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NCERT Class 11 Mathematics Chapter 2 Relations and Functions
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Relations and Functions
Chapter – 2
| Exercise 2.1 |
1. If (x/3 + 1, y – 2/3) = (5/3, 1/3) find the values of x and y.
Ans:
2. If the set A has 3 elements and the set B = {3, 4, 5} , then find the number of elements in (A×B).
Ans: It is given that set A has 3 elements and the elements of set B are 3, 4, and 5.
⇒ Number of elements in set B = 3
Number of elements in (A x B)
= (Number of elements in A) x (Number of elements in B)
= 3 x 3 = 9
Thus, the number of elements in (A x B) is 9.
3. If G = {7, 8} and H = {5, 4, 2} find G×H and H×G.
Ans: G = {7, 8} and H = {5, 4, 2}
We know that the Cartesian product P x Q of two non-empty sets P and Q is defined as
P x Q = {(p, q): P∈ P, q ∈Q}
∴G⨯H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}
HxG = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}.
4. State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.
(i) If P = {m, n} and Q = {n, m} then P×Q = {(m, n), (n, m)}.
Ans: False
If P = {m, n} and Q = {n, m}, then
P x Q = {(m, m), (m, n), (n, m), (n, n)}.
(ii) If A and B are non-empty sets, then A×B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B.
Ans: True.
(iii) If A = {1, 2} B = {3, 4} then A ×(B ∩ ∅)= ∅.
Ans: True.
5. If A = {- 1, 1} find A × A × A.
Ans: It is known that for any non-empty set A, A x A x A is defined as
A x A x A = {(a, b, c): a, b, c ∈ A}
It is given that A = {- 1, 1}
∴ A ⨯ A ⨯ A = {(- 1, – 1, – 1), (- 1, – 1, 1), (- 1, 1, – 1), (- 1, 1, 1) , (1, -1, -1), (1, -1, 1), (1, 1, -1), (1, 1, 1)}.
6. If A×B = {(a, x), (a, y), (b, x), (b, y)} Find A and B.
Ans: It is given that A x B = {(a, x), (a, y), (b, x), (b, y)}
We know that the Cartesian product of two non-empty sets P and Q is defined as P x Q = {(p. q): p ∈ P, q ∈Q}
∴ A is the set of all first elements and B is the set of all second elements.
Thus, A = {a, b} and B = {x, y}.
7. Let A = {1, 2} B = {1, 2, 3, 4} C = {5, 6} and D = {5, 6, 7, 8} . Verify that.
(i) A × (B ∩ C) = (A × B) ∩ (A × C).
Ans: To verify: A x (B ∩ C) = (A x B) ∩ (A x C)
We have B ∩ C = {1, 2, 3, 4} ∩ {5,6} = ∅
∴ L.H.S.=A x (B ∩ C)= A x ∅ = ∅
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)}
∴ R .H.S.=(A x B) ∩ (A x C) = ∅
∴ L .H.S.= R.H.S
Hence, A x (B ∩ C) = (A x B) ∩ (A x C).
(ii) A × C is a subset of B × D.
Ans: To verify: A x C is a subset of B x D
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)}
We can observe that all the elements of set A x C are the elements of set B x D.
Therefore, A x C is a subset of B x D.
8. Let A = {1, 2} and B = {3, 4} . Write A × B How many subsets will A × B have? List them.
Ans: A = {1, 2} and B = {3, 4}
∴ A x B = {(1, 3), (1, 4), (2, 3), (2, 4)}
⇒n(A x B) = 4
We know that if C is a set with n(C) = m, then n[P(C)] = 2ᵐ.
Therefore, the set A x 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)}.
9. Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A×B
find A and B, where x, y and z are distinct elements.
Ans: It is given that n(A) = 3 and n(B) = 2; and (x, 1), (y, 2), (z, 1) are in A x B.
We know that A = Set of first elements of the ordered pair elements of A x B
B = Set of second elements of the ordered pair elements of A x B.
∴ x, y, and zare the elements of A; and 1 and 2 are the elements of B.
Since n(A) = 3 and n(B) = 2 , it is clear that A = {x, y, z) and B = {1, 2}.
10. The Cartesian product A × A has 9 elements among which are found (-1, 0) and (0,1). Find the set A and the remaining elements of A × A.
Ans: We know that if n(A) = p and n(B) = q; then n(A × B) = pq.
∴ n(A × A) = n(A) × n(A)
It is given that n(A × A) = 9
∴ n(A) × n(A) = 9
⇒n(A) = 3
The ordered pairs (-1, 0) and (0, 1) are two of the nine elements of A x A.
We know that A ⨯ A= {(a, a): a ∈A). Therefore, -1, 0, and 1 are elements of A.
Since n(A) = 3 it is clear that A = {- 1, 0, 1}
The remaining elements of set A x A are (-1,-1), (-1, 1), (0, -1), (0, 0), (1,-1), (1, 0), and (1, 1).
| Exercise 2.2 |
1. Let A = {1, 2, 3 ,…,14} Define a relation R from A to A by R = {(x, y): 3x – y = 0 where x, y ∈ A}. Write down its domain, codomain and range.
Ans: The relation R from A to A is given as
R = {(x, y): 3x – y = 0 where x, y ∈ A}
i.e., R = {(x, y): 3x = y, where x, y ∈ A}
∴ R = {(1, 3), (2, 6), (3, 9), (4, 12)}
The domain of R is the set of all first elements of the ordered pairs in the relation.
∴ Domain of R = {1, 2, 3, 4}
The whole set A is the codomain of the relation R.
∴ Codomain of R = A = {1, 2, 3, …, 14}
The range of R is the set of all second elements of the ordered pairs in the relation.
∴ Range of R = {3, 6, 9, 12}.
2. Define a relation R on the set N of natural numbers by R= {(x, y) : y = x + 5 ,x is a natural number less than 4 ; x, y ∈ N} Depict this relationship using roster form. Write down the domain and the range.
Ans: R= {(x, y): y = x + 5 , xis a natural number less than 4, x, y ∈ N}
The natural numbers less than 4 are 1, 2, and 3.
∴ R = {(1, 6), (2, 7), (3, 8)}
The domain of R is the set of all first elements of the ordered pairs in the relation.
∴ Domain of R = {1, 2, 3}
The range of R is the set of all second elements of the ordered pairs in the relation.
∴ Range of R = {6, 7, 8}.
3. A = {1, 2, 3, 5} and B = {4, 6, 9} Define a relation R from A to B by R = {(x, y)}:the difference between x and y is odd; x ∈ A, y ∈ B}. Write R in roster form.
Ans: A = {1, 2, 3, 5} and B = {4, 6, 9}
R = {(x, y): the difference between xand yis odd; x ∈ A, y ∈B}
∴ R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}.
4. The Fig2.7 shows a relationship between the sets P and Q. Write this relation.
What is its domain and range?
(i) in set-builder form.
Ans: R = {(x, y): y = x- 2; x ∈P} or R={(x, y): y = x – 2 for x = 5, 6, 7}.
(ii) roster form.
Ans: R = {(5, 3), (6, 4), (7, 5)}
Domain of R = {5, 6, 7}
Range of R = {3, 4, 5}.
5. Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by.
{(a, b) : a, b ∈ A b is exactly divisible by a}.
(i) Write R in roster form.
Ans: R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)}.
(ii) Find the domain of R.
Ans: Domain of R = {1, 2, 3, 4, 6}.
(iii) Find the range of R.
Ans: Range of R = {1, 2, 3, 4, 6}.
6. Determine the domain and range of the relation R defined by R = {(x, x + 5) x ∈ {0, 1, 2, 3, 4, 5}}.
Ans: R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}
∴R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}
∴ Domain of R = {0, 1, 2, 3, 4, 5}
Range of R = {5, 6, 7, 8, 9, 10}.
7. Write the relation R={(x, x³) : x is a prime number less than 10} in roster form.
Ans: R = {(x, x³): x is a prime number less than 10}
The prime numbers less than 10 are 2, 3, 5, and 7.
∴R = {(2, 8), (3, 27), (5, 125), (7, 343)}.
8. Let A = {x, y, z} and B = {1, 2} . Find the number of relations from A to B.
Ans: It is given that A = {x, y, z} and B = {1, 2}
∴ A × B = {(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)}
Since n(A × B) = 6, the number of subsets of A × B is 2⁶.
Therefore, the number of relations from A to B is 2⁶.
9. Let R be the relation on Z defined by R = {(a,b): a, b ∈ Z a – b is an integer}. Find the domain and range of R.
Ans: R = {(a, b): a, b ∈ Z, a – bis an integer}
It is known that the difference between any two integers is always an integer.
∴ Domain of R = Z
Range of R = Z.
| Exercise 2.3 |
1. Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.
(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}.
Ans: {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}
Since 2, 5, 8, 11, 14, and 17 are the elements of the domain of the given relation having their unique images, this relation is a function.
Here, domain = {2, 5, 8, 11, 14, 17) and range = {1}.
(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}.
Ans: {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
Since 2, 4, 6, 8, 10, 12, and 14 are the elements of the domain of the given relation having their unique images, this relation is a function.
Here, domain = {2, 4, 6, 8, 10, 12, 14) and range = {1, 2, 3, 4, 5, 6, 7}.
(iii) {(1, 3),(1,5),(2,5).
Ans: {(1, 3), (1, 5), (2, 5)}
Since the same first element i.e., 1 corresponds to two different images i.e., 3 and 5, this relation is not a function.
2. Find the domain and range of the following real functions:
(i) f(x) = – |x|
Ans: f(x) = – |x|
f(- 1) = – |- 1| = – 1
f(o) = – |0| = 0
f(1) = – |1| = – 1
Here, x can be any real number, but f(x) will always be negative or zero.
Therefore,
Domain of function = R (All real numbers)
Range of the function = Negative real numbers
(ii) f(x) = √9 – x²
Ans: f(x) = √9 – x²
9 – x² ≥ 0
x² < 9
x < ±3
x ∈ [- 3, 3]
f(- 3) = √9 – (- 3)² = 0 (Real number)
f(0) = √9 – 0 = 3 (Real number)
f(3) = √9 – (3) = 0 (Real number)
Here,
– 3 ≤ x ≤ 3
0 ≤ f(x) ≤ 3
Therefore,
Domain of function = [- 3, 3]
Range of the function = [0, 3]
3. A function f is defined by f(x) = 2x – 5. Write down the values of
(i) f (0).
Ans: f(0) = 2 × 0 – 5 = 0 – 5 = – 5.
(ii) f (7).
Ans: f(7) = 2 × 7 – 5 = 14 – 5 = 9.
(iii) f (-3).
Ans: f(- 3) = 2 × (- 3) – 5 = – 6 – 5 = – 11.
4. The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by t(C) Find:
(i) t (0).
Ans:
(ii) t(28).
Ans:
(iii) t(-10).
Ans:
(iv) The value of C, when t(C) = 212.
Ans:
5. Find the range of each of the following functions.
(i) f(x) = 2 – 3x, x ∈ R, x > 0.
Ans: Given x > 0
⇒ 3x > 0
⇒ – 3x < 0
⇒ 2 – 3x < 2
⇒ f(x) < 2
∴ Range of f = (-∞, 2)
(ii) f(x) = x² + 2, x is a real number.
Ans: f(x) = x² + 2 x, is a real number
The values of f(x) for various values of real numbers xcan be written in the tabular form as
| x | 0 | ±0.3 | ±0.8 | ±1 | ±2 | ±3 | …. |
| f(x) | 2 | 2.09 | 2.64 | 3 | 6 | 11 | ……….. |
Thus, it can be clearly observed that the range of fis the set of all real numbers greater than 2.
i.e., range of f = [2, ∞)
Alter:
Let x be any real number.
Accordingly.
x² ≥ 0
⇒ x² + 2 ≥ 0 + 2
⇒ x² + 2 ≥ 2
⇒ f(x) ≥ 2
∴ Range of f = [2, ∞)
(iii) f(x) = x, x is a real number.
Ans: f(x) = x, x is a real number
It is clear that the range of f is the set of all real numbers.
∴ Range of f = R.
| Miscellaneous Exercise on Chapter 2 |
Ans: The relation fig defined as
It is observed that for
0 ≤ x < 3, f(x) = x²
3 < x ≤ 10 f(x) = 3x
Also, at x = 3 f(x) = 3² = 9 or f(x) = 3 × 3 = 9
i.e., at x = 3 f(x) = 9
Therefore, for 0 ≤ x ≤ 10 the images of f(x) are unique.
Thus, the given relation is a function.
The relation gis defined as
It can be observed that for x = 2 g(x) = 2² = 4 and g(x) = 3 × 2 = 6
Hence, element 2 of the domain of the relation corresponds to two different images i.e., 4 and 6. Hence, this relation is not a function.
Ans:
Ans: It can be seen that function f is defined for all real numbers except at x = 6 and x = 2
Hence the domain of f is R – {2, 6}
4. Find the domain and the range of the real function f defined by f(x) = √(x – 1) .
Ans: The given real function is f(x) = √x – 1
It can be seen that √x – 1 is defined for (x – 1) ≥ 0
i.e., f(x) = √(x – 1) is defined for x ≥ 1
Therefore the domain of f is the set of all real numbers greater than or equal to
1 i.e.,
the domain of f = [1, ∞)
As X ≥ 1
⇒ (x – 1) ≥ 0
⇒ √x – 1 ≥ 0
f(x) ≥ 0
Therefore the range of f is the set of all real numbers greater than or equal to o i.e., the range of f = [0, ∞)
5. Find the domain and the range of the real function f defined by f(x) = |x – 1|.
Ans: The given real function isf(x) = |x-1|.
It is clear that |x-1| is defined for all real numbers.
∴ Domain of f= R
Also, for x ∈ R, |x-1| assumes all real numbers.
Hence, the range of fis the set of all non-negative real numbers.
Ans:
The range of f is the set of all second elements. It can be observed that all these elements are greater than or equal to 0 but less than 1.
[Denominator is greater numerator]
Thus, range of f = [0, 1)
7. Let f, g : R → R be defined, respectively by f(x) = x + 1 g(x) = 2x – 3 Find f + g, f – g and f/g.
Ans: f, g : R → R is defined as
f(x) = x + 1
g(x) = 2x – 3
Now, (f + g)(x) = f(x) + g(x) = (x + 1) + (2x – 3) = 3x – 2
∴ (f + g)(x) = 3x – 2
Now, (f – g)(x) = f(x) – g(x) = (x + 1) – (2x – 3) = x + 1 – 2x + 3 = – x + 4
∴ (f – g)(x) = – x + 4
8. Let f= {(1,1), (2,3), (0,-1), (-1, -3)} be a function from Z to Z defined by f(x) = ax + b, for some integers a, b. Determine a, b.
Ans: f = {(1, 1), (2, 3), (0, – 1), (- 1, – 3)}
f(x) = ax + b
(1, 1) ∈ f
⇒ f(1) = 1
⇒ a x 1 + b = 1
⇒ a + b = 1
(0, – 1) ∈ f
⇒ f(0) = – 1
⇒ a x 0 + b = – 1
⇒ b = – 1
On substituting b = – 1 in a + b = 1 we obtain a+(-1)=1 ⇒a = 1 + 1 = 2 .
Thus, the respective values of a and bare 2 and -1.
9. Let R be a relation from N to N defined by R = {(a, b) : a, b ∈ N and a=b²}. Are the following true?
Justify your answer in each case.
(i) (a,a) ∈ R, for all a ∈ N.
Ans: It can be seen that 2 ∈ N;however, 2 ≠ 2² = 4.
Therefore, the statement “(a, a) ∈ R, for all a ∈ N” is not true.
(ii) (a,b) ∈ R, implies (b,a) ∈ R.
Ans: It can be seen that (9, 3) ∈ N because 9, 3 ∈ N and 9 = 3².
Now, 3 ≠ 9² = 81; therefore, (3, 9) ∠”° N
Therefore, the statement “(a, b) ∈ R, implies (b, a) ∈ R” is not true.
(iii) (a,b) ∈ R, (b,c) ∈ R implies (a,c) ∈ R.
Ans: It can be seen that (16, 4) ∈ R, (4, 2) ∈ R because 16, 4, 2 ∈ N and 16 = 4² and 4 = 2².
Now, 16 ≠ 2² = 4; therefore, (16, 2)∠ “° N
Therefore, the statement “(a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R” is not true.
10. Let A={1,2,3,4}, B = {1,5,9,11,15,16} and f= {(1,5), (2,9), (3,1), (4,5), (2,11)} Are the following true?
(i) f is a relation from A to B Justify your answer in each case.
Ans: A relation from a non-empty set A to a non-empty set B is a subset of the Cartesian product A x B.
It is observed that fig a subset of A x B.
Thus, fig a relation from A to B.
(ii) f is a function from A to B.
Ans: Since the same first element i.e., 2 corresponds to two different images i.e., 9 and 11, relation f is not a function.
11. Let f be the subset of Z × Z defined by f = {(ab, a + b): a, b ∈ Z}. Is f a function from Z to Z? Justify your answer.
Ans: The relation fis defined as f= {(ab, a + b):a, b ∈ Z}
We know that a relation from a set A to a set B is said to be a function if every element of set A has unique images in set B.
Since 2, 6, -2, – 6 ∈ Z ( 2 × 6, 2 + 6), ( – 2 × -6,-2 +(-6)) ∈ f
i.e., (12, 8), (12, – 8) ∈ f
It can be seen that the same first element i.e., 12 corresponds to two different images i.e., 8 and -8. Thus, relation fis not a function.
12. Let A = {9,10,11,12,13} and let f: A→N be defined by f (n) = the highest prime factor of n. Find the range of f.
Ans: A = {9, 10, 11, 12, 13}
f: A →Nis defined as
f(n) = The highest prime factor of n
Prime factor of 9 = 3
Prime factors of 10 = 2, 5
Prime factor of 11 = 11
Prime factors of 12 = 2, 3
Prime factor of 13 = 13
∴ f(9) = The highest prime factor of 9 = 3
f(10) = The highest prime factor of 10 = 5
f(11) = The highest prime factor of 11 = 11
f(12) = The highest prime factor of 12 = 3
f(13) = The highest prime factor of 13 = 13
The range of fis the set of all f(n) where n ∈ A.
∴Range of f = {3, 5, 11, 13}.
