Mathematics solutions 2
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Solutions for Class 12 Maths – Chapter 1 – Relations and Functions Ex 1.1

  • Question 2

    • Relation R in the set of Real Numbers

    R = {(a, b) : a ≤ b2}

    • Reflexive: No
    • Symmetric: No
    • Transitive: No

    Explanation:

    The relation R is not reflexive because for any real number a, (a, a) is not in R since a is not less than or equal to a2.

    It is not symmetric because if (a, b) is in R, where a ≤ b2, it does not imply that (b, a) is in R since a is not necessarily less than or equal to b2.

    It is not transitive because if (a, b) and (b, c) are in R, where a ≤ b2 and b ≤ c2, it does not imply that (a, c) is in R since a is not necessarily less than or equal to c2.

  • Question 3

    • Relation R in the set {1, 2, 3, 4, 5, 6}

    R = {(a, b) : b = a + 1}

    • Reflexive: No
    • Symmetric: Yes
    • Transitive: Yes

    Explanation:

    The relation R is not reflexive because for any element a in the set, (a, a+1) is not in R since a+1 is not equal to a.

    It is symmetric because if (a, b) is in R, where b = a + 1, then (b, a) is also in R since a = b - 1, which satisfies the condition.

    It is transitive because if (a, b) and (b, c) are in R, where b = a + 1 and c = b + 1, then (a, c) is also in R since c = a + 2 = (a + 1) + 1.

  • Question 4

    • Relation R in the set of Real Numbers

    R = {(a, b) : a ≤ b}

    • Reflexive: Yes
    • Symmetric: No
    • Transitive: Yes

    Explanation:

    The relation R is reflexive because for any real number a, (a, a) is in R since a ≤ a.

    It is not symmetric because if (a, b) is in R, where a ≤ b, it does not imply that (b, a) is in R since b may not necessarily be less than or equal to a.

    It is transitive because if (a, b) and (b, c) are in R, where a ≤ b and b ≤ c, then (a, c) is also in R since a ≤ c.

  • Question 5

    • Relation R in the set of Real Numbers

    R = {(a, b) : a ≤ b3}

    • Reflexive: Yes
    • Symmetric: No
    • Transitive: Yes

    Explanation:

    The relation R is reflexive because for any real number a, (a, a) is in R since a ≤ a3.

    It is not symmetric because if (a, b) is in R, where a ≤ b3, it does not imply that (b, a) is in R since b may not necessarily be less than or equal to a3.

    It is transitive because if (a, b) and (b, c) are in R, where a ≤ b3 and b ≤ c3, then (a, c) is also in R since a ≤ c3.

  • Question 6

    • Relation R

    Symmetry:

    1. Show that the relation R is symmetric

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

    Since (1, 2) exists in R, (2, 1) must also exist.

    Thus, R is symmetric.

    Reflexivity:

    2. Show that the relation R is not reflexive

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

    Neither (1, 1) nor (2, 2) exists in R.

    Thus, R is not reflexive.

    Transitivity:

    3. Show that the relation R is not transitive

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

    Given (1, 2) and (2, 1), (1, 1) should exist for transitivity.

    However, (1, 1) does not exist in R.

    Thus, R is not transitive.

  • Question 7

    • Equivalence Relation for Book Pages

    Let A be the set of all books in the library of a college.

    Given relation R = {(x, y) : x and y have the same number of pages}.

    To show that R is an equivalence relation:

    Reflexivity:

    For any book x, it must have the same number of pages as itself.

    So, (x, x) exists for all x in A.

    Symmetry:

    If book x has the same number of pages as book y, then book y also has the same number of pages as book x.

    So, if (x, y) exists, then (y, x) also exists.

    Transitivity:

    If book x has the same number of pages as book y, and book y has the same number of pages as book z, then book x also has the same number of pages as book z.

    So, if (x, y) and (y, z) exist, then (x, z) also exists.

    Conclusion:

    Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

  • Question 8

    • Equivalence Relation in Set A

    Let A = {1, 2, 3, 4, 5}.

    Given relation R = {(a, b) : |a – b| is even}.

    To show that R is an equivalence relation:

    Reflexivity:

    For any element a in A, |a - a| = 0, which is even.

    So, (a, a) exists for all a in A.

    Symmetry:

    If |a - b| is even, then |b - a| is also even.

    So, if (a, b) exists, then (b, a) also exists.

    Transitivity:

    If |a - b| is even and |b - c| is even, then |a - c| is also even.

    So, if (a, b) and (b, c) exist, then (a, c) also exists.

    Relation in Subsets:

    All elements of {1, 3, 5} are related to each other:

    • For (1, 3): |1 - 3| = 2 (even)
    • For (1, 5): |1 - 5| = 4 (even)
    • For (3, 5): |3 - 5| = 2 (even)

    All elements of {2, 4} are related to each other:

    • For (2, 4): |2 - 4| = 2 (even)

    No element of {1, 3, 5} is related to any element of {2, 4}.

  • Question 9

    • Relation R (ii)

    Let A = {x ∈ Z : 0 ≤ x ≤ 12}.

    Given relation R = {(a, b) : a = b}.

    To show that R is an equivalence relation:

    Reflexivity:

    For any element a in A, a = a.

    So, (a, a) exists for all a in A.

    Symmetry:

    If a = b, then b = a.

    So, if (a, b) exists, then (b, a) also exists.

    Transitivity:

    If a = b and b = c, then a = c.

    So, if (a, b) and (b, c) exist, then (a, c) also exists.

    Elements Related to 1:

    Only 1 is related to itself.

  • Question 10

    • 1. Symmetric but neither reflexive nor transitive example: Relation R = {(1, 2), (2, 1), (2, 3)} This relation is symmetric because for every (a, b) in R, (b, a) is also in R. It is not reflexive because there is no (a, a) pair in R. It is not transitive because although (1, 2) and (2, 3) are in R, (1, 3) is not.

    2. Transitive but neither reflexive nor symmetric example: Relation R = {(1, 2), (2, 3)} This relation is transitive because for every (a, b) and (b, c) in R, (a, c) is also in R. It is not reflexive because there is no (a, a) pair in R. It is not symmetric because although (1, 2) is in R, (2, 1) is not.

    3. Reflexive and symmetric but not transitive example: Relation R = {(1, 1), (2, 2), (1, 2), (2, 1)} This relation is reflexive because every element is related to itself. It is symmetric because for every (a, b) in R, (b, a) is also in R. It is not transitive because although (1, 2) and (2, 1) are in R, (1, 1) is not.

    4. Reflexive and transitive but not symmetric example: Relation R = {(1, 1), (2, 2), (1, 2), (2, 1), (3, 3)} This relation is reflexive because every element is related to itself. It is transitive because for every (a, b) and (b, c) in R, (a, c) is also in R. It is not symmetric because although (1, 2) is in R, (2, 1) is not.

    5. Symmetric and transitive but not reflexive example: Relation R = {(1, 2), (2, 1), (2, 3), (3, 2)} This relation is symmetric because for every (a, b) in R, (b, a) is also in R. It is transitive because for every (a, b) and (b, c) in R, (a, c) is also in R. It is not reflexive because there is no (a, a) pair in R.

  • Question 11

    • Equivalence Relation in Plane

    Let A be the set of points in a plane.

    Given relation R = {(P, Q) : distance of point P from the origin is same as distance of point Q from the origin}.

    To show that R is an equivalence relation:

    Reflexivity:

    For any point P in the plane, the distance of P from the origin is equal to itself.

    So, (P, P) exists for all points P in A.

    Symmetry:

    If the distance of point P from the origin is the same as the distance of point Q from the origin, then vice versa.

    So, if (P, Q) exists, then (Q, P) also exists.

    Transitivity:

    If the distance of point P from the origin is the same as the distance of point Q from the origin, and the distance of point Q from the origin is the same as the distance of point R from the origin, then the distance of point P from the origin is also the same as the distance of point R from the origin.

    So, if (P, Q) and (Q, R) exist, then (P, R) also exists.

    Set of Points Related to a Point P ≠ (0, 0):

    The set of all points related to a point P ≠ (0, 0) forms a circle passing through P with the origin as the center.

  • Question 12

    • Equivalence Relation in Triangles

    Let A be the set of all triangles.

    Given relation R = {(T1, T2) : T1 is similar to T2}.

    To show that R is an equivalence relation:

    Reflexivity:

    Every triangle is similar to itself.

    So, (T, T) exists for all triangles T in A.

    Symmetry:

    If triangle T1 is similar to triangle T2, then triangle T2 is also similar to triangle T1.

    So, if (T1, T2) exists, then (T2, T1) also exists.

    Transitivity:

    If triangle T1 is similar to triangle T2, and triangle T2 is similar to triangle T3, then triangle T1 is also similar to triangle T3.

    So, if (T1, T2) and (T2, T3) exist, then (T1, T3) also exists.

    Triangles Related:

    Among the given triangles:

    • T1 with sides 3, 4, 5 is similar to T2 with sides 5, 12, 13 (both are right-angled triangles).
    • T1 with sides 3, 4, 5 is similar to T3 with sides 6, 8, 10 (both are scaled versions of the same right-angled triangle).
    • T2 with sides 5, 12, 13 is similar to T3 with sides 6, 8, 10 (both are scaled versions of the same right-angled triangle).
  • Question 13

    • Equivalence Relation in Polygons

    Let A be the set of all polygons.

    Given relation R = {(P1, P2) : P1 and P2 have the same number of sides}.

    To show that R is an equivalence relation:

    Reflexivity:

    Every polygon has the same number of sides as itself.

    So, (P, P) exists for all polygons P in A.

    Symmetry:

    If polygon P1 has the same number of sides as polygon P2, then polygon P2 also has the same number of sides as polygon P1.

    So, if (P1, P2) exists, then (P2, P1) also exists.

    Transitivity:

    If polygon P1 has the same number of sides as polygon P2, and polygon P2 has the same number of sides as polygon P3, then polygon P1 has the same number of sides as polygon P3.

    So, if (P1, P2) and (P2, P3) exist, then (P1, P3) also exists.

    Polygons Related to Right Angle Triangle T(3, 4, 5):

    The set of all elements related to the right angle triangle T with sides 3, 4, and 5 consists of all other right angle triangles with sides having the same ratios, as well as polygons with the same number of sides.

  • Question 14

    • Equivalence Relation in Lines

    Let L be the set of all lines in the XY plane.

    Given relation R = {(L1, L2) : L1 is parallel to L2}.

    To show that R is an equivalence relation:

    Reflexivity:

    Every line is parallel to itself.

    So, (L, L) exists for all lines L in L.

    Symmetry:

    If line L1 is parallel to line L2, then line L2 is also parallel to line L1.

    So, if (L1, L2) exists, then (L2, L1) also exists.

    Transitivity:

    If line L1 is parallel to line L2, and line L2 is parallel to line L3, then line L1 is also parallel to line L3.

    So, if (L1, L2) and (L2, L3) exist, then (L1, L3) also exists.

    Lines Related to y = 2x + 4:

    The set of all lines related to the line y = 2x + 4 consists of all lines parallel to the line y = 2x + 4.

  • Question 15

    • The correct answer is:

    (A) R is reflexive and symmetric but not transitive.

    Explanation:

    • Reflexive: (1, 1), (2, 2), (3, 3), and (4, 4) are present in R, indicating reflexivity.
    • Symmetric: (1, 2) is in R, but (2, 1) is not, so R is not symmetric.
    • Transitive: (1, 2) and (2, 2) are in R, but (1, 2) and (2, 2) together do not imply (1, 2), showing lack of transitivity.
  • Question 16

    • The correct answer is:

    (B) (3, 8) ∈ R

    Explanation:

    • In the relation R = {(a, b) : a = b – 2, b > 6}, for an ordered pair (a, b) to be in R, a must be equal to b - 2, and b must be greater than 6.
    • Substituting b = 8, we get a = 8 - 2 = 6.
    • So, (3, 8) satisfies the condition a = b - 2 and b > 6, hence (3, 8) ∈ R.