RD Sharma Class 11 Chapter 2 Relations solutions

Download Pdf of RD Sharma Class 11 Chapter 2 RELATIONS solutions

Much of mathematics is about finding a pattern – a recognisable link between quantities that change. In our daily life, we come across many patterns that characterise relations such as brother and sister, father and son, teacher and student. In mathematics also, we come across many relations such as number m is less than number n, line l is parallel to line m, set A is a subset of set B. In all these, we notice that a relation involves pairs of objects in certain order. In this Chapter, we will learn how to link pairs of objects from two sets and then introduce relations between the two objects in the pair. Finally, we will learn about special relations which will qualify to be functions. The concept of function is very important in mathematics since it captures the idea of a mathematically precise correspondence between one quantity with the other.

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RD Sharma Class 11 Chapter 1 SETS solutions

Download Pdf of RD Sharma Class 11 Chapter 1 SETS solutions

The concept of set serves as a fundamental part of the present day mathematics. Today this concept is being used in almost every branch of mathematics. Sets are used to define the concepts of relations and functions. The study of geometry, sequences, probability, etc. requires the knowledge of sets. The theory of sets was developed by German mathematician Georg Cantor (1845-1918). He first encountered sets while working on “problems on trigonometric series”. In this Chapter, we discuss some basic definitions and operations involving sets.

We shall say that a set is a well-defined collection of objects.

The following points may be noted :

1.  Objects, elements and members of a set are synonymous terms.
2.  Sets are usually denoted by capital letters A, B, C, X, Y, Z, etc.
3.  The elements of a set are represented by small letters a, b, c, x, y, z, etc

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RD Sharma Solutions For Class 10 Math Chapter 1 Real Numbers Exercise 1.1

 Q1. If a and b are two odd positive integers such that a > b, then prove that one of the two numbers \(\frac{a+b}{2} \) and \(\frac{a-b}{2} \) is odd and the other is even. 

Solution: We know that any odd positive integer is of the form 4q+1 or, 4q+3 for some whole number q.    Now that it’s given a > b 

So, we can choose a= 4q+3 and b= 4q+1. 

∴ \( \frac{a+b}{2}=\frac{\left[ (4q+3)+(4q+1) \right]}{2} \)  

⇒ \(\Rightarrow \frac{a+b}{2}=\frac{8q+4}{2} \)(a+b)/2 = (8q+4)/2 ⇒ (a+b)/2 = 4q+2 = 2(2q+1) which is clearly an even number. 

Now, doing (a-b)/2 

⇒ \(\Rightarrow \frac{a-b}{2}=\frac{\left[ (4q+3)-(4q+1) \right]}{2} \)  

⇒ (a-b)/2 = (4q+3-4q-1)/2 

⇒ (a-b)/2 = (2)/2 

⇒ (a-b)/2 = 1 which is an odd number. 

Hence, one of the two numbers (a+b)/2 and (a-b)/2 is odd and the other is even. 

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