x² +y² = 25
y = 土\(\sqrt { 25-{ x }^{ 2 } } \)
x = 0 ⇒ y = 士5
y = 0 ⇒ x = 土5
(0, 5), (0, -5) ∈R
x = 3 ⇒ y = 土4
x = -3 ⇒ y = 土4
Domain of R = {0, 3, -3, -4, 4, -5, 5}
Domain of R^-1 {0, 3, -3, -4, 4, -5, 5}

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 R = {(a, b): a < b²} where a, b ∈ R
reflexivity: We know that \(\left(1\over 2\right)\le\left(1\over 2\right)^2\)is not true
\(⇒\ \left({1\over 2},{1\over 2}\right)∉R\)
⇒ R is not reflexive
Symmetry: We know that -1< 3² but 3 ≰ (-1)² is not true
⇒ (-1, 3) ∈ R but (3, -1)∉R
∴ R is not symmetric
Transitive: We observe that 2< (-3)² and -3< (1)² but 2 ≰ (1)² is not true
⇒ (2, -3) ∈ R and (-3, 1) ∈ R but (2, 1) ∉ R
⇒ R is not transitive
∴ R is neither reflexive nor symmetric nor transitive.

Given f(x) = 3x - 4
Let y = 3x - 4 ⇒ y + 4 = 3x
\(⇒ x={y+4\over 3}\)
Let g(y) = \(y+4\over 3\)
Now gof(n) = g(f(n)) = g(3\(\times\) -4) = \({3x-4+4\over 3}={3x\over 3}=x\)
and fog(y) = f(g(y)) = \(f\left(y+4\over 4\right)=3\left(y+4\over 3\right)-4=y+4-4=y\)
Thus, gof(x) = I_x and fog (y) = I_y
This implies that f and g are bijections and inverses to each other
Hence f is bijection and \(f^{-1} (x)={y+4\over 3}\)
Replacing y by x, we get f^-1 (x) = \(\frac { x+4 }{ 3 } \)

Hence, the graph of y = f^-1(x) is the reflection of the graph of f in y = x

Let A = {1, 2,3,4,8} = {5, 2,3,6}
A and B have two elements in common.
Number of elements common to A \(\times\) B and B x A = 2 \(\times\)2 = 2²
Similarly here we have 17² element in common

\(n[(A \times B) \cup(A \times C)]=n(A) \times n(B \cup C)\)
= 2 \(\times\) 3 = 6

It is onto not one-one

\(\text { Since } \sin 30^{\circ}=\frac{1}{2}\)
\(\sin 150^{\circ}=\frac{1}{2}\)

Given R = {(a, b) : a ≤ b} where a, b ∈ R.
Reflexivity : For any a ∈ R, a ≤ a
⇒ (a, a) ∈ R
⇒ R is reflexive
Symmetry : For 2≤3 ⇒ (2,3) ∈ R
but (3,2) ∉ R(ஃ 3 ≰ 2)
R is not symmetric.
Transitivity : Let (a,b) ∈ R and (b,c) ∈ R
⇒  Let (a,b) ∈ R and (b,c) ∈ R
⇒ a≤b and b≤c
⇒ a≤c ⇒ (a,c) ∈ R
ஃ R is transitive.
Hence, R is reflexive and transitive but not symmetric.

Given relation is aRb if a is not a sister of b.
Let a, b, C \(\in \)  A.
Reflexivity : aRa \(\Rightarrow\)  a is not a sister of a
\(\therefore\) R is reflexive.
Symmetricity: aRb \(\Rightarrow\)  bRa
a is not a sister of b \(\Rightarrow\) b is not a sister of a.
\(\therefore\) R is symmetric.
Transitivity : aRb and bRC \(\Rightarrow\) aRC
a is not a sister of b, b is not a sister of C [Eg : Mother is not a sister of daughter, daughter is not a sister of chithi, but mother is a sister of chithi.]
\(\Rightarrow\) a is a sister of C.
\(\therefore\) R is not transitive.
\(\therefore\) R is reflexive, symmetric and but not transitive.

Given relation is "aRb if a is not a sister of b". and a, b, c \(\in \) A.
Reflexivity: aRa \(\Rightarrow\) a is not a sister of a
\(\therefore\)  R is reflexive.         
Symmetric : aRb \(\Rightarrow\) bRa
a is not a sister of b \(\Rightarrow\) b is not a sister of a
\(\therefore\) R is not symmetric.
Transitivity: aRb and bRC \(\Rightarrow\) aRC
a is not a sister of b and b is not a sister of C. [Eg : Mother is not a sister of daughter, daughter is not a sister of chithi, but mother is a sister of chithi.]
\(\Rightarrow\) a is not a sister of C.
\(\therefore\)  R not is transitive.

Since A\(\times\) A contains 16 elements, then A must have 4 elements
\(\Rightarrow\) n(A) = 4.
The elements of A \(\times\) A are (1, 3) and (0, 2)
\(\therefore\) The possibilities of elements of A are {0, 1, 2, 3}

Let C = {x \(\in \) Z : x is even and < 10}
\(\Rightarrow\) C = {2, 4, 6, 8} 
\(\Rightarrow\) C is a finite set.

Let B = {x\(\in \)N : x is an odd prime number}
\(\Rightarrow\) B = {3, 5, 7, 11, .......}
\(\Rightarrow\) B is an infinite set.

Given that : f(x) =\(\frac { 1 }{ \sqrt { x-5 } } \)
Here, it is clear that/ex) is real when x - 5 > 0 ⇒ x > 5
Hence, the domain = \((5,\infty)\)
Now to find the range put
\(f(x)=y=\frac { 1 }{ \sqrt { x-5 } } \)
\(⇒ \sqrt { x-5 } =\frac { 1 }{ y } \Rightarrow x-5=\frac { 1 }{ { y }^{ 2 } } \)
\(⇒ x=\frac { 1 }{ { y }^{ 2 } } +5\)
For x \(\in \) \((5,\infty),\) y \(\in \) R⁺.
Hence, the range of f = R⁺

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.

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

We know that n(A\(\cup\) B) = n(A-B)+n(B-A)+n(A\(\cap\)B) if A and B are not disjoint.
\(\Rightarrow\) n(A-B)+n(B-A) = n(A\(\cup\)B) - n(A\(\cap\)B)
\(\Rightarrow\) n(A\(\Delta \)B) = 10-3
\(\Rightarrow\) \(\therefore\) n(A\(\Delta \)B) = 7
\(\therefore\) n[P(A\(\Delta \)B)] = 2⁷= 128