Samacheer kalvi 10th Maths - Relations and Functions Ex 1.6

10th Maths Book Back Question and Answers – Chapter 1 Exercise 1.6:

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Samacheer Kalvi 10th Maths Book Back Answers – Ex 1.6 Relations and Functions

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Chapter 1 Ex 1.6 Relations and Functions

1.If n(A × B) = 6 and A = {1, 3} then n(B) is
(1) 1
(2) 2
(3) 3
(4) 6
Solu.:
(3) 3
Hint:
If n(A × B) = 6
A = {1, 1}, n(A) = 2
n(B) = 3

2.A = {a, b,p}, B = {2, 3}, C = {p, q, r, s)
then n[(A ∪ C) × B] is ………….
(1) 8
(2) 20
(3) 12
(4) 16
Solu.:
(3) 12
Hint: A ∪ C = [a, b, p] ∪ [p, q, r, s]
= [a, b, p, q, r, s]
n (A ∪ C) = 6
n(B) = 2
∴ n [(A ∪ C)] × B] = 6 × 2 = 12

3.If A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8} then state which of the following statement is true.
(1) (A × C) ⊂ (B × D)
(2) (B × D) ⊂ (A × C)
(3) (A × B) ⊂ (A × D)
(4) (D × A) ⊂ (B × A)
Solu.:
(1) (A × C) ⊂ (B × D)]
Hint:
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)}
∴ (A × C) ⊂ B × D it is true

4. If there are 1024 relations from a set A = {1, 2, 3, 4, 5} to a set B, then the number of elements in B is ………………….
(1) 3
(2) 2
(3) 4
(4) 8
Solu.:
(2) 2
Hint: n(A) = 5
n(A × B) = 10
(consider 1024 as 10)
n(A) × n(B) = 10
5 × n(B) = 10
n(B) = 105 = 2
n(B) = 2

5. The range of the relation R = {(x, x²)|x is a prime number less than 13} is
(1) {2, 3, 5, 7}
(2) {2, 3, 5, 7, 11}
(3) {4, 9, 25, 49, 121}
(4) {1, 4, 9, 25, 49, 121}
Solu.:
(3) {4, 9, 25, 49, 121}]
Hint:
R = {(x, x²)/x is a prime number < 13}
The squares of 2, 3, 5, 7, 11 are
{4, 9, 25, 49, 121}

6. If the ordered pairs (a + 2,4) and (5, 2a + 6) are equal then (a, b) is ………
(1) (2, -2)
(2) (5, 1)
(3) (2, 3)
(4) (3, -2)
Solu.:
(4) (3, -2)
Hint:

The value of a = 3 and b = -2

7. Let n(A) = m and n(B) = n then the total number of non-empty relations that can be defined from A to B is
(1) mⁿ
(2) n^m
(3) 2^mn– 1
(4) 2^mn
Solu.:
(4) 2^mn
Hint:
n(A) = m, n(B) = n
n(A × B) = 2^mn

8. If {(a, 8),(6, b)} represents an identity function, then the value of a and 6 are respectively
(1) (8,6)
(2) (8,8)
(3) (6,8)
(4) (6,6)
Solu.:
(1) (8,6)
Hint: f = {{a, 8) (6, 6)}. In an identity function each one is the image of it self.
∴ a = 8, b = 6

9. Let A = {1, 2, 3, 4} and B = {4, 8, 9, 10}. A function f : A → B given by f = {(1, 4),(2, 8),(3, 9),(4, 10)} is a
(1) Many-one function
(2) Identity function
(3) One-to-one function
(4) Into function
Solu.:
(3) One-to one function
Hint:
A = {1, 2, 3, 4), B = {4, 8, 9,10}

10. If f(x) = 2x² and g (x) = 13x, Then fog is

Solu.:
(3) 29×2
Hint:
f(x) = 2x²
g(x) = 13x
fog = f(g(x)) = f(13x)=2(13x)2
= 2 × 19×2=29×2

11. If f: A → B is a bijective function and if n(B) = 7, then n(A) is equal to …………..
(1) 7
(2) 49
(3) 1
(4) 14
Solu.:
(1) 7
Hint:
n(B) = 7
Since it is a bijective function, the function is one – one and also it is onto.
n(A) = n(B)
∴ n(A) = 7

12. Let f and g be two functions given by f = {(0, 1), (2, 0), (3, -4), (4, 2), (5, 7)} g = {(0, 2), (1, 0), (2, 4), (-4, 2), (7, 0)} then the range of fog is
(1) {0, 2, 3, 4, 5}
(2) {-4, 1, 0, 2, 7}
(3) {1, 2, 3, 4, 5}
(4) {0, 1, 2}
Solu.:
(4) {0, 1, 2}
Hint:
gof = g(f(x))
fog = f(g(x))
= {(0, 2),(1, 0),(2, 4),(-4, 2),(7, 0)}
Range of fog = {0, 1, 2}

13. Let f (x) = 1+x2−−−−−√ then ………………..
(1) f(xy) = f(x) f(y)
(2) f(xy) > f(x).f(y)
(3) f(xy) < f(x). f(y)
(4) None of these
Solu.:
(3) f(xy) < f(x) . f(y)

14. If g = {(1, 1),(2, 3),(3, 5),(4, 7)} is a function given by g(x) = αx + β then the values of α and β are
(1) (-1, 2)
(2) (2, -1)
(3) (-1, -2)
(4) (1, 2)
Solu.:
(2) (2,-1)
Hint:
g(x) = αx + β
α = 2
β = -1
g(x) = 2x – 1
g(1) = 2(1) – 1 = 1
g(2) = 2(2) – 1 = 3
g(3) = 2(3) – 1 = 5
g(4) = 2(4) – 1 = 7

15. f(x) = (x + 1)³ – (x – 1)³ represents a function which is …………….
(1) linear
(2) cubic
(3) reciprocal
(4) quadratic
Solu.:
(4) quadratic
Hint: f(x) = (x + 1)³ – (x – 1)³
[using a³ – b³ = (a – b)³ + 3 ab (a – b)]
= (x + 1 – x + 1)³ + 3(x + 1) (x – 1)
(x + 1 – x + 1)
= 8 + 3 (x² – 1)²
= 8 + 6 (x² – 1)
= 8 + 6x² – 6
= 6x² + 2
It is quadratic polynomial

Chapter 1

Exercise 1 Relations and Functions Unit

If the ordered pairs (x² – 3x, y² + 4y) and (-2, 5) are equal, then find x and y.
Solu.:
(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.
Solu.:
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−−−−−√−−−−−−−−−−√−−−−−−−−−−−−−−−√
Solu.:
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?
Solu.:
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.
Solu.:

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.
Solu.:

10. Write the domain of the following real functions

Solu.:
(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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01 Apr 2022