NCERT Solutions for CBSE Class 12th Maths - Chapter 1: Relations and Functions (Exercise 1.1) are available here. You can also download these solutions with the help of download link given at the end of this article. Here, solutions from question number 9 to 16 are available. Solutions from question number 1 to 8 are available in **Part 1. **Most of the questions given below are based on the concepts related to Mathematical relation and their types.

These concepts are important for CBSE Class 12th Maths board exam 2018 and other competitive exams like WBJEE, JEE Mains, UPSEE etc. In order to score well in exams, students must study these NCERT Solutions thoroughly.

*NCERT Solutions for Class 12 Maths - Chapter 1: Relations and Functions (Exercise 1.1 – Question number 9 to 16) are given below*

**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 9:**

**( i)**

**NCERT Exemplar Class 12 Mathematics – Chapter 1 Relations and Functions**

(*ii*)

**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 10:**

**( i)**

Let *A* = {5, 6, 7}

R = {(5, 6), (6, 5)}

R is not reflexive [because (5, 5), (6, 6), (7, 7) ∉ R]

Now,

(6, 5) ∈ R & (5, 6) ∈ R

Thus, R is symmetric

R is not transitive [since (5, 5) ∉ R]

R is symmetric but not reflexive or transitive.

**( ii)**

R = {(*a*, *b*): *a* < *b*}

For any *a *∈ R,

(*a*, *a*) ∉ R, *a* = *a*

Thus, R is not reflexive.

Now,

(1, 2) ∈ R (Since 1 < 2)

But, 2 is not less than 1

Hence, (2, 1) ∉ R

Thus, R is not symmetric.

Now,

Let (*a*, *b*), (*b*, *c*) ∈ R

*a* < *b* and *b* < *c*

*a* < *c*

⇒ (*a*, *c*) ∈ R

Hence, R is transitive

Hence, R is transitive but not reflexive and symmetric.

**( iii)**

Let *A* = {4, 6, 8}

*A* = {(4, 4), (6, 6), (8, 8), (4, 6), (6, 4), (6, 8), (8, 6)}

R is reflexive [as (4, 4), (6, 6), (8, 8)} ∈ R]

R is symmetric [As (*a*, *b*) ∈ R ⇒ (*b*, *a*) ∈ R]

R is not transitive [As (4, 6), (6, 8) ∈ R, but (4, 8) ∉ R]

Thus, R is reflexive and symmetric but not transitive.

**( iv)**

R = {*a*, *b*): *a*^{3} ≥ *b*^{3}}

(*a*, *a*) ∈ R as *a*^{3} = *a*^{3}

Thus, R is reflexive

Now,

(2, 1) ∈ R (as 2^{3} ≥ 1^{3})

But,

(1, 2) ∉ R (as 1^{3} < 2^{3})

Hence, 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

Thus, R is transitive

Hence, R is reflexive and transitive but not symmetric.

**( v)**

Let *A* = {−5, −6}

R = {(−5, −6), (−6, −5), (−5, −5)}

R is not reflexive [Since (−6, −6) ∉ R]

R is symmetric [Since (−5, −6) ∈ R and (−6, −5} ∈ R]

(−5, −6), (−6, −5) ∈ R

(−5, −5) ∈ R

Hence, R is transitive

Thus, R is symmetric and transitive but not reflexive.

**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 11:**

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

We know that,

The distance of point P from the origin is always the same as the distance of the same point P from the origin.

(P, P) ∈ R

Hence, R is reflexive

Let (P, Q) ∈ R

The distance of point P from the origin = the distance of point Q from the origin

The distance of point Q from the origin = the distance of point P from the origin

(Q, P) ∈ R

Hence, R is symmetric

Now,

Let (P, Q), (Q, S) ∈ R

The distance of points Q and S from the origin is the same.

(P, S) ∈ R [As the distance of points P and S from the origin is the same]

Hence, R is transitive

Thus, R is an equivalence relation.

If O = (0, 0)

OP = *k*

Set of all points related to P is at a distance of *k* from the origin.

Thus, this set of points forms a circle with the centre as the origin and this circle passes through point P.

**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 equivalence relation. Consider three right angle 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 12:**

**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 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 13:**

R = {(*P*_{1}, *P*_{2}): *P*_{1} and *P*_{2} have same the number of sides}

(*P*_{1}, *P*_{1}) ∈ R [Since the same polygon has the same number of sides with itself]

Therefore, R is reflexive

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

Thus, 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

Therefore, R is transitive.

Thus, 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.

Thus, the set of all elements in *A* related to triangle *T* is the set of all triangles.

**Question 14:** Let L be the set of all 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 lines related to the line *y* = 2*x* + 4.

**Solution 14:**

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

Any line, *L*_{1} is parallel to itself i.e., (*L*_{1}, *L*_{1}) ∈ R

Hence, R is reflexive

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

Hence, R is symmetric

Now,

Let (*L*_{1}, *L*_{2}), (*L*_{2}, *L*_{3}) ∈R

*L*_{1} is parallel to *L*_{2}

*L*_{2} is parallel to *L*_{3}

⇒ *L*_{1} is parallel to *L*_{3.}

Hence, R is transitive.

Thus, R is an equivalence relation.

The set of all lines related to the line *y* = 2*x* + 4 is the set of all lines that are parallel to the line *y* = 2*x* + 4.

*y* = 2*x* + 4

Slope of above line, *m* = 2

We know that,

Parallel lines have the same slopes.

The line parallel to the given line,

*y* = 2*x* + *c*, where *c* ∈ R

Thus, the set of all lines related to the given line is *y* = 2*x* + *c*, where *c* ∈ R.

**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 15:**

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

We know that,

(*a*, *a*) ∈ R**,** for every *a* ∈ {1, 2, 3, 4}

Hence, R is reflexive

Also,

(1, 2) ∈ R, but (2, 1) ∉ R

Thus, R is not symmetric

We can see that,

(*a*, *b*), (*b*, *c*) ∈ R

⇒ (*a*, *c*) ∈ R for all *a*, *b*, *c* ∈ {1, 2, 3, 4}

Hence, R is transitive

Thus, R is reflexive and transitive but not symmetric.

The correct answer is B.

**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 16:**

R = {(*a*, *b*): *a *= *b* − 2, *b* > 6}

Now,

*b* > 6, (2, 4) ∉ R

Also,

3 ≠ 8 − 2, (3, 8) ∉ R

And,

8 ≠ 7 − 2

Thus, (8, 7) ∉ R

Now,

Consider (6, 8)

We know 8 > 6 and also, 6 = 8 − 2

Thus, (6, 8) ∈ R

The correct answer is C.** **

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