Very Important Previous Year Geography Difference between questions with Answers
Saturday, January 29, 2022
Saturday, January 22, 2022
10th samacheer kalvi maths chapter 1 Relations and Functions, Exercise 1.1 questions with answer
Example 1.1
If A = {1,3,5} and B = {2,3} then,
(i) find A x B and B x A.
(ii) Is AxB = BxA? If not why?
(iii) Show that n(BxA) = n(A) x n(B)
Solutions
Given, A = {1,3,5} and B = {2,3}
(i) We need to find AxB and BxA
so, A x B = {1,3,5} x {2,3}
= {(1,2),(1,3),(3,2),(3,3),(5,2),(5,3)} ---------(1)
B x A = {2,3} x {1,3,5}
= {(2,1),(2,3),(2,5),(3,1),(3,3),(3,5)} ---------(2)
(ii) We need to check AxB = BxA, since from equation (1) and (2),
AxB ≠ BxA
Since cross products are not commutative.
(iii) n(BxA) means number of elements in (BxA), similarly n(AxB) means number of elements in (AxB).
n(AxB) = 6 (from equation (1)) ----------(3)
n(BxA) = 6 (from equation (2)) ----------(4)
From equation (3) and (4),
n(AxB) = n(BxA)
Hence Verified
Example 1.2
If AxB = {(3,2),(3,4),(5,2),(5,4)} then find A and B
Solution
Given, AxB = {(3,2),(3,4),(5,2),(5,4)}
From this we can understand that first coordinate of sets gives us a value of A and second coordinate gives us the values of B.
Thus, A = {3,5} and B = {2,4}
Example 1.3
Let A = {x∈N|1<x<4}, B = {x ∈W| 0≤ x <2} and C = {x ∈N| x<3}
Then verify that
(i) A x(BUC) = (AxB) U (AxC)
(ii) Ax(B∩C) = (AxB)∩(AxC)
Solution
Given, A = {x∈N|1<x<4},
So, A = {2,3}
Given, B = {x ∈W| 0≤ x <2}
So, B = {0,1}
Given, C = {x ∈N| x<3}
So, C = {1,2}
(i)A x(BUC) = (AxB) U (AxC)
LHS = A x(BUC)
BUC = {0,1} U {1,2}
("U" means we need to join both the sets, if same number repeated in both its enough to write one time)
BUC = {0,1,2}
Ax(BUC) = {2,3} x {0,1,2}
Ax(BUC) = {(2,0),(2,1),(2,2),(3,0),(3,1),(3,2)} ------------(1)
RHS = (AxB) U (AxC)
(AxB) = {2,3} x {0,1}
= {(2,0),(2,1),(3,0),(3,1)}
(AxC) = {2,3} x {1,2}
= {(2,1),(2,2),(3,1),(3,2)}
(AxB) U (AxC) = {(2,0),(2,1),(3,0),(3,1)} U {(2,1),(2,2),(3,1),(3,2)}
("U" means we need to join both the sets, if same number repeated in both its enough to write one time)
(AxB) U (AxC) = {(2,0),(2,1),(2,2),(3,0),(3,1),(3,2)} ------------(2)
From equation (1) and (2)
LHS = RHS
Equation (1) = Equation (2)
A x(BUC) = (AxB) U (AxC)
Hence verified
(ii) Ax(B∩C) = (AxB)∩(AxC)
Solution
LHS = Ax(B∩C) (we need to solve brackets first)
B∩C = {0,1} ∩ {1,2}
"∩" means we need to select elements that is present in both sets
So, B∩C = {1} (Now we know both A and B∩C)
Ax(B∩C) = {2,3} x {1}
= {(2,1),(3,1)} ------------------(1)
RHS = (AxB)∩(AxC)
(AxB) = {2,3} x {0,1}
= {(2,0),(2,1),(3,0),(3,1)}
(AxC) = {2,3} x {1,2}
= {(2,1),(2,2),(3,1),(3,2)}
(AxB)∩(AxC) = {(2,0),(2,1),(3,0),(3,1)} ∩ {(2,1),(2,2),(3,1),(3,2)}
= {(2,1),(3,1)} ---------------------(2)
From equation (1) and (2)
LHS = RHS
Equation (1) = Equation (2)
Ax(B∩C) = (AxB)∩(AxC)
Hence Verified
Exercise 1.1
1.Find A x B, A x A and B x A
(i) A = {2,-2,3} and B = {1,-4}
Solution
A x B = {2,-2,3} x {1,-4}
= {(2,1),(2,-4),(-2,1),(-2,-4),(3,1),(3,-4)}
A x A = {2,-2,3} x {2,-2,3}
= {(2,2),(2,-2),(2,3),(-2,2),(-2,-2),(-2,3),(3,2),(3,-2),(3,3)}
B x A = {1,-4} x {2,-2,3}
= {(1,2),(1,-2),(1,3),(-4,2),(-4,-2),(-4,3)}
(ii) A = B = {p,q}
Solution
Since both A and B are same sets we get same answer for all questions
A x B = {p,q} x {p,q}
= {(p,p),(p,q),(q,p),(q,q)}
A x A = {(p,p),(p,q),(q,p),(q,q)}
B x A = {(p,p),(p,q),(q,p),(q,q)}
(iii) A = {m,n}; B = ЁЭЬЩ
Solution
Here B is a null set. That means that's equal to 0.
A x B = {m,n} x {} (Since B = ЁЭЬЩ, we can put like this )
= {} (Multiplication with null set will gives you a again a null set)
A x A = {m,n} x {m,n}
= {(m,m),(m,n),(n,m),(n,n)}
B x A = {} x {m,n}
= {} (Multiplication with null set will gives you a again a null set)
2.Let A = {1,2,3} and B = {x | x is a prime number less than 10}. Find A x B and B x A.
Solution
Given, A = {1,2,3}
B = {Prime numbers less than 10}
B = {2,3,5,7}
A x B = {1,2,3} x {2,3,5,7}
= {(1,2),(1,3),(1,5),(1,7),(2,2),(2,3),(2,5),(2,7),(3,2),(3,3),(3,5),(3,7)}
B x A = {2,3,5,7} x {1,2,3}
= {(2,1),(2,2),(2,3),(3,1),(3,2),(3,3),(5,1),(5,2),(5,3),(7,1),(7,2),(7,3)}
3.If B x A = {(-2,3),(-2,4),(0,3),(0,4),(3,3),(3,4)}, find A and B
Solution
Given, B x A = {(-2,3),(-2,4),(0,3),(0,4),(3,3),(3,4)}
From this we can understand that first coordinate of sets gives us a value of B and second coordinate gives us the values of A.
Thus, B = {-2,0,3}
A = {3,4}
4.If A = {5,6}, B = {4,5,6}, C = {5,6,7} show that A x A = (B x B) ∩ (C x C).
Solution
Given, A = {5,6}
B = {4,5,6}
C = {5,6,7}
To prove A x A = (B x B) ∩ (C x C)
LHS = A x A
A x A = {5,6} x {5,6}
= {(5,5),(5,6),(6,5),(6,6)} --------------(1)
RHS = (B x B) ∩ (C x C)
B x B = {4,5,6} x {4,5,6}
= {(4,4),(4,5),(4,6),(5,4),(5,5),(5,6),(6,4),(6,5),(6,6)}
C x C = {5,6,7} x {5,6,7}
= {(5,5),(5,6),(5,7),(6,5),(6,6),(6,7),(7,5),(7,6),(7,7)}
(B x B) ∩ (C x C) = {(4,4),(4,5),(4,6),(5,4),(5,5),(5,6),(6,4),(6,5),(6,6)} ∩ {(5,5),(5,6),(5,7),(6,5),(6,6),(6,7),(7,5),(7,6),(7,7)}
"∩" means we need to select elements that is present in both sets
= {(5,5),(5,6),(6,5),(6,6)} ----------------(2)
From equation (1) and (2)
LHS = RHS
Equation (1) = Equation (2)
A x A = (B x B) ∩ (C x C)
Hence Verified
5.Given A = {1,2,3}, B = {2,3,5}, C = {3,4} and D = {1,3,5}, Check if (A∩C) x (B∩D) = (AxB)∩(CxD) is true?
Solution
LHS = (A∩C) x (B∩D)
(A∩C) = {1,2,3} ∩ {3,4}
"∩" means we need to select elements that is present in both sets
= {3}
(B∩D) = {2,3,5} ∩ {1,3,5}
"∩" means we need to select elements that is present in both sets
= {3,5}
(A∩C) x (B∩D) = {3} x {3,5}
= {(3,3),(3,5)} ---------------------(1)
RHS = (AxB)∩(CxD)
(AxB) = {1,2,3} x {2,3,5}
= {(1,2),(1,3),(1,5),(2,2),(2,3),(2,5),(3,2),(3,3),(3,5)}
(CxD) = {3,4} x {1,3,5}
= {(3,1),(3,3),(3,5),(4,1),(4,3),(4,5)}
(AxB)∩(CxD) = {(1,2),(1,3),(1,5),(2,2),(2,3),(2,5),(3,2),(3,3),(3,5)} ∩ {(3,1),(3,3),(3,5),(4,1),(4,3),(4,5)}
"∩" means we need to select elements that is present in both sets
= {(3,3),(3,5)} -----------------------(2)
From equation (1) and (2)
LHS = RHS
Equation (1) = Equation (2)
(A∩C) x (B∩D) = (AxB)∩(CxD)
Hence Verified
6.Given A = {x∈W | x<2}, B = {x∈N |1<x≤ 4} and C = {3,5}. Verify that
(i) A x (BUC) = (AxB)U(AxC)
Solution
Given, A = {x∈W | x<2}
A = {0,1}
B = {x∈N |1<x≤4}
B = {2,3,4}
C = {3,5}
LHS = A x (BUC) (First we need to solve brackets)
BUC = {2,3,4} U {3,5}
("U" means we need to join both the sets, if same number repeated in both its enough to write one time)
= {2,3,4,5} (Now we know both A and BUC )
AxBUC = {0,1} x {2,3,4,5}
= {(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)} ------------------(1)
RHS = (AxB)U(AxC)
("U" means we need to join both the sets, if same number repeated in both its enough to write one time)
AxB = {0,1} x {2,3,4}
= {(0,2),(0,3),(0,4),(1,2),(1,3),(1,4)}
AxC = {0,1} x {3,5}
= {(0,3),(0,5),(1,3),(1,5)}
(AxB)U(AxC) = {(0,2),(0,3),(0,4),(1,2),(1,3),(1,4)} U {(0,3),(0,5),(1,3),(1,5)}
= {(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)} ------------------(2)
From equation (1) and (2)
LHS = RHS
Equation (1) = Equation (2)
A x (BUC) = (AxB)U(AxC)
Hence Verified
(ii) A x (B∩C) = (AxB)∩(AxC)
Solution
LHS = A x (B∩C)
(B∩C) = {2,3,4} ∩ {3,5}
"∩" means we need to select elements that is present in both sets
= {3}
A x (B∩C) = {0,1} x {3}
= {(0,3),(1,3)} --------------------(1)
RHS = (AxB)∩(AxC)
(AxB) = {0,1} x {2,3,4}
= {(0,2),(0,3),(0,4),(1,2),(1,3),(1,4)}
(AxC) = {0,1} x {3,5}
= {(0,3),(0,5),(1,3),(1,5)}
(AxB)∩(AxC) = {(0,2),(0,3),(0,4),(1,2),(1,3),(1,4)}∩ {(0,3),(0,5),(1,3),(1,5)}
"∩" means we need to select elements that is present in both sets
= {(0,3),(1,3)} ----------------(2)
From equation (1) and (2)
LHS = RHS
Equation (1) = Equation (2)
A x (B∩C) = (AxB)∩(AxC)
Hence Verified
(iii) (AUB)xC = (AxC)U(BxC)
Solution
LHS = (AUB)xC
(AUB) = {0,1} U {2,3,4}
("U" means we need to join both the sets, if same number repeated in both its enough to write one time)
= {0,1,2,3,4}
(AUB)xC = {0,1,2,3,4} x {3,5}
= {(0,3),(0,5),(1,3),(1,5),(2,3),(2,5),(3,3),(3,5),(4,3),(4,5)} ---------(1)
RHS = (AxC)U(BxC)
(AxC) = {0,1} x {3,5}
= {(0,3),(0,5),(1,3),(1,5)}
(BxC) = {2,3,4} x {3,5}
= {(2,3),(2,5),(3,3),(3,5),(4,3),(4,5)}
(AxC)U(BxC) = {(0,3),(0,5),(1,3),(1,5)} U {(2,3),(2,5),(3,3),(3,5),(4,3),(4,5)}
("U" means we need to join both the sets, if same number repeated in both its enough to write one time)
= {(0,3),(0,5),(1,3),(1,5),(2,3),(2,5),(3,3),(3,5),(4,3),(4,5)} ---------(2)
From equation (1) and (2)
LHS = RHS
Equation (1) = Equation (2)
(AUB)xC = (AxC)U(BxC)
Hence Verified
7.Let A = The set of all natural numbers less than 8, B = The set of all prime numbers less than 8, C = The set of even prime number. Verify that
(i)(A∩B)xC = (AxC) ∩ (BxC)
Solution
Given, A = The set of all natural numbers less than 8 (Natural numbers = 1 to infinity)
So, A = {1,2,3,4,5,6,7}
B = The set of all prime numbers less than 8 (Prime numbers are divided only by 1 and number itself)
So, B = {2,3,5,7}
C = The set of even prime number (The only even prime number is 2)
So, C = {2}
LHS = (A∩B)xC
(A∩B) = {1,2,3,4,5,6,7} ∩ {2,3,5,7}
"∩" means we need to select elements that is present in both sets
= {2,3,5,7}
(A∩B)xC = {2,3,5,7} x {2}
= {(2,2),(3,2),(5,2),(7,2)} ---------(1)
RHS = (AxC) ∩ (BxC)
(AxC) = {1,2,3,4,5,6,7} x {2}
= {(1,2),(2,2),(3,2),(4,2),(5,2),(6,2),(7,2)}
(BxC) = {2,3,5,7} x {2}
= {(2,2),(3,2),(5,2),(7,2)}
(AxC) ∩ (BxC) = {(1,2),(2,2),(3,2),(4,2),(5,2),(6,2),(7,2)} ∩ {(2,2),(3,2),(5,2),(7,2)}
"∩" means we need to select elements that is present in both sets
= {(2,2),(3,2),(5,2),(7,2)} ----------(2)
From equation (1) and (2)
LHS = RHS
Equation (1) = Equation (2)
(A∩B)xC = (AxC) ∩ (BxC)
Hence Verified
(ii) Ax(B-C) = (AxB) - (AxC)
Solution
Given, A = {1,2,3,4,5,6,7}
B = {2,3,5,7}
C = {2}
LHS = Ax(B-C)
(B-C) = {2,3,5,7} - {2}
("-" means we need to remove C elements from B)
= {3,5,7}
Ax(B-C) = {1,2,3,4,5,6,7} x {3,5,7}
= {(1,3),(1,5),(1,7),(2,3),(2,5),(2,7),(3,3),(3,5),(3,7),(4,3),(4,5),(4,7),(5,3),(5,5),(5,7),(6,3),(6,5),(6,7), (7,3),(7,5),(7,7)}----------------(1)
RHS = (AxB) - (AxC)
(AxB) = {1,2,3,4,5,6,7} x {2,3,5,7}
= {(1,2),(1,3),(1,5),(1,7),(2,2),(2,3),(2,5),(2,7),(3,2),(3,3),(3,5),(3,7),(4,2),(4,3),(4,5),(4,7),(5,2), (5,3),(5,5),(5,7),(6,2),(6,3),(6,5),(6,7),(7,2),(7,3),(7,5),(7,7)} (AxC) = {1,2,3,4,5,6,7} x {2}
= {(1,2),(2,2),(3,2),(4,2),(5,2),(6,2),(7,2)}
(AxB) - (AxC) = {(1,2),(1,3),(1,5),(1,7),(2,2),(2,3),(2,5),(2,7),(3,2),(3,3),(3,5),(3,7),(4,2),(4,3),(4,5), (4,7),(5,2),(5,3),(5,5),(5,7),(6,2),(6,3),(6,5),(6,7),(7,2),(7,3),(7,5),(7,7)} -
{(1,2),(2,2),(3,2),(4,2),(5,2),(6,2),(7,2)}
("-" means we need to remove C elements from B)
= {(1,3),(1,5),(1,7),(2,3),(2,5),(2,7),(3,3),(3,5),(3,7),(4,3),(4,5),(4,7),(5,3),(5,5),(5,7),(6,3),(6,5), (6,7), (7,3),(7,5),(7,7)}----------------(2)
From equation (1) and (2)
LHS = RHS
Equation (1) = Equation (2)
Ax(B-C) = (AxB) - (AxC)
Hence Verified
Subscribe to:
Comments (Atom)