NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions
Table of Contents
Exercise 1.1
Question 1:
Determine whether each of the following relations are reflexive, symmetric and transitive:
(i) Relation R in the set A = {1, 2, 3, …, 13, 14} defined as R = {(x, y) : 3x – y = 0}
(ii) Relation R in the set N of natural numbers defined as R = {(x, y) : y = x + 5 and x < 4}
(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y) : y is divisible by x}
(iv) Relation R in the set Z of all integers defined as R = {(x, y) : x – y is an integer}
(v) Relation R in the set A of human beings in a town at a particular time given by
(a) R = {(x, y) : x and y work at the same place}
(b) R = {(x, y) : x and y live in the same locality}
(c) R = {(x, y) : x is exactly 7 cm taller than y}
(d) R = {(x, y) : x is wife of y}
(e) R = {(x, y) : x is father of y}
Answer
(i) A = {1, 2, 3 … 13, 14}
R = {(x, y): 3x − y = 0}
∴R = {(1, 3), (2, 6), (3, 9), (4, 12)}
R is not reflexive since (1, 1), (2, 2) … (14, 14) ∉ R.
Also, R is not symmetric as (1, 3) ∈R, but (3, 1) ∉ R. [3(3) − 1 ≠ 0]
Also, R is not transitive as (1, 3), (3, 9) ∈R, but (1, 9) ∉ R. [3(1) − 9 ≠ 0]
Hence, R is neither reflexive, nor symmetric, nor transitive.
(ii) R = {(x, y): y = x + 5 and x < 4} = {(1, 6), (2, 7), (3, 8)}
It is seen that (1, 1) ∉ R.
∴R is not reflexive. (1, 6) ∈R But,(6, 1) ∉ R.
∴R is not symmetric.
Now, since there is no pair in R such that (x, y) and (y, z) ∈R, then (x, z) cannot belong to R.
∴ R is not transitive. Hence, R is neither reflexive, nor symmetric, nor transitive.
(iii) A = {1, 2, 3, 4, 5, 6} R = {(x, y): y is divisible by x}
We know that any number (x) is divisible by itself. (x, x) ∈R
∴R is reflexive.
Now, (2, 4) ∈R [as 4 is divisible by 2]
But, (4, 2) ∉ R. [as 2 is not divisible by 4]
∴R is not symmetric.
Let (x, y), (y, z) ∈ R.
Then, y is divisible by x and z is divisible by y.
∴z is divisible by x. ⇒ (x, z) ∈R
∴R is transitive.
Hence, R is reflexive and transitive but not symmetric.
(iv) R = {(x, y): x − y is an integer}
Now, for every x ∈ Z, (x, x) ∈R as x − x = 0 is an integer.
∴R is reflexive.
Now, for every x, y ∈ Z if (x, y) ∈ R, then x − y is an integer.
⇒ −(x − y) is also an integer.
⇒ (y − x) is an integer. ∴ (y, x) ∈ R
∴R is symmetric.
Now, Let (x, y) and (y, z) ∈R, where x, y, z ∈ Z.
⇒ (x − y) and (y − z) are integers.
⇒ x − z = (x − y) + (y − z) is an integer.
∴ (x, z) ∈R
∴R is transitive.
Hence, R is reflexive, symmetric, and transitive.
(v) (a) R = {(x, y): x and y work at the same place} (x, x) ∈ R
∴ R is reflexive.
If (x, y) ∈ R, then x and y work at the same place.
⇒ y and x work at the same place.
⇒ (y, x) ∈ R. ∴R is symmetric.
Now, let (x, y), (y, z) ∈ R
⇒ x and y work at the same place and y and z work at the same place.
⇒ x and z work at the same place. ⇒ (x, z) ∈R
∴ R is transitive.
Hence, R is reflexive, symmetric, and transitive.
(b) R = {(x, y): x and y live in the same locality}
Clearly (x, x) ∈ R as x and x is the same human being.
∴ R is reflexive.
If (x, y) ∈R, then x and y live in the same locality.
⇒ y and x live in the same locality.
⇒ (y, x) ∈ R
∴R is symmetric.
Now, let (x, y) ∈ R and (y, z) ∈ R.
⇒ x and y live in the same locality and y and z live in the same locality.
⇒ x and z live in the same locality.
⇒ (x, z) ∈ R
∴ R is transitive. Hence, R is reflexive, symmetric, and transitive.
(c) R = {(x, y): x is exactly 7 cm taller than y}
Now, (x, x) ∉ R Since human being x cannot be taller than himself.
∴R is not reflexive.
Now, let (x, y) ∈R.
⇒ x is exactly 7 cm taller than y.
Then, y is not taller than x. ∴ (y, x) ∉R
Indeed if x is exactly 7 cm taller than y, then y is exactly 7 cm shorter than x.
∴R is not symmetric.
Now, Let (x, y), (y, z) ∈ R.
⇒ x is exactly 7 cm taller thany and y is exactly 7 cm taller than z.
⇒ x is exactly 14 cm taller than z . ∴(x, z) ∉R
∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
(d) R = {(x, y): x is the wife of y}
Now, (x, x) ∉ R Since x cannot be the wife of herself.
∴R is not reflexive.
Now, let (x, y) ∈ R
⇒ x is the wife of y.
Clearly y is not the wife of x.
∴(y, x) ∉ R Indeed if x is the wife of y, then y is the husband of x.
∴ R is not symmetric.
Let (x, y), (y, z) ∈ R
⇒ x is the wife of y and y is the wife of z.
This case is not possible.
Also, this does not imply that x is the wife of z.
∴(x, z) ∉ R ∴R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
(e) R = {(x, y): x is the father of y} (x, x) ∉ R
As x cannot be the father of himself.
∴R is not reflexive.
Now, let (x, y) ∈R. ⇒ x is the father of y.
⇒ y cannot be the father of y.
Indeed, y is the son or the daughter of y.
∴(y, x) ∉ R
∴ R is not symmetric.
Now, let (x, y) ∈ R and (y, z) ∈ R.
⇒ x is the father of y and y is the father of z.
⇒ x is not the father of z.
Indeed x is the grandfather of z.
∴ (x, z) ∉ R
∴R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions
Question 2:
Show that the relation R in the set R of real numbers, defined as R = {(a, b) : a ≤ b2} is neither reflexive nor symmetric nor transitive.
Answer
R = {(a, b): a ≤ b 2 } , Relation R is defined as the set of real numbers.
It can be observed that
\(\left(\frac{1}{2}, \frac{1}{2}\right) \in R. \text{Since} \frac{1}{2} > \left(\frac{1}{2}\right)^2 = \frac{1}{4}\)
∴R is not reflexive.
Now, (1, 4) ∈ R as 1 < 4 2
But, 4 is not less than 1 2 .
∴(4, 1) ∉ R
∴R is not symmetric.
Now, (3, 2), (2, 1.5) ∈ R (as 3 < 2 2 = 4 and 2 < (1.5) 2 = 2.25)
But, 3 > (1.5) 2 = 2.25
∴(3, 1.5) ∉ R
∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions
Question 3:
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive.
Answer
Let A = {1, 2, 3, 4, 5, 6}.
A relation R is defined on set A as: R = {(a, b): b = a + 1}
∴R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}
We can find (a, a) ∉ R, where a ∈ A.
For instance, (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) ∉ R
∴R is not reflexive.
It can be observed that (1, 2) ∈ R, but (2, 1) ∉ R.
∴R is not symmetric.
Now, (1, 2), (2, 3) ∈ R
But, (1, 3) ∉ R
∴R is not transitive
Hence, R is neither reflexive, nor symmetric, nor transitive.
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions
Question 4:
Show that the relation R in R defined as R = {(a, b) : a ≤ b}, is reflexive and transitive but not symmetric.
Answer
R = {(a, b); a ≤ b}
Clearly (a, a) ∈ R as a = a.
∴R is reflexive.
Now, (2, 4) ∈ R (as 2 < 4)
But, (4, 2) ∉ R as 4 is greater than 2.
∴ R is not symmetric.
Now, let (a, b), (b, c) ∈ R.
Then, a ≤ b and b ≤ c ⇒ a ≤ c ⇒ (a, c) ∈ R
∴R is transitive.
Hence,R is reflexive and transitive but not symmetric.
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions
Question 5:
Check whether the relation R in R defined by R = {(a, b) : a ≤ b3} is reflexive, symmetric or transitive.
Answer
It is observed that \(\left(\frac{1}{2}, \frac{1}{2}\right) \notin \)R as \(\frac{1}{2} > \left(\frac{1}{2}\right)^3 = \frac{1}{8}.\)
∴R is not reflexive.
Now,
\((1, 2) \in \) R as \(1 < 2^3 = 8)\)
But, (2, 1) \(\notin\) R as (1 < 2^3 = 8)
∴R is not symmetric.
We have, \(\left(3, \frac{3}{2}\right), \left(3, \frac{3}{2}, \frac{6}{5}\right) \in\) R as 3 < \(\left(\frac{3}{2}\right)^3\) and \(\frac{3}{2} < \left(\frac{6}{5}\right)^3\).
But \(\left(3, \frac{6}{5}\right) \notin\) R as 3 < \(\left(\frac{6}{5}\right)^3 \).
∴R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions
Question 6:
Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.
Answer
Let A = {1, 2, 3}.
A relation R on A is defined as R = {(1, 2), (2, 1)}.
It is seen that (1, 1), (2, 2), (3, 3) ∉R.
∴ R is not reflexive.
Now, as (1, 2) ∈ R and (2, 1) ∈ R, then R is symmetric.
Now, (1, 2) and (2, 1) ∈ R
However, (1, 1) ∉ R
∴ R is not transitive.
Hence, R is symmetric but neither reflexive nor transitive.
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions
Question 7:
Show that the relation R in the set A of all the books in a library of a college, given by R = {(x, y) : x and y have same number of pages} is an equivalence relation.
Answer
Set A is the set of all books in the library of a college.
R = {x, y): x and y have the same number of pages}
Now, R is reflexive since (x, x) ∈ R as x and x has the same number of pages.
Let (x, y) ∈ R
⇒ x and y have the same number of pages.
⇒ y and x have the same number of pages.
⇒ (y, x) ∈ R ∴R is symmetric.
Now, let (x, y) ∈R and (y, z) ∈ R.
⇒ x and y and have the same number of pages and y and z have the same number of pages.
⇒ x and z have the same number of pages.
⇒ (x, z) ∈ R
∴R is transitive.
Hence, R is an equivalence relation.
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions
Question 8:
Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b) : |a – b| is even}, is an equivalence relation.
Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.
Answer
A = {1, 2, 3, 4, 5} and R = {(a,b) : |a-b| is even} , then R = {(1, 3), (1, 5), (3, 5), (2, 4)}
It is clear that for any element a ∈ A , we have|a-a| = 0 (which is even).
∴R is reflexive.
Let (a,b) ∈ R.
⇒ |a-b| is even
⇒ |-(a-b) | = |b-a| is also even
⇒ (b,a) ∈ R
∴R is symmetric.
Now, let (a,b) ) ∈ R and (b, c) ∈ R.
⇒ |a-b| is even and |b -c| is even
⇒ (a – b) is even and (b – c) is even
⇒ (a – c) = (a – b) + (b – c) is even [Sum of two even integer is even]
⇒ |a – c | is even
∴R is transitive.
Hence, R is an equivalence relation.
Now, all elements of the set {1, 3, 5} are related to each other as all the elements of this subset are odd.
Thus, the modulus of the difference between any two elements will be even.
Similarly, all elements of the set {2, 4} are related to each other as all the elements of this subset are even.
Also, no element of the subset {1, 3, 5} can be related to any element of {2, 4} as all elements of {1, 3, 5} are odd and all elements of {2, 4} are even.
Thus, the modulus of the difference between the two elements (from each of these two subsets) will not be even.
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions
Question 9:
Show that each of the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by
(i) R = {(a, b) : |a – b| is a multiple of 4}
(ii) R = {(a, b) : a = b}
is an equivalence relation. Find the set of all elements related to 1 in each case.
Answer
(a) A ={x in Z : 0 <= <= 12} = {0,1,2,3,4,5,6,7,8,9,10,11,12}
R = {(a,b):|a-b| is a multiple of 4}
For any element a ∈ A, we have (a,a ) ∈ R as |a – a| = 0 is a multiple of 4.
∴R is reflexive.
Now, let (a, b) ∈ R
⇒ |a- b| is a multiple of 4.
⇒ |-(a-b)|=|b -a| is a multiple of 4
⇒ (b, a ) ∈ R
∴R is symmetric.
Now, let (a,b), (b,c) ∈ R.
⇒ |a – b| is multiple of 4 and |b – c| is a multiple of 4
⇒ (a-b) is a multiple of 4 and (b-c) is a multiple of 4
⇒ (a-c) = (a – b) + (b – c) is a multiple of 4
⇒ (a,c)∈ R.
∴ R is transitive.
Hence, R is an equivalence relation.
The set of elements related to 1 is {1, 5, 9}
since |1 – 1| = 0 is a multiple of 4 |5 – 1| = 4 is a multiple of 4 |9 – 1| = 8 is a multiple of 4
(b) R = {(a,b): a = b}
For any element a ∈ A, we have ( a,a ) ∈ R, since a = a.
∴R is reflexive.
Now, let (a,b ) ∈ R and (b, c) ∈ R.
⇒ a = b and b = c
⇒ a = c
⇒ (a,c) ∈ R.
∴ R is transitive.
Hence, R is an equivalence relation.
The elements in R that are related to 1 will be those elements from set A which are equal to 1.
Hence, the set of elements related to 1 is {1}.
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions
Question 10:
Give an example of a relation. Which is
(i) Symmetric but neither reflexive nor transitive.
(ii) Transitive but neither reflexive nor symmetric.
(iii) Reflexive and symmetric but not transitive.
(iv) Reflexive and transitive but not symmetric.
(v) Symmetric and transitive but not reflexive.
Answer
(i) Let A = {5, 6, 7}. Define a relation R on A as R = {(5, 6), (6, 5)}.
Relation R is not reflexive as (5, 5), (6, 6), (7, 7) ∉ R.
Now, as (5, 6) ∈ R and also (6, 5) ∈ R, R is symmetric.
⇒ (5, 6), (6, 5) ∈ R, but (5, 5) ∉ R
∴R is not transitive.
Hence, relation R is symmetric but not reflexive or transitive.
(ii)Consider a relation R in R defined as: R = {(a, b): a < b}
For any a ∈ R, we have (a, a) ∉ R since a cannot be strictly less than a itself.
In fact, a = a.
∴ R is not reflexive.
Now, (1, 2) ∈ R (as 1 < 2)
But, 2 is not less than 1.
∴ (2, 1) ∉ R ∴ R is not symmetric. N
ow, let (a, b), (b, c) ∈ R. ⇒ a < b and b < c ⇒ a < c ⇒ (a, c) ∈ R
∴R is transitive.
Hence, relation R is transitive but not reflexive and symmetric.
(iii)Let A = {4, 6, 8}.
Define a relation R on A as: A = {(4, 4), (6, 6), (8, 8), (4, 6), (6, 4), (6, 8), (8, 6)}
Relation R is reflexive since for every a ∈ A, (a, a) ∈R i.e., (4, 4), (6, 6), (8, 8)} ∈ R.
Relation R is symmetric since (a, b) ∈ R ⇒ (b, a) ∈ R for all a, b ∈ R.
Relation R is not transitive since (4, 6), (6, 8) ∈ R, but (4, 8) ∉ R.
Hence, relation R is reflexive and symmetric but not transitive.
(iv) Define a relation R in R as: R = {a, b): a 3 ≥ b 3 }
Clearly (a, a) ∈ R as a 3 = a 3 . ∴R is reflexive.
Now, (2, 1) ∈ R (as 2 3 ≥ 1 3 ) But, (1, 2) ∉ R (as 1 3 < 2 3 )
∴ R is not symmetric.
Now, Let (a, b), (b, c) ∈ R. ⇒ a 3 ≥ b 3 and b 3 ≥ c 3 ⇒ a 3 ≥ c 3
⇒ (a, c) ∈ R
∴R is transitive.
Hence, relation R is reflexive and transitive but not symmetric.
(v) Let A = {−5, −6}.
Define a relation R on A as: R = {(−5, −6), (−6, −5), (−5, −5)}
Relation R is not reflexive as (−6, −6) ∉ R.
Relation R is symmetric as (−5, −6) ∈ R and (−6, −5}∈R.
It is seen that (−5, −6), (−6, −5) ∈ R. Also, (−5, −5) ∈ R.
∴The relation R is transitive.
Hence, relation R is symmetric and transitive but not reflexive.
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions
Question 11:
Show that the relation R in the set A of points in a plane given by R = {(P, Q) : distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.
Answer
R = {(P, Q): distance of point P from the origin is the same as the distance of point Q from the origin}
Clearly, (P, P) ∈ R since the distance of point P from the origin is always the same as the distance of the same point P from the origin.
∴R is reflexive.
Now, Let (P, Q) ∈ R.
⇒ The distance of point P from the origin is the same as the distance of point Q from the origin.
⇒ The distance of point Q from the origin is the same as the distance of point P from the origin.
⇒ (Q, P) ∈ R
∴R is symmetric.
Now, Let (P, Q), (Q, S) ∈ R.
⇒ The distance of points P and Q from the origin is the same and also, the distance of points Q and S from the origin is the same.
⇒ The distance of points P and S from the origin is the same.
⇒ (P, S) ∈ R ∴R is transitive.
Therefore, R is an equivalence relation.
The set of all points related to P ≠ (0, 0) will be those points whose distance from the origin is the same as the distance of point P from the origin.
In other words, if O (0, 0) is the origin and OP = k, then the set of all points related to P is at a distance of k from the origin.
Hence, this set of points forms a circle with the centre as the origin and this circle passes through point P.
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions
Question 12:
Show that the relation R defined in the set A of all triangles as R = {(T1, T2) : T1 is similar to T2}, is equivalence relation. Consider three right angle triangles T1 with sides 3, 4, 5, T2 with sides 5, 12, 13 and T3 with sides 6, 8, 10. Which triangles among T1 , T2 and T3 are related?
Answer
R = {(T 1 , T 2) : T 1 is similar to T 2 }
R is reflexive since every triangle is similar to itself.
Further, if (T 1 , T 2 ) ∈ R, then T 1 is similar to T 2 .
⇒ T 2 is similar to T 1 . ⇒ (T 2 , T 1 ) ∈R
∴R is symmetric.
Now, Let (T 1 , T 2 ), (T 2 , T 3 ) ∈ R.
⇒ T 1 is similar to T 2 and T 2 is similar to T 3 .
⇒ T 1 is similar to T 3 . ⇒ (T 1 , T 3 ) ∈ R
∴ R is transitive.
Thus, R is an equivalence relation.
Now, we can observe that: Therefore, 3/6 = 4/8 = (5/10 = 1/2)
∴The corresponding sides of triangles T 1 and T 3 are in the same ratio.
Then, triangle T 1 is similar to triangle T 3 .
Hence, T 1 is related to T 3 .
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions
Question 13:
Show that the relation R defined in the set A of all polygons as R = {(P1, P2) 😛1 and P2 have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?
Answer
R = {(P 1 , P 2 ): P 1 and P 2 have same the number of sides}
R is reflexive since (P 1 , P 1 ) ∈ R as the same polygon has the same number of sides with itself.
Let (P 1 , P 2 ) ∈ R.
⇒ P 1 and P 2 have the same number of sides.
⇒ P 2 and P 1 have the same number of sides.
⇒ (P 2 , P 1 ) ∈ R
∴R is symmetric.
Now, Let (P 1 , P 2 ), (P 2 , P 3 ) ∈ R.
⇒ P 1 and P 2 have the same number of sides.
Also, P 2 and P 3 have the same number of sides.
⇒ P 1 and P 3 have the same number of sides.
⇒ (P 1 , P 3 ) ∈ R ∴R is transitive.
Hence, R is an equivalence relation.
The elements in A related to the right-angled triangle (T) with sides 3, 4, and 5 are those polygons which have 3 sides (since T is a polygon with 3 sides).
Hence, the set of all elements in A related to triangle T is the set of all triangles.
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions
Question 14:
Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L1, L2) : L1 is parallel to L2}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.
Answer
R = {(L 1 , L 2 ): L 1 is parallel to L 2 }
R is reflexive as any line L 1 is parallel to itself i.e., (L 1 , L 1 ) ∈ R.
Now, Let (L 1 , L 2 ) ∈ R.
⇒ L 1 is parallel to L 2 .
⇒ L 2 is parallel to L 1 .
⇒ (L 2 , L 1 ) ∈ R
∴ R is symmetric.
Now, Let (L 1 , L 2 ), (L 2 , L 3 ) ∈R.
⇒ L 1 is parallel to L 2 .
Also, L 2 is parallel to L 3 .
⇒ L 1 is parallel to L 3 .
∴R is transitive.
Hence, R is an equivalence relation.
The set of all lines related to the line y = 2x + 4 is the set of all lines that are parallel to the line y = 2x + 4.
Slope of line y = 2x + 4 is m = 2
It is known that parallel lines have the same slopes.
The line parallel to the given line is of the form y = 2x + c, where c ∈R.
Hence, the set of all lines related to the given line is given by y = 2x + c, where c ∈ R.
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions
Question 15:
Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4,4), (1, 3), (3, 3), (3, 2)}. Choose the correct answer.
(A) R is reflexive and symmetric but not transitive.
(B) R is reflexive and transitive but not symmetric.
(C) R is symmetric and transitive but not reflexive.
(D) R is an equivalence relation.
Answer
R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}
It is seen that (a, a) ∈ R, for every a ∈{1, 2, 3, 4}.
∴ R is reflexive. It is seen that (1, 2) ∈ R, but (2, 1) ∉ R.
∴R is not symmetric.
Also, it is observed that (a, b), (b, c) ∈ R
⇒ (a, c) ∈ R for all a, b, c ∈ {1, 2, 3, 4}.
∴ R is transitive.
Hence, R is reflexive and transitive but not symmetric.
The correct answer is B.
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions
Question 16:
Let R be the relation in the set N given by R = {(a, b) : a = b – 2, b > 6}. Choose the correct answer.
(A) (2, 4) ∈ R (B) (3, 8) ∈ R (C) (6, 8) ∈ R (D) (8, 7) ∈ R
Answer
R = {(a, b): a = b − 2, b > 6}
Now, since b > 6, (2, 4) ∉ R
Also, as 3 ≠ 8 − 2, (3, 8) ∉ R
And, as 8 ≠ 7 − 2 (8, 7) ∉ R
Now, consider (6, 8).
We have 8 > 6 and also, 6 = 8 − 2.
∴(6, 8) ∈ R The correct answer is C.
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions
Exercise 1.2
Question 1:
Show that the function f : R∗ → R∗ defined by f(x) = \(\frac{1}{x}\) is one-one and onto, where R∗ is the set of all non-zero real numbers. Is the result true, if the domain R∗ is replaced by N with co-domain being same as R∗?
Answer
It is given that f : R∗ → R∗ defined by f(x) = \(\frac{1}{x}\) .
one-one:
f (x) = f (y)
\(\Rightarrow\) \(\frac{1}{x}\) = \(\frac{1}{y}\)
\(\Rightarrow\) x = y
\(\therefore\) f is one-one.
onto:
It is clear that for y \(\in\) R; there exists x = \(\frac{1}{y} \in R\), such that
f (x) = \(\frac{1}{\left(\frac{1}{y}\right)}\) = y.
\(\therefore\) f is onto.
Thus, the given function (f) is one-one and onto.
Now, g (x) = \(\frac{1}{x}\).
g (x1) = g (x2) \(\Rightarrow\) \(\frac{1}{x_1}\) = \(\frac{1}{x_2}\) \(\Rightarrow\) x1 = x2
\(\therefore\) g is one-one.
Question 2:
Check the injectivity and surjectivity of the following functions:
(i) f : N → N given by f(x) = x2
(ii) f : Z → Z given by f(x) = x2
(iii) f : R → R given by f(x) = x2
(iv) f : N → N given by f(x) = x3
(v) f : Z → Z given by f(x) = x3
Answer
(i) f: N → N is given by, f(x) = x 2
It is seen that for x, y ∈N, f(x) = f(y) ⇒ x 2 = y 2 ⇒ x = y.
∴f is injective.
Now, 2 ∈ N. But, there does not exist any x in N such that f(x) = x 2 = 2.
∴ f is not surjective.
Hence, function f is injective but not surjective.
(ii) f: Z → Z is given by, f(x) = x 2
It is seen that f(−1) = f(1) = 1, but −1 ≠ 1. ∴ f is not injective. Now,−2 ∈ Z.
But, there does not exist any element x ∈Z such that f(x) = x 2 = −2. ∴ f is not surjective.
Hence, function f is neither injective nor surjective.
(iii) f: R → R is given by, f(x) = x 2
It is seen that f(−1) = f(1) = 1, but −1 ≠ 1. ∴ f is not injective. Now,−2 ∈ R.
But, there does not exist any element x ∈ R such that f(x) = x 2 = −2. ∴ f is not surjective.
Hence, function f is neither injective nor surjective.
(iv) f: N → N given by, f(x) = x 3
It is seen that for x, y ∈N, f(x) = f(y) ⇒ x 3 = y 3 ⇒ x = y. ∴f is injective. Now, 2 ∈ N.
But, there does not exist any element x in domain N such that f(x) = x 3 = 2. ∴ f is not surjective
Hence, function f is injective but not surjective.
(v) f: Z → Z is given by, f(x) = x 3 I
It is seen that for x, y ∈ Z, f(x) = f(y) ⇒ x 3 = y 3 ⇒ x = y. ∴ f is injective.
Now, 2 ∈ Z. But, there does not exist any element x in domain Z such that f(x) = x 3 = 2.
∴ f is not surjective. Hence, function f is injective but not surjective.
Question 3:
Prove that the Greatest Integer Function f : R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
Answer
f: R → R is given by, f(x) = [x]
It is seen that f(1.2) = [1.2] = 1, f(1.9) = [1.9] = 1. ∴ f(1.2) = f(1.9), but 1.2 ≠ 1.9.
∴ f is not one-one.
Now, consider 0.7 ∈ R.
It is known that f(x) = [x] is always an integer.
Thus, there does not exist any element x ∈ R such that f(x) = 0.7.
∴ f is not onto.
Hence, the greatest integer function is neither one-one nor onto.
Question 4:
Show that the Modulus Function f : R \(\rightarrow\) R, given by f(x) = | x |, is neither oneone nor onto, where | x | is x, if x is positive or 0 and | x | is – x, if x is negative.
Answer
: R \(\rightarrow\) R, given by \(f(x) = | x | = \begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0 \end{cases}
\)
It is seen that f (-1) = |-1| = 1, f (1) = |1| = 1.
\(\therefore\) f (-1) = f (1), but -1 \(\neq\) 1.
\(\therefore\) f is not one-one.
Now, consider – 1 \(\in\) R.
It is known that f (x) = |x| is always non-negative.
\(\therefore\) f is not onto.
Hence, the modulus function is neither one-one nor onto.
Question 5:
Show that the Signum Function f : R → R, given by
\[
f(x) = \begin{cases}
1, & \text{if } x > 0 \\
0, & \text{if } x = 0 \\
1, & \text{if } x < 0
\end{cases}
\]
is neither one-one nor onto.
Answer
f : R → R, given by
\[
f(x) = \begin{cases}
1, & \text{if } x > 0 \\
0, & \text{if } x = 0 \\
1, & \text{if } x < 0
\end{cases}
\]
It is seen that f(1) = f(2) = 1, but 1 ≠ 2. ∴ f is not one-one.
Now, as f(x) ) takes only 3 values (1, 0, or −1) for the element −2 in co-domain R , there does not exist any x in domain R such that f(x) = −2. ∴ f is not onto.
Hence, the signum function is neither one-one nor onto.
Question 6:
Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.
Answer
It is given that A = {1, 2, 3}, B = {4, 5, 6, 7}.
f: A → B is defined as f = {(1, 4), (2, 5), (3, 6)}.
∴ f (1) = 4, f (2) = 5, f (3) = 6
It is seen that the images of distinct elements of A under f are distinct.
Hence, function f is one-one.
Question 7:
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) f : R \(\rightarrow\) R defined by f (x) = 3 – 4x
(ii) f : R \(\rightarrow\) R defined by f (x) = 1 + x2
Answer
(i) f : R \(\rightarrow\) R defined by f (x) = 3 – 4x
Let x1 , x2 \(\in\) R such that f (x1) = f (x2)
\(\Rightarrow\) 3 – x1 = 3 – 4x2 \(\Rightarrow\) -4x1 = – 4x2 \(\Rightarrow\) x1 = x2
\(\therefore\) f is one-one.
For any real number (y) in R, there exists \(\frac{3 – y}{4}\) in R such that
f \(\left(\frac{3 – y}{4}\right) = 3 – 4\left(\frac{3 – y}{4}\right) = y\)
\(\therefore\) f is onto.
Hence, f is biective.
(ii) f : R \(\rightarrow\) R defined by f (x) = 1 + x2
Let x1 , x2 \(\in\) R such that f (x1) = f (x2)
\(\Rightarrow\) 1 + \(x_1^2\) = 1 + \(x_2^2\) \(\Rightarrow\) \(x_1^2\) = \(x_2^2 \Rightarrow\) x1 = \(\pm\) x2
\(\therefore\) f (x1) = f (x2) does not imply that x1 = x2
For example f (1) = f (-1) = 2
\(\therefore\) f is not one-one.
Consider an element -2 in co-domain in R.
It is seen that f (x) = 1 + x2 is positive for all x \(\in\) R.
Thus, there does not exist any x in domain R such that f (x) = -2.
\(\therefore\) f is not onto.
Hence, f is neither one-one nor onto.
Question 8:
Let A and B be sets. Show that f : A × B → B × A such that f(a, b) = (b, a) is bijective function.
Answer
f : A /(\times\) B \(\rightarrow\) B \(\times\) A is defined as f (a, b) = (b, a)
Let (a1, b1), (a2, b2) \(\in\) A \(\times\) B such that f (a1, b1) = f (a2, b2)
\(\Rightarrow\) (b1, a1) = (b2, a2)
\(\Rightarrow\) b1 = b2 and a1 = a2
\(\Rightarrow\) (a1, b1) = (a2, b2)
\(\therefore\) f is one-one.
Now, let (b, a) \(\in\) B \(\times\) A be any element.
Then, there exists (a, b) \(\in\) A \(\times\) B such that f (a, b) = (b, a).
\(\therefore\) f is onto.
Hence, f is bijective.
Question 9:
Let f : N → N be defined by
\[
f(x) = \begin{cases}
\frac{n + 1}{2}, & \text{if n is odd } \\
\frac{n}{2}, & \text{if n is even}
\end{cases}
\text{for all} n \in \mathbb{N}
\]
State whether the function f is bijective. Justify your answer.
Answer
\[
f(x) = \begin{cases}
\frac{n + 1}{2}, & \text{if n is odd } \\
\frac{n}{2}, & \text{if n is even}
\end{cases}
\text{for all} n \in \mathbb{N}
\]
For n = 1, 2
f (1) = \(\frac{n + 1}{2} = \frac{1 + 1 }{2}\) = 1 and f (2) = \(\frac{n}{2} = \frac{2 }{2}\) = 2
f (1) = f (2) but 1 \(\neq\) 2.
\(\therefore\) f is not one-one.
For a natural number, ‘a’ in co-doamin N
If n is odd
\(\therefore\) n = 2r + 1 for some r \(\in \) N. Then, there exists 4r + 1 \(\in\) N such that
f (4r + 1) = \(\frac{4r + 1 + 1}{2}\) = 2r + 1
If n is even
\(\therefore\) n = 2r for some r \(\in \) N. Then, there exists 4r \(\in\) N such that
f (4r) = \(\frac{4r}{2}\) = 2r
\(\therefore\) f is onto.
Hence, f is not a biective function.
Question 10:
Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by
\(f(x) = (\frac{x – 2}{x – 3})\). Is f one-one and onto? Justify your answer.
Answer
For x 1, x 2 \(\in\) A with f (x 1) = f (x 2)
\(\Rightarrow\) \(\frac{x_1 – 2}{x_1 – 3}\) = \(\frac{x_2 – 2}{x_2 – 3}\)
\(\Rightarrow\) (x 1 – 2)(x 2 – 3) = (x 1 – 3)(x 2 – 2)
\(\Rightarrow\) x 1x 2 – 3x 1 – 2x 2 + 6 = x 1x 2 – 2x 1 – 3x 2 + 6
\(\Rightarrow\) -3x 1 – 2x 2 = -2x 1 – 3x 2
\(\Rightarrow\) – x 1 = – x 2 \(\Rightarrow\) x 1 = x 2
\(\therefore\) f (x 1) = f (x 2) \(\Rightarrow\) x 1 = x 2
\(\therefore\) f is one-one.
Now let y = f (x) = \(\frac{x – 2}{x – 3} \Rightarrow\) (x – 3)y = x – 2
\(\Rightarrow\) xy – 3y = x – 2 \(\Rightarrow\) x(y – 1) = 3y – 2
\(\Rightarrow\) x = \(\frac{3y – 2}{y – 1}\)
\(\Rightarrow\) For every y \(\neq\) 1, i.e., when y \(\in\) co-domain B, there exists
\(\;\) x = \(\frac{x – 2}{x – 3} \in\) = R – {3} such that
\(\;\) f (x) = \(\frac{x – 2}{x – 3} = \frac{\frac{3y – 2}{y – 1} – 2}{\frac{3y – 2}{y – 1} – 3}\)
\(\quad\) = \(\frac{3y – 2 – 2y + 2}{3y – 2 – 3y + 3}\) = y
\(\therefore\) f is onto.
Question 11:
Let f : R → R be defined as f(x) = x4. Choose the correct answer.
(A) f is one-one onto (B) f is many-one onto
(C) f is one-one but not onto (D) f is neither one-one nor onto.
Answer
f: R→ R is defined as f(x) = x 4 .
Let x, y ∈ R such that f(x) = f(y).
⇒ x 4 = y 4 ⇒ x = ±y ∴f(x 1 ) = f ( x2 ) does not imply that x 1 = x 2
For instance, f(1) = f(-1) = 1
∴ f is not one-one.
Consider an element 2 in co-domain R .
It is clear that there does not exist any x in domain R such that f(x) = 2.
∴ f is not onto. Hence, function f is neither one-one nor onto.
The correct answer is D.
Question 12:
Let f : R → R be defined as f(x) = 3x. Choose the correct answer.
(A) f is one-one onto (B) f is many-one onto
(C) f is one-one but not onto (D) f is neither one-one nor onto
Answer
f: R → R be defined as f(x) = 3x.
Let x, y ∈ R such that f(x) = f(y).
⇒ 3x = 3y
⇒ x = y
∴ f is one – one
Also, for any real number (y) in co-domain R, there exists y/3 in R such that f(y/3) = 3(y/3) = y
∴ f is onto.
Hence, function f is one-one and onto.
The correct answer is A.
Miscellaneous Exercise on Chapter 1
Question 1:
Show that the function f : R → {x ∈ R : – 1 < x < 1} defined by \(f(x) = \frac{x}{1 + |x|},\) ,x ∈ R is one one and onto function.
Answer
f : R → {x ∈ R : – 1 < x < 1} defined by \(f(x) = \frac{x}{1 + |x|},\) ,x ∈ R
Suppose f(x) = f(y), where x, y ∈ R.
\(\Rightarrow \frac{x}{1 + x} = \frac{y}{1 + y} \Rightarrow\) x – xy = y + xy
\(\Rightarrow\) 2xy = x – y
Since x is positive and y is negative.
x > y \(\Rightarrow\) x – y > 0
But, 2xy is negative.
Then, 2xy \(\neq\) x – y.
Thus, the case of x being positive and y being negative can be ruled out.
When x and y are both positive,
\(f(x) = f(y) \Rightarrow \frac{x}{1+x} = \frac{y}{1+y} \Rightarrow x + xy = y +xy \Rightarrow x = y.\)
When x and y are both negative,
\(f(x) = f(y) \Rightarrow \frac{x}{1 – x} = \frac{y}{1 – y} \Rightarrow x – xy = y – xy \Rightarrow x = y.\)
\(\therefore\) f is one-one.
Now, let y \(\in\) ∈ R such that -1<y<1.
If x is negative, then there exists x = \(\frac{y}{1 + y} \in R\) such that
\(f(x) = f\left(\frac{y}{1 + y}\right) = \frac{(\frac{y}{1+y})}{1 + \frac{y}{1+y}} = \frac{\frac{y}{1+y}}{1 + \frac{-y}{1+y}} = \frac{y}{1 + y – y} = y\)
If x is positive, then there exists x = \(\frac{y}{1 – y} \in R\) such that
\(f(x) = f\left(\frac{y}{1 – y}\right) = \frac{(\frac{y}{1-y})}{1 + \frac{y}{1-y}} = \frac{\frac{y}{1-y}}{1 + \frac{y}{1-y}} = \frac{y}{1 – y + y} = y\)
\(\therefore\) f is onto.
Hence, f is one-one and onto.
Question 2
Show that the function f : R → R given by f(x) = x3 is injective.
Answer
f: R → R is given as f(x) = x 3 .
Suppose f(x) = f(y), where x, y ∈ R. ⇒ x 3 = y 3 … (1)
Now, we need to show that x = y.
Suppose x ≠ y, their cubes will also not be equal. ⇒ x 3 ≠ y 3
However, this will be a contradiction to (1). ∴ x = y
Hence, f is injective.
Question 3:
Given a non empty set X, consider P(X) which is the set of all subsets of X.
Define the relation R in P(X) as follows:
For subsets A, B in P(X), ARB if and only if A ⊂ B. Is R an equivalence relation on P(X)? Justify your answer.
Answer
Since every set is a subset of itself, ARA for all A ∈ P(X). ∴R is reflexive.
Let ARB ⇒ A ⊂ B. This cannot be implied to B ⊂ A.
For instance, if A = {1, 2} and B = {1, 2, 3}, then it cannot be implied that B is related to A.
∴ R is not symmetric.
Further, if ARB and BRC, then A ⊂ B and B ⊂ C.
⇒ A ⊂ C ⇒ ARC
∴ R is transitive.
Hence, R is not an equivalence relation since it is not symmetric.
Question 4:
Find the number of all onto functions from the set {1, 2, 3,……, n} to itself.
Answer
Onto functions from the set {1, 2, 3, … ,n} to itself is simply a permutation on n symbols 1, 2, …, n.
Thus, the total number of onto maps from {1, 2, … , n} to itself is the same as the total number of permutations on n symbols 1, 2, …, n, which is n!
Question 5:
Let A = {– 1, 0, 1, 2}, B = {– 4, – 2, 0, 2} and f, g : A → B be functions defined by f(x) = x2 – x, x ∈ A and g( x) = \(2|x – \frac{1}{2}| – 1\) x ∈ A. Are f and g equal? Justify your answer.
(Hint: One may note that two functions f : A → B and g : A → B such that f(a) = g(a) ∀ a ∈ A, are called equal functions).
Answer
It is given that A = {– 1, 0, 1, 2}, B = {– 4, – 2, 0, 2}
Also, it is given that f, g: A\(\rightarrow\)B are defined by f(x) = x2 – x, x ∈ A and g(x) = \(2|x – \frac{1}{2}| – 1\) x ∈ A.
f(-1) = 2 and g(-1) = 2
\(\Rightarrow\) f(-1) = g(-1)
f(0) = 0 and g(0) = 0
\(\Rightarrow\) f(0) = g(0)
f(1) = 0 and g(1) = 0
\(\Rightarrow\) f(1) = g(1)
f(2) = 2 and g(2) = 2
\(\Rightarrow\) f(2) = g(2)
Hence, the functions f and g are equal.
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions
Question 6:
Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is
(A) 1 (B) 2 (C) 3 (D) 4
Answer
It is clear that 1 is reflexive and symmetric but not transitive.
Therefore, option (A) is correct.
Question 7:
Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is
(A) 1 (B) 2 (C) 3 (D) 4
Answer
2, Therefore, option (B) is correct.
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions
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