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NCERT Solutions for Class 11th Maths Chapter-2 “Relations and Functions”


Exercise 2.1 : Solutions of Questions on Page Number : 33
Q1 :If, find the values of x and y.
Answer :
It is given that.Since the ordered pairs are equal, the corresponding elements will also be equal.
Therefore, and∴ x= 2 and y= 1


Q2 :If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A x B)?
Answer :
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.


Q3 :If G = {7, 8} and H = {5, 4, 2}, find G x H and H x G.
Answer :
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 x H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}
H x G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}


Q4 :State whether each of the following statement 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 x Q = {(m, n), (n, m)}.
(ii) If A and B are non-empty sets, then A x B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B.
(iii) If A = {1, 2}, B = {3, 4}, then A x (B ∩ Φ) = Φ.
Answer :
(i)
False
If P = {m, n} and Q = {n, m}, then
P x Q = {(m, m), (m, n), (n, m), (n, n)}
(ii) True
(iii) True


Q5 :If A = {-1, 1}, find A x A x A.
Answer :
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 x A x 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)}


Q6 :If A x B = {(a, x), (a, y), (b, x), (b, y)}. Find A and B.
Answer :
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}


Q7 :Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that
(i) A x (B ∩C) = (A x B) ∩(A x C)
(ii) A x C is a subset of B x D
Answer :
(i)
To verify: A x (B ∩C) = (A x B) ∩(A x C)
We have B ∩C = {1, 2, 3, 4} ∩{5, 6} = Φ
∴L.H.S. = A x (B ∩C) = A x Φ = Φ
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) = Φ
∴L.H.S. = R.H.S
Hence, A x (B ∩C) = (A x B) ∩(A x C)
(ii) 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.


Q8 :Let A = {1, 2} and B = {3, 4}. Write A x B. How many subsets will A x B have? List them.
Answer :
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)}


Q9 :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 x B, find A and B, where x, y and z are distinct elements.
Answer :
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 z are 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}.


Q10 :The Cartesian product A x A has 9 elements among which are found (-1, 0) and (0, 1). Find the set A and the remaining elements of A x A.
Answer :
We know that if n(A) = p and n(B) = q, then n(A x B) = pq.
∴ n(A x A) = n(A) x n(A)
It is given that n(A x A) = 9
∴ n(A) x 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 x 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 : Solutions of Questions on Page Number : 35

Q1 :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.
Answer :
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 codomainof 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}


Q2 :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.
Answer :
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}


Q3 :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.
Answer :
A = {1, 2, 3, 5} and B = {4, 6, 9}
R = {(x, y): the difference between x and y is odd; x ∈A, y ∈B}
∴R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}


Q4 :The given figure shows a relationship between the sets P and Q. write this relation
(i) in set-builder form (ii) in roster form.
What is its domain and range?

Answer :
According to the given figure, P = {5, 6, 7}, Q = {3, 4, 5}
(i) R = {(x, y): y = x- 2; x ∈P} or R = {(x, y): y = x- 2 for x= 5, 6, 7}
(ii) R = {(5, 3), (6, 4), (7, 5)}
Domain of R = {5, 6, 7}
Range of R = {3, 4, 5}


Q5 :Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by
{(a, b): a, b ∈A, bis exactly divisible by a}.
(i) Write R in roster form
(ii) Find the domain of R
(iii) Find the range of R.
Answer :
A = {1, 2, 3, 4, 6}, R = {(a, b): a, b ∈ A, b is exactly divisible by a}
(i) 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) Domain of R = {1, 2, 3, 4, 6}
(iii) Range of R = {1, 2, 3, 4, 6}


Q6 :Determine the domain and range of the relation R defined by R = {(x, x+ 5): x ∈{0, 1, 2, 3, 4, 5}}.
Answer :
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}


Q7 :Write the relation R = {(x, x3): x is a prime number less than 10} in roster form.
Answer :
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)}


Q8 :Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.
Answer :
It is given that A = {x, y, z} and B = {1, 2}.
∴ A x B = {(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)}
Since n(A x B) = 6, the number of subsets of A x B is 26.
Therefore, the number of relations from A to B is 26.


Q9 :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.
Answer :
R = {(a, b): a, b ∈ Z, a – b is 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 :Solutions of Questions on Page Number:44

Q1 :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)}
(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
(iii) {(1, 3), (1, 5), (2, 5)}
Answer :
(i) {(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)}
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)}
Since the same first element i.e., 1 corresponds to two different images i.e., 3 and 5, this relation is not a function.


Q2 :Find the domain and range of the following real function:
(i) f(x) = -|x| (ii)

Answer :
(i) f(x) = -|x|, x ∈R
We know that |x| =Since f(x) is defined for x ∈ R, the domain of f is R.
It can be observed that the range of f(x) = -|x| is all real numbers except positive real numbers.
∴ The range of f is (-∞, 0].
(ii)Since is defined for all real numbers that are greater than or equal to -3 and less than or equal to 3, the domain of f(x) is {x :-3 ≤ x ≤ 3} or [-3, 3].
For any value of x such that -3 ≤ x≤ 3, the value of f(x) will lie between 0 and 3.
∴The range of f(x) is {x: 0 ≤ x ≤ 3} or [0, 3].


Q3 :A function f is defined by f(x) = 2x- 5. Write down the values of
(i) f(0)    (ii) f(7)     (iii) f(-3)
Answer :
The given function is f(x) = 2x- 5.
Therefore,
(i) f(0) = 2 x 0 – 5 = 0 – 5 = -5
(ii) f(7) = 2 x 7 – 5 = 14 – 5 = 9
(iii) f(-3) = 2 x (-3) – 5 = – 6 – 5 = -11


Q4 :The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by.
Find   (i) t(0)   (ii) t(28)     (iii) t(-10)   (iv) The value of C, when t(C) = 212
Answer :
The given function is.Therefore,

(i)

(ii)

(iii)

(iv) It is given that t(C) = 212

Thus, the value of t,when t(C) = 212, is 100.


Q5 :Find the range of each of the following functions.
(i) f(x) = 2 – 3x, x ∈ R, x> 0.
(ii) f(x) = x2+ 2, x, is a real number.
(iii) f(x) = x, xis a real number
Answer :
(i)
f(x) = 2 -3x, x ∈ R, x> 0
The values of f(x) for various values of real numbers x> 0 can be written in the tabular form as

x 0.01 0.1 0.9 1 2 2.5 4 5
f(x) 1.97 1.7 -0.7 -1 -4 -5.5 -10 -13 ..

 

Thus, it can be clearly observed that the range of fis the set of all real numbers less than 2.
i.e., range of f= (-∞, 2)
Alter:
Let x > 0
⇒3x > 0
⇒ 2 -3x< 2
⇒ f(x) < 2
∴Range of f = (-∞, 2)
(ii) 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 f is 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
It is clear that the range of f is the set of all real numbers.
∴ Range of f = R


Exercise Miscellaneous : Solutions of Questions on Page Number : 46

Q1 :The relation f is defined byThe relation g is defined byShow that f is a function and g is not a function.
Answer :
The relation f is 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 g is 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 gcorresponds to two different images i.e., 4 and 6. Hence, this relation is not a function.


Q2 :If f(x) = x2, find  

Answer :


Q3 :Find the domain of the function

Answer :
The given function is 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}.


Q4 :Find the domain and the range of the real function f defined by 

Answer :
The given real function is It can be seen that is defined for (x -1) ≥0.
i.e., 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 ⇒Therefore, the range of f is the set of all real numbers greater than or equal to 0 i.e., the range of f= [0,∞).


Q5 :Find the domain and the range of the real function f defined by f(x) = |x- 1|.
Answer :
The given real function is f(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 f is the set of all non-negative real numbers.


Q6 :Let

be a function from R into R. Determine the range of f.

Answer :

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)


Q7 :Let f, g: R→R be defined, respectively by f(x) = x + 1, g(x) = 2x – 3. Find f+ g, f-g and Answer :
f, g: R R is defined as f(x) = x + 1, g(x) = 2x – 3
(f+ g) (x) = f(x) + g(x) = (x + 1) + (2x – 3) = 3x – 2
∴(f + g) (x) = 3x – 2
(f – g) (x) = f(x) – g(x) = (x + 1) – (2x – 3) = x+ 1 – 2x+ 3 = -x+ 4
∴ (f – g) (x) = -x+ 4


Q8 :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.
Answer :
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 b are 2 and -1.


Q9 :Let R be a relation from N to N defined by R = {(a, b): a, b ∈ N and a = b2}. Are the following true?
(i) (a, a) ∈ R, for all a ∈ N
(ii) (a, b) ∈ R, implies (b, a) ∈ R
(iii) (a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R.
Answer :
R = {(a, b): a, b ∈ N and a = b2}
(i) 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) 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) 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.


Q10 :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 (ii) f is a function from A to B.
Justify your answer in each case.
Answer :
A = {1, 2, 3, 4} and B = {1, 5, 9, 11, 15, 16}
∴A x B = {(1, 1), (1, 5), (1, 9), (1, 11), (1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11), (2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11), (3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4, 11), (4, 15), (4, 16)}
It is given that f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}
(i) 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 f is a subset of A x B.
Thus, f is a relation from A to B.
(ii) Since the same first element i.e., 2 corresponds to two different images i.e., 9 and 11, relation f is not a function.


Q11 :Let f be the subset of Z x Z defined by f = {(ab, a+ b): a, b ∈ Z}. Is fa function from Z to Z: justify your answer.
Answer :
The relation f is defined as f = {(ab, a+ b): a, b ∈ Z}
We know that a relation f 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 x 6, 2 + 6), (-2 x -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 f is not a function.


Q12 :Let A = {9, 10, 11, 12, 13} and let f: A → Nbe defined by f(n) = the highest prime factor of n. Find the range of f.
Answer :
A = {9, 10, 11, 12, 13}
f: A → N is 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 f is the set of all f(n), where n ∈A.
∴Range of f= {3, 5, 11, 13}