NCERT Solutions for Maths Class 12 Exercise 1.1

# NCERT Solutions for Maths Class 12 Exercise 1.1

Hello Students! In this post, you will find the complete NCERT Solutions for Maths Class 12 Exercise 1.1.

## NCERT Solutions for Maths Class 12 Ex 1.1

### Maths Class 12 Ex 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}

Solution:
(i) Relation R in the set A = {1, 2, 3, …. 13, 14} defined as R = {(x, y): 3x – y = 0}
(a) Putting y = x, we get 3x – x ≠ 0
R is not reflexive.
(b) If 3x – y = 0, then 3y – x ≠ 0
R is not symmetric.
(c) If 3x – y = 0, 3y – z = 0,then 3x – z ≠ 0
R is not transitive.

(ii) Relation R in the set N of natural numbers defined by R = {(x, y): y = x + 5 and x < 4}
(a) Putting y = x, we get x ≠ x + 5
R is not reflexive.
(b) If y = x + 5, then x ≠ y + 5
R is not symmetric.
(c) If y = x + 5, z = y + 5, then z ≠ x + 5
R is not transitive.

(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y): y is divisible by x}
(a) Putting y = x, we get, x is divisible by x
R is reflexive.
(b) If y is divisible by x, then x is not divisible by y, when x ≠ y
R is not symmetric.
(c) If y is divisible by x and z is divisible by y, then z is divisible by x.

For example, if 8 is divisible by 4 and 16 is divisible by 8, then 16 is divisible by 4 R is transitive.

(iv) Relation R in the set Z of all integers defined as R = {(x, y): x – y is an integer}
(a) Putting y = x, we get, x – x = 0, which is an integer
R is reflexive.
(b) If x – y is an integer, then y – x is also an integer
R is symmetric.
(c) If x – y is an integer and y – z is an integer, then x – z is also an integer
R is transitive.

(v) Relation R in the set of human beings in a town at a particular time is given.
(a) R = {(x, y): x and y work at the same place}.

It is reflexive as x works at the same place.

It is symmetric since x and y or y and x work at the same place.
If x, y work at the same place and y, z work at the same place, then x, z also work at the same place. Therefore, R is transitive.

(b) R = {(x, y): x and y live in the same locality}
It is reflexive as x lives in the same locality.

It is symmetric since x and y or y and x live in the same locality.
If x, y live in the same locality and y, z live in the same locality, then x, z also live in the same locality. Therefore, R is transitive.

(c) R = {(x, y): x is exactly 7 cm taller than y}.

It is not reflexive because x cannot be 7 cm taller than x.

It is not symmetric because x is exactly 7 cm taller than y but y cannot be exactly 7 cm taller than x.

It is not transitive because if x is exactly 7 cm taller than y and y is exactly 7 cm taller than z, then x is not exactly 7 cm taller than z.

(d) R = {(x, y): x is wife of y}
It is not reflexive because x cannot be wife of x.

It is not symmetric because x is wife of y but y is not wife of x.
It is transitive because if x is a wife of y then y is a male and cannot be the wife of anybody else. Here, (x, y)
R but (y, z) do not belong to R. Therefore, by the special case, it is transitive.

(e) R = {(x, y): x is a father of y}
It is not reflexive because x cannot be the father of himself.

It is not symmetric because x is a father of y but y cannot be the father of x.
It is not transitive because x is a father of y and y is a father of z but x cannot be the father of z.

### Maths Class 12 Ex 1.1 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.

Solution:
(i) R is not reflexive since a is not less than or equal to a² for all a
R, for example, 1/3 is not less than or equal to 1/9.
(ii) R is not symmetric since a ≤ b² but b is not less than or equal to a², for example, 1 < 2² but 2 is not less than 1².
(iii) R is not transitive because for a ≤ b² and b ≤ c², a is not less than or equal to c², for example, 3 ≤ (-2)², -2 ≤ (-1)², but 3 is not less than or equal to (-1)².

### Maths Class 12 Ex 1.1 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.

Solution:
(i) R is not reflexive because a ≠ a + 1.
(ii) R is not symmetric because if b = a + 1, then a ≠ b + 1
(iii) R is not transitive because if b = a + 1 and c = b + 1, then c ≠ a + 1.

### Maths Class 12 Ex 1.1 Question 4.

Show that the relation R in R defined as R = {(a, b): a ≤ b}, is reflexive and transitive but not symmetric.

Solution:
(i) R is reflexive because if b = a, then a ≤ a
a = a is true.
(ii) R is not symmetric because a ≤ b but b is not less than or equal to a, for example, 3 ≤ 5, but 5 is not less than or equal to 3.
(iii) R is transitive because if a ≤ b and b ≤ c, then a ≤ c, for example, 3 ≤ 5, 5 ≤ 6
3 ≤ 6.

### Maths Class 12 Ex 1.1 Question 5.

Check whether the relation R in R defined by R = {(a, b): a ≤ b³} is reflexive, symmetric or transitive.

Solution:
(i) R is not reflexive since a is not less than or equal to a3 for all a
R, for example, 1/3 is not less than (1/3)3.
(ii) R is not symmetric since a ≤ b3 but b is not less than or equal to a3, for example, 1 < 23 but 2 is not less than 13.
(iii) R is not transitive because for a ≤ b3 and b ≤ c3, a is not less than or equal to c3, for example, 3 ≤ (-2)3, -2 ≤ (-1)3, but 3 is not less than or equal to (-1)3.

### Maths Class 12 Ex 1.1 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.

Solution:
(i) Since (1, 1), (2, 2), (3, 3) do not belong to relation R.
R is not reflexive.
(ii) Since (1, 2) and (2, 1) belong to R. Therefore, R is symmetric.
(iii) Since (1, 2) and (2, 1) belong to R but (1, 1) does not belong to R. Therefore, R is not transitive.

### Maths Class 12 Ex 1.1 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.

Solution:
(i) The number of pages in the same book remains the same.
Therefore, the relation R is reflexive.
(ii) If the book x has the same number of pages as the book y, then the book y has the same number of pages as the book x.
Therefore, the relation R is symmetric.
(iii) If the books x and y have the same number of pages and the books y and z have also the same number of pages, then the books x and z have the same number of pages.
Therefore, the relation R is transitive.

Thus, the relation R is an equivalence relation.

### Maths Class 12 Ex 1.1 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}.

Solution:
We have the set A = {1, 2, 3, 4, 5} and the relation R = {(a, b): |a – b| is even}
Therefore, R = {(1, 3), (1, 5), (3, 5), (2, 4)}
(a) (i) Suppose a is an element of the set A. Then |a – a| = 0, which is even.
Therefore, R is reflexive.
(ii) If |a – b| is even, then |b – a| is also even. For example, if |1 – 3| is even, then |3 – 1| is also even.
Therefore, R is symmetric.
(iii) If |a – b| and |b – c| are even, then |a – c| is also even. For example, if |1 – 3| and |3 – 5| are even, then |1 – 5| is  also even.
Therefore, R is transitive.

Hence, R is an equivalence relation.

(b) All the elements of {1, 3, 5} are related to each other.
Because |1 – 3| = 2, |3 – 5| = 2, |1 – 5| = 4. All are even numbers.
Similarly, all the elements of {2, 4} are related to each other.

Because |2 – 4| = 2, which is an even number.
Now, no element of set {1, 3, 5} is related to any element of set {2, 4}.

Because |1 – 2| = 1, |1 – 4| = 3, |3 – 2| = 1, |3 – 4| = 1, |5 – 2| = 3, |5 – 4| = 1 are not even numbers.

### Maths Class 12 Ex 1.1 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.

Solution:
The set A = {x
Z: 0 ≤ x ≤ 12} = {0, 1, 2, ….., 12}
(i) R = {(a, b): |a – b| is a multiple of 4}
R = {(1, 5), (1, 9), (2, 6), (2, 10), (3, 7), (3, 11), (4, 8), (4, 12), (5, 9), (6, 10), (7, 11), (8, 12), (0, 0), (1, 1), (2, 2), ….. (12, 12)}
(a) For any element a
A, we have (a, a) R. For example, (0, 0), (1, 1), (2, 2), ….., (12, 12) R.
R is reflexive.
(b) Here, (1, 5)
R and (5, 1) R, i.e., (a, b) and (b, a) both belong to R.

Therefore, R is symmetric.
(c) Here, (1, 5)
R and (5, 9) R implies (1, 9) R. It means for a, b, c A, (a, b) R and (b, c) R implies (a, c) R. This shows when |a – b| and |b – c| are both multiples of 4, then |a – c| is also a multiple of 4. Therefore, R is transitive.

R is an equivalence relation.

The set of all elements related to 1 is {1, 5, 9}.

(ii) R = {(a, b): a = b}
R = {(0, 0), (1, 1), (2, 2), ….., (12, 12)}
(a) For any element a
A, we have a = a, it means (a, a) R
Therefore, R is reflexive.
(b) Again if (a, b)
R, then (b, a) also R. It means if a = b, then b = a.
Therefore, R is symmetric.
(c) If (a, b)
R and (b, c) R, then (a, c) R. It means if a = b and b = c, then a = c.
Therefore, R is transitive.

The set of all elements related to 1 is {1}.

### Maths Class 12 Ex 1.1 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.

Solution:
(i) Let A = {straight lines in a plane}
R = {(a, b): a is perpendicular to b}

(a) A line a is not perpendicular to itself. Therefore, R is not reflexive.

(b) If a is perpendicular to b, then b is also perpendicular to a. Therefore, R is symmetric.

(c) If a is perpendicular to b and b is perpendicular to c, then a is not perpendicular to c. Therefore, R is not transitive.

Hence, R is symmetric but neither reflexive nor transitive.

(ii) Let A = {real numbers}

R = {(a, b): a > b}
(a) An element is not greater than itself.
Therefore, R is not reflexive.
(b) If a > b, then b is not greater than a. Therefore, R is not symmetric.
(c) If a > b and b > c, then a > c. Therefore, R is transitive.
Hence, R is transitive but neither reflexive nor symmetric.

(iii) Let A = {1, 2, 3}

R = {(a, b): |a – b| is either 0 or 1}
R = {(1, 1), (2, 2), (3, 3) (1, 2), (2, 1), (2, 3), (3, 2)}

(a) (1, 1), (2, 2), (3, 3) R, therefore, R is reflexive.
(b) (1, 2), (2, 1), (1, 3), (3, 1)
R, therefore, R is symmetric.
(c) Since (1, 2)
R and (2, 3) R but (1, 3) R. Therefore, R is not transitive.

Hence, R is reflexive and symmetric but not transitive.

(iv) Let A = {1, 2, 3}

R = {(a, b): a ≤ b}

R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}
(a) (1, 1), (2, 2), (3, 3)
R, therefore, R is reflexive.
(b) (1, 2)
R, but (2, 1) R, therefore, R is not symmetric.
(c) (1, 2)
R and (2, 3) R implies (1, 3) R, therefore, R is transitive.

Hence, R is reflexive and transitive but not symmetric.

(v) Let A = {1, 2, 3}

R = {(a, b): 0 < |a – b| ≤ 2}

R = {(1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)}
(a) Since (1, 1), (2, 2), (3, 3)
R, therefore, R is not reflexive.
(b) Since (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)
R, therefore, R is symmetric.
(c) Since (1, 2)
R and (2, 3) R implies (1, 3) R, therefore, R is transitive.

Hence, R is symmetric and transitive but not reflexive.

### Maths Class 12 Ex 1.1 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.

Solution:
Let O be the origin. Then the relation R is defined as
R = {(P, Q): OP = OQ}
(i) R is reflexive. For any distance OP, OP = OP. Therefore, R is reflexive.
(ii) R is symmetric. If OP = OQ, then OQ = OP. Therefore, R is symmetric.
(iii) R is transitive. Let OP = OQ and OQ = OR, then OP = OR. Therefore, R is transitive.
Hence, the relation R is an equivalence relation.
For all points P ≠ (0, 0), OP = constant.

Therefore, P lies on a circle with origin as centre.

### Maths Class 12 Ex 1.1 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?

Solution:
(i) In a set A of all triangles, the relation R is defined as

R = {(T1, T2): T1 is similar to T2}
(a) R is reflexive. A triangle T is similar to itself. Therefore, (T, T)
R for all T A.
Therefore, R is reflexive.
(b) R is symmetric. If triangle T1 is similar to triangle T2, then triangle T2 is similar to triangle T1. Therefore, R is symmetric.
(c) R is transitive. If triangle T1 is similar to triangle T2 and triangle T2 is similar to triangle T3, then triangle T1 is similar to triangle T3. Therefore, R is transitive.
Hence, R is an equivalence relation.

(ii) According to the proportionality theorem, two triangles are similar, if their sides are proportional. Now, sides 3, 4, 5 of triangle T1 are proportional to the sides 6, 8, 10 of triangle T3.
Therefore, triangle T1 is related to triangle T3.

### Maths Class 12 Ex 1.1 Question 13.

Show that the relation R defined in the set A of all polygons as R = {(P1, P2): P1 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?

Solution:
Let n be the number of sides of polygon P1. Then, the relation R is defined as
R = {(P1, P2): P1 and P2 are n-sided polygons}
(i) (a) R is reflexive. Any polygon P1 has n sides. Therefore, R is reflexive.
(b) R is symmetric. If P1 has n sides, P2 also has n sides and if P2 has n sides, P1 also has n sides. Therefore, R is symmetric.
(c) R is transitive. If P1, P2 are n-sided and P2, P3 are n-sided polygons, then P1 and P3 are also n-sided polygons. Therefore, R is transitive.

Hence, R is an equivalence relation.
(ii) The set of all elements in A related to the right angle triangle T with sides 3, 4, and 5 = {all the triangles in a plane}

### Maths Class 12 Ex 1.1 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.

Solution:
It is given that the set L = {all the lines in XY plane}

R = {(L1, L2): L1 is parallel to L2}
(i) (a) R is reflexive. The line L1 is parallel to itself. Therefore, R is reflexive.
(b) R is symmetric. If L1 is parallel to L2, then L2 is parallel to L1. Therefore, R is symmetric.
(c) R is reflexive. If L1 is parallel to L2 and L2 is parallel to L3, then L1 is parallel to L3. Therefore, R is transitive.
Hence, R is an equivalence relation.
(ii) The set of all lines related to the line y = 2x + 4 is y = 2x + c, where c
R, because y = 2x + c is always parallel to y = 2x + 4.

### Maths Class 12 Ex 1.1 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.

Solution:
(b) R is reflexive and transitive but not symmetric.
(i) (1, 1), (2, 2), (3, 3), (4, 4)
R. Therefore, R is reflexive.

(ii) (1, 2) R but (2, 1) does not belong to R. Therefore, R is not symmetric.

(iii) (1, 2), (2, 2) R implies (1, 2) R. (1, 3), (3, 3) R implies (1, 3) R. Again, (1, 3), (3, 2) R implies (1, 2) R. Therefore, R is transitive.

### Maths Class 12 Ex 1.1 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

Solution:
(c) (6, 8)
R satisfies the condition a = b – 2, because if b = 8, a = 8 – 2 = 6 and b > 6.

If we take b = 7, a = 7 – 2 = 5 and (5, 7) is not given in the option.
Therefore, option (c) is correct.