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Since x – x = 0 is multiple of 2, (x, x) ∈ R

Therefore reflexive.

If y – x is a multiple of 2 then x – y is also a multiple of 2. Therefore (x, y) ∈ R ⇒ (y, x) ∈ R. Hence symmetric.

If y – x is a multiple of 2 and z-y is a multiple of 2, then their sum y – x + z – y = z – x is a multiple of 2.

Therefore (x, y), (y, z) ∈ R ⇒ (x, z) ∈ R

Hence transitive.

Therefore R is an equivalance relation.

According to the question,

P = {x: x < 3, x ∈N}, Q = {x : x ≤ 2, x ∈W} where W is the set of whole numbers

P = {1, 2}

Q = {0, 1, 2}

Now

(P∪Q) = {1, 2}∪{0, 1, 2} = {0, 1, 2}

And,

(P∩Q) = {1, 2}∩{0, 1, 2} = {1, 2}

We need to find the Cartesian product of (P∪Q) = {0, 1, 2} and (P∩Q) = {1, 2}

So,

(P∪Q) × (P∩Q) = {0, 1, 2} × {1, 2}

= {(0, 1), (0, 2), (1, 1), (1, 2), (2, 1), (2, 2)}

Hence, the Cartesian product = {(0, 1), (0, 2), (1, 1), (1, 2), (2, 1), (2, 2)}

We have, R₃ = {(x, |x)) | x is real number}

Clearly, domain of R₃ = R

Now, x ∈ R and |x| ≥ 0 .

Range of R₃ is [0,∞)

(i) R = {(4, 2) (6, 2) (8, 2) (6, 3) (9, 3) (8, 4)} 

(ii) Domain of R = {4, 6, 8, 9} 

(iii) Range of R = {2, 3, 4} 

(iv) R^-1 = \(\frac{1}{R}\) {(2, 4) (2, 6) (2, 8) (3, 6) (3, 9) (4, 8)}

According to the question,

A = {x: x ∈ W, x < 2}, B = {x : x ∈N, 1 < x < 5} C = {3, 5}; W is the set of whole numbers

A = {x: x ∈ W, x < 2} = {0, 1}

B = {x : x ∈N, 1 < x < 5} = {2, 3, 4}

(i)

(B∩C) = {2, 3, 4} ∩ {3, 5}

(B∩C) = {3}

A × (B∩C) = {0, 1} × {3} = {(0, 3), (1, 3)}

Hence, the Cartesian product = {(0, 3), (1, 3)}

(ii)

(B∪C) = {2, 3, 4} ∪ {3, 5}

(B∪C) = {2, 3, 4, 5}

A × (B∪C) = {0, 1} × {2, 3, 4, 5} = {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)}

Hence, the Cartesian product = {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)}

According to the question,

R₃ = {(x, |x|) |x is a real number} is a relation

Domain of R₃ consists of all the first elements of all the ordered pairs of R₃, i.e., x,

It is also given that x is a real number,

So, Domain of R₃ = R

Range of R contains all the second elements of all the ordered pairs of R₃, i.e., |x|

It is also given that x is a real number,

So, |x| = |R|

⇒ |x|≥0,

i.e., |x| has all positive real numbers including 0

Hence,

Range of R₃ = [0, ∞)

The given statement is true

Explanation:

Given A × B = {(a, x), (a, y), (b, x), (b, y)}

So

Set A will be set of first element of ordered pairs in A x B

So, A = {a, b}

And B will be set of first element of ordered pairs in A x B

So B = {x, y}

According to the question, A = {1, 2, 3, 4, 5}, S = {(x, y) : x ∈A, y ∈A}

(i) x + y = 5

So, we find the ordered pair such that x + y = 5, where x and y belongs to set A = {1, 2, 3, 4, 5},

1 + 1 = 2≠5

1 + 2 = 3≠5

1 + 3 = 4≠5

1 + 4 = 5⇒ the ordered pair is (1, 4)

1 + 5 = 6≠5

2 + 1 = 3≠5

2 + 2 = 4≠5

2 + 3 = 5⇒ the ordered pair is (2, 3)

2 + 4 = 6≠5

2 + 5 = 7≠5

3 + 1 = 4≠5

3 + 2 = 5⇒ the ordered pair is (3, 2)

3 + 3 = 6≠5

3 + 4 = 7≠5

3 + 5 = 8≠5

4 + 1 = 5⇒ the ordered pair is (4, 1)

4 + 2 = 6≠5

4 + 3 = 7≠5

4 + 4 = 8≠5

4 + 5 = 9≠5

5 + 1 = 6≠5

5 + 2 = 7≠5

5 + 3 = 8≠5

5 + 4 = 9≠5

5 + 5 = 10≠5

Therefore, the set of ordered pairs satisfying x + y = 5 = {(1,4), (2,3), (3,2), (4,1)}.

(ii) x + y < 5

So, we find the ordered pair such that x + y<5, where x and y belongs to set A = {1, 2, 3, 4, 5}

1 + 1 = 2<5 ⇒ the ordered pairs is (1, 1)

1 + 2 = 3<5 ⇒ the ordered pairs is (1, 2)

1 + 3 = 4<5 ⇒ the ordered pairs is (1, 3)

1 + 4 = 5

1 + 5 = 6>5

2 + 1 = 3<5 ⇒ the ordered pairs is (2, 1)

2 + 2 = 4<5 ⇒ the ordered pairs is (2, 2)

2 + 3 = 5

2 + 4 = 6>5

2 + 5 = 7>5

3 + 1 = 4<5 ⇒ the ordered pairs is (3, 1)

3 + 2 = 5

3 + 3 = 6>5

3 + 4 = 7>5

3 + 5 = 8>5

4 + 1 = 5

4 + 2 = 6>5

4 + 3 = 7>5

4 + 4 = 8>5

4 + 5 = 9>5

5 + 1 = 6>5

5 + 2 = 7>5

5 + 3 = 8>5

5 + 4 = 9>5

5 + 5 = 10>5

Therefore, the set of ordered pairs satisfying x + y< 5 = {(1,1), (1,2), (1,3), (2, 1), (2,2), (3,1)}.

(iii) x + y > 8

So, we find the ordered pair such that x + y>8, where x and y belongs to set A = {1, 2, 3, 4, 5}

1 + 1 = 2<8

1 + 2 = 3<8

1 + 3 = 4<8

1 + 4 = 5<8

1 + 5 = 6<8

2 + 1 = 3<8

2 + 2 = 4<8

2 + 3 = 5<8

2 + 4 = 6<8

2 + 5 = 7<8

3 + 1 = 4<8

3 + 2 = 5<8

3 + 3 = 6<8

3 + 4 = 7<8

3 + 5 = 8

4 + 1 = <8

4 + 2 = 6<8

4 + 3 = 7<8

4 + 4 = 8

4 + 5 = 9>8, so one of the ordered pairs is (4, 5)

5 + 1 = 6<8

5 + 2 = 7<8

5 + 3 = 8

5 + 4 = 9>8, so one of the ordered pairs is (5, 4)

5 + 5 = 10>8, so one of the ordered pairs is (5, 5)

Therefore, the set of ordered pairs satisfying x + y > 8 = {(4, 5), (5, 4), (5,5)}.

We have,

R₂ = {(x, y) | x and y are integers and x² + y² – 64}

So, we get,

x² = 0 and y² = 64 or x² = 64 and y² = 0

x = 0 and y = ±8 or x = ±8 and y = 0

Therefore, R₂ = {(0, 8), (0, –8), (8,0), (–8,0)}

According to the question,

R₁ = {(x, y) | y = 2x + 7, where x ∈R and – 5 ≤ x ≤ 5} is a relation

The domain of R₁ consists of all the first elements of all the ordered pairs of R₁, i.e., x,

It is also given – 5 ≤ x ≤ 5.

Therefore,

Domain of R₁ = [–5, 5]

The range of R contains all the second elements of all the ordered pairs of R₁, i.e., y

It is also given y = 2x + 7

Now x ∈ [–5,5]

Multiply LHS and RHS by 2,

We get,

2x ∈ [–10, 10]

Adding LHS and RHS with 7,

We get,

2x + 7 ∈ [–3, 17]

Or, y ∈ [–3, 17]

So,

Range of R₁ = [–3, 17]

The given statement is true.

Explanation:

Now

Cartesian product of set A = {1, 2, 3} and B = {3, 4} is

A×B = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}

Cartesian product of set A = {1, 2, 3} and C = {4, 5, 6} is

A×C = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

Now,

(A×B)∪(A×C) is union of set A×B and set A×C elements, so

(A×B)∪(A×C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}

This is the required Cartesian product.

We have, R₁ = {(x, y)|y = 2x + 7, where x∈ R and -5 ≤x ≤ 5}

Domain of R₁ = {-5 ≤ x ≤ 5, x ∈ R} = [-5, 5]

x ∈ [-5, 5]

=> 2x ∈ [-10,10]

=>2x + 7∈ [-3, 17]

Range is [-3, 17]

We have, R₂ = {(x, y) | x and y are integers and x² + y² = 64}

Clearly, x² = 0 and y² = 64 or x² = 64 and y² = 0

x = 0 and y = ±8

or, x = ±8 and y = 0

R = {(0, 8), (0, -8), (8,0), (-8,0)}

The given statement is false.

Explanation: 

The set P = {1, 2} is given

⇒ n(P) = 2

Now we need to find P×P×P,

So number of elements in P×P×P, will be

n(P×P×P) = n(P)×n(P)×n(P) = 2× 2× 2 = 8

But given P×P×P set has just 4 elements; hence it is not the set of P×P×P.

The set of P×P×P is

P×P×P = {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)}

We have, R = {(x,y):x,y∈ W, x² + y² = 25}

= {(0,5), (3,4), (4, 3), (5,0)}

Domain of R = Set of first element of ordered pairs in R = {0,3,4, 5}

Range of R = Set of second element of ordered pairs in R = {5,4, 3, 0}

The given statement is false.

Explanation:

given R = {(x, y) : y = x – 5, x, y ∈Z}

This means set R contains numbers such that y = x-5, so

So when x = 5, y becomes

y = x-5 = 5-5 = 0

So corresponding y will be 0,

So (5,2) does not belong to R.

It can be observed that,

x + y + z + w = 360º (Complete angle)

It is given that,

x + y = z + w

∴ x + y + x + y = 360º

2(x + y) = 360º

x + y = 180º

Since x and y form a linear pair, AOB is a line.

Given, A = {2, 4, 6, 9} and B = {4, 6, 18, 27, 54} and 

R = {(a, b): a ∈ A, b ∈ B, a is a 

factor of b and a<b} 

Roster form 

R = {(2, 4), (2, 6), (2, 18), (2, 54), (6, 18), (6, 54), (9, 18), (9, 27), (9, 54)} 

Domain of R = {2, 6, 9} 

Range of R = {4, 6, 18, 27, 54}

We have, A = {1,2, 3,4, 5}, S= {(x,y) : x ∈ A,y∈ A}

(i) The set of ordered pairs satisfying x + y= 5 is {(1,4), (2,3), (3,2), (4,1)}

(ii) The set of ordered pairs satisfying x+y < 5 is {(1,1), (1,2), (1,3), (2, 1), (2,2), (3,1)}

(iii) The set of ordered pairs satisfying x +y > 8 is {(4, 5), (5,4), (5, 5)}.

Since,  A = {2, 4, 6, 9}

B = {4, 6, 18, 27, 54},

we have to find a set of ordered pairs (a, b) such that a is factor of b and a < b.

Since, 2 is a factor of 4 and 2 < 4.

So, (2, 4) is one such ordered pair.

Similarly, (2, 6), (2, 18), (2, 54) are other such ordered pairs. Thus the required set of ordered pairs is

{(2, 4), (2, 6), (2, 18), (2, 54), (6, 18), (6, 54,), (9, 18), (9, 27), (9, 54)}.

Given R = {(x, y): 3x -y = 0, x, y ∈ A}

= {(1, 3), (2, 6), (3, 9), (4,12)}

Domain = {1, 2, 3, 4}

Co-domain = A Range = {3, 6, 9,12}

Radius of 1^st circle = 19 cm

Radius of 2^nd circle = 9 cm

Circumference of 1^st circle = 2π(19) = 38π cm

Circumference of 2^nd circle = 2π (9) = 18π cm

Let radius of required circle = R cm

Circumference of required circle = 2πR = c₁+c₂

2πR = 38π + 18π

2πR = 56π

R = 28 cms

Area of required circle = πr²

=22/7 x 28 x 28

=2464 cm²

We have R x R = {(x, y) : x, y ∈ R } which represents the coordinates of all the points in two dimensional space and R x R x R = {(x, y, z) x,y,z ∈ R } which represents the coordinates of all the points in three-dimensional space.

Grass eating animals that performs the rumination are called ruminants.

Maharshi Bharadwaj wrote “Yantra Sarvaswa’.

(i) To verify: A × (B ∩ C) = (A × B) ∩ (A × C)
We have B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Φ
∴ L.H.S. = A × (B ∩ C) = A × Φ = Φ
A × B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
∴ R.H.S. = (A × B) ∩ (A × C) = Φ
∴ L.H.S. = R.H.S
Hence, A × (B ∩ C) = (A × B) ∩ (A × C)
(ii) To verify: A × C is a subset of B × D
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
A × D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8),
(3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}
We can observe that all the elements of set A × C are the elements of set
B × D. Therefore, A × C is a subset of B × D.

The process of bringing back the cud in to the mouth in small lumps chews it again is called rumination.

Shakmuni wrote ‘Kaalganana’.

Partially digested food in the rumen of grass eating animals is called cud.

It was addressed as ‘Ajarbhar’ by the scholars.

A = {1, 2} and B = {3, 4}
∴ A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
⇒ n(A × B) = 4
We know that if C is a set with n(C) = m, then n[P(C)] = 2^m.
Therefore, the set A × B has 2⁴ = 16 subsets. These are
Φ, {(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)}, {(1, 3), (1, 4)}, {(1, 3), (2, 3)},
{(1, 3), (2, 4)}, {(1, 4), (2, 3)}, {(1, 4), (2, 4)}, {(2, 3), (2, 4)},
{(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1, 3), (2, 3),
(2, 4)}, {(1, 4), (2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3), (2, 4)}

A {1, 2, 3}, B = {3, 4}, C = {4, 5, 6}, B ∩ C = {4} 

(i) A × (B ∩ C) 

= {1, 2, 3} × {4} 

= {(1, 4), (2, 4), 3, 4)} 

(ii) (A × B) ∩ (A × C) 

= {(1, 2, 3} × {3, 4)} ∩ ({1, 2, 3} × {4, 5, 6}) 

= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)} 

∩ {(1, 4), 1, 5), (1, 6), (2, 4), (2, 5), 

(2, 6), (3, 4), (3, 5), (3, 6)} 

= {(1, 4), (2, 4), (3, 4)}

Given A = {1, 2} and B = {3, 4} 

A x B = {(1, 3), (1,4), (2, 3), (2, 4)} 

∴n (A x B) = 4 

Number of subsets of A x B = 24 =16 

Subsets of A x B are: A x B, φ, {(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)}, {(1, 3), (1, 4)}, {(1, 3), (2, 3)}, {(1, 3), (2,4)}, {(1,4), (2, 3)}, {(1, 4), (2, 4)} {(2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1,4), (2, 3), (2, 4)}, {(2, 3), (2,4), (1, 3)}.

(4) 3

A ∩ C = {a,b,c} ∩ {a, e, i, o, u}

= {a}

n(A ∩ C) = 1

n[(A ∩ C) x B] = n(A ∩ C) x n(B)

= 1 x 3

= 3

A x B = {(a, x) (a, y) (b, x) (b, y) (c, x) (c, y)}

A = {a, b, c}

B = {x,y}

A = {1,2,3}, B = {2,3,5}, C = {3,4} D = {1,3,5} 

A ∩ C = {1,2,3} ∩ {3,4} 

= {3} 

B ∩ D = {2,3, 5} ∩ {1,3,5} 

= {3,5} 

(A ∩ C) x (B ∩ D) = {3} x {3,5} 

= {(3, 3)(3, 5)} … (1) 

A x B = {1,2,3} × {2,3,5} 

= {(1,2) (1,3) (1,5) (2, 2) (2, 3) (2, 5) (3, 2) (3, 3) (3, 5)} 

C x D = {3,4} × {1,3,5} 

= {(3,1) (3, 3) (3, 5) (4,1) (4, 3) (4, 5)} 

(A x B) ∩ (C x D) = {(3, 3) (3, 5)} … (2)

From (1) and (2) we get 

(A ∩ C) x (B ∩ D) = (A x B) ∩ (C x D) 

This is true.

A = {5, 6}, B = {4, 5, 6}, C = {5, 6, 7} 

A x A = {(5, 5), (5, 6), (6, 5), (6, 6)} … (1) 

B x B = {(4, 4), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)} … (2) 

C x C = {(5, 5), (5, 6), (5, 7), (6, 5), (6, 6), (6, 7), (7, 5), (7, 6), (7, 7)} … (3) 

(B x B) ∩ (C x C) = {(5, 5), (5,6), (6, 5), (6,6)} … (4) 

(1) = (4)

A x A = (B x B) ∩ (C x C) 

It is proved.

A = {a, b} and B – {x, y}

A = {4,5,6}; B = {1,2} C = {0,1}

B ∩ C = {1,2} ∩ {0, 1}

= {1}

A x (B ∩ C) = {4,5,6} x {1}

= {(4, 1) (5, 1) (6, 1)} …. (1)

A x B = {4,5,6} x {1,2}

= {(4, 1) (4, 2) (5, 1) (5, 2) (6, 1) (6, 2)}

A x C = {4,5,6} x {0, 1}

= {(4,0) (4,1) (5,0) (5, 1) (6, 0) (6, 1)}

(A x B) ∩ (A x C) = {(4, 1) (5, 1) (6, 1)} … (2)

From (1) and (2) we get

A x (B ∩ C) = (A x B) ∩ (A x C)

(i) B∩C={4} 

A x (B∩C) = (1,4), (2, 4), (3,4)} 

(ii) A x B = {(1, 3), (1,4), (2, 3), (2, 4), (3, 3), (3, 4)} 

A x C = {(1, 4), (1, 5) (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)} 

(A x B)∩(A x C)= {(1, 4), (2, 4), (3, 4)} 

(iii) B ∪ C={3,4, 5, 6} 

∴ A x (B ∪ C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)} 

(iv) A x B = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)} 

A x C = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)} 

(A x B)∪(A x C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6) (3, 3), (3, 4), (3, 5), (3,6)}.

A = set of first elements = {p, m}

B = set of second elements = {q, r}

(i) B ∩ C = { } 

∴ A x (B ∩ C) = φ ………….. (1)

A x B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2,4)} 

A x C = {(1, 5), (1,6), (2, 5) (2,6)} 

∴ (A x B) ∩ (A x C) = φ ………………. (2) 

From (1) and (2), we get 

A x (B∩C) = (A x B) ∩(A x C) 

(ii) A x C = {(1, 5), (1,6), (2, 5), (2, 6)} 

B x D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}. 

Clearly every elements of A x C is an element of B x D. A x C ⊂B x D.

C) Gymnosperm

The correct answer is  Angiosperms.

Gills is the respiratory organs present in Pisces.

The correct answer is Sponges.