Tamil Nadu Board Class 10 Solutions for Mathematics – Sets And Functions (English Medium)
Exercise 1.1
Solution 1:
In the above Venn diagram, the circle inside (shaded) represents set A which is a subset of set B represented by the bigger circle.
Hence, every element in set A belongs to B.
A ∪ B = {z | z ∈ A or z ∈ B}
Hence, A ∪ B = B.
Solution 2:
Given A ⊂ B.
From the diagram above, every element in set A (shaded small circle) belongs to set B (bigger circle).
A ∩ B = {z | z ∈ A and z ∈ B}
Hence, A ∩ B = A
As we know, A \ B = {x | x ∈ A and x ∉ B}
Hence, A \ B = ϕ
Solution 3:
i. P \ R = {a, b, c} \ {a, e, f, s}
P \ R = {b, c}
ii. Q ∩ R = {g, h, x, y} ∩ {a, e, f, s}
Q ∩ R = ϕ
iii. (P ∩ Q) = {a, b, c} ∩ {g, h, x, y}
= ϕ
R \ (P ∩ Q) = {a, e, f, s} \ ϕ
= {a, e, f, s}
Solution 4:
i. (B ∩ C) = {2,4,6} ∩ {1,2,3,4,5,6}
= {2,4,6}
A U (B ∩ C) = {4,6,7,8,9} U {2,4,6}
= {2, 4, 6, 7, 8, 9}
ii. (B U C) = {2,4,6} U {1,2,3,4,5,6}
= {1,2,3,4,5,6}
A ∩ (B UC) = {4,6,7,8,9} ∩ {1,2,3,4,5,6}
= {4,6}
iii. (C \ B) = {1,2,3,4,5,6} \ {2,4,6}
= {1,3,5}
A \ (C \ B) = {4,6,7,8,9} \ {1,3,5}
= {4,6,7,8,9}
Solution 5:
Commutative Property of set union is as given below.
A U B = B U A
A U B = {a, x, y, r, s} U {1,3,5,7, – 10}
= { a, x, y, r, s, 1,3,5,7, – 10} ……(1)
B U A = {1,3,5,7, – 10} U {a, x, y, r, s}
= { a, x, y, r, s, 1,3,5,7, – 10} …….(2)
Since, (1)=(2)
A U B = B U A. Hence proved.
Solution 6:
A ∩ B = {l, m, n, o, 2, 3, 4, 7} ∩ {2, 5, 3, – 2, m, n, o, p}
= {m,n,o,2,3} ……..(1)
B ∩ A = {2, 5, 3, – 2, m, n, o, p} ∩ {l, m, n, o, 2, 3, 4, 7}
= {m,n,o,2,3} ………(2)
Since, (1)=(2)
A ∩ B = B ∩ A. Hence proved.
Solution 7:
A = {2, 3, 7}
B = {6,7,8,9,10,11,12}
C = {1,4,5,6}
B U C = {6,7,8,9,10,11,12} U {1,4,5,6}
= {1,4,5,6, 7,8,9,10,11,12}
A U (B U C) = {2,3,7} U {1,4,5,6,7,8,9,10,11,12}
= {1,2,3,4,5,6, 7,8,9,10,11,12} ……(1)
A U B = {2,3,7} U {6,7,8,9,10,11,12}
= {2,3, 6,7,8,9,10,11,12}
(A U B) U C = {2,3, 6,7,8,9,10,11,12} U {1,4,5,6}
= {1,2,3,4,5,6, 7,8,9,10,11,12} .……(2)
Hence, From (1) and (2),
A U (B U C)= (A U B) U C
Solution 8:
P ∩ (Q ∩ R) = (P ∩ Q) ∩ R
(Q ∩ R) = {a,e,i,o,u} ∩ {a,c,e,g}
= {a,e}
P ∩ (Q ∩ R) = {a,b,c,d,e} ∩ {a,e}
= {a,e} ………….(1)
(P ∩ Q) = {a,b,c,d,e} ∩ {a,e, i,o,u}
= {a,e}
(P ∩ Q) ∩ R = {a,e} ∩ {a,c,e,g}
= {a,e} ………….(2)
Since, (1)=(2)
P ∩ (Q ∩ R) = (P ∩ Q) ∩ R
Solution 9:
(B \ C) = {6,10,12,18,24} \ {7,10,12,14,21,28}
= {6,18,24}
A \ (B \ C) = {5,10,15, 20} \ {6,18,24}
= {5,10,15,20} …………(1)
(A \ B) = {5,10,15, 20} \ {6,10,12,18,24}
= {5,15,20}
(A \ B) \ C = {5,15,20} \ {7,10,12,14,21,28}
= {5,15,20} ………….(2)
From (1) and (2), we get that,
A \ (B \ C) ≠ (A \ B) \ C
Solution 10:
(B \ C) = {- 2, – 1,0} \ {- 6, – 4, – 2}
= {-1,0}
A \ (B \ C) = {- 5, – 3, – 2, – 1} \ {-1,0}
= {-5,-3,-2} ……….(1)
(A \ B) = {- 5, – 3, – 2, – 1} \ {- 2, – 1,0}
= {-5,-3}
(A \ B) \ C = {-5,-3} \ {- 6, – 4, – 2}
= {-5,-3} ……….(2)
From (1) and (2), we conclude that,
A \ (B \ C) ≠ (A \ B) \ C
i.e Set difference operation is not associative.
Solution 11(i):
(B ∩ C) = {- 1,- 2, 3,4,5,6} ∩ {- 1, 2,3,4,5,7}
= {-1,3,4,5}
A U (B ∩ C) = {- 3,- 1, 0, 4,6,8,10} U {-1,3,4,5}
= {-3,-1,0, 3,4,5,6,8,10} ……….(1)
(A U B) = {- 3,- 1, 0, 4,6,8,10} U {- 1,- 2, 3,4,5,6}
= {- 3,-2,- 1, 0,3, 4,5,6,8,10}
(A U C) = {- 3,- 1, 0, 4,6,8,10} U {- 1, 2,3,4,5,7}
= {- 3,- 1, 0,2,3, 4,5,6,7,8,10}
(A U B) ∩ (A U C) = {- 3,-2,- 1, 0,3, 4,5,6,8,10} ∩ {- 3,- 1, 0,2,3,4,5,6,7,8,10}
= {-3,-1,0,3,4,5,6,8,10} ………(2)
Hence, from (1) and (2),
A U(B ∩ C) = (A U B) ∩ (A U C)
Solution 11(ii):
(B U C) = {- 1,- 2, 3,4,5,6} U {- 1, 2,3,4,5,7}
= {-1,-2,2,3,4,5,6,7}
A ∩ (B U C) = {- 3,- 1, 0, 4,6,8,10} ∩ {-1,- 2,2,3,4,5,6,7}
= {-1,4,6} …………(1)
(A ∩ B) = {- 3,- 1, 0, 4,6,8,10} ∩ {- 1,- 2, 3,4,5,6}
= {-1,4,6}
(A ∩ C) = {- 3,- 1, 0, 4,6,8,10} ∩ {- 1, 2,3,4,5,7}
= {- 1, 4}
(A ∩ B) U (A ∩ C) = {-1,4,6} U {- 1, 4}
= {-1,4,6} ………(2)
Hence, from (1) and (2),
A ∩ (B U C)= (A ∩ B)U(A ∩ C)
Solution 11(iii):
To show: A U (B ∩ C) = (A U B) ∩ (A U C)
Hence proved.
Solution 11(iv):
To show: A ∩ (B ∪ C)= (A ∩ B) U (A ∩ C)
Hence proved.
Solution 11(iv):
To show: A ∩ (B ∪ C)= (A ∩ B) U (A ∩ C)
Hence proved.
Exercise 1.2
Solution 1:
Solution 2:
Solution 3(i):
Solution 3(ii):
Solution 3(iii):
Solution 3(iv):
Solution 3(v):
Solution 3(vi):
Solution 3(vii):
Solution 4:
From Fig 3 and Fig 4 , we can say that
(A ∩ B) ∪ (A \ B)= A
Solution 5:
(A ∩ B)= {4,8,16,20,24,28}
(A ∪ B)’ will contain all elements other than that in (A ∪ B)
Hence, (A ∪ B)’ = {12}
OR
A’ = {4,12,20,28}
B’ = {8,12,24}
Using De Morgan’s law for complementation, we get
(A ∪ B)’ = A’ ∩ B’
= {4,12,20,28} ∩ {8,12,24}
(A ∪ B)’ = {12}
(A ∩ B) = {16}
Hence, (A ∩ B)’ will contain all elements other than that in (A ∩ B).
(A ∩ B)’ = {4,8,12,20,24,28}
OR
(A ∩ B)’ = A’ ∪ B’
= {4,12,20,28} ⋃ {8,12,24}
(A ∩ B)’ ={4,8,12,20,24,28}
Solution 6:
De Morgan’s laws for complementation, we get
(A ∪ B)’ = A’ ∩ B’ and
(A ∩ B)’ = A’ ∪ B’
(A ∪ B) = {a, b, f, g} ∪ {a, b, c}
= {a,b,c,f,g}
(A ∪ B)’ = {d,e,h} ……….(1)
(A ∩ B) = {a, b, f, g} ∩ {a, b, c}
(A ∩ B) = {a,b}
(A ∩ B)’ = {c,d,e,f,g,h} ………..(2)
A’ = {c,d,e,h}
B’ = {d,e,f,g,h}
A’ ∩ B’ = {c,d,e,h} ∩ {d,e,f,g,h}
= {d,e,h} ………(3)
A’ ∪ B’ = {c,d,e,h} ∪ {d,e,f,g,h}
= {c,d,e,f,g,h} …………(4)
From (1) and (3) , we can say that
(A ∪ B)’ = A’ ∩ B’
And From (2) and (4) , we can say that
(A ∩ B)’ = A’ ∪ B’
Hence, De Morgan’s laws of complementation are verified.
Solution 7:
De Morgan’s laws for set difference says that
For any three sets A,B and C, we have
A \ (B ∪ C)= (A \ B) ∩ (A \ C)
A \ (B ∩ C)= (A \ B) ∪ (A \ C)
Let us prove them
(B ∪ C) = {1, 2, 5, 7} ∪ {3,9,10,12,13}
= {1,2,3,5,7,9,10,12,13}
A \ (B ∪ C) = {11,15} ………..(1)
(A \ B) = {3,9,11,13,15}
(A \ C) = {1,5,7,11,15}
(A \ B) ∩ (A \ C) = {3,9,11,13,15} ∩ {1,5,7,11,15}
(A \ B) ∩ (A \ C)= {11,15} …………(2)
From (1) and (2) , we can say that
A \ (B ∪ C)= (A \ B) ∩ (A \ C)
(B ∩ C) = {1, 2, 5, 7} ∩ {3,9,10,12,13}
= Ф
A \ (B ∩ C)= {1, 3, 5, 7, 9,11,13,15}\ Ф
= {1, 3, 5, 7, 9,11,13,15} ………(3)
(A \ B) = {1, 3, 5, 7, 9,11,13,15} \ {1, 2, 5, 7}
= {3,9,11,13,15}
(A \ C) = {1, 3, 5, 7, 9,11,13,15} \ {3,9,10,12,13}
= {1,5,7,11,15}
(A \ B) ∪ (A \ C) = {3,9,11,13,15} ∪ {1,5,7,11,15}
= {1, 3, 5, 7, 9,11,13,15} ……….(4)
From (3) and (4) , we can say that
A \ (B ∩ C)= (A \ B) ∪ (A \ C)
Thus, we have verified De Morgan’s laws for set difference.
Solution 8:
(B ∩ C) = {1, 5, 10, 15, 20, 30} ∩ {7, 8, 15, 20, 35, 45, 48}
= {15,20}
A \ (B ∩ C) = {10, 15, 20, 25, 30, 35, 40, 45, 50} \ {15,20}
= {10,25,30,35,40,45,50} …….(1)
(A \ B) = {10, 15, 20, 25, 30, 35, 40, 45, 50}\{1, 5, 10, 15, 20, 30}
= {25,35,40,45,50}
(A \ C) = {10, 15, 20, 25, 30, 35, 40, 45, 50}\{7, 8, 15, 20, 35, 45, 48}
= {10,25,30,40,50}
(A \ B) ∪ (A \ C) = {25,35,40,45,50} ∪ {10,25,30,40,50}
= {10,25,30,35,40,45,50} ……..(2)
Hence from (1) and (2), we can say that
A \ (B ∩ C) = (A \ B) ∪ (A \ C)
Solution 9(i):
From Fig 2 and Fig 5,
A ∪ (B ∩ C)= (A ∪ B) ∩ (A ∪ C)
Solution 9(ii):
From Fig 2 and Fig 5,
A ∩ (B ∪ C)= (A ∩ B) ∪ (A ∩ C)
Solution 9(iii):
From Fig 2 and Fig 5,
(A ∪ B)’= A’ ∩ B’
Solution 9(iv):
From Fig 2 and Fig 5,
A \ (B ∪ C) = (A \ B) ∩ (A \ C)
Exercise 1.3
Solution 1:
Using the formula of Cardinality of sets, we get
n(A ∪ B)= n(A)+ n(B)- n(A ∩ B)
Hence, n(A ∪ B) = 200+300-100
= 400
Also, (A ∪ B)’ = (A’ ∩ B’) …….(1)
n(A ∪ B)’ = n(U)- n(A ∪ B)
= 700-400
n(A ∪ B)’= 300 ………(2)
Hence, n(A’ ∪ B’) = 300 (From (1) and (2))
Solution 2:
Using the formula of Cardinality of sets, we get
n(A ∪ B)= n(A)+ n(B)- n(A ∩ B)
Hence,
410 = 285 + 195 – n(A ∩ B)
n(A ∩ B) = 70
Also, (A ∩ B)’ = (A’ ∪ B’) ………(1)
n(A ∩ B)’ = n(U)- n(A ∩ B)
= 500-70
n(A ∩ B)’= 430 ………..(2)
Hence, n(A’ ∪ B’) = 430 (From (1) and (2))
Solution 3:
Using the formula of Cardinality of sets, we get
n(A ∪ B ∪ C)= n(A)+ n(B)+ n(C)- n(A ∩ B)- n(B ∩ C)- n(A ∩ C)+n(A ∩ B ∩ C)
= 17 + 17 + 17 – 7 – 6 – 5 + 2
n(A ∪ B ∪ C) = 35
Solution 4(i):
Given A = {4,5,6}, B = {5,6,7,8} and C = {6,7,8,9}
LHS = n(A ∪ B ∪ C)
= n[{4,5,6} ∪ {5,6,7,8} ∪ {6,7,8,9}]
= n[{4,5,6,7,8,9}]
LHS = 6 …………(1)
(A ∩ B) = {4,5,6} ∩ {5,6,7,8}
= {5,6}
(B ∩ C) = {5,6,7,8} ∩ {6,7,8,9}
= {6,7,8}
(A ∩ C) = {4,5,6} ∩ {6,7,8,9}
= {6}
(A ∩ B ∩ C) = {4,5,6} ∩ {5,6,7,8} ∩ {6,7,8,9}
= {6}
Hence
n(A) = 3, n(B) = 4, n(C) = 4, n(A ∩ B) = 2, n(B ∩ C) = 3, n(A ∩ C)= 1, n(A ∩ B ∩ C) = 1
RHS = n(A)+ n(B)+ n(C)- n(A ∩ B)- n(B ∩ C)- n(A ∩ C)+ n(A ∩ B ∩ C)
= 3 + 4 + 4 -2 – 3 – 1 + 1
RHS = 6 ……………(2)
Hence, From (1) and (2), we get
LHS=RHS
Hence proved that n(A ∪ B ∪ C) = n(A)+ n(B)+ n(C)- n(A ∩ B)- n(B ∩ C)- n(A ∩ C)+ n(A ∩ B ∩ C)
Solution 4(ii):
Given A = {a,b,c,d,e}, B = {x, y, z} and C = {a,e,x}
LHS = n(A ∪ B ∪ C)
= n[{a,b,c,d,e} ∪ {x,y, z} ∪ {a,e,x}]
= n[{a,b,c,d,e,x,y,z}]
LHS = 8 ……….(1)
(A ∩ B) = {a,b,c,d,e} ∩ {x,y, z}
= ϕ
(B ∩ C) = {x,y, z} ∩ {a,e,x}
= {x}
(A ∩ C) = {a,b,c,d,e} ∩ {a,e,x}
= {a,e}
(A ∩ B ∩ C) = {a,b,c,d,e} ∩ {x, y, z} ∩ {a,e,x}
= ϕ
Hence,
n(A) = 5, n(B) = 3, n(C) = 3, n(A ∩ B) = 0, n(B ∩ C) = 1, n(A ∩ C)= 2, n(A ∩ B ∩ C) = 0
RHS = n(A)+ n(B)+ n(C)- n(A ∩ B)- n(B ∩ C)- n(A ∩ C)+ n(A ∩ B ∩ C)
= 5 + 3 + 3 – 0 – 1 – 2 + 0
RHS = 8 …………(2)
Hence, From (1) and (2), we get
LHS=RHS
Hence proved that n(A ∪ B ∪ C) = n(A)+ n(B)+ n(C)- n(A ∩ B)- n(B ∩ C)- n(A ∩ C)+ n(A ∩ B ∩ C)
Solution 5:
Given : n(C) = 60, n(P) = 40, n(B) = 30, n(P ∩ C) = 15, n(P ∩ B) = 10, n(B ∩ C) = 5, n(P ∩ C ∩ B) = 0
We can get the following Venn diagram from the data,
n(P ∪ B ∪ C) = n(P)+ n(B)+ n(C)- n(P ∩ C)- n(P ∩ B)- n(B ∩ C)+ n(P ∩ B ∩ C)
= 40 +30+ 60 – 15 – 10 – 5 + 0
= 100
From the Venn diagram, the number of students who enrolled in atleast any one of the three subjects equals= 15+15+40+10+5+15= 100
Solution 6:
Let us consider there are 100 people in town. n(U) = 100
Hence, number of people speaking Tamil n (T)= 85
number of people speaking English n (E)= 40
number of people speaking Hindi n (H)= 20
Also, n(E ∩ T )= 32, n(T ∩ H)= 13, n(E ∩ H)= 10
To find: n(T ∩ E ∩ H)
Solution:
Let x be the number of people who can speak all the
three languages.
We get the following Venn diagram for the data
Also we know that,
n(T ∪ E ∪ H) = n(T)+ n(E)+ n(H)- n(T ∩ E)- n(E ∩ H)- n(T ∩ H)+ n(T ∩ E ∩ H)
100 = 85+40+20-32-13-10+x
x = 10
Hence, the percentage of people who can speak all the three languages is 10%.
Solution 7:
Given: n(U) = 170, n(T) = 115, n(R) = 110, n(M) = 130
Also, n(T ∩ M) = 85, n(T ∩ R) = 75, n(R ∩M)= 95 , n (R ∩ M ∩ T) = 70
We get the following Venn diagram for the data:
From the diagram above, we get the following:
(i) Number of people using only radio = 10
(ii) Number of people using only television = 25
(iii) Number of people using only television and magazine but not radio = 15
Solution 8:
Given: n(U) = 4000, n(F) = 2000, n(T) = 3000, n(H) = 500
n(F ∩ T) = 1500, n(F ∩ H)= 300, n(T ∩ H)=200, n(T ∩ H ∩ F)=50
We get the following Venn diagram for the data:
(i) Number of people who do not know any of the three languages is 450
(ii) Number of people who know atleast one of the three languages is (4000-450) = 3550
(iii) Number of people knowing only two languages is (1450+150+250)= 1850
Solution 9:
Given: n(U) = 120, n(F) = 93, n(K) =63, n(C) = 45,n(F ∩ K)= 45, n(K ∩ C) = 24, n(C ∩ F)= 27
We assume that all use at least one mode of cooking from the above given three modes.
Hence, n(U)=n(K ∪ C ∪ F)=120
To find: n(F ∩ K ∩ C)
Solution:
We know that,
n(K ∪ C ∪ F) = n(F)+ n(K)+ n(C)- n(F ∩ K)- n(K ∩ C) – n(F ∩ C)+ n(F ∩ K ∩C)
120 = 93+63+45-45-24-27+ n(F ∩ K ∩ C)
Hence,
n(F ∩ K ∩ C)= 15
15 families use firewood, kerosene and cooking gas.
Exercise 1.4
Solution 1:
A function f is meant to provide a unique output y for every input value of x.
In arrow diagram (i), 3 out of 4 elements of P have a unique image in Q. But one element ‘c’ does not have it. Hence it is not a function.
In arrow diagram (ii), every element in L has a unique image in M. Hence it is a function.
Solution 2:
Domain of F= {1,2,4,5,3}
Range of F = {3,5,7,9,1}
Solution 3:
(i) Both 12 and 14 have the same image 3 for the function.
Hence, the given function is not one-one.
The function is also not onto, since 0 ∈ B but there is no x ∈ A which is associated with 0 in B.
Hence, the function is neither one-one nor onto.
(ii) Range of f2={1}
The function f2: A ⟶B defined by f2(x) = 1 for every x ∈ A is a constant function.
(iii) Range of f3= {0,1,2,3,5} = B
f3: A⟶B is one-one and onto because f3 maps distinct elements of A into distinct images in B and every element in B is an image of some element in A.
Solution 4:
i. R1 = {(1,3),(3,5),(5,7)}
Hence, not every element in X has a unique image in Y. Hence it is not a function.
ii. R2 = { (1, 1), (2, 1), (3, 3), (4, 3), (5, 5) }
R2 is a function since every element of X has a unique image in Y.
Since both 1 and 2 of set X are related to 1 of set Y. The function is not one-one.
Range of R2= {1,3,5}. 7 and 9 are not images of any element in the domain. Hence, the function is not onto.
iii. R3 = { (1, 1), (1, 3), (3, 5), (3, 7), (5, 7) }
For the above relation, 1 is related to both 1 and 3 of set X. This violates the definition of a function since every element should be related to a unique element in a function. Hence, the given relation is not a function.
iv. R4 = { (1, 3), (2, 5), (4, 7), (5, 9), (3, 1) }
R4 is a function of the type one to one and onto. Since, every element in X has a unique image in Y, no two elements of X have the same image in Y and range of R4 ={3,5,7,9,1} = B
Solution 5:
Let the Identity function R be defined as A → A. The following diagram shows an identity function.
For identity function, every element is related to itself.
Hence, a = -2, b = -5, c = 8, d = -1
Solution 6:
From the given data, we get
f= {(-1, -0.5),(-1,-1),(1,1),(2,0.5)}
Range of f = {-0.5,-1, 1, 0.5}. Since, range has elements which are not in set A, it is not a function from A to A.
Solution 7:
Range of f = {7,4,9,6,2,3}= B
Hence, it is onto function. Also, every element of A is associated with unique element in B and no two element of A are associated with same element of B. Hence it is a one-to-one function too. Hence, it is both a one-to-one and onto function.
Solution 8:
Pre-images of 2 are 12 and 14.
Pre-images of 3 are 13 and 15.
Solution 9:
Given that, f(x) = 2x – 1
f(5)=2 × 5-1= 9 and
f(8) = 2 × 8-1= 15
Hence, a = 9 and b = 15
Solution 10:
i. Given that f = {( x,y) : y = 3 – 2x, x ∈ A, y ∈ B }
Hence the elements of f are , f = {(5,-7),(6,-9),(7,-11),(8,-13)}
ii. co-domain is B which is { -11, 4, 7, -10,-7, -9,-13}
iii. Range is the images of A in B present in the function = {-7,-9,-11,-13}
iv. It is a one-to-one function, since no element in B is associated with more than one element in A. It is not onto since the range set is not equal to the co-domain.
Solution 11:
(i) The given graph represents a function as a vertical line cuts the graph at one point A.
(ii) The given graph represents a function as a vertical line cuts the graph at one point X.
(iii) The given graph does not represent a function as a vertical line cuts the graph at two points A and B.
(iv) The given graph does not represent a function as a vertical line cuts the graph at two points P and Q.
(v) The given graph represents a function as a vertical line cuts the graph at one point A.
Solution 12:
Given that f = { (-1, 2), (- 3, 1), (-5, 6), (- 4, 3) }.
(i) Table form:
x | -1 | -3 | -5 | -4 |
f(x) | 2 | 1 | 6 | 3 |
(ii) Arrow diagram:
Solution 13:
From the given data , f(x) = {(6,1),(9,2),(15,4),(18,5),(21,6)}
(i) Arrow diagram:
(ii) Set of ordered pairs
f(x) = {(6,1),(9,2),(15,4),(18,5),(21,6)}
(iii) Table
x | 6 | 9 | 15 | 18 | 21 |
f(x) | 1 | 2 | 4 | 5 | 6 |
(iv) Graph
Solution 14:
(i) Arrow diagram
(ii) Set of ordered pairs
f(x) = {(4,3), (6,4), (8,5),(10,6)}
(iii) Table
x | 4 | 6 | 8 | 10 |
f(x) | 3 | 4 | 5 | 6 |
Solution 15:
[-3,7) = {x ∈ A: -3 ≤ x < 7}={-3,-2,-1,0,1,2,3,4,5,6}
(i) f(5) = 2(5)-3 = 7
f(6) = 2(6)-3 = 9
f(5)+ f(6) = 7+9=16
(ii) f(1) = 4(1)2-1 = 3
f(-3) = 4(-3)2-1 = 35
f(1)- f(-3) = 3 – 35 =-32
(iii) f(-2) =4(-2)2-1 = 15
f(4)= 3(4)-2 = 10
f(-2)- f(4)= 15-10 = 5
(iv) f(3) = 3(3)-2 = 7
f(-1) = 4(-1)2-1 = 3
f(6) = 2(6)-3 = 9
f(1) = 4(1)2-1 = 3
Solution 16:
(i) f(-4) = -4+5 = 1
f(2) = 2+5 = 7
2 f (- 4) + 3 f (2) = 2(1) + 3(7) = 23
(ii) f (- 7) = (-7)2+2(-7)+1= 49-14+1 = 36
f (- 3) = -3+5 = 2
f (- 7) – f (- 3) = 36 – 2 = 34
Exercise 1.5
Solution 1:
Answer : A
Solution 2:
Answer : C
Solution 3:
Answer : C
Solution 4:
A\B represents the set which has all elements in set A and not in set B.
A \ B = {p, q, r, s} \ {r, s, t, u} = {p, q}
Answer : A
Solution 5:
If n(A) = m
n[ p(A)] = 2m
Hence,
2m=64
m= 6
n(A) = 6
Answer : A
Solution 6:
Distributive law of sets.
A ∩(B ∪ C) = (A ∩ B ) ∪ (A ∩ C)
Answer : B
Solution 7:
Both are disjoint sets. Hence, there intersection is an empty set.
Answer : A
Solution 8:
A \ B is the set of elements in A but not in B. Option B has A ∩B on the right hand side which is the set of elements common to A and B. Hence, option B is not correct. Other options give the same set as A \ B.
Answer : B
Solution 9:
The De Morgan’s law of set difference states that
B \ (A ∪ C) = (B \ A) ∩ (B \ C)
Answer : B
Solution 10:
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
40 = 20 + 30 – n(A ∩ B)
n(A ⋂ B) = 10
Answer: B
Solution 11:
For an identity function f(x) = x for all ‘x’.
Hence, for the pair (x, 2), we get x=2.
For the pair (4, y), we get y=4.
Answer : A
Solution 12:
For a constant function, every element in domain is associated with a constant element.
Here, 7 is related to 11. Since it’s a constant function, 5 should also be related to 11. Hence, a=11.
Answer : B
Solution 13:
(-1)x is 1, when ‘x’ is an even natural number.
(-1)x is -1, when ‘x’ is an odd natural number.
Hence, range set is {1, -1}
Answer : C
Solution 14:
In the given function, ‘3’ is associated with two elements of domain, 6 and 5.
Answer : D
Solution 15:
All elements of set A are associated with a unique element of set B. Hence, f is a function.
No element of set B is associated with two or more elements of set A. Hence, the function is one-one.
The range set of the function is {-1, 2, 1, 5, 9}.
Since range set is not equal to set B, the function is not onto.
Answer : A
Solution 16:
For the given diagram, the element ‘2’ of set C is associated with two elements of set D. This violates the definition of a function. Hence, the relation represented by the diagram is not a function.
Answer : D
Solution 17:
f(5) = 5 – 2=3
f(6) = 6 – 2=4
f(7) = 7 – 2=5
Hence, range of ‘f’ is {3, 4, 5}
Answer : D
Solution 18:
f(-4)=(-4)2 + 5=21
Answer : B
Solution 19:
If the range of a function is a singleton set, then all elements in the domain are related to a single element in co-domain. Hence, it is a constant function.
Answer : A
Solution 20:
A bijective function is both one-one and onto. Hence, every element in the domain is related to exactly one and distinct element of co-domain. Also, co-domain set is equal to the range set. Hence, for a bijective function, number of elements in domain set is equal to number of elements in range set.
Hence, if n(A) = 5, then n(B)=5.
Answer : C