{"id":29217,"date":"2022-03-28T17:00:10","date_gmt":"2022-03-28T11:30:10","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=29217"},"modified":"2022-03-28T17:13:11","modified_gmt":"2022-03-28T11:43:11","slug":"ncert-solutions-for-class-12-maths-chapter-1-ex-1-1","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-1-ex-1-1\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.1"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> Chapter 1 Relations and Functions Ex 1.1 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-1-ex-1-1\/<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions 1.1<\/h2>\n

\"NCERT<\/p>\n

Class 12 Maths Chapter 1 Exercise 1.1 Question 1.<\/strong>
\nDetermine whether each of the following relations are reflexive, symmetric and transitive.
\ni. Relation R in the set A= {1, 2, 3,… 13, 14} defined as R = {(x, y): 3x – y = 0}
\nii. Relation R in the set N of natural numbers defined as R = {(x, y): y = x + 5 and x < 4}
\niii. Relation R in the set A= (1, 2, 3, 4, 5, 6} as R = {(x, y): y is divisible by x}
\niv. Relation R in the set Z of all integers defined as R = {(x, y): x -y is an integer}
\nv. Relation R in the set A of human beings in a town at a particular time given by
\na. R = {(x, y) : x and y work at the same place}
\nb. R= {(x, y) : x and y live in the same locality}
\nc. R = {(x, y) : x is exactly 7 cm taller than y}
\nd. R = {(x, y) : x is wife of y}
\ne. R = {(x, y) : x is father of y}
\nSolution:
\ni. Reflexive R = {(1, 3),(2, 6), (3, 9), (4, 12)}
\n(1.1) \u2208 R
\n\u2234 R is not reflexive.<\/p>\n

Symmetric
\n(1, 3) \u2208 R but (3, 1) \u2209 R
\n\u2234 R is not symmetric.<\/p>\n

Transitive
\n(1, 3) \u2208 R and (3,9) \u2208 R but (1, 9) \u2209 R
\n\u2234 R is not transitive.
\nHence R is neither reflexive, nor symmetric nor transitive.<\/p>\n

ii. R = {(1, 6),(2, 7), (3, 8)}
\n(1, 1) \u2209 R
\n\u2234 R is not reflexive.
\n(1, 6) \u2208 R but (6, 1) \u2209 R
\n\u2234 R is not symmetric. In R there does not exist ordered pairs of the form (x, y) and (y, z)
\n\u2234 R is transitive.<\/p>\n

iii. R = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2.2) , (2,4), (2,6), (3,3), (3,6), (4,4), (5,5), (6,6)}
\nReflexive
\n(x, x) \u2208 R for all x \u2208 A, since x is divisible by x.
\n\u2234 R is reflexive<\/p>\n

Symmetric
\n(1,2) \u2208 R but (2, 1) \u2208 R
\n\u2234R is not symmetric.<\/p>\n

Transitive
\nIf y is divisible by x and z is divisible by y, then z is divisible by x. i.e., (x, y) \u2208 R
\n(y, z) \u2208R \u21d2 (x, z) \u2208R for all x, y, z \u2208 A
\n\u2234R is transitive<\/p>\n

iv. Reflexive
\nx – x = 0, which is an integer
\ni.e., (x, x)\u2208 R for every x \u2208 Z
\n\u2234 R is reflexive<\/p>\n

Symmetric
\n(x, y) \u2208 R
\n\u21d2 x – y is an integer
\n\u21d2 y – x is an integer
\n\u21d2 (y, x) \u2208 R
\n\u2234 R is symmetric
\nTransitive
\n(x, y) \u2208 R, (y, z) \u2208 R \u21d2 x – y is an integer and y – z is an integer.
\n\u21d2 (x – y) + (y – z) is an integer
\n\u21d2 x – z is an integer
\n\u21d2 (x, z) \u2208 R for all x, y, z \u2208 Z
\n\u2234 R is transitive.<\/p>\n

v. a. Reflexive
\n(x, x) \u2208 R for every x \u2208 A, since x and x work at the same place.
\n\u2234R is reflexive.<\/p>\n

Symmetric
\nLet (x, y) \u2208 R
\n\u21d2 x and y work at the same place
\n\u21d2 y and x work at the same place
\n\u21d2 (y, x) \u2208 R for all x, y \u2208 A
\n\u2234 R is symmetric<\/p>\n

Transitive
\nLet (x, y), (y, z) \u2208 R
\n\u21d2 x and y work at the same place;
\ny and z work at the same place.
\n\u21d2 x and z work at the same place
\n\u21d2 (x, z) \u2208 R
\n\u2234 R is transitive
\nHence R is reflexive, symmetric and transitive.<\/p>\n

b. Reflexive, symmetric and transitive
\n(Similar to v.a)<\/p>\n

c. Reflexive
\nSince x cannot 7 cm taller than x, (x, x) \u2209 R.
\n\u2234 R is not reflexive<\/p>\n

Symmetric
\n(x, y) \u2208 R \u21d2 x is exactly 7 cm taller than y
\n\u21d2 y cannot be exactly 7 cm taller than x.
\n\u21d2 (y, x) \u2209 R
\n\u2234 R is not symmetric<\/p>\n

Transitive
\nLet (x, y),(y, z) \u2208 R \u21d2 x is exactly 7 cm
\ntaller than y and y is exactly 7 cm taller than z.
\n\u21d2 x is exactly 14 cm taller than z.
\n\u21d2 (x, z) \u2209 R \u2234 R is not transitive.<\/p>\n

d. Reflexive
\n(x, x) \u2209 R as x cannot be the wife of x.
\n\u2234 R is not reflexive
\nSymmetric
\nLet (x, y) \u2208 R \u21d2 x is the wife of y
\n\u21d2 y is not the wife of x \u21d2 (y, x) \u2209 R
\n\u2234 R is not symmetric<\/p>\n

Transitive
\nLet (x, y) \u2208 R \u21d2 x is the wife of y \u21d2 y
\ncannot be the wife of any z
\n\u21d2 (y, z) \u2209 R
\nThat is, there does not exist any (x, y) and (y, z) in R.
\n\u2234 R is transitive<\/p>\n

e. Reflexive
\n(x, x) \u2209 R as x cannot be the father of x.
\n\u2234 R is not reflexive<\/p>\n

Symmetric
\n(x, y) \u2208 R \u21d2 x is the father of y.
\n\u21d2 y is not the father of x
\n\u21d2 (y, x) \u2209 R
\n\u2234 R is not symmetric<\/p>\n

Transitive
\nLet (x, y), (y, z) \u2208 R \u21d2 x is the father of y
\nand y is the father of z
\n\u21d2 x cannot be the father of z
\n\u21d2 (x, z) \u2209 R
\n\u2234 R is not transitive<\/p>\n

\"NCERT<\/p>\n

Chapter 1 Math Class 12 Question 2.<\/strong>
\nShow that the relation R in the set R of real numbers, defined as R= {(a, b); a<b\u00b2} is neither reflexive nor symmetric nor transitive.
\nSolution:
\nReflexive
\nLet a = \\(\\frac { 1 }{ 2 }\\). Then a\u00b2 = \\(\\frac { 1 }{ 4 }\\)
\n\u21d2 \\(\\frac { 1 }{ 2 }\\) is not less than or equal to (\\(\\frac { 1 }{ 2 }\\))\u00b2
\n\u21d2 (\\(\\frac { 1 }{ 2 }\\), \\(\\frac { 1 }{ 2 }\\)) \u2209 R
\n\u2234 R is not reflexive.<\/p>\n

Symmetric
\nSince 1 \u2264 2\u00b2, (1, 2) \u2208 R
\nBut 2 is not less than or equal to 1\u00b2
\n\u21d2 (2, 1) \u2208 R
\ni.e., (1, 2) \u2208 R and (2, 1) \u2209 R
\n\u2234 R is not symmetric<\/p>\n

Transitive
\n1 \u2264 (- 2)\u00b2 and – 2 \u2264 (0)\u00b2
\nBut 1 is not less than or equal to 0\u00b2
\ni.e., (1, – 2), (- 2, 0) \u2208 R but (1, 0) \u2209 R
\n\u2234 R is not transitive<\/p>\n

Class 12 Maths Ex 1.1 Question 3.<\/strong>
\nCheck 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.
\nSolution:
\nR={(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}
\n(1, 1) \u2209 R
\n\u2234 R is not reflexive
\n(1, 2) \u2208 R and (2, 1) \u2209 R
\n\u2234 R is not symmetric
\n(3, 4) \u2208 R , (4, 5) \u2209 R but (3, 5) \u2209 R
\n\u2234 R is not transitive<\/p>\n

Class 12 Ex 1.1 Question 4.<\/strong>
\nShow that the relation R in R defined as R = {(a, b); a < b), is reflexive and transitive but not symmetric.
\nSolution:
\nReflexive
\n(a, a) \u2208 R as a < a \u2234 R is reflexive
\nSymmetric
\n(1, 2) \u2208 R \u21d2 1 \u2264 2 \u21d2 2 \u2264 1
\n\u21d2 (2, 1) \u2209 R
\n\u2234 R is not symmetric<\/p>\n

Transitive
\n(a, b), (b, c) \u2208 R a \u2264 b , b \u2264 c
\n\u21d2 a \u2264 c which is true
\n\u2234 R is transitive<\/p>\n

Maths Class 12 Chapter 1 Exercise 1.1 Question 5.<\/strong>
\nCheck whether the relation R in R defined by R = {(a,b); a \u2264 b\u00b3} is reflexive, symmetric or transitive.
\nSolution:
\nReflexive
\nLet a = \\(\\frac { 1 }{ 2 }\\). Then \\(\\frac { 1 }{ 2 }\\) is not less than (\\(\\frac { 1 }{ 2 }\\))\u00b3.
\ni.e., (\\(\\frac { 1 }{ 2 }\\), \\(\\frac { 1 }{ 2 }\\)) \u2209 R
\n\u2234 R is not reflexive.<\/p>\n

Symmetric
\nSince 1 \u2264 2\u00b3, (1, 2) \u2208 R
\nBut 2 is not less than or equal to L3<\/sup>.
\n\u21d2 (2, 1) \u2209 R
\ni.e., (1, 2) \u2208 R and (2, 1) \u2209 R
\n\u2234 R is not symmetric<\/p>\n

Transitive
\n9 \u2264 3\u00b3 and 3 \u2264 2\u00b3, i.e., (9, 3), (3, 2) \u2208 R
\nBut 9 is not less than or equal to 2\u00b3
\ni.e., (9, 2) \u2209 R
\ni.e., (9, 3),(3, 2) \u2208 R but (9, 2) \u2209 R
\n\u2234 R is not transitive.<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nShow that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.
\nSolution:
\n(1, 1) \u2209 R \u2234 R is not reflexive
\nSymmetric
\n(1, 2), (2, 1) \u2208 R \u21d2 (2, 1), (1, 2) \u2208 R
\n\u2234 R is symmetric
\nTransitive
\n(1, 2), (2, 1) \u2208 R but (1, 1) \u2209 R
\n\u2234 R is not transitive<\/p>\n

Question 7.
\nShow 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.
\nSolution:
\nReflexive
\n(x, x) \u2208 R, since book x and x have same number of pages.
\n\u2234 R is reflexive<\/p>\n

Symmetric
\nIf (x, y) \u2208 R, then books x and y have same number of pages i.e., books y and x have same number of pages. Hence (y, x) \u2208 R.
\n\u2234 R is symmetric<\/p>\n

Transitive
\nLet (x, y, (y, z) \u2208 R \u21d2 books x, y and books y, z have same number of pages.
\n\u21d2 books x and z have same number of pages.
\n\u21d2 (x, z) \u2208 R
\n\u2234 R is symmetric
\nR is trarisitive R is reflexive, symmetric and transitive. Hence R is an equivalence relation.<\/p>\n

\"NCERT<\/p>\n

Question 8.
\nShow 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}.
\nSolution:
\nR= {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (5, 1), (5, 3),(5, 5)}
\nReflexive
\nSince |a – a| = 0 which is even (a, a) \u2208 R
\n\u2234 R is reflexive<\/p>\n

Symmetric
\nLet (d, b) \u2208 R \u21d2 |a – b| is even
\n\u21d2 |b – a| is even
\n\u21d2 (b, a) \u2208 R for all a, b \u2208 A
\n\u2234 R is symmetric<\/p>\n

Transitive
\nLet (a, b), (b, c) \u2208 R
\n\u21d2 |a – b| is even and |b – c| is even
\n\u21d2 |a – c| is even
\n\u2234 R is transitive
\nHence R is an equivalence relation.
\nThe elements of {1, 3, 5} are related to each other, since |1 – 1|, |1 – 3|, |1 – 5| etc. are even numbers.
\nThe elements of {2, 4} are related to each other, since |2 – 2|, |2 – 4|, |4 – 2|, |4 – 4| are even numbers.
\nNo elements of {1, 3, 5} is related to any element of {2, 4}, since the absolute value of the difference of any element from {1, 3, 5} with any element from {2, 4} is not even.<\/p>\n

Question 9.
\nShow that each of the relation R in the set A= {x \u2208 Z : 0 \u2264 x \u2264 12}, given by
\ni. R = {(a, b): |a – b| is a multiple of 4}
\nii. R = {(a, b): a = b} is an equivalence relation. Find the set of all elements related to 1 in each case.
\nSolution:
\ni. Reflexive
\n|a – a| = 0 is a multiple of 4, a \u2208 A
\n\u21d2 (a, a) \u2208 R
\n\u2234 R is reflexive<\/p>\n

Symmetric
\nLet (a, b) \u2208R \u21d2 |a – b| is a multiple of 4
\n\u21d2 |b – a| is a multiple of 4
\n\u21d2 (b, a) \u2208R
\n\u2234 R is symmetric<\/p>\n

Transitive
\nLet (a, b), (b, c) \u2208 R
\n\u21d2 |a – b| and |b – c| are multiple of 4
\n\u21d2 a- b and b – c are multiple of 4
\nNow a – c = a – b + b – c
\n= Multiple of 4 + multiple of 4
\n= Multiple of 4
\n\u2234 |a – c| is a multiple of 4 \u21d2 (a, c) \u2208R
\ni.e., (a, b), (b, c) \u2208 R \u21d2 (a, c) \u2208 R
\n\u2234R is transitive
\nHence R is an equivalence relation.
\nThe set of elements related to 1 is {1, 5,9}<\/p>\n

ii. Reflexive
\nLet a \u2208 A. Then a = a
\n\u21d2 (a, a) \u2208R for all a \u2208 A
\n\u2234 R is reflexive<\/p>\n

Symmetric
\nLet (a, b) \u2208 R \u21d2 a – b
\n\u21d2 b = a
\n\u21d2 (b, a) \u2208 R
\n\u2234 R is symmetric<\/p>\n

Transitive
\nLet {a, b), (b, c) \u2208 R \u21d2 a = b and b = c
\n\u21d2 a = c
\n\u21d2 (a, c) \u2208 R
\n\u2234 R is transitive.
\nHence R is an equivalence relation.
\nThe set of elements related to 1 is {1}<\/p>\n

\"NCERT<\/p>\n

Question 10.
\nGive an example of a relation, which is
\ni. Symmetric but neither reflexive nor transitive.
\nii. Transitive but neither reflexive nor symmetric.
\niii. Reflexive and symmetric but not transitive’
\niv. Reflexive and transitive but not symmetric.
\nv. Symmetric and transitive but not reflexive.
\nSolution:
\nLet A= {1, 2, 3}
\ni. Let R= {(1, 2), (2, 1)}
\nReflexive
\nR is not reflexive, since (1, 1) \u2209 R
\nSymmetric
\nR is symmetric, since (1, 2), (2, 1) \u2208 R
\n\u21d2 (2, 1),(1, 2) \u2208 R
\nTransitive
\nR is not transitive, since (1, 2), (2, 1) \u2208R but (1, 1) \u2209 R
\n\u2234 R is symmetric, but neither reflexive nor transitive.<\/p>\n

ii. Let R= {(1,2), (1,3), (3, 2)}
\nReflexive
\nR is not reflexive, since (1, 1) \u2209 R
\nSymmetric
\nR is not symmetric, since(1, 2) \u2208 R, but (2, 1) \u2209 R
\nTransitive
\nR is transitive, since (1, 3), (3, 2) \u2208 R and
\n(1, 2) \u2208 R
\n\u2234 R is transitive but neither reflexive nor symmetric.<\/p>\n

iii. Let R = {(1, 1), (2, 2), (3, 3), (1, 2),
\n(2,1) , (2,3), (3,2)}
\nReflexive
\nR is reflexive, since (1,1), (2,2), (3,3) \u2208 R
\nSymmetric
\nR is symmetric, since (1, 2) \u2208 R
\n\u21d2 (2, 1) \u2208 R(2, 3) \u2208 R \u21d2 (3, 2) \u2208R
\nTransitive
\nR is not transitive, since (1, 2), (2, 3) \u2208 R but(1, 3) \u2209 R
\n\u2234 R is reflexive and symmetric but not transitive.<\/p>\n

iv. Let R= {(1, 1), (2, 2), (3, 3), (1, 2)}
\nReflexive
\nR is reflexive, since (1, 1), (2, 2), (3, 3) \u2208 R
\nSymmetric
\nR is not symmetric, since (1, 2) \u2208 R but (2, 1) \u2209 R
\nTransitive
\nR is transitive, since (1, 1), (1, 2) \u2208 R and , (1, 2) \u2208 R<\/p>\n

v. Let R= {(1, 2), (2, 1), (1, 1), (2, 2)}
\nReflexive
\nR is not reflexive, since (3, 3) \u2209 R
\nSymmetric
\nR is symmetric, since (1, 2) \u2208 R \u21d2 (2, 1) \u2208 R
\nTransitive
\nR is transitive, since (1, 2), (2, 1) \u2208 R \u21d2 (1, 1) \u2208 R
\nand (2, 1 )(1, 2) \u2208 R \u21d2 (2, 2) \u2208 R<\/p>\n

Question 11.
\nShow 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 \u2260 (0, 0) is the circle passing through P with origin as centre.
\nSolution:
\nLet O be the origin
\nThen R = {(P, Q) : OP = OQ}
\nReflexive
\nLet P be a point in the plane Then OP = OP
\n\u21d2 (P, P) \u2208 R for all P
\n\u2234 R is reflexive<\/p>\n

Symmetric
\nLet (P, Q) \u2208 R
\n\u21d2 OP = OQ
\n\u21d2 OQ = OP
\n\u21d2 (Q, P) \u2208 R
\n\u2234 R is symmetric.<\/p>\n

Transitive
\nLet (P, Q), (Q, S) \u2208 R
\n\u21d2 OP = OQ and OQ = OS
\n\u21d2 OP = OS
\n\u21d2 (P, S) \u2208 R
\n\u2234 R is transitive
\nHence R is an equivalence relation.
\nLet P \u2260 (0,0) be a point in the plane. Consider the circle with centre at origin and radius OP. Then the set of points on this circle are related to P since the distance from the origin to any point on the circle is OP.
\nHence the set of points related to P is the circle passing through P with origin as centre.<\/p>\n

\"NCERT<\/p>\n

Question 12.
\nShow that the relation R defined in the set A of all triangles as R = {(T1<\/sub>, T2<\/sub>) : T1<\/sub> is similar to T2<\/sub>}, is an equivalence relation. Consider three right angled triangles T1<\/sub> with sides 3,4, 5, T2<\/sub> with sides 5, 12, 13 and T3<\/sub> with sides 6, 8,10. Which triangles among T1<\/sub>, T1<\/sub> and T3<\/sub> are related?
\nSolution:
\nR = {(T1<\/sub>, T2<\/sub>): T1<\/sub> is similar to T2<\/sub>
\nReflexive
\nLet T \u2208 A
\nSince T is similar to T, (T, T) \u2208 R
\n\u2234 R is reflexive<\/p>\n

Symmetric
\nLet (T1<\/sub>,T2<\/sub>) \u2208 R
\n\u21d2 T1<\/sub> is similar to T2<\/sub>
\n\u21d2 T2<\/sub> is similar to T1<\/sub>
\n\u21d2 (T2<\/sub>,T1<\/sub>) \u2208 R
\n\u2234 R is symmetric<\/p>\n

Transitive
\nLet (T1<\/sub>, T2<\/sub>), (T2<\/sub>, T3<\/sub>) \u2208 R
\n\u21d2 T1<\/sub> is similar to T2<\/sub> and T2<\/sub> is similar to T3<\/sub>
\n\u21d2 T1<\/sub> is similar to T3<\/sub>
\n\u21d2 (T1<\/sub>, T3<\/sub>) \u2208 R
\n\u2234 R is transitive
\nHence R is an equivalence relation.
\nTwo triangles are similar if their sides are proportional. The sides 3, 4, 5 of triangle T1<\/sub> is proportional to the sides 6, 8, 10 of triangle T3<\/sub>
\n(\\(\\frac { 3 }{ 6 }\\) = \\(\\frac { 4 }{ 8 }\\) = \\(\\frac { 5 }{ 10 }\\))
\ni. e., the sides of and T1<\/sub> are proportional. Hence T1<\/sub> is related to T3<\/sub>.<\/p>\n

Question 13.
\nShow that the relation R defined in the set A of all polygons as R = {(P1<\/sub>, P2<\/sub>): P1<\/sub> and P2<\/sub> 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?
\nSolution:.
\nR = {(P1<\/sub>, P2<\/sub>): P1<\/sub> and P2<\/sub> have same number of sides}
\nReflexive
\nLet P \u2208 A
\nSince P and P have same number of sides,
\n(P, P) \u2208 R for all P.
\n\u2234 R is reflexive<\/p>\n

Symmetric
\nLet (P1<\/sub>, P2<\/sub>) \u2208 R
\n\u21d2 Number of sides of P1<\/sub> = Number of sides of P2<\/sub>
\n\u21d2 Number of sides of P1<\/sub> = Number of sides of P1<\/sub>
\n\u21d2 (P2<\/sub>, P1<\/sub>) \u2208 R
\n\u2234 R is symmetric<\/p>\n

Transitive
\nLet (P1<\/sub>, P2<\/sub>), (P2<\/sub>, P3<\/sub>) \u2208 R
\n\u21d2 P1<\/sub>, P2<\/sub> have same number of sides and P2<\/sub>, P3<\/sub> have same number of sides.
\n\u21d2 P1<\/sub>, P3<\/sub> have same number of sides
\n\u21d2 (P1<\/sub>, P3<\/sub>) \u2208 R
\n\u2234 R is transitive
\nHence R is an equivalence relation.
\nThe right triangle with sides 3, 4, 5 is a polygon having 3 sides.
\n\u2234 T he set of elements of A related to T is the set of triangles in the plane.<\/p>\n

\"NCERT<\/p>\n

Question 14.
\nLet L be the set of all lines in XY plane and R be the relation in L defined as R = {(L1<\/sub>, L2<\/sub>) : L1<\/sub> is parallel to L2<\/sub>}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.
\nSolution:.
\nReflexive
\n(L1<\/sub>, L1<\/sub>) \u2208R since the line L1<\/sub> is parallel to itself.
\n\u2234 R is reflexive<\/p>\n

Symmetric
\n(L1<\/sub>, L2<\/sub>) \u2208 R \u21d2 L1<\/sub> is parallel to L2<\/sub>
\n\u21d2 L2<\/sub> is parallel to L1<\/sub> \u21d2 (L2<\/sub>, L1<\/sub>) \u2208 R
\n\u2234 R is symmetric.<\/p>\n

Transitive
\n(L1<\/sub>, L2<\/sub>) \u2208R, (L2<\/sub>, L3<\/sub>) \u2208 R
\nL1<\/sub>, L2<\/sub> are parallel and L2<\/sub>, L3<\/sub> are parallel
\n\u21d2 L1<\/sub> and L3<\/sub> are parallel. \u21d2 (L1<\/sub>, L3<\/sub>) \u2208 R
\n\u2234 R is transitive.
\nHence R is an equivalence relation.
\nThe set of all lines related to the line y = 2x + 4 is the line y = 2x + c where C \u2208 R.<\/p>\n

Question 15.
\nLet 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.
\na. R is reflexive and symmetric but not transitive.
\nb. R is reflexive and transitive but not symmetric.
\nc. R is symmetric and transitive but not reflexive.
\nd. R is an equivalence relation.
\nSolution:
\nAnswer:
\nb. R is reflexive and transitive but not symmetric.
\n(1, 2) \u2208 R but (2, 1) \u2209 R
\n\u2234 R is not symmetric<\/p>\n

Question 16.
\nLet R be the relation in the set N given by R = {{a, b) : a = b – 2, b > 6}. Choose the correct answer.
\na. (2, 4) \u2208 R
\nb. (3, 8) \u2208 R
\nc. (6, 8) \u2208 R
\nd. (8, 7) \u2208 R
\nSolution:
\nAns. c
\nIf b = 8, a = 6 – 2 = 8 – 2 = 6
\n\u2234 (6, 8) \u2208 R is correct.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.1 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-1-ex-1-1\/ NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions 1.1 Class 12 Maths Chapter 1 Exercise 1.1 Question 1. Determine whether each of the following relations are …<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.1<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.1 - MCQ Questions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-1-ex-1-1\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.1 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.1 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-1-ex-1-1\/ NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions 1.1 Class 12 Maths Chapter 1 Exercise 1.1 Question 1. 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