10th Maths Book Back Question and Answers – Exercise 1.1:
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Samacheer Kalvi 10th Maths Book Back Answers – Ex 1.1 Relations and Functions
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Chapter 1:
Exercise 1.1 Relations and Functions
1.Find A × B, A × A and B × A
(i) A = {2,-2,3} and B = {1,-4}
(ii) A = B = {p,q]
(iii) A= {m,n} ; B = (Φ)
Solu.:
(i) A = {2,-2,3}, B = {1,-4}
A × B = {(2, 1), (2, -4), (-2, 1), (-2, -4), (3,1) , (3,-4)}
A × A = {(2, 2), (2,-2), (2, 3), (-2, 2), (-2, -2), (-2, 3), (3, 2), (3, -2), (3,3) }
B × A = {(1, 2), (1, -2), (1, 3), (-4, 2), (-4, -2), (-4,3)}
(ii) A = B = {(p,q)]
A × B = {(p, p), {p, q), (q, p), (q, q)}
A × A = {(p, p), (p, q), (q, p), (q, q)}
B × A = {(p,p), {p, q), (q, p), (q, q)}
(iii) A = {m,n} × Φ
A × B = { }
A × A = {(m, m), (m, n), (n, m), (n, n)}
B × A = { }
2.Let A= {1,2,3} and B = {× | x is a prime number less than 10}. Find A × B and B × A.
Solu.:
A = {1,2,3}, B = {2, 3, 5, 7}
A × B = {1,2,3} × {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 × A = {2, 3, 5, 7} × {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 × A = {(-2, 3),(-2, 4),(0, 3),(0, 4),(3, 3), (3, 4)} find A and B.
Solu.:
B × A ={(-2, 3), (-2, 4), (0, 3), (0, 4), (3, 3), (3, 4)}
A = {3, 4), B = { -2, 0, 3}
4.If A= {5, 6}, B = {4, 5 ,6}, C = {5, 6, 7}, Show that A × A = (B × B) ∩ (C × C)
Solu.:
A ={5,6}, B = {4,5,6}, C = {5, 6,7}
A × A = {5, 6} × {5,6}
= {(5, 5) (5, 6) (6, 5) (6, 6)} ….(1)
B × B = {4, 5, 6} × {4, 5, 6}
= {(4, 4)(4, 5)(4, 6)(5, 4)(5, 5) (5, 6) (6, 4)(6, 5) (6, 6)}
C × C = {5,6,7} × {5,6,7}
= {(5, 5)(5, 6)(5, 7)(6, 5)(6, 6) (6, 7)(7, 5)(7, 6) (7, 7)}
(B × B) ∩ (C × C) = {(5, 5)(5, 6)(6, 5)(6, 6)} ….(2)
From (1) and (2) we get
A × A = (B × B) ∩ (C × C)
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) = (A × B) ∩ (C × D) is true?
Solu.:
LHS = {(A∩C) × (B∩D)
A ∩C = {3}
B ∩D = {3, 5}
(A ∩ C) × (B ∩ D) = {(3, 3) (3, 5)} ………….. (1)
RHS = (A × B) ∩ (C × D)
A × B = {(1, 2), (1, 3), (1, 5), (2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5)}
C × D = {(3, 1), (3, 3), (3, 5), (4, 1), (4, 3), (4, 5)}
(A × B) ∩ (C × D) = {(3, 3), (3, 5)} …(2)
∴ (1) = (2) ∴ It is true.
6.Let A = {x ∈ W | x < 2}, B = {x ∈ N | 1 < 1 < × < 4} and C = {3,5}. Verify that
(i) A × (B ∪ C) = (A × B) ∪ (A × C)
(ii) A × (B ∩ C) = (A × B) ∩ (A × C)
(iii) (A ∪ B) × C = (A × C) ∪ (B × C)
Solu.:
(i) A = {0, 1}
B = {2,3,4}
C = {3,5}
(i) A × (B ∪ C) = (A × B) ∪ (A × c)
B ∪ C = {2, 3,4} ∪ {3,5}
= {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)} ….(1)
A × B = {0, 1} × {2,3,4}
= {(0,2) (0,3) (0,4) (1,2) (1,3) (1,4) }
A × C = {0, 1} × {3, 5}
{(0, 3) (0, 5) (1,3) (1,5)}
(A × B) ∪ (A × C) = {(0, 2) (0, 3) (0, 4) (0, 5) (1, 2)(1, 3)(1, 4)(1, 5)} ….(2)
From (1) and (2) we get
A × (B ∪ C) = (A × B) ∪ (A × C)
(ii) A × (B n C) = (A × B) n (A × C)
B ∩ C = {2,3,4} ∩ {3,5}
= {3}
A × (B ∩ C) = {0, 1} × {3}
= {(0,3) (1,3)} ….(1)
A × B = {0,1} × {2,3,4}
= {(0, 2) (0, 3) (0, 4) (1,2) (1,3) (1,4)}
A × C = {0,1} × {3,5}
{(0, 3) (0, 5) (1,3) (1,5)}
(A × B) n (A × C) = {(0, 3) (1, 3)} ….(2)
From (1) and (2) we get
A × ( B n C) = (A × B) n (A × C)
(iii) (A ∪ B) × C = (A × C) ∪ (B × C)
A ∪ B = {0, 1} ∪ {2,3,4}
= {0,1, 2, 3, 4}
(A ∪ B) × C = {0, 1,2, 3,4} × {3,5}
= {(0, 3) (0, 5) (1, 3) (1, 5)(2, 3) (2, 5) (3, 3)(3, 5) (4, 3)(4, 5)} ….(1)
A × C = {0, 1} × {3,5}
= {(0,3) (0,5) (1,3) (1,5)}
B × C = {2,3,4} × {3,5}
= {(2,3) (2,5) (3,3) (3,5)(4,3)(4,5)}
(A × C) ∪ (B × C) = {(0, 3) (0, 5) (1, 3) (1, 5) (2, 3)(2, 5) (3, 3) (3, 5) (4, 3) (4, 5)} ….(2)
From (1) and (2) we get
(A ∪ B) × C = (A × C) ∪ (B × C)
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 numbers. Verify that
(i) (A ∩ B) × c = (A × C) ∩ (B × C)
(ii) A × (B – C ) = (A × B) – (A × C)
A = {1, 2, 3, 4, 5, 6, 7}
B = {2, 3, 5, 7}
C = {2}
Solu.:
(i)(A ∩ B) × C = (A × c) ∩ (B × C)
LHS = (A ∩ B) × C
A ∩ B = {2, 3, 5, 7}
(A ∩ B) × C = {(2, 2), (3, 2), (5, 2), (7, 2)} ………… (1)
RHS = (A × C) ∩ (B × C)
(A × C) = {(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (7, 2)}
(B × C) = {2, 2), (3, 2), (5, 2), (7, 2)}
(A × C) ∩ (B × C) = {(2, 2), (3, 2), (5, 2), (7, 2)} ……….. (2)
(1) = (2)
∴ LHS = RHS. Hence it is verified.
(ii) A × (B – C) = (A × B) – (A × C)
LHS = A × (B – C)
(B – C) = {3,5,7}
A × (B – C) = {(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 = (A × B) – (A × C)
(A × B) = {(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)}
(A × C) = {(1, 2), (2, 2),(3, 2),(4, 2), (5, 2), (6, 2), (7, 2)}
(A × B) – (A × C) = {(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)
(1) = (2) ⇒ LHS = RHS.
Hence it is verified.
Chapter 1
Exercise: 1 Relation and Functions
1. If the ordered pairs (x2 – 3x, y2 + 4y) and (-2, 5) are equal, then find x and y.
Sol:
(x2 – 3x, y2 + 4y) = (-2, 5)
x2 – 3x = -2
x2 – 3x + 2 = 0
2. The cartesian product A × A has 9 elements among which (-1, 0) and (0,1) are found. Find the set A and the remaining elements of A × A.
Sol:
n(A × A) = 9
n(A) = 3
A = {-1,0,1}
A × A = {-1, 0, 1} × {-1, 0, 1}
A × A = {(-1,-1)(-1, 0) (-1, 1)
(0, -1) (0, 0) (0, 1)
(1,-1) (1, 0) (1, 1)}
The remaining elements of A × A =
{(-1, -1) (-1, 1) (0, -1) (0, 0) (1,-1) (1,0) (1,1)}
3. Given that
(i) f(0)
(ii) f(3)
(iii) f(a + 1) in terms of a.(Given that a > 0)
Solu.:
(i) f(0) = 4
(ii) f(3) = 3−1−−−−√=2–√
(iii) f(a + 1) = a+1−1−−−−−−−−√=a−−√
4. Let A = {9,10,11,12,13,14,15,16,17} and let f : A → N be defined by f(n) = the highest prime factor of n ∈ A. Write f as a set of ordered pairs and find the range of f.
Solu.:
A= {9, 10, 11, 12, 13, 14, 15, 16, 17}
f: A → N
f(x) = the highest prime factor n ∈ A
f = {(9, 3) (10, 5) (11, 11) (12, 3) (13, 13) (14, 7) (15, 5) (16, 2) (17, 17)}
Range of f = {3, 5, 11, 13, 7, 2, 17}
= {2, 3, 5, 7, 11, 13, 17}
5. Find the domain of the function f(x) = 1+1−1−x2−−−−−√−−−−−−−−−−√−−−−−−−−−−−−−−−√
Sol:
f(x) = 1+1−1−x2−−−−−√−−−−−−−−−−√−−−−−−−−−−−−−−−√
Domain of f(x) = {-1, 0, 1}
(x2 = 1, -1, 0, because 1−x2−−−−−√ should be +ve, or 0)
6. If f (x) = x2, g(x) = 3x and h(x) = x – 2, Prove that (f o g)o h = f o(g o h).
Solu.:
f(x) = x2 ; g(x) = 3x and h(x) = x – 2
L.H.S. = (fog) oh
fog = f[g(x)]
= f(3x)
= (3x)2 = 9x2
(fog) oh = fog[h(x)]
= fog (x – 2)
= 9(x – 2)2
= 9[x2 – 4x + 4]
= 9x2 – 36x + 36 ….(1)
R.H.S. = fo(goh)
goh = g [h(x)]
= g(x – 2)
= 3(x – 2)
= 3x – 6
fo(goh) = fo [goh (x)]
= f(3x – 6)
= (3x – 6)2
= 9x2 – 36x + 36 ….(2)
From (1) and (2) we get
L.H.S. = R.H.S.
(fog) oh = fo {goh)
7. A = {1, 2} and B = {1, 2, 3, 4} , C = {5, 6} and D = {5, 6, 7, 8} . Verify whether A × C is a subset of B × D?
Sol:
A = {1, 2), B = (1, 2, 3, 4)
C = {5, 6}, D = {5, 6, 7, 8)
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
B × 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)}
(A × C) ⊂ (B × D) It is proved.
8. If f(x) = x−1x+1, x ≠ 1 show that f(f(x)) = −1x, Provided x ≠ 0.
Sol:
Hence it is proved.
9. The function/and g are defined by f(x) = 6x + 8; g(x) = x−23.
(i) Calculate the value of gg(12)
(ii) Write an expression for g f(x) in its simplest form.
Sol:
10. Write the domain of the following real functions
Sol:
(i) f(x) = 2x+1x−9
The denominator should not be zero as the function is a real function.
∴ The domain = R – {9}
(ii) p(x) = −54×2+1
The domain is R.
(iii) g(x) = x−2−−−−−√
The domain = [2, ∝)
(iv) h(x) = x + 6
The domain is R.
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