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 angle 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 and T3 are related?

Class 12, NCERT Chapter 1, Exercise 1.1, Q12

It is given that the relation R defined in the set A of all triangles as

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

Now, R is reflexive as every triangle is similar to itself.

Now, if (T1, T2) ϵ R, then T1 is similar to T2.

⇒ T2 is similar to T1.

⇒ (T1, T2) ϵ R

Therefore, R is symmetric.

Now, if (T1, T2), (T2, T3) ϵ R,

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

⇒ T1 is similar to T3.

⇒ (T1, T3) ϵ R

Therefore, R is transitive.

Therefore, R is an equivalence relation.

Now, we can see that,

Therefore, the corresponding sides of triangles T1 and T3 are in the same ratio.

Thus, triangle T1 is similar to triangle T3.

Therefore, T1 is related to T3.

Comments

In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer. (i) f : R → R defined by f(x) = 3 – 4x (ii) f : R → R defined by f(x) = 1 + x^2

Class 12, NCERT Chapter 1, Exercise 1.2, Q7 (i) Here f(x) = 3 – 4x Then f(x 1 ) = 3 – 4x 1 and f (x 2 ) = 3 – 4x 2 Now f(x 1 ) = f (x 2 ) ⇒ 3 – 4x 1 = 3 – 4x 2 ⇒ – 4x 1 = –4x 2 ⇒ x 1 = x 2 ∴ f(x 1 ) = f(x 2 ) & x 1 = x 2 So, f is one-one function. Let f(x) = y ∈ R Then y=3 - 4x ⇒ x= 3-y/4 (ii) Let x 1 = 2 and x 2 = –2 ∈ R Here f(x) = 1 + x 2 Then f(x 1 ) = f (2) = 1 + (2) 2 = 5 and f (x 2 ) = f(–2) = 1 + (–2) 2 = 5 ∴ f (x 1 ) = f(x 2 ) but x 1 ≠ x 2 So, f is not one-one function. Let f(x) = – 2 ⇒R Then 1 + x 2 = –2 ⇒ x 2 = –3 = ± √–3 ∈R So, f is not onto function.

Let f : R → R be defined as f(x) = x 4 . Choose the correct answer. (A) f is one-one onto (B) f is many-one onto (C) f is one-one but not onto (D) f is neither one-one nor onto.

Class 12, NCERT Chapter 1, Exercise 1.2, Q11 f : R → R is defined as f (x) = x 4 Let x , y ∈ R such that f ( x ) = f ( y ). ∴ does not imply that x 1 =x 2 . For instance, f (1) = f (-1) = 1 ∴ f is not one-one. Consider an element 2 in co-domain R . It is clear that there does not exist any x in domain R such that f ( x ) = 2. ∴ f is not onto. Hence, function f is neither one-one nor onto. The correct answer is D

Let L be the set of all lines in a plane and R be the relation in L defined as R = {(L1 , L2 ) : L1 is perpendicular to L2 }. Show that R is symmetric but neither reflexive nor transitive.

RELATIONS AND FUNCTIONS Class 12, NCERT Chapter 1, Example 3 Solution Figure R is not reflexive, as a line L1 can not be perpendicular to itself, i.e., (L1 , L1 ) ∉ R. R is symmetric as (L1 , L2 ) ∈ R ⇒ L1 is perpendicular to L2 ⇒ L2 is perpendicular to L1 ⇒ (L2 , L1 ) ∈ R. R is not transitive. Indeed, if L1 is perpendicular to L2 and L2 is perpendicular to L3 , then L1 can never be perpendicular to L3 . In fact, L1 is parallel to L3 , i.e., (L1 , L2 ) ∈ R, (L2 , L3 ) ∈ R but (L1 , L3 ) ∉ R. Example1 Example2 Example3 ←you are here

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