Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 1 Relations and Functions Ex 1.1 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 1 Relations and Functions 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,â€¦.13, 14} defined as R={(x, y):3x â€“ 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 â€“ 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, ….. , 14) defined as

R = {(x, y) : 3x – y = 0}

(a) Put y = x, 3x – x â‰ 0 â‡’ R is not reflexive.

(b) If 3x – y = 0, then 3x – x = 0 â‡’ R is not symmetric.

(c) If 3x – y = 0, 3y – z = 0, then 3x – z â‰ 0 â‡’ R is not transitive.

(ii) Relation in tire set N of natural number is defined by

R = {(x, y): y – x + 5 and x < 4}

(a) Putting y = x, x â‰ x + 5 â‡’ R is not reflexive.

(b) Putting y = x + 5, then x â‰ y + 5 â‡’ R is not symmetric.

(c) If y – x + 5, z = y + 5, then z â‰ x + 5 â‡’ R is not transitive.

(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y): x is divisible by y}

(a) Putting y = x, x is divisible by x â‡’ R is reflexive.

(b) If y is divisible by x, then x is not divisible by y, when x â‰ y â‡’ R is not symmetric.

(c) If x is divisible by y and y is divisible by z, then x is divisible by z,

e.g., 4 is divisible by 2 and 2 is divisible by 1

â‡’ 4 is divisible by 1 â‡’ R is transitive.

(iv) Relation R in Z of all integers defined as R = {(x, y): x – y is an integer}

(a) x – x = 0 is an integer â‡’ R is reflexive.

(b) x – y is an integer, so is x â‡’ R is symmetric.

(c) x – y is an integer, y – z is an integer and x is also an integer â‡’ R is transitive.

(v) R is a set of human beings in a town at a particular time given by

(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.

It is transitive since if x, y work at the same place and y, z work at the same place, then x and z also work at the same place.

(b) R: f(x, y): x and y live in the same locality}

With similar reasoning as in part (a), R is reflexive, symmetric and transitive.

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

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

It is not symmetric : If x is exactly 7 cm taller than y, then y cannot be exactly 7 cm taller than x.

It is not transitive : If x is exactly 7 cm taller than y and if 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}

R is not reflexive : x cannot be wife of x.

R is not symmetric : x is wife of y but y is not wife of x.

R is not transitive : if x is a wife of y, then y cannot be the wife of anybody else.

(e) R = {(x, y): x is father of y}

It is not reflexive : x cannot be father of himself.

It is not symmetric : x is father of y but y cannot be the father of x.

It is not transitive : If x is father of y and y is father of z, then x cannot be the father of z.

Question 2.

Show that the relation R in the set R of real numbers, defined as

R = {(a, b) : a â‰¤ b^{2}}, is neither reflexive, nor symmetric, nor transitive.

Solution:

(i) R is not reflexive âˆµ a is not less than or equal to a^{2} for all a âˆˆ R, e.g., \(\frac{1}{2}\) is not less than \(\frac{1}{4}\).

(ii) R is not symmetric, since if a â‰¤ b^{2}, then b is not less than or equal to a^{2}, e.g., 2 < 5^{2} but 5 is not less than or equal to 2^{2}.

(iii) R is not transitive : Here, also if a â‰¤ b^{2}, b â‰¤ c^{2}, then a is not less than or equal to c^{2}, e.g., 2 < (- 2)^{2}, – 2 < (-1)^{2}. But 2 is not less than (-1)^{2}.

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 and transitive.

Solution:

(i) R is not reflexive : a â‰ a + 1.

(ii) R is not symmetric : If b = a + 1, then a â‰ b + 1.

(iii) R is not transitive :If b = a + 1, c = b + 1, then c â‰ a + 1.

Question 4.

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

Solution:

R = {(a, b): a â‰¤ b}

(i) R is reflexive. Replacing b by a, a â‰¤ a â‡’ a = a is true.

(ii) R is not symmetric : a < b and b < a which is not true.

e.g., 2 < 3, but 3 is not less than 2.

(iii) R is transitive : If a â‰¤ b and b â‰¤ c, then a â‰¤ c.

e.g. 2 < 3, 3 < 4 â‡’ 2 < 4.

Question 5.

Check whether the relation R, defined by R = {(a, b) : a â‰¤ b^{3}} is reflexive, symmetric or transitive.

Solution:

(i) a â‰¤ a^{3} is not true. e.g. \(\frac{1}{2}\) is not less than (\(\frac{1}{2}\))^{3}

âˆ´ R is not reflexive.

(ii) If a â‰¤ b^{3}, then b is not less than or equal a3, e.g. 1 â‰¤ 2^{3}, but 2 is not less than 1^{3}.

â‡’ R is not symmetric.

(iii) If a < b^{3} and b < c^{3}, then but a is not necessarily less than c^{3}.

e.g. a = 7, b = 2, c = 1.5

a < b^{3}, âˆµ 7 < 2^{3} = 8; b < c^{3}, âˆµ 2 < (1.5)^{3} = 3.375

But a > c^{3}, âˆµ 7 > (1.5)^{3} = 3.375

âˆ´ R is not Transitive

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) (1, 1), (2, 2), (3, 3) do not belong to relation R.

âˆ´ R is not reflexive.

(ii) It is symmetric because (1, 2) and (2, 1) belong to R.

(iii) There are only two elements 1 and 2 in this relation and there is no third element c in it. â‡’ R is not transitive.

Question 7.

Show that the relation R in the set A of all books in a library of a college, given by R = {(x, y): x and y have the same number of pages} is an equivalence relation.

Solution:

(i) The number of pages in a book remain the same.

â‡’ Relation R is reflexive.

(ii) The book x has the same number of pages as the book y.

â‡’ Book y has the same number of pages as the book x.

â‡’ The relation R is symmetric.

(iii) Books x and y have the same number of pages.

â‡’ Also books y and z have the same number of pages.

â‡’ Books x and z also have the same number of pages.

â‡’ R is transitive also.

Thus, R is an equivalence relation.

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:

A = {1, 2, 3, 4, 5} and R = {(a, b): | a – b | is even}

R = {(1, 3), (1, 5), (3, 5), (2, 4), (3, 1), (5, 1), (5, 3), (4, 2)}

(a) (i) Let us take any element a of set A.

Then, | a – a | = 0 is even,

â‡’ R is reflexive.

(ii) If | a – b | is even, then | b – a | is also even, where R = {{a, b): | a – b | is even}

â‡’ R is symmetric.

(iii) Further a – c = a – b + b – c

If | a – b | and | b – c | are even, then their sum | a – b + b – c | is also even.

â‡’ | a – c | is even, R is transitive.

Hence, R is an equivalence relation.

(b) Elements of {1, 3, 5} are related to each other.

Since | 1 – 3 | = 2, | 3 – 5| = 2, | 1 – 5 | = 4. All are even numbers.

â‡’ Elements of {1, 3, 5} are related to each other. Similarly, elements of {2, 4} are related to each other, since | 2 – 4 | = 2 is an even number. No element of set {1, 3, 5} is related to any element of {2,4}, because | 2 – 1 | is not even, | 2 – 3 | is not even, etc.

Question 9.

Show that the relation R in the set A = [x âˆˆ Z : 0 â‰¤ x â‰¤ 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 âˆˆ Z : 0 â‰¤ x â‰¤ 12} = {0, 1, 2, ………, 12}

(i) R = {(a, b): | a – b | is a multiple of 4}

|a – b | = 4k or b = a + 4k.

âˆ´ 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), (0, 4), (4, 0), (0, 8), (8, 0), (0, 12), (12, 0), (5,1), (9,1), (6, 2), (10, 2), (7, 3), (11, 3), (8, 4), (12,4), (9, 5), (10,6), (11, 7), (12, 8)}

(a) a – a = 0 = 4k, where k = 0 â‡’ (a, a) âˆˆ R.

âˆ´ R is reflexive.

(b) If | a – b | = 4k, then | b – a | = 4k i.e., (a, b) and (b, a) both belong to R. âˆ´ R is symmetric.

(c) a – c = a – b + b – c.

When a – b and b – c both are multiples of 4, then a – c is also a multiple of 4. This shows if (a, b), (b, c) âˆˆ R, then (a, c) also âˆˆ R.

âˆ´ R is an equivalence relation.

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

(ii) R – {(a, b) : a = b} = {(0, 0), (1, 1), (2, 2), ……. ,(12, 12)}

(a) a = a â‡’ (a, a) âˆˆ R âˆ´R is reflexive.

(b) Again if (a, b) âˆˆ R, then (b, a) also âˆˆR.

Since a – b and (a, b) âˆˆ R â‡’ R is symmetric.

(c) If (a, b) âˆˆR, then (b, c) âˆˆ R â‡’ a = b = c

âˆ´ a = c â‡’ (a, c) âˆˆ R. Hence R is transitive.

Set related to 1 is {1}.

Question 10.

Give examples of relations which are

(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:

Let A = set of straight lines in a plane.

(i) R : {(a, b): a is perpendicular to b]

Let a and b be two perpendicular lines.

(a) If line a is perpendicular to b, then b is perpendicular to a â‡’ R is symmetric.

(b) But a is not a perpendicular to itself.

âˆ´ R is not reflexive.

(c) If a is perpendicular to b and & is perpendicular to c, then a is not perpendicular to c.

âˆ´R is not transitive.

Thus, R is symmetric but neither reflexive nor transitive.

(ii) Let A = set of real numbers and R = {(a, b): a > b}

(a) An element is not greater than itself.

âˆ´ R is not reflexive.

(b) If a > b, then b is not greater than a.

â‡’ R is not symmetric.

(c) If a > b and b> c, then a > c.

Thus, R is transitive.

Hence, R is transitive but neither reflexive nor symmetric.

(iii) The relation R in the set {1, 2, 3}, is given by R = [(a, b) : a + b â‰¤ 4}

R = {(1, 1), (1, 2), (2, 1), (1, 3), (3, 1), (2, 2)}

Here (1,1), (2,2) âˆˆ R â‡’ R is reflexive.

(1, 2), (2, 1), (1, 3), (3, 1) âˆˆ R â‡’ R is symmetric.

But it is not transitive, since (2, 1) âˆˆ R and (1, 3) âˆˆ R does not imply (2, 3) âˆˆ R.

(iv) The relation R in the set {1, 2, 3}, given by

R = {(a, b): a < b} = (1, 1), (1, 2), (2, 2), (3, 3), (2, 3), (1, 3)

(a) (1, 1), (2, 2), (3, 3) âˆˆ R â‡’ R is reflexive.

(b) (1, 2) âˆˆ R, but (2, 1) âˆ‰ R â‡’ R is not symmetric.

(c) (1,2) âˆˆ R, (2,3) âˆˆ R. Also, (1, 3) âˆˆ R

â‡’ R is transitive.

(v) The relation R in the set {1,2, 3}, given by R = [(a, b) : 0 < | a – b | â‰¤ 2}

= {(1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)} .

(a) R is not reflexive,

since, (1, 1), (2, 2), (3, 3) do not belong to R.

(b) R is symmetric.

âˆµ (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2) âˆˆ R

(c) R is transitive because (1, 2) âˆˆ R, (2, 3) âˆˆ R and also (1, 3) âˆˆ R.

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 the same as the distance of Q from the origin}, is an equivalence relation. Further, show that the set of all points related to P â‰ (0, 0) is the circle passing through P with origin as the centre.

Solution:

Let O be the origin, then the relation R = {(P, Q): OP = OQ}

(i) R is reflexive. Take any distance OP.

We know OP = OP. âˆ´ R is reflexive.

(ii) R is symmetric. If OP = OQ, then OQ = OP â‡’ R is symmetric.

(iii) R is transtive. If OP = OQ and OQ = OR, then OP = OR. âˆ´ R is transitive.

Hence, R is an equivalence relation.

Since OP = k (consonant) â‡’ P lies on a circle with centre

at the origin and radius k.

Question 12.

Show that the relation R, defined in the set A of all triangles as R = {(T_{1}, T_{2}) : T_{1} is similar to T_{2}}, is an equivalence relation. Consider three right triangles T_{1} with sides 3, 4, 5; T_{2} with sides 5, 12, 13 and T_{3} with sides 6, 8, 10. Which triangles among T_{1}, T_{2} and T_{3} are related?

Solution:

(i) In a set of triangles, R = {(T_{1}, T_{2}): T_{1} is similar to T_{2}},

(a) A triangle T_{1} is similar to itself. R is reflexive.

(b) If triangle T_{1} is similar to triangle T_{2}, then T_{2} is similar to triangle T_{1}. âˆ´ R is symmetric.

(c) Let T_{1} is similar to triangle T_{2} and T_{2} to T_{3}. Then, triangle T_{1} is similar to triangle T_{3}. âˆ´ R is transitive.

Hence, R is an equivalence relation.

(ii) Two triangles are similar if their sides are proportional, blow sides 3, 4, 5 of traiangle T_{1} are proportional to the sides 6, 8, 10 of triangle T_{3}. Therefore, T_{1} is related to T_{3}.

Question 13.

Show that the relation R, defined in the set A of all polygons as R = {(P_{1}, P_{2}) : P_{1} and P_{2} have the 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 a polygon P_{1}.

R = {(P_{1}, P_{2}): P_{1} and P_{2} are n sided polygons}.

(i) (a) Any polygon P_{1} has n sides â‡’ R is reflexive.

(b) If P_{1} has n sides, P_{2} also has n sides, then if P_{2} has n sides, P_{1} also has n sides.

â‡’ R is symmetric.

(c) Let P_{1}, P_{2}; P_{2}, P_{3} are pairs of n sided polygons. Thus, P_{1} and P_{3} are also n sided polygons.

â‡’ R is transitive.

Hence, R is an equivalence relation.

(ii) The set A = set of all the triangles in a plane.

Question 14.

Let L be the set of all the lines in XY-plane and R be the relation in L defined as R = {(L_{1}, L_{2}) : L_{1} is parallel to L_{2}}. Show that R is an equivalence relation. Find the set of all the lines related to the line y = 2x + 4.

Solution:

L = Set of all the lines in XY-plane.

R = {(L_{1}, L_{2}): L_{1} is parallel to L_{2}}

(i) (a) L_{1} is parallel to itself â‡’ R is reflexive.

(b) L_{1} is parallel to L_{2} â‡’ L_{2} is parallel to L_{1}.

â‡’ R is symmetric.

(c) Let L_{1} is parallel to L_{2} and L_{2} is parallel to L_{3}. Obviously, L_{1} is parallel to L_{3}.

â‡’ R is transitive.

Hence, R is an equivalence relation.

(ii) Set of parallel lines related to y = 2x + 4 is y = 2x + c, where c is an arbitrary constant.

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:

Let R be the relation with set {1,2, 3,4}, given by R = {(1,2), (2,2), (1,1), (4,4), (1,3), (3,3), (3,2)}.

(a) (1, 1), (2, 2), (3, 3), (4, 4) âˆˆ R â‡’ R is reflexive.

(b) (1, 2) âˆˆ R but (2, 1) does not belongs to R.

âˆ´ R is not symmetric.

(c) If (1, 3) âˆˆ R and (3, 2) âˆˆ R, then (1, 2) also âˆˆ R.

âˆ´ R is transitive.

â‡’ Part (b) is the correct answer.

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) âˆˆ R

(B) (3, 8) âˆˆ R

(C) (6, 8) âˆˆ R

(D) (8, 7) âˆˆ R

Solution:

Part (C) satisfies the condition

that a = b – 2, i.e., b = 8 – 2 and b > 6. Here, b = 8.

â‡’ Part (C) is the correct answer.