∈ Rand (b, c) ∈ R
Then (a, b) ∈ R and (b, c) ∈ R
⇒ |a - b| is a multiple of 4 and Ib - c| is a multiple of 4.
⇒ |a - b| = 4⋋ and |b - c| = 4μ for some ⋋ μ∈ N
⇒ a-b = ±4 -and b-c = ±4μ for some ⋋, μ ∈ N
⇒ a-c = ±4⋋ ±4μ for some ⋋, μ ∈ N
⇒ |a - c| is a multiple of 4.
⇒ (a-c)∈R
∴ R is transitive.
Hence, R is an equivalence relation.

B ⋂ C = {4, 6}
A\(B ⋂ C) = {0, 1, 2, 3}
A\B = {0,2}
A\B = {0, 1, 3}
(A\B) U (A\C)= {a, 1,2, 3}
From (1) and (2), A\(B ∩ C) = (A\B) U (A\C)
Venn diagram:

From (ii) and (v), we get
A\(B ∩ C) = (A\B) U (A\C)

Venn diagram

From (ii) and (iv), it is clear that A ∩ (B ∩ C) = (A ∩ B) ∩ C.

Given A = {a, b, c, d}, B = {a, c, e}, C = {a, e}
B∩C = {a, c}
A∩(B∩C) = {a}
A ∩ B = {a, e}
(A ∩ B) ∩ C = {a}
From (1) and (2), it is clear that A ∩ (B ∩ C) = (A ∩ B) ∩ C.

(i) y = sin (-x)

Let y = sin x.
Then sin(-x) is the reflection of the graph of sin x, about y-axis.
(ii) y = -sin(-x)

-sin(-x) is the reflection of the graph of sin(-x) about the x-axis.
(iii) \(y=sin\left( {\pi\over 2}+x\right)\)
Let y = sinx.
Then \(sin\left( {\pi\over 2}+n\right)\)causes the shift to the left for \(\pi\over 2\) unit to the sin x curve.
(iv) \(y=sin\left({\pi\over 2}-x \right)\)

Let y = sin x. Then \(sin\left( {\pi\over 2}-n\right)\) causes the shift to the left for \(\pi\over 2\) unit to the sin (-x) curve.

Given f(x) = 1.23x where x represents the number of American dollars.
and g(y) = 50.50y where y represents the number of Singapore dollars.

To convert American dollars to Indian rupees, we have to find out go f(x)
∴ go f(x) = g(f(x))
= g(1.23x)
= 50.50[1.23x]
= 62.115x
∴ The function for exchange rate of American dollars in terms of Indian rupee is g o f (x) = 62.115x.

\(\mathrm{f}(x)=\left\lfloor\begin{array}{lll} x & -x \mid, \mathrm{f}(x) \end{array}=\left\lfloor\begin{array}{ll} x & -x \end{array}\right.\right.\)
\(f(0) =0-0=0 \)
\(f(6.5) =6-6.5=|-0.5|=0.5 \)
\(f(-7.2) =8-7.2=0.8 \)
\(\therefore \text { Range is }[0,1)\)

\(\text { If } 4 \in X \text { then }(4,4) \notin \mathrm{R}\)
R is not reflexive
Symmetric can be easily checked

Let X = (a,b,c)
Then R = Universal relation
= {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)}.
It is transitive

\(\text { If } x \in(-\infty,-2), \text { Let } x=-3\)
\(\text { then } f(x)=|-5|+|1|=6=-2 x\)
\(\text { If } x \in(-2,-2), \text { Let } x=0 \text { then }\)
\(f(x)=|0-2|+|0+2|=4\)
\(\text { If } x \in(2, \infty), \text { Let } x=4\)
\(\text { then } f^{}(x)=|2|+|6|=8=2 x\)
\(\therefore \mathrm{f}(x)=\left\{\begin{array}{rl} -2 x & x \in(-\infty,-2] \\ 4 & x \in(-2,2] \\ 2 x & x \in(2, \infty) \end{array}\right.\)

\(\text { Since }\left|x^{2}+y^{2}\right|<2, x, y \text { must be } 0,1,-1 \text {. }\)

\(\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R} \text { is defined by }\)
\(\mathrm{f}(x)=1-|x|\)
\(\text { The range is }(-\infty, 1] \text { as } f(-\infty)=-\infty\)
\(f(0)=1\)
\(f(\infty) =-\infty\)

It is not a function since it has two images

It is clear that the curve can be obtained from that of y = sin x using translation and dilation. So first we draw y = sin x. From that it is easy to draw the curve y = sin (x - 1), then draw y = 2 sin (x - 1) and finally y = 2 sin (x - 1) + 3.

\(f(x)={1\over 2}x^2\) causes the graph of the function \(f(x)=x^2\) stretches towards x-axis since the multiplying factor is \({1\over 2}\) which is less than one. \(f(x)=2x^2\)causes the graph of the function \(f(x)=x^2\) compresses towards the y-axis that is, moves away from the x-axis since the multiplying factor is 2 which is greater than one.

For the curve \(f(x)=\sqrt{x},\) we have \(f(-x)=\sqrt{-x}\) and hence \(f(-x)=\sqrt{-x}\) where x < 0, is the reflection of \(f(x)=\sqrt{x}\) about y-axis.

We have \(f(x) =\frac { 1 }{ 2-sin\ 3x } \)
-1 ≤ sin 3x ≤ 1 for all x \(\in\) R
⇒ -1 ≤  - sin 3x ≤  1 for all x \(\in\) R
⇒ 1 ≤  2 - sin 3x ≤  3 for all x \(\in\) R
⇒ 2 - sin 3x ≠ 0 for all x \(\in\) R
⇒ f(x) = \(\frac { 1 }{ 2-sin\ 3x } \) is defined for all x \(\in\) R
Hence, domain (f) =R
Range of f: As discused above
1 ≤ 2 - sin 3x ≤ 3 for all x \(\in\) R
⇒ \(\frac { 1 }{ 3 } \le \frac { 1 }{ 2-sin\ 3x } \)-sin3x ≠ 0 for all x \(\in\) R
⇒ \(\frac { 1 }{ 3 } \)≤ f(x) ≤1  for all x \(\in\) R.
⇒ f(x) \(\in\) R [1/3, 1]
Hence, range (f) = [1/3, 1]

Clearly, g o f = {(1, 1), (2, 2), (3, 2)}. But f o g is not defined because the range of g = {1, 2, 4} is not contained in the domain of f = {1, 2, 3}.

To check whether compositions can be defined, let us find the domain and range of these functions.
Domain of f = {1, 2, 3}, Range of f = {2, 4}, Domain of g = {2, 3, 4} and Range of g = {1, 2}. Since the range of f is contained in the domain of g we can define g o f, so as to find the image of 1 under g o f, we first find the image of 1 under f and then its image under g. The image of 1 under f is 2 and its image under g is 1. So (g o f) (1) = g(f(1)) = g(2) = 1.
Similarly we find that (g o f) (2) = 1 and (g o f) (3) = 2. So g o f = {(1, 1), (2, 1), (3, 2)}.
Similarly f o g = {(2, 2), (3, 2), (4, 2)}.

If x < -3 or x > 3, then x² will be greater than 9 and hence 9 - x² will become negative which has no square root in R.
So x must lie on the interval [- 3, 3].
Also if \(x\ge-1\) or \(x\le 1,\) then x²-1 will become negative or zero. If it is negative, x² - 1 has no square root in R.  If it is zero, f is not defined. So, x must lie outside [- 1, 1]. 
That is x must lie on \(( -\infty,-1 ]\cup[1,\infty),\) Combining these two conditions, the largest possible domain for f is \([-3,3]\cap((-\infty, -1)\cup(1,\infty)).\)  That is \([-3,-1)\cup(1, 3].\)

Given w(x) = 0.35x
Since x represents the number of men, it will take only positive integers.
\(\therefore\)  W : W \(\rightarrow\)  R⁺
Hence the domain is the set of whole numbers. 

(i) Since f, g, h are functions from R \(\rightarrow\) R,
(f + g)o h: R \(\rightarrow\) R and f o h + g o h: R\(\rightarrow\) R. For any x \(\in \) R,
[(f + g)oh](x) = (f + g)(h(x) - f(h(x)) + g(h(x)) = f o h(x) + g o h(x)
(f + g)o h = f o h + g o h
(ii) Also fo(g + h) = f[(g + h)(x)] for any x \(\in \)R
= f[g(x) + h(x)] = f(g(x) + f(h(x)) = f o g(x) + f o h(x)
\(\therefore\) fo(g + h) = fog(x) + f o h(x).

Let S be the set of all triangles in a plane and let R be the relation on S defined by (Δ₁, Δ₂) ∈ R⇔ triangle Δ₁ is congruent to triangle Δ₂.
Reflexivity : For each triangle Δ ∈ S, we have Δ = Δ ⇒ (Δ, Δ) ∈ R for all Δ ∈ S.
⇒ R is reflexive on S.
Symmetry : Let Δ₁ Δ₂ ∈ S such that (Δ₁, Δ₂) ∈ R. Then (Δ₁ Δ₂) ∈ R ⇒ (Δ₁, Δ₂) ∈ R
∴ R is symmetric on S
Transitivity : Let Δ₁, Δ₂, Δ₃, ∈ S such that (Δ₁, Δ₂) ∈ R and (Δ₂, Δ₃) ∈ R
Then (Δ₁, Δ₂) ∈ Rand (Δ₂, Δ₃) ∈ R ⇒ (Δ₁, Δ₃) ∈ R.
∴ R is transitive on S.
Since R is reflexive, symmetric and transitive, R is an equivalence relation.

Given R is the set of all real numbers. Then R x R is the set of all ordered pairs (x, y) where x, y ∈ R.
R \(\times\)R = {(x,y): x, y ∈ R}
Clearly R \(\times\)R is the set of all points in xy-plane. Now R \(\times\)R \(\times\)R = {(x, y, z) : x,y, x ∈ R}.
∴ R \(\times\)R \(\times\)R represents the set of all points in space.

The Venn diagram of A∩B∩C is as follows.

The shaded region is A' U (A ∩ B) or (A\B)'

The shaded region is (A ∩ B) U (A ∩ B)

We have A = A ⋂ U
A = A ⋂ (B ⋂ B)'    [∵ B U B' =U]
A = (A ⋂ B) U (A ⋂ B')    [⋂ is distribute over union]
A = (A ⋂ B) U Φ    [given A ⋂ B' = Φ]
A = A ⋂ B
A ⊂ B. Hence proved

{x ∈ R: 0 < x < 1} in an infinite set since any interval has got infinite number of elements

Set of concentric circles in a plane is an infinite set since number of concentric circles in a plane is infinite

Given A = {0}  ...(1)
B = {x : x > 15 and x < 5} ⇒ B = Ф  ...(2)
C = {x : x-5 = 0} ⇒ B = {5}   ...(3)
D = { x : x² = 25} ⇒ D = {-5, 5}  ...(4)
E = {x : x is an integral positive root of x² - 2x - 15 = 0}
⇒ E = {5}  ...(5)
From (3) and (5), clearly C = E.
Hence C and E are equal sets

Range of cosine function is -1 \(\le \)cos x \(\le \) 1
\(\Rightarrow\) -2 \(\le \) 2 cos x \(\le \) 2        (Multiplied by 2)
\(\Rightarrow\) -2 -1 \(\le \) 2 cos x -1 \(\le \) 2-1
\(\Rightarrow\) -3 \(\le \) 2 cos x-1 \(\le \) 1
\(\Rightarrow \frac { -1 }{ 3 } >\frac { 1 }{ 2cosx-1 } >\frac { 1 }{ 1 } \)
\(\Rightarrow \frac { -1 }{ 3 } f(x)>1\)
\(\therefore \ Range \) \(=\left(-\infty,-\frac{1}{3}\right] \cup[1, \infty)\)

Let f = {(1, b) (2, b) (3, c) (4, e)}
Different elements in A does not have different images in B
∴ f is not one- one
Now, Co-domain = {a, b, e, d}, Range = {b, e}
Co-domain ≠ range
∴ f is not onto. Hence f is neither one - one and nor onto.

Let P be the set of all triangles in a plane R is defined as aRb if a is similar to b.
Let a, b, c ∊ P.

Reflexivity: aRa ⇒ a is similar to a for all a ∊ P.
∴ R is reflexive
Symmetricity: aRb ⇒ bRa
a is similar to b ⇒ b is similar to a for all a, b ∊ p.
Transitivity: aRb, and bRC ⇒ aRC.
a is similar to b and b is similar to C
⇒ a is similar to C.

Hence R is an equivalence relation.

Given f : R ⇾ R, defined by f(x) = cos x
We know f(0) = cos 0 = 1
and f(2π) = cos 2π = 1
∴ f(0) = f(2π) ⇒ 0 ≠ (2π)
∴ f is not one-one.
Since the values of cos x tie between -1 and 1, the range of f(x) is not equal to its co-domains.
∴ f is not onto.
Hence, f is neither one-one nor onto.

Let f and g denote the identity function and the modulus function.
Then f : R ⟶ R is defined as f(x) = x and g : R ⟶ R is defined as g(x) = |x|
g(x) = 0 ⇒ |x| = 0 ⇒ x =0
∴ The quotient of f by g is \({f\over g}:R-\{0\}⟶R\) and is defined as
 \(\left(f\over g \right)(x )={f(x)\over g(x)}={x\over |x|}=\begin{cases} {x\over x}=1\ if\ xx>0 \\{x\over -x}=-1\ if\ x<0 \end{cases}\)

Given n(p(A)) = 1024 = 2¹⁰
\(\Rightarrow\) n(A) = 10      [\(\therefore\) if n(A) = n, then n(p(A)) = 2ⁿ]
n(p(B)) = 32 = 2⁵
\(\Rightarrow\) n(B) = 5.
We know that,
n(A\(\cup\) B) = n(A) + n(B) - n(A\(\cap\)B)
\(\Rightarrow\) 15 = 10+5 - (A\(\cap\)B)
\(\Rightarrow\) n(A\(\cap\)B) = 0.

B-A = {4,5,6,7}
(B-A)\(\cup\)C = {3,4,5,6,7,9}....(1)
B\(\cup\)C = {3, 4, 5, 6, 7,9}
A-C = {1, 2}
(B\(\cup\)C) - (A-C) = {3,4,5,6,7,9} ....(2)
From (1) and (2), (B-A) \(\cup\) C = (B\(\cup\)C) - (A-C)
Hence verified

A = { x | x is a whole number less than or equal to 7 } is the set-bilder form of A.

We have f(x) = \(\frac { { x }^{ 2 }-9 }{ x-3 } \)
Domain of f : Clearly f(x) is not defined for x - 3 = 0 i.e. x = 3. Therefore, Domain (f) = R- {3}
Range of f: Let f(x) = y. Then,
f(x) = y ⇒ \(\frac { { x }^{ 2 }-9 }{ x-3 } \)=y ⇒ x+3 = y
it follows from the above relation that y takes all real values except 6 when x takes values in the ser R - {3}. Therefore, Range (f) = R {6}.

A = {1, 2, 3, 4, 5,6, 7, 8, 9}

(i) f( m) = f( n) \(\Rightarrow\) m² = n² \(\Rightarrow\) m = n since \(m,\ n\in N.\) Thus f is one-to-one. But, non-perfect square elements in the co-domain do not have pre-images and hence not onto.
(ii) Two different elements in the domain have same images and hence f is not one-to-one. Clearly the range of f is a proper subset of R. Thus it is not onto.

(i) If f(n) = f(m), then n + 2 = m + 2 and hence m = n. Thus f is one-to-one. As 1 has no pre-image, this function is not onto.
(ii) As above, this function is one-to-one. If m is in the co-domain, then m − 2 is in the domain and f(m − 2) = (m − 2) + 2 = m; thus m has a pre-image and hence this function is onto.

(i) \(\rho\) is not reflexive because (3, 3) is not in \(\rho\). As (1, 1) and (2, 2) are in \(\rho,\) it is enough to include the pair (3, 3) to \(\rho\) so as to make it reflexive.
(ii) \(\rho\) is not symmetric because (1, 2) is in p, but (2, 1) is not in \(\rho.\)  It is enough to include the pair (2, 1) to \(\rho\) so as to make it symmetric. It is enough to remove the pair (1, 2) from p so as to make it symmetric.
(iii) \(\rho\) is not transitive because (3, 1) and (1,3) are in,\(\rho\) but (3, 3) is not in \(\rho.\) To make it transitive we have to include (3, 3) in \(\rho\) . Even after including (3, 3), the relation is not transitive because (3, 1) and (I, 2) are in \(\rho\) , but (3, 2) is not in \(\rho\). To make it transitive we have to include (3, 2) also in \(\rho\) , Now it becomes transitive. So (3, 3) and (3, 2) are to be included into so as to make \(\rho\) transitive. But if we remove (3, 1) from \(\rho\), then it becomes transitive.
(iv) We have seen that
i) to make \(\rho\) reflexive, we have to include (3, 3);
ii) to make \(\rho\) symmetric, we have to include (2, 1);
iii) and to make \(\rho\) transitive, we have to include (3, 3) and (3, 2).
To make \(\rho\) as an equivalence relation we have to include all these pairs. So after including the pairs the relation becomes {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (3, 2)}.
But this relation is not symmetric because (3,2) is in the relation and (2,3) is not in the relation. So we have to include (2,3) also. Now the new relation becomes {(1, 1); (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (3, 2), (2, 3)}.
It can be seen that this relation is reflexive, symmetric and transitive and hence it is an equivalence relation. Thus we have to include (3, 3), (2, 1), (3, 2) and (2, 3) to \(\rho\) so as to make it an equivalence relation.

If A = {a, b, c} and f = {(a, c) (b, c) (c, b)} : (f : A \(\rightarrow\)A).
Given f : A ⇾ A

This is a function. Since different elements of A does not have different images in A.
∴ f is not one-one.
Here
Co-domain = {a, b, c}
But Range = {b, c}
f is not onto since co-domain ≠ Range.

Given f(1) = a, f(2) = b, f(3) = c
∴ f = {(1, a) (2, b) (3, c)}...(1)
Clearly f is a bijection and invertible
∴  f^-1 = {(a, 1) (b, 2) (c, 3)}
⇒ (f^-1)^-1 = {(1, a) (2, b) (3, c)}...(2)
From (1) and (2), it is clear that f = (f^-1)^-1

f(-1) = (-1)² and f(1) = 1²
f(-1) = f(1) but -1 ≠ 1
ஃ f is one-one
Also -1 in the co-domain R is not having any image in the domain R.
ஃ f is not onto.

Yes, since 4.7 is a point in between 3 and 5

From (ii) and (v), (AUB)' = A' ∩ B'

Given A⊂B

(i) From the diagram A ∩ B =A
(ii) A\B = Ø

Given P = {1, 2, 3, 5, 7, 11, 13}
and Q = {2, 4, 6, 8, 10, 12, 14}
Now P⋂Q = {2}
Since P⋂Q ≠ Ø, the given sets are not disjoint.

Given A = {5, 10, 15, 20, 25, 30}
B = {1, 3, 7, 10, 12, 15, 18, 25}
Now AUB = {1, 3, 5, 7, 10, 12, 15, 18, 20, 25, 30}