# Plus Two Math's Solution Ex 1.1 Chapter1 Relations and Function

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Here is the solution for Exercise 1.1 Chapter 1 Relation and Function of NCERT plus two maths. Here we have given a detailed explanation of each and every exercise so that students can understand the concepts easily without any difficulty. The solution to each and every question is provided here so you can solve them by yourself if you don’t get the answer here.

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 Board SCERT, Kerala Text Book NCERT Based Class Plus Two Subject Math's Textbook Solution Chapter Chapter 1 Exercise Ex 1.1 Chapter Name Relations and Function Category Plus Two Kerala

## Kerala Syllabus Plus Two Math's Textbook Solution Chapter  1 Relation and Function Exercises 1.1

### Chapter  1  Relations and Function Solution

Kerala plus two maths NCERT textbooks, we provide complete solutions for the exercise and answers provided at the end of each chapter. We also cover the entire syllabus given by the Board of secondary education, Kerala state.

### Chapter  1  Relations and Function Exercise   1.1

Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as

R = {(ab): b = a + 1} is reflexive, symmetric or transitive.

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

$\therefore$ R is not reflexive as (1,1)..........(6,6) etc are not belonging to R

for each 'a' element of 'A'; '(a, a)' is not an element of R

$\therefore$  R is not symmetric

'(a, b)' is an element of 'R' but (b, a) does not belong to R

$\therefore$  R is not transitive also

For some 'a, b, c ' element of 'R'; '(a, b)' element of 'R'; '(b, c)' element of 'R'; but '(a, c)' is not an element of 'R'

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 = {(xy): 3x − y = 0}

(ii) Relation R in the set N of natural numbers defined as

R = {(xy): y = x + 5 and x < 4}

(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as

R = {(xy): y is divisible by x}

(iv) Relation R in the set Z of all integers defined as

R = {(xy): x − y is as integer}

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

(a) R = {(xy): and y work at the same place}

(b) R = {(xy): x and y live in the same locality}

(c) R = {(xy): is exactly 7 cm taller than y}

(d) R = {(xy): x is wife of y}

(e) R = {(xy): x is father of y}

(i) A = {1, 2, 3 … 13, 14}

R = {(xy): 3x − y = 0}

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

R is not reflexive since (1, 1), (2, 2) … (14, 14) ∉ R.

Also, R is not symmetric as (1, 3) ∈R, but (3, 1) ∉ R. [3(3) − 1 ≠ 0]

Also, R is not transitive as (1, 3), (3, 9) ∈R, but (1, 9) ∉ R.

[3(1) − 9 ≠ 0]

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

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

It is seen that (1, 1) ∉ R.

∴R is not reflexive.

(1, 6) ∈R

But,

(6, 1) ∉ R.

∴R is not symmetric.

Now, since there is no pair in R such that (xy) and (yz) ∈R, then (xz) cannot belong to R.

∴ R is not transitive.

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

(iii) A = {1, 2, 3, 4, 5, 6}

R = {(xy): y is divisible by x}

We know that any number (x) is divisible by itself.

(xx) ∈R

∴R is reflexive.

Now,

(2, 4) ∈R [as 4 is divisible by 2]

But,

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

∴R is not symmetric.

Let (xy), (yz) ∈ R. Then, y is divisible by x and z is divisible by y.

z is divisible by x.

⇒ (xz) ∈R

∴R is transitive.

Hence, R is reflexive and transitive but not symmetric.

(iv) R = {(xy): x − y is an integer}

Now, for every x ∈ Z, (xx) ∈R as x − x = 0 is an integer.

∴R is reflexive.

Now, for every xy ∈ Z if (xy) ∈ R, then x − y is an integer.

⇒ −(x − y) is also an integer.

⇒ (y − x) is an integer.

∴ (yx) ∈ R

∴R is symmetric.

Now,

Let (xy) and (yz) ∈R, where xyz ∈ Z.

⇒ (x − y) and (y − z) are integers.

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

∴ (xz) ∈R

∴R is transitive.

Hence, R is reflexive, symmetric, and transitive.

(v) (a) R = {(xy): x and y work at the same place}

(xx) ∈ R

∴ R is reflexive.

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

⇒ y and x work at the same place.

⇒ (yx) ∈ R.

∴R is symmetric.

Now, let (xy), (yz) ∈ 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.

⇒ (xz) ∈R

∴ R is transitive.

Hence, R is reflexive, symmetric, and transitive.

(b) R = {(xy): x and y live in the same locality}

Clearly (xx) ∈ R as x and x is the same human being.

∴ R is reflexive.

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

⇒ y and x live in the same locality.

⇒ (yx) ∈ R

∴R is symmetric.

Now, let (xy) ∈ R and (yz) ∈ 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 = {(xy): x is exactly 7 cm taller than y}

Now,

(xx) ∉ R

Since human being cannot be taller than himself.

∴R is not reflexive.

Now, let (xy) ∈R.

⇒ x is exactly 7 cm taller than y.

Then, y is not taller than x.

∴ (yx) ∉R

Indeed if x is exactly 7 cm taller than y, then y is exactly 7 cm shorter than x.

∴R is not symmetric.

Now,

Let (xy), (yz) ∈ R.

⇒ x is exactly 7 cm taller thanand y is exactly 7 cm taller than z.

⇒ x is exactly 14 cm taller than z.

∴(xz) ∉R

∴ R is not transitive.

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

(d) R = {(xy): x is the wife of y}

Now,

(xx) ∉ R

Since x cannot be the wife of herself.

∴R is not reflexive.

Now, let (xy) ∈ R

⇒ x is the wife of y.

Clearly, y is not the wife of x.

∴(yx) ∉ R

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

∴ R is not symmetric.

Let (xy), (yz) ∈ R

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

This case is not possible. Also, this does not imply that x is the wife of z.

∴(xz) ∉ R

∴R is not transitive.

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

(e) R = {(xy): x is the father of y}

(xx) ∉ R

As x cannot be the father of himself.

∴R is not reflexive.

Now, let (xy) ∈R.

⇒ x is the father of y.

⇒ y cannot be the father of y.

Indeed, y is the son or the daughter of y.

∴(yx) ∉ R

∴ R is not symmetric.

Now, let (xy) ∈ R and (yz) ∈ R.

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

⇒ x is not the father of z.

Indeed x is the grandfather of z.

∴ (xz) ∉ R

∴R is not transitive.

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

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

R = {(ab): a ≤ b2} is neither reflexive nor symmetric nor transitive.

R = {(ab): a ≤ b2}

It can be observed that

∴R is not reflexive.

Now, (1, 4) ∈ R as 1 < 42

But, 4 is not less than 12.

∴(4, 1) ∉ R

∴R is not symmetric.

Now,

(3, 2), (2, 1.5) ∈ R

(as 3 < 22 = 4 and 2 < (1.5)2 = 2.25)

But, 3 > (1.5)2 = 2.25

∴(3, 1.5) ∉ R

∴ R is not transitive.

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

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

R = {(ab); a ≤ b}

Clearly (aa) ∈ R as a.

∴R is reflexive.

Now,

(2, 4) ∈ R (as 2 < 4)

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

∴ R is not symmetric.

Now, let (ab), (bc) ∈ R.

Then,

a ≤ b and b ≤ c

⇒ a ≤ c

⇒ (ac) ∈ R

∴R is transitive.

Hence, R is reflexive and transitive but not symmetric.

Check whether the relation R in R defined as R = {(ab): a ≤ b3} is reflexive, symmetric, or transitive.

R = {(ab): â‰¤ b3}

It is observed that

âˆ´ R is not reflexive.

Now,

(1, 2) âˆˆ R (as 1 < 23 = 8)

But,

(2, 1) âˆ‰ R (as 2 > 13 = 1)

âˆ´ R is not symmetric.

We have

But

âˆ´ R is not transitive.

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

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.

Let A = {1, 2, 3}.

A relation R on A is defined as R = {(1, 2), (2, 1)}.

It is seen that (1, 1), (2, 2), (3, 3) ∉R.

∴ R is not reflexive.

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

Now, (1, 2) and (2, 1) ∈ R

However,

(1, 1) ∉ R

∴ R is not transitive.

Hence, R is symmetric but neither reflexive nor transitive.

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

Set A is the set of all books in the library of a college.

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

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

Let (xy) ∈ R ⇒ x and y have the same number of pages.

⇒ y and x have the same number of pages.

⇒ (yx) ∈ R

∴R is symmetric.

Now, let (xy) ∈R and (yz) ∈ R.

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

⇒ x and z have the same number of pages.

⇒ (xz) ∈ R

∴R is transitive.

Hence, R is an equivalence relation.

Show that the relation R in the set A = {1, 2, 3, 4, 5} given by

, 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}.

A = {1, 2, 3, 4, 5}

It is clear that for any element a ∈A, we have (which is even).

∴R is reflexive.

Let (ab) ∈ R.

∴R is symmetric.

Now, let (ab) ∈ R and (bc) ∈ R.

⇒ (ac) ∈ R

∴R is transitive.

Hence, R is an equivalence relation.

Now, all elements of the set {1, 3, 5} are related to each other as all the elements of this subset are odd. Thus, the modulus of the difference between any two elements will be even.

Similarly, all elements of the set {2, 4} are related to each other as all the elements of this subset are even.

Also, no element of the subset {1, 3, 5} can be related to any element of {2, 4} as all elements of {1, 3, 5} are odd and all elements of {2, 4} are even. Thus, the modulus of the difference between the two elements (from each of these two subsets) will not be even.

Show that each of the relation R in the set, given by

(i)

(ii)

is an equivalence relation. Find the set of all elements related to 1 in each case.

(i)

For any element a ∈A, we have (aa) ∈ R as is a multiple of 4.

∴R is reflexive.

Now, let (ab) ∈ R ⇒ is a multiple of 4.

⇒ (ba) ∈ R

∴R is symmetric.

Now, let (ab), (bc) ∈ R.

⇒ (ac) ∈R

∴ R is transitive.

Hence, R is an equivalence relation.

The set of elements related to 1 is {1, 5, 9} since

(ii) R = {(ab): a = b}

For any element a ∈A, we have (aa) ∈ R, since a = a.

∴R is reflexive.

Now, let (ab) ∈ R.

⇒ a = b

⇒ b = a

⇒ (ba) ∈ R

∴R is symmetric.

Now, let (ab) ∈ R and (bc) ∈ R.

⇒ a = b and b = c

⇒ a = c

⇒ (ac) ∈ R

∴ R is transitive.

Hence, R is an equivalence relation.

The elements in R that are related to 1 will be those elements from set A which are equal to 1.

Hence, the set of elements related to 1 is {1}.

Given 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.

(i) Let A = {5, 6, 7}.

Define a relation R on A as R = {(5, 6), (6, 5)}.

Relation R is not reflexive as (5, 5), (6, 6), (7, 7) ∉ R.

Now, as (5, 6) ∈ R and also (6, 5) ∈ R, R is symmetric.

(5, 6), (6, 5) ∈ R, but (5, 5) ∉ R

∴R is not transitive.

Hence, relation R is symmetric but not reflexive or transitive.

(ii)Consider a relation R in defined as:

R = {(ab): a < b}

For any an ∈ R, we have (aa) ∉ R since a cannot be strictly less than itself. In fact, a = a.

∴ R is not reflexive.

Now,

(1, 2) ∈ R (as 1 < 2)

But, 2 is not less than 1.

∴ (2, 1) ∉ R

∴ R is not symmetric.

Now, let (ab), (bc) ∈ R.

⇒ a < b and b < c

⇒ a < c

⇒ (ac) ∈ R

∴R is transitive.

Hence, relation R is transitive but not reflexive and symmetric.

(iii)Let A = {4, 6, 8}.

Define a relation R on A as:

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

Relation R is reflexive since for every a ∈ A, (aa) ∈R i.e., (4, 4), (6, 6), (8, 8)} ∈ R.

Relation R is symmetric since (ab) ∈ R ⇒ (ba) ∈ R for all ab ∈ R.

Relation R is not transitive since (4, 6), (6, 8) ∈ R, but (4, 8) ∉ R.

Hence, relation R is reflexive and symmetric but not transitive.

(iv) Define a relation R in R as:

R = {ab): a3 ≥ b3}

Clearly (aa) ∈ R as a3 = a3.

∴R is reflexive.

Now,

(2, 1) ∈ R (as 23 ≥ 13)

But,

(1, 2) ∉ R (as 13 < 23)

∴ R is not symmetric.

Now,

Let (ab), (bc) ∈ R.

⇒ a3 ≥ b3 and b3 ≥ c3

⇒ a3 ≥ c3

⇒ (ac) ∈ R

∴R is transitive.

Hence, relation R is reflexive and transitive but not symmetric.

(v) Let A = {−5, −6}.

Define a relation R on A as:

R = {(−5, −6), (−6, −5), (−5, −5)}

Relation R is not reflexive as (−6, −6) ∉ R.

Relation R is symmetric as (−5, −6) ∈ R and (−6, −5}∈R.

It is seen that (−5, −6), (−6, −5) ∈ R. Also, (−5, −5) ∈ R.

∴The relation R is transitive.

Hence, relation R is symmetric and transitive but not reflexive.

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 center.

It is given that

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

Now, it is clear that

(P, P) ∈  R since the distance of point P from the origin is always the same as the distance of the same point P from the origin.

Therefore, R is reflexive.

Now, Let us take (P, Q) ∈ R,

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

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

⇒ (Q, P)∈ R

Therefore, R is symmetric.

Now, Let (P, Q), (Q, S) ∈  R

⇒ The distance of point P and Q from the origin is always the same as the distance of the same point Q and S from the origin.

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

⇒ (P,S) ∈  R

Therefore, R is transitive.

Therefore, R is an equivalence relation.

The set of all points related to P ≠ (0,0) will be those points whose distance from the origin is the same as the distance of point P from the origin.

So, if O(0,0) is the origin and OP = k, then the set of all points related to P is at a distance of k from the origin.

Therefore, this set of points forms a circle with the center as the origin, and this circle passes through

Show that the relation R defined in the set A of all triangles as R = {(T1T2): T1 is similar to T2}, is an equivalence relation. Consider three right angle triangles T1 with sides 3, 4, 5, T2 with sides 5, 12, 13, and T3 with sides 6, 8, 10. Which triangles among T1T2, and T3 are related?

R = {(T1T2): T1 is similar to T2}

R is reflexive since every triangle is similar to itself.

Further, if (T1T2) ∈ R, then T1 is similar to T2.

⇒ T2 is similar to T1.

⇒ (T2T1) ∈R

∴R is symmetric.

Now,

Let (T1T2), (T2T3) ∈ R.

⇒ T1 is similar to T2 and T2 is similar to T3.

⇒ T1 is similar to T3.

⇒ (T1T3) ∈ R

∴ R is transitive.

Thus, R is an equivalence relation.

Now, we can observe that:

∴The corresponding sides of triangles T1 and T3 are in the same ratio.

Then, triangle T1 is similar to triangle T3.

Hence, T1­ is related to T3.

Show that the relation R defined in the set A of all polygons as R = {(P1P2): P1 and P2 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?

R = {(P1P2): P1 and P2 have same the number of sides}

R is reflexive since (P1P1) ∈ R as the same polygon has the same number of sides with itself.

Let (P1P2) ∈ R.

⇒ P1 and P have the same number of sides.

⇒ P2 and P1 have the same number of sides.

⇒ (P2P1) ∈ R

∴R is symmetric.

Now,

Let (P1P2), (P2P3) ∈ R.

⇒ P1 and P2 have the same number of sides. Also, P2 and P3 have the same number of sides.

⇒ P1 and P3 have the same number of sides.

⇒ (P1P3) ∈ R

∴R is transitive.

Hence, 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 that have 3 sides (since T is a polygon with 3 sides).

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

Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L1L2): L1 is parallel to L2}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.

R = {(L1L2): L1 is parallel to L2}

R is reflexive as any line L1 is parallel to itself i.e., (L1L1) ∈ R.

Now,

Let (L1L2) ∈ R.

⇒ L1 is parallel to L2.

⇒ L2 is parallel to L1.

⇒ (L2L1) ∈ R

∴ R is symmetric.

Now,

Let (L1L2), (L2L3) ∈R.

⇒ L1 is parallel to L2. Also, L2 is parallel to L3.

⇒ L1 is parallel to L3.

∴R is transitive.

Hence, R is an equivalence relation.

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

Slope of line y = 2x + 4 is m = 2

It is known that parallel lines have the same slopes.

The line parallel to the given line is of the form y = 2x + c, where c ∈R.

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

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.

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

It is seen that (aa) ∈ R, for every a ∈{1, 2, 3, 4}.

∴ R is reflexive.

It is seen that (1, 2) ∈ R, but (2, 1) ∉ R.

∴R is not symmetric.

Also, it is observed that (ab), (bc) ∈ R ⇒ (ac) ∈ R for all abc ∈ {1, 2, 3, 4}.

∴ R is transitive.

Hence, R is reflexive and transitive but not symmetric.

The correct answer is B.

Let R be the relation in the set given by R = {(ab): b − 2, > 6}. Choose the correct answer.

(A) (2, 4) ∈ R (B) (3, 8) ∈R (C) (6, 8) ∈R (D) (8, 7) ∈ R

R = {(ab): b − 2, b > 6}

Now, since b > 6, (2, 4) ∉ R

Also, as 3 ≠ 8 − 2, (3, 8) ∉ R

And, as 8 ≠ 7 − 2

(8, 7) ∉ R

Now, consider (6, 8).

We have 8 > 6 and also, 6 = 8 − 2.

∴(6, 8) ∈ R

The correct answer is C.

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