Relations and Functions NCERT Solutions

NCERT Exercise 1.1 Q: 10 - 16

Question:10. Give an example of a relation. Which is

(i)Symmetric but neither reflexive nor transitive

Solution:

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

R is not reflexive because (1, 1)∉ R.

R is symmetric because (I, 2)∈ R

⇒ (2, 1) ∈R.

R is not transitive because (1, 2)∈ R and (2, 1)∈ R but (1, 1)∉ R.

(ii) Transitive but neither reflexive nor symmetric

Solution:

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

R is not reflexive because (2, 2)∉ R, (1, 1)∉ R, and (3, 3) ∉ R

R is not symmetric because (1, 2)∈ R but (2, 1)∉ R

R is transitive because (1, 2)∈ and (2, 3) ∈ R ⇒ (1, 3)∈ R

(iii)Reflexive and symmetric but not transitive

Solution:

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

R is reflective because (1, 1), (2, 2), (3, 3) ∈ R

R is Symmetric because (1, 2)∈ R ⇒(2, 1) ∈ R.

R is not transitive because (1, 2) ∈ R and (2, 3) ∈ R but (1, 3) ∉ R.

(iv) Reflexive and transitive but not symmetric

Solution:

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

R is reflective because (1, 1), (2,2), (3, 3), (4, 4), (5, 5) &isnin;R.

R is not symmetric because (1, 2)∈ R but (2, 1) ∉ R.

R is transitive because (1, 2) ∈ R and (2, 3 )∈ R

⇒ (1,3) ∈ R

(v) Symmetric and transitive but not reflexive.

Solution:

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

R is not reflexive because (3, 3)∉ R

R is symmetric because (2, 3)∈ R

⇒ (3, 2) ∈R.

R is transitive because (1, 3) ∈ R and (3, 1) ∈ R

⇒ (1, 1) ∈ R

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 original, 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 original as centre.

Solution:

Let O be the origin

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

={(P, Q) :OP =OQ}

Now let OP =x then (x, x) ∈ R since OP = OP.

So, R is reflexive relation.

Let OP = OQ = x than (x, x) ∈ R ⇒ (x, x) ∈ R.

So, R is symmetric relation.

Let OP = OQ =x and OQ =OR =x then

(x, x )∈ R and (x, x) ∈ R ⇒ (x, x) ∈ R.

So, R is transitive relation.

Thus, R is an equivalence relation.

The distance of all points related to P, from the origin is same as OP. Since a circle is the locus of all points equidistant from a fixed point (here 0), The set of the points related to P is a circle passing through P with O as the fixed point (centre).

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

Solution:

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

= {(T₁, T₂) : T₁ ~ T₂} where T₁, T₂ ∈ A

We know that every triangle is similar to itself.

⇒ {( T₁, T₂ ∈ R) for all T_{1 ∈ A}}

So, R is reflexive relation.

{( T₁, T₂): ∈ R Where T₁, T₂ ∈ A}

T₁ ~ T₂ then T₂ ~ T₁ ⇒ (T₂, T₁) ∈ R

So, R is symmetric relation.

(T₁, T₂) ∈ R and (T₂, T₃) ∈ R

Where, T₁, T₂, T₃ ∈ A

T₁ ~ T₂ and T₂ ~ T₃

⇒ (T₁, T₃) ∈ R

So, R is transitive relation.

Thus, R is an equivalence relation

We know that two triangles are similar if their sides are proportional. We see that sides of triangle T₁ are proportional to the sides of triangle T₃

Thus, T₁ is related to T₃

Question: 13. Show that the relation R defined in the set A of all polygons as

Solution: Here

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

Where, P₁, P₂ ∈ A

Now every polygon has same number of sides to itself.

⇒ (P₁, P₂) ∈ R for all P P₁ ∈ A

So, R is reflexive relation.

(P₁, P₂) ∈ R where P₁, P₂ ∈ A

⇒ number of sides in P₁ = number of sides in P₂

⇒ number of sides in P₂ = number of sides in P₁

⇒ (P₁, P₂) ∈ R

So, R is symmetric relation.

(P₁, P₂) ∈ R and (P₂, P₃) ∈ R

Where, P₁, P₂, P₃ ∈ A

⇒ number of sides in P₁ = number of sides in P₂

And, number of sides in P₂ = number of sides in P₃

⇒ number of sides in P₁ = number of sides in P₃

⇒ (P₁, P₃) ∈ R

So, R is transitive relation.

Thus R is an equivalence relation.

The set A is set of all the triangles.

Question: 14. Let L be the set of all lines in 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.

Solution: Here, R = {(L₁, L₂ : L₁ is parallel to L₂}

Where, L₁, L₂ ∈ L

Since, every line is parallel to itself.

⇒ L₁, L₂ ∈ R for all L₁ ∈ L

So, R is reflexive relation.

If (L₁, L₂) ∈ R where L₁, L₂ ∈ L

⇒ L₁ is parallel to L₂

⇒ L₂ is parallel to L₁

⇒ (L₂, L₁ ∈ R

So, R is symmetric relation.

(L₁, L₂ ∈ R and L₂, L₃ ∈ R

Where, L₁, L₂, L₃ ∈ L

⇒ L₁ is parallel to L₂

⇒ L₂ is parallel to L₃

⇒ L₁ is parallel to L₃

⇒ (L₁, L₃) ∈ R

So, R is transitive relation.

Thus, R is an equivalence relation.

The set of parallel lines related to the line y = 2x + 4 is y = 2x + c where c is any arbitrary constant.

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:

Here A={ 1, 2, 3, 4}

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

R is reflexive because (1, 1), (2, 2), (3, 3), (4, 4) ∈ R

R is not symmetric because (1, 2)∈ R but (2, 1) ∉ R.

R is transitive because (3, 3), (3, 2), (3, 2) ∈ R, (1, 2), (2, 2), (1, 2) ∈ R, etc.

Thus answer is (B) R is reflexive and transitive but not symmetric.

Question: 16. Let R be the relation in the set N given by R = {(a, b) :a = b – 2, b > 6 is correct answer.

(A) (2, 4)∈R

(B)(3, 8)∈ R

(C) (6, 8) ∈ R

(D) (8, 7) ∈ R

Solution:

(A) Is not the answer because 4 < 6

(B) does not satisfy the equation a = b – 2

(C) Satisfy the equation because 6= 8 – 2 and 8 > 6

(D) does not satisfy the equation a = b – 2

Thus the answer is (C) (6, 8) ∈

12-math-home

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