Given U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {1, 3, 5, 7, 9}
B = {2, 3, 5, 9, 10}
A' = {2, 4, 6, 8, 10}
B' = {1, 4, 6, 7, 8}
A'UB' = {1, 2, 4, 6, 7, 8, 10}

\(\left\{ x:\frac { x-4 }{ x+2 } =3,x\in R-\{ -2\} \right\} \)
Let D = \(\left\{ x:\frac { x-4 }{ x+2 } =3,x\in R-\{ -2\} \right\} \)
\(\Rightarrow\) D = {x: x - 4 = 3x + 6, x \(\in \) R}
\(\Rightarrow\) D = {x:-4-6 = 3x-x, x  ​​​​​​\(\in \) R}
\(\Rightarrow\) D = {x:2x = -10, x \(\in \) R}
\(\Rightarrow\) D = {x:x = -5, x\(\in \)R}
\(\Rightarrow\) D = {-5}

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

\(\mathrm{T}: x-y \text { is an integer } \Rightarrow x \mathrm{R} y\)
\(\text { i) } x-x=0 \text { is an integer }\)
\(\therefore \text { T is reflexive. }\)
\(\text { ii) }(x-y) \text { is an integer } \Rightarrow y-x \text { is also an integer }\)
\(\therefore \mathrm{T} \text { is symmetry }\)
iii) If (x - y)is an integer and y- z is also an integer, by adding
x - z is also an integer.
\(\therefore \mathrm{T} \text { is transitive }\)
Thus, T is an equivalence relation.
\(\mathrm{S}: \mathrm{y}=x+1 \Rightarrow x \mathrm{~S} y\)
\(\text { i) } x=x+1 \Rightarrow x S x \text { is not true. }\)
\(\therefore S \text { is not reflexive. }\)
Hence T is an equivalence relation but S is not an equivalence relation

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

It is onto not one-one

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

For any [-1, 0], we have
fog(x) = f(g(x)) = f([x]) = f( -1) = -1
gof(x) = g(f(x)) = g(-1) = [-1] = -1
⇒ fog(x) = gof(x) for all x ∈ [-1,0)
Hence fog and gof coincide on [-1, 0)

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.

(i) y = -x³

Let f(x) = x³
Since y = -f(x), this is the reflection of the graph off about the x-axis.
(ii)  y = x³+1

Let f(x) = x³
Since y = f(x) + 1, this is the graph of f(x) shifts to the upward for one unit.
(iii) y=x³-1

Let f(x) = x³
Since y = f(x) -1, this is the graph of (x) shifts to the downward for one unit.
(iv) y = (x + 1)³

Let f(x) = x³
y = (x + 1)³, causes the graph of f(x) shifts to the left for one unit.

The natural number.
A simple cipher is to assign a natural number to each alphabet.
This is, a is represented by 1, b is represented by 2, ....., z is represented by 26.
This correspondence can be written as the set of ordered pairs {(a, 1), (b, 2), (z, 26)}.
This set of ordered pairs of relation.
The domain of the relation is {a, b, ...., z} and the range is {1, 2, ..... 26 }.

As (a, a) \(\in\) R for all a \(\in\) S, R is reflexive.
There is no pair (a, b) in R such that (b, a) \(\notin\) R. In other words, for every pair (a, b) \(\in\) R, (b, a) is also in R. Thus R is symmetric.
We cannot find two pairs (a, b) and (b, c) in R, such that (a, c) \(\notin\) R. Thus the statement "R is not transitive" is not true; therefore, the statement "R is transitive" is true; hence R is transitive. Since R is reflexive, symmetric and transitive. Thus, this relation is an equivalence relation.

Clearly x ∈ A ∩ B
⇒ x ∈ A and x ∈ B
⇒ x = 3n and x = 4n, n ∈ Z
⇒ x is a multiple of 3 and x is a multiple of 4
⇒ x is a multiple of 3 and 4 both
⇒ x is a multiple of 12
⇒ x = 12n, n ∈ Z
Hence A ∩ B= {x : x = 12n, n∈Z}

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.

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.

Given A\(\times\)B = {(x, 1)(y, 2) (z, 1)}
Since n(A) = 3 and n(B) = 2,
A \(\times\) B will have 6 elements.
The remaining elements of A \(\times\) B will be (x, 2) (y, 1) (z, 2)
\(\therefore\) A\(\times\) B = {(x,1)(y,2)(z,1)(x,2)(y,1)(z,2)}
\(\therefore\) A = {x, y, z} and B = {1, 2}

Step 1 :
Draw the Graph of y = x²

Step 2 :
The graph of y =(x -1)², shifts to the right for one unit.
Step 3 :
The graph of y = 3 (x - 1)², compresses towards the Y-axis that is moves away from the X-axis since the multiplying factor is 3 which is greater than 1.
Step 4:
The graph of y = 3 (x - 1)²+ 5, causes the shift to the upward for 5 units.