Note Book

  Welcome To Class : XI Subject : Maths Chapter : 1 Exercise : 1.3 Que 1 : Suppose that 120 students are studying in 4 sections of eleventh standard in a school. Let A denote the set of students and B denote the set of the sections. Define a relation from A to B as “x related toy if the student x belongs to the section y”. Is this relation a function? What can you say about the inverse relation? Explain your answer. Solution :   A = {set of students in 11th standard} B = {set of sections in 11th standard} R : A ➝ B ⇒ x related to y ⇒ Every students in eleventh Standard must in one section of the eleventh standard. ⇒ It is a function. Inverse relation cannot be a function since every section of eleventh standard cannot be related to one student in eleventh standard. Que 2 : Write the values of f at – 4, 1, -2, 7, 0 if Solution : f(-4) = -(-4) + 4         = 4+4         = 8 f(1) = 1 – 1²         = 0 f(-2) = (-2)² ...

  Welcome To Class        : XI Subject   : Maths Chapter  : 1 Exercise : 1.2 Que 1 : Discuss the following relations for reflexivity, symmetricity and transitivity: (i) The relation R defined on the set of all positive integers by “mRn if m divides n”. (ii) Let P denote the set of all straight lines in a plane. The relation R defined by “lRm if l is perpendicular to m”. (iii) Let A be the set consisting of all the members of a family. The relation R defined by “aRb if a is not a sister of b”. (iv) Let A be the set consisting of all the female members of a family. The relation R defined by “aRb if a is not a sister of b”. (v) On the set of natural numbers the relation R defined by “xRy if x + 2y = 1”. (i) Solution : S = {set of all positive integers} (a) mRm ⇒ ‘m’ divides’m’ ⇒ reflexive (b) mRn ⇒ m divides n but nRm ⇒ n does not divide m (i.e.,) mRn ≠ nRm It is not symmetric  (c) mRn ⇒ nRr as n divides r It is transitive. (ii) Solution ...

Welcome To Class       : XI Subject   : Maths Chapter  : 1 Exercise : 1.1 Que 1 : Write the following in roster form: (i)   {x ∈ N : x² < 121 and x is a prime}; Solution : x is a Prime Number; x² < 121 Let 2² = 4, 3² = 9 , 5² = 25, 7² = 49, (i.e) x²<121 So We Take 2,3,5,7 A = {2, 3, 5, 7} (ii) The set of all positive roots of the equation (x – 1)(x + 1)(x2 – 1) = 0; Solution : Let x-1= 0;        x+1 = 0;        x²+1 = 0 x = 1;              x = -1;             x = ±1 Let "B"  as the the set of all positive roots of the equation   (x – 1)(x + 1)(x2 – 1)  B = { 1, -1, ±1} Only Positive Roots So We Take B = {1} B = {1} (iii) {x ∈ N : 4x + 9 < 52}; Solution :              4x + 9 < 52 Add (-9) On Both Side        4x + 9 – 9 < 52 – 9       ...