NCERT Solutions for CBSE Class 12th Maths - Chapter 1: Relations and Functions (Exercise 1.1) are given below. Here, you will get solutions from question number 1 to 8. Most of the questions given below are based on the concepts of Mathematical relations and their types i.e., reflexive, symmetric, transitive etc.

These concepts are important for CBSE Class 12th Maths board exam and other competitive exams like JEE Mains, WBJEE etc.

*NCERT Solutions for Class 12 Maths - Chapter 1: Relations and Functions (Exercise 1.1 – Question number 1 to 8) are given below*

**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*): 3*x* – *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*}

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

**( 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 1:**

**( i)**

*A*= {1, 2, 3,...13, 14}

R = {(*x*,* y*): 3*x* ‒ *y* = 0}

Hence,

R = {(1, 3), (2, 6), (3, 9), (4, 12)}

R is not reflexive [because (1, 1), (2, 2) ... (14, 14) ∉ R]

R is also not symmetric [(1, 3) ∈R, but (3, 1) ∉ R. {3(3) − 1 ≠ 0}]

Also, R is not transitive as well [(1, 3), (3, 9) ∈R, but (1, 9) ∉ R]

[3(1) − 9 ≠ 0]

Thus, we can say that R is neither reflexive, nor symmetric, nor transitive.

**( ii)**

R = {(*x*, *y*): *y* = *x* + 5 and *x* < 4} = {(1, 6), (2, 7), (3, 8)}

We know that

(1, 1) ∉ R

Hence, R is not reflexive.

(1, 6) ∈R

But,

(6, 1) ∉ R

Hence, R is not symmetric.

Now, we have seen that

There is no pair in R such that (*x*, *y*) and (*y*, *z*) ∈R, then (*x*, *z*) ∉ R

Hence, R is not transitive.

Thus, R is neither reflexive, nor symmetric, nor transitive.

**( iii)**

*A* = {1, 2, 3, 4, 5, 6}

R = {(*x*, *y*): *y* is divisible by *x*}

We know that,

Let an every number x is divisible by itself.

Thus, (*x*, *x*) ∈R

Hence, R is reflexive.

Also,

(2, 4) ∈ R [since 4 is divisible by 2]

But,

(4, 2) ∉ R [since 2 is not divisible by 4]

Hence, R is not symmetric.

Let (*x*, *y*), (*y*, *z*) ∈ R

Hence, *z* is divisible by *x*

(*x*, *z*) ∈R

Therefore, R is transitive.

Hence, R is reflexive and transitive but not symmetric.

**( iv)**

R = {(*x*, *y*): *x* − *y* is an integer}

*x* ∈ **Z**, (*x*, *x*) ∈R [because *x* − *x* = 0 is an integer]

Hence, R is reflexive.

*x*, *y* ∈ **Z** if (*x*, *y*) ∈ R, then *x* − *y* is an integer.

⇒ − (*x* − *y*) is also an integer.

⇒ (*y* − *x*) is an integer.

Hence, (*y*, *x*) ∈ R

Therefore, R is symmetric.

Now,

Let (*x*, *y*) and (*y*, *z*) ∈R, where *x*, *y*, *z* ∈ **Z**

(*x* − *y*) and (*y* − *z*) are integers.

⇒ *x** *− *z* = (*x* − *y*) + (*y* − *z*) is an integer.

(*x*, *z*) ∈R

Therefore, R is transitive.

Thus, R is reflexive, symmetric, and transitive.

**CBSE Class 12 Mathematics Syllabus 2017 – 2018**

**( v)**

(*a*)

R = {(*x*, *y*): *x* and *y* work at the same place}

(*x*, *x*) ∈ R

Hence, R is reflexive.

If (*x*, *y*) ∈ R, then *x* and *y* work at the same place.

*y* and *x* work at the same place.

(*y*, *x*) ∈ R

Therefore, R is symmetric.

Let (*x*, *y*), (*y*, *z*) ∈ R

*x* and *y* work at the same place and *y* and *z* work at the same place.

*x* and *z* work at the same place.

⇒ (*x*, *z*) ∈R

Hence, R is transitive.

Thus, R is reflexive, symmetric, and transitive.

(*b*)

R = {(*x*, *y*): *x* and *y* live in the same locality}

(*x*, *x*) ∈ R [since *x* and *x* is the same human being]

Hence, R is reflexive

If (*x*, *y*) ∈R, then *x* and *y* live in the same locality

*y* and *x* live in the same locality.

⇒ (*y*, *x*) ∈ R

Hence, R is symmetric.

Let (*x*, *y*) ∈ R and (*y*, *z*) ∈ R.

*x* and *y* live in the same locality and *y* and *z* live in the same locality.

*x* and *z* live in the same locality.

⇒ (*x,* *z*) ∈ R

∴ R is transitive.

Hence, R is reflexive, symmetric, and transitive.

(*c*)

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

(*x*, *x*) ∉ R [as human being *x** *cannot be taller than himself]

Therefore, R is not reflexive.

Let (*x*, *y*) ∈R

*x* is exactly 7 cm taller than *y*.

Then, *y* is not taller than *x*.

Thus, (*y*, *x*) ∉R

Therefore, R is not symmetric.

Now,

Let (*x*, *y*), (*y*, *z*) ∈ R

*x* is exactly 7 cm taller than *y** *and *y* is exactly 7 cm taller than z.

*x* is exactly 14 cm taller than *z** *.

Hence, (*x*, *z*) ∉R

Thus, R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

(*d*) R = {(*x*, *y*): *x* is the wife of *y*}

(*x*, *x*) ∉ R [As *x* cannot be the wife of herself]

Thus, R is not reflexive.

Let (*x*, *y*) ∈ R

*x* is the wife of *y.*

We can see that *y* is not the wife of *x*.

Hence, (*y*, *x*) ∉ R

if *x* is the wife of *y*, then *y* is the husband of *x*.

∴ R is not transitive.

Let (*x*, *y*), (*y*, *z*) ∈ R

*x* is the wife of *y* and *y* is the wife of *z*.

This case is not possible.

∴(*x*, *z*) ∉ R

Hence, R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

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

(*x*, *x*) ∉ R [Since, *x* cannot be the father of himself]

Therefore, R is not reflexive.

Let (*x*, *y*) ∈R

*x* is the father of *y.*

*y* cannot be the father of *y.*

*y* is the son or the daughter of *y.*

∴(*y*, *x*) ∉ R

Hence, R is not symmetric.

Let (*x*, *y*) ∈ R and (*y*, *z*) ∈ R.

*x* is the father of *y* and *y* is the father of *z*.

*x* is not the father of *z*.

*x* is the grandfather of *z*.

∴ (*x*, *z*) ∉ R

Thus, R is not transitive.

Therefore, R is neither reflexive, nor symmetric, nor transitive.

**Question 2:** Show that the relation R in the set R of real numbers, defined as R = {(*a*, *b*): *a* ≤ *b*^{2}} is neither reflexive nor symmetric nor transitive.

**Solution 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 or transitive.

**Solution 3:**

Let *A* = {1, 2, 3, 4, 5, 6}.

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

Hence,

R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

(*a*, *a*) ∉ R, where *a *∈ A

(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) ∉ R

∴R is not reflexive.

We can see that,

(1, 2) ∈ R, but (2, 1) ∉ R

Hence, R is not symmetric

Now, (1, 2), (2, 3) ∈ R

But,

(1, 3) ∉ R

Hence, R is not transitive

Thus, R is neither reflexive, nor symmetric, nor transitive.

**Question 4:** Show that the relation R in R defined as R = {(*a*, *b*) : *a* ≤ *b*}, is reflexive and transitive but not symmetric.

**Solution 4:**

R = {(*a*, *b*); *a* ≤ *b*}

We know that

(*a*, *a*) ∈ R as *a *= *a*

Thus, R is reflexive

Now,

(2, 4) ∈ R (as 2 < 4)

But, (4, 2) ∉ R as 4 is greater than 2.

Hence, R is not symmetric.

Let (*a*, *b*), (*b*, *c*) ∈ R.

Then,

*a* ≤ *b* and *b* ≤ *c*

*a* ≤ *c*

⇒ (*a*, *c*) ∈ R

Hence, R is transitive

Therefore, R is reflexive and transitive but not symmetric.

**Question** **5:** Check whether the relation R in R defined by R = {(*a*, *b*) : *a* ≤ *b*^{3}} is reflexive, symmetric or transitive.

**Solution 5:**

**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 6:**

Let *A* = {1, 2, 3}.

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

We know that,

(1, 1), (2, 2), (3, 3) ∉ R

Hence, R is not reflexive.

Now,

As (1, 2) ∈ R and (2, 1) ∈ R, then R is symmetric.

(1, 2) and (2, 1) ∈ R

Also,

(1, 1) ∉ R

Thus, R is not transitive.

Therefore, R is symmetric but neither reflexive nor transitive.

**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 7:**

Set *A* = Set of all books in the library of a college

R = {*x*, *y*): *x* and *y* have the same number of pages}

Now,

R is reflexive since (*x*, *x*) ∈ R as *x* and *x* has the same number of pages.

Let (*x*, *y*) ∈ R

*x* and *y* have the same number of pages.

*y* and *x* have the same number of pages.

⇒ (*y*, *x*) ∈ R

Hence, R is symmetric.

Let (*x*, *y*) ∈R and (*y*, *z*) ∈ R

*x* and *y* and have the same number of pages and *y* and *z* have the same number of pages.

Therefore, *x* and *z* have the same number of pages

⇒ (*x*, *z*) ∈ R

Hence, R is transitive.

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 8:**

**Download NCERT Solutions for Class 12 Maths: Chapter 1 Relations and Functions in PDF format**

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