Correct the mistakes if any 5 ∉ ( 0 , 1 ) .

Given statement is 5 ∉ 0 , 1 If a and b are two real numbers such that a < b , then an open interval a , b is the set of all real numbers that lies between a and b excluding both the end points. i.e., a , b = x ∈ R : a < x < b Here, 0 , 1 is the set of all real numbers that lies between 0 and 1 excluding both the end-points. So, 5 is not an element of 0 , 1 , and we write it as 5 ∉ ( 0 , 1 ) Thus, the given statement is true i.e., it has no mistakes.

Correct the mistakes if any - 1 ∉ ( - ∞ , - 1 ) .

Given statement is - 1 ∉ ( - ∞ , - 1 ) If a and b are two real numbers such that a < b , then an open interval a , b is the set of all real numbers that lies between a and b excluding both the end points. i.e., a , b = x ∈ R : a < x < b Here ( - ∞ , - 1 ) is the set of all real numbers that lies between - ∞ and - 1 excluding both the end-points. So, - 1 is not an element of ( - ∞ , - 1 ) , and we write it as - 1 ∉ ( - ∞ , - 1 ) Thus, the given statement is true i.e., it has no mistakes.

Correct the mistakes if any 3 ∈ [ 1 , ∞ ) .

If a and b are any two real numbers such that a < b then the interval [ a , b ) represent all the numbers between a and b , including a but not b i.e., [ a , b ) = x ∈ R : a ≤ x < b Clearly, [ 1 , ∞ ) is the set of all real numbers that lies between 1 and ∞ including 1 but not ∞ . Therefore, an element 3 belongs to the interval [ 1 , ∞ ) . And, we write it as 3 ∈ [ 1 , ∞ ) . Hence, 3 ∈ [ 1 , ∞ ) is correct and no correction is needed.

Correct the mistakes if any 3 ∈ [ 3 , ∞ ) .

If a and b are any two real numbers such that a < b then the interval [ a , b ) represent all the numbers between a and b , including a but not b i.e., [ a , b ) = x ∈ R : a ≤ x < b Clearly, [ 3 , ∞ ) is the set of all real numbers that lies between 3 and ∞ , including 3 but not ∞ . Therefore, an element 3 belongs to the interval [ 3 , ∞ ) . And, we write it as 3 ∈ [ 3 , ∞ ) . Hence, 3 ∈ [ 3 , ∞ ) is correct and no correction is needed.

Correct the mistakes if any 2 ∈ ( - ∞ , 2 ) .

Given statement is 2 ∈ ( - ∞ , 2 ) If a and b are two real numbers such that a < b , then an open interval a , b is the set of all real numbers that lies between a and b excluding both the end points. i.e., a , b = x ∈ R : a < x < b Here ( - ∞ , 2 ) is the set of all real numbers that lies between - ∞ and 2 excluding both the end-points. So, 2 is not an element of ( - ∞ , 2 ) , and we write it as 2 ∉ ( - ∞ , 2 ) Thus, the given statement is false and the correction is 2 ∉ ( - ∞ , 2 ) .

Let 0 < a < b < c . Then a 2 c < a 2 + b 2 + c 2 a + b + c < c 2 a .

Given, that, we have: 0 < a < b < c ⇒ c > b > a > 0 ⇒ c 2 > b 2 > a 2 > 0 Now, a < b < c ⇒ 3 a < a + b + c < 3 c ⇒ 1 3 c < 1 a + b + c < 1 3 a → 1 And a 2 < b 2 < c 2 ⇒ 3 a 2 < a 2 + b 2 + c 2 < 3 c 2 → 2 From 1 and 2 , we get: 3 a 2 3 c < a 2 + b 2 + c 2 a + b + c < 3 c 2 3 a ⇒ a 2 c < a 2 + b 2 + c 2 a + b + c < c 2 a Hence, proved the inequality.

E M B I B E

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 17 - Real Number System>EXERCISES 1(b)

Odisha Board Solutions for Chapter: Real Number System, Exercise 2: EXERCISES 1(b)

Author:Odisha Board

Odisha Board Mathematics Solutions for Exercise - Odisha Board Solutions for Chapter: Real Number System, Exercise 2: EXERCISES 1(b)

Attempt the free practice questions on Chapter 17: Real Number System, Exercise 2: EXERCISES 1(b) with hints and solutions to strengthen your understanding. Bureau's Higher Secondary Elements of Mathematics Vol.1 solutions are prepared by Experienced Embibe Experts.