NCERT 12 Maths Relations and Functions Chapter 1 – Exercise

NCERT Solutions for Class 12 Maths and Chapter Wise Solutions, NCERT 12 Maths Relations and Functions Chapter 1 – Exercise.

Relations and Functions

Q 1. Determine whether each of the following relations are reflexive, symmetric and transitive:

  1. Relation R in the set A = {1,2,3, ….., 13, 14} defined as R = {(x, y): 3x-y=0}
  2. Relation R in the set N of natural numbers defined as R = {(x, y): y = x+5 and x<4}
  3. Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y): y is divisible by x}
  4. Relation R in the set Z of all integers defined as R = {(x, y): x − y is as integer}
  5. 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}

Solutions 1. (a) Given that, A = {1, 2, 3, 4,….., 12, 13, 14} and R is = {(x,y) : 3x – y = 0} for reflexive relation (x, x) ϵ R ∀ x ϵ A

  If y = x, 3x – y = 0 → 2x ≠ 0, R is not reflexive.

→ (x, x) ∉R.

 

(b) Given R = {(x, y) : 3x-y=0} for symmetric relation (x,y) ϵ R →(y, x)ϵ R

If (x, y) ϵR → 3x – y = 0, then 3y-x ≠ 0 →(y, x) ∉ R, so R is not symmetric.

For example, if x =1, y = 3, 3×1–3=0, 3×3-1 = 9-1 =8≠ 0

 

(c) For transitive, if (x, y) ϵ R →3x-y=0, (y,z) ϵ R→ 3y-z = 0, then 3x-z≠ 0. So R is not transitive

For example, when x=1, y=3, z=9, then 3×1-3=0, 3×3-9=0, 3×1-9≠ 0

 

2. Here, A = N, the set of natural numbers and

R = {(x, y): y = x+5, x<4}

= {(x, x+5) : x ϵ N and x<4} = {(1, 6), (2, 7), (3, 8)}

a. For reflexive (x, x) ϵ R∀ x putting y=x, x ≠y +5 → (1, 1) ∉ So, R is not reflexive.

 

b. For symmetrical (x, y) ϵ R → (y, x) ϵ R putting y=x+5, then x ≠y+5→ (1, 6) ϵ R but (6, 1) ∉ So, R is not symmetric.

 

c .For transitivity (x, y) ϵ R, (y, z) ϵ R→ (x, z) ϵ R if y = x+5, z=y+5, then z≠ x+5. Since, (1, 6) ϵ R and there is no order pair in R which has 6 as the first element same in the case for (2, 7) and (3, 8). So, R is not transitive.

 

3.  (a) For reflexive, we know that x is divisible by x for all x ϵ A.

  (x, x) ϵ R for all x ϵ R. So, R is reflexive.

 

(b) For symmetry, we observe that 6 is divisible by 2. This
means that (2, 6) ϵ R but (6, 2) ∉R. So, R is not symmetric.

 

(c)For transitivity, let (x, y) ϵ R and (y, z) ϵ R, then z is divisible by x.

→                          (x, z) ϵ R

For example, 2 is divisible by 1, 4 is divisible by 2.

So, 4 is divisible by 1. So, R is transitive.

4. (a) For reflexive put y=x, x-x=0 which is an integer for all x ϵ Z. So, R is reflexive on Z.

 

(b) For symmetry, let (x, y) ϵ R, then (x-y) is an integer λ and also

y-x=-λ                              [λ ϵZ = -λ ϵZ]

  y-x is an integer (y, x) ϵ R. So, R is symmetric.

 

(c) For transitivity let (x, y) ϵ R and (y, z) ϵ R

  x-y= integer and y-z= integers, then x-z is also an integer

   (x, z) ϵ R. So, R is transitive.

 

5. (a) R is reflexive, symmetric and transitive obviously.

 

(b) R is reflexive, symmetric and transitive obviously.

 

(c) Here, R is not reflexive as x is not 7 cm taller than x

R is not symmetric as if x is exactly 7cm taller than y, then y cannot be 7cm taller than x and R is not transitive as if x is exactly 7 cm taller than y and y is exactly 7 cm taller than z, then x is exactly 14 cm taller than z.

 

(d) Here, R is not reflexive ; as x is not wife of x, R is not symmetric as x is wife of y, then y is husband (not wife) of x and R is transitive as transitivity is not contradicted in this case. Whenever, (x, y) ϵ R, then (y, z) ∉R for any z as if x is wife of y, then y is a male and a male cannot be a wife.

 

(e) Here, R is not reflexive; as x cannot be father of x. For any x, R is not symmetric as if x is father of y, then y cannot be father of x. R is not transitive as if x is father of y and y is father of z, then x is grandfather (not father) of z.

 

Q 2. Show that the relation R in the set R of real numbers, defined as R = {(a, b) : a ≤ b2} is neither reflexive nor symmetric nor transitive.

Sol. We have R = {(a, b) : a≤b2}, where a, b ϵ R

For reflexivity, we observe that 1/2 ≤ (1/2)2 is not true.

So, R is not reflexive as (1/2, 1/2) ∉ R.

For symmetry, we observe that -1 ≤ 32 but 3 ≰ (-1)2

(-1, 3) ϵ R but (3-1) ∉ R

So, R is not symmetric

For transitivity,  we observe that 2 ≤ (-3)2 and -3 ≤ (1)2 but 2 ≰ (1)2

(2, -3) ϵ R and (-3, 1) ϵ R but (2, 1) ∉ R. So, R is not transitive.

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

 

Q 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.

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

A relation R is defined on set A is

R = {(a, b): b = a+1}. Therefore, R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

Now, 6  ϵ A but (6, 6) ∉ R.

Therefore, R is not reflexive.

It can be observed that (1, 2) ∈ R, but (2, 1) ∉ R.

 R is not symmetric.

Now, (1, 2), (2, 3) ∈ 

But, (1, 3) ∉ R

∴R is not transitive

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

 

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

Sol. 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.

 

Q 5. Check whether the relation R in R defined by R = {(a, b) : a ≤ b3} is reflexive, symmetric or transitive.

Sol. Given that R = {(a, b) : a ≤ b3 }

relations and functions

Therefore, R is not transitive

Hence, R is neither reflexive nor symmetric nor transitive

 

Q 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.

Sol. Given that 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.

Therefore, R is not reflexive

Now, (1, 2) ϵ R and (2, 1)ϵ R. So, R is symmetric

Now (1, 2) and (2, 1) ϵ R but (1, 1) ∉ R. So, R is not transitive.

Hence, R is symmetric but neither reflexive nor transitive

 

Q 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.

Sol. Here, set A is the set of all books in the library of a college.

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

Now, R is reflexive, since {x, x} ϵ R as x and x has the same number of pages.

Let {x, y} ϵ R

→ x and y have the same number of pages.

→ y and x have the same number of pages.

→ (y, x)ϵ R. So, R is symmetric. Now, let (x, y) ϵ R and (y, z) ϵ R.

→ x and y 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 → (x, z) ϵ R

Therefore, R is transitive. Hence, R is an equivalence relation.

 

Q 8. Show that the relation R in the set A ={1, 2, 3, 4, 5} is 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}.

Sol. Given that A = {1, 2, 3, 4, 5}, R = {(a, b) : |a-b| is even}

Let a ϵ A → |a-a| = 0 (which is even), ∀ a

So, R is reflexive

Let (a, b) ϵ R → |a-b| is even →|-(b-a)| = |b-a| is also even

→ (b, a) ϵR. So, R is symmetric

Now, Let (a, b) ϵR and (b, c)ϵR

→ |a-b| is even and |b-c| is even

→ (a-b) is even and (b-c) is even

→ (a-c) = (a-b) +(b-c) is even [sum of two even integers is even]

→ |a-c| is even → (a, c)ϵR

So, 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 set are odd. So, 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 set are even. So, the modulus of the difference between any two elements will be even.

Also, no element of the {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. So, the modulus of the modules of the difference between the two elements (from each of these two subsets) will not be even.

 

Q 9.Show that the relation R in the set S = {x ϵ Z : 0≤ x≤ 12}, given by

i. R = {(a, b) : |a-b| is a multiple of 4}

ii. R = {(a, b) ; a=b}, are equivalence relations. Find the set of all elements related to 1 in the each case.

Sol. Given that A = {x ϵ Z : 0 ≤ x ≤ 12}

= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

i. R = {(a, b) : |a-b| is a multiple of 4}

For any element of a ϵ A, we have (a, a) ϵ R as |a-a| = 0 is a multiple of 4. Therefore R is reflexive.

Now, let (a, b)ϵR

→ |a-b| is a multiple of 4,  |b-a| = |-(a-b)| = |a-b| is a multiple of 4

→(b, a) ϵ R. Therefore, R is symmetric. Now let (a, b), (b, c) ϵR

So, |a-b| and |b-c| is multiple of 4.

→|a-c| is also multiple of 4 (a, c)ϵR

Therefore, R is transitive.

Hence, R is an equivalence relation.

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

|1-1| =0 is multiple of 4

|5-1| =4 is a multiple of 4 and

|9-1| = 8 is a multiple of 4

 

ii. R = {(a, b) :a=b}

For any element a ϵ A, we have (a, a)ϵR, as a=a

Therefore, R is reflexive

Now, let (a, b)ϵR →(b, a)ϵR

a=b → b=a

(b, a)ϵ R

Therefore, R is symmetric

Now, let (a, b)ϵR and (b, c)ϵR

a=b and b=c →(a, c)ϵR. Therefore, 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}.

 

Q 10. Given example of relation which are

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

Sol. i. Let A ={5, 6, 7}

Defined 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)ϵR and also (6, 5) ϵR. So, R is symmetric

(5, 6), (6, 5) ϵR but (5, 5) ∉ R. Therefore, R is not transitive.

Hence, relation R is symmetric but neither reflexive nor transitive

 

ii. Consider a relation R in R defined as

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

For any a ϵ R, we have (a, a) ∉ R, since a cannot be strictly less than a itself. Therefore, R is not reflexive.    [a=a]

Now, (1, 2)ϵ R (as 1<2)

But (2, 1) ∉ R as 2 is not less than 1.

Therefore, R is not symmetric.

Now, let (a, b), (b, c) ϵ R

→ a< b and b <c  → a< c → (a, c) ϵ R

Therefore, R is transitive.

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

 

iii. Let A = {4, 6, 8}

Defined 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 ϵ A, (a, a)ϵ R, i.e., (4, 4), (6, 6), (8, 8) ϵ R

Relation R is symmetric since, (a, b) ϵ R → (b, a) ϵ R for all a, b ϵ R.

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

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

 

iv. Define a relation R in R as:

R = {(a, b) : a3 ≥ b3}

Clearly (a, a) ϵ R as a3 = a3. Therefore, R is reflexive.

Now, (2, 1)ϵ R ( as 23≥13). But (1, 2)∉ R (as 13<23)

Therefore, R is not symmetric.

Now, let (a, b), (b, c) ϵ R

→ a3 ≥ b3 and b3 ≥ c3→(a, c)ϵR. Therefore, 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)ϵ R and (-6, -5)ϵR

And (-5, -6), (-6, -5) ϵ R→ (-5, -5) ϵ R

Therefore, the relation R is transitive.

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

 

Q 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.

Sol. Here, R = {(P, Q) : distance of point P from the origin is same as the distance of point Q from the origin}. Clearly (P, P) ϵR, since the distance of point P from the origin is always the same as the distance of the same point from the origin.

Therefore, R is reflexive.

Now, let (P, Q) ϵR.

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

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

(Q, P)ϵ R

Therefore, R is symmetric.

Now, let (P, Q), (Q, S) ϵ R

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

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

→ (P, S) ϵ 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 points P from the origin.

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

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

 

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

Sol.  Here, R ={( T1, T2): T1 is similar to T2}

R is reflexive, since every triangle is similar to itself.

Further, if (T1, T2) ϵR, then T1 is similar to T2

T2 is similar to T1 → (T2, T1)ϵR

Therefore, R is symmetric.

Now, let (T1, T2), T2, T3)ϵR

→ T1, is similar to T2 and Tis  similar to T3

→ → (T1, T3) ϵ R

Therefore, R is transitive. Thus, R is an equivalence relation.

Now, we can observe that 3/6 = 4/8 = 5/10 = (1/2)

Therefore, 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.

 

Q 13. Show that the relation R, defined in the set A of all polygons as R = {(P1, P2) : P1 and P2 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?

Sol. Here, R = {(P1, P2) : P1 and P2  have same 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 which 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.

 

Q 14. Let L be the set of all lines XY-plane and R be the relation in L defined as R = {(L1, L2): 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.

Sol. Here, R = {(L1, L2): 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.

 

Q 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.i. R is reflexive and symmetric but not transitive

ii. R is reflexive and transitive but not symmetric

iii. R is symmetric and transitive but not reflexive

iv. R is an equivalence relation

Sol. Here, R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}

Since, (a, a)ϵ R, for every a ϵ {1, 2, 3, 4}. Therefore, R is reflexive.

Now, since (1, 2) ϵR but (2, 1)∉R. Therefore, R is not symmetric.

Also, it is observed that (a, b), (b, c) ϵR.

→ (a, c)ϵ R. For all a, b, c ϵ {1, 2, 3, 4}

Therefore, R is transitive. Hence, R is reflexive and transitive but not symmetric. Thus, the correct answer is (b).

 

Q 16. Let R is the relation in the set N given by R = {(a, b) : a=b-2, b> 6}. Choose the correct answer

a. (2, 4) ϵ R

b. (3, 8) ϵ R

c. (6, 8) ϵ R

d. (8, 7) ϵ R

Sol. Given R = {(a, b) : a=b-2, b> 6}

Now, since, b> 6 so (2, 4) ∉R. Also, as 3≠8-2 so (3, 8) ∉R and as 8≠7 – 2

Therefore, (8, 7)∉R

Now, consider (6, 8). We have 8>6 and also, 6= 8-2

Therefore, (6, 8) ϵ R. Thus, the correct answer is (c)

means that (2, 6) R but (6, 2) R. So, R is not symmetric.

(c)For transitivity, let (x, y) R and (y, z) R, then z is divisible by x.

(x, z) R

For example, 2 is divisible by 1, 4 is divisible by 2

So, 4 is divisible by 1. So, R is transitive.

(iv) (a) For reflexive put y = x, x-x=0 which is an integer for all x Z. So, R is reflexive on Z.

Updated: September 22, 2020 — 2:21 pm

Leave a Reply

Your email address will not be published.