PHYSICS-15-10- 11th (PQRS) SOLUTION

https://docs.google.com/document/d/1iMddMC_00m0Bh68j_7iORZCw_rSO-iz2/edit?usp=sharing&ouid=109474854956598892099&rtpof=true&sd=true sets and their Algebraic properties, Relations, Equivalence relations, M a pping, one-one, Into a nd Onto ma ppings, C om position of mappings The concept of set is the basis of the modern Mathematics. It is widely used in various branches of Mathematics. `Set' was used for the first time by a German Mathematician 'George-cantor'. He defined set as "Any collection into a whole of definite and distinct objects of our intuition or thought". This definition of set was discussed and modified to the most acceptable form as "A set is any collection of distinct and distinguishable objects of our intuition or thought". In this chapter, the emphasis is on developing the graphical approach among students while solving a problem, from the very beginning. The use of Venn diagrams makes many problems very simple and it should be put to use as frequently as possible. The concept of relation is very useful to understand a function. If function or not as a function can only be understood the concept of relation is clear. 'aRb' means 'a' is R-related to b', where R may be any given relation between a & b. The concept of function lays the foundation of the study of the most important branch 'calculus' of mathematics. The word 'function' is derived from a Latin word meaning 'operation'. Function is also called mapping. SET - “Set is a well-defined collection of distinct objects” The objects of a set have a common property. An object having this property belongs to this set and another object not possessing this property does not belong to that set. For example, the collection of books written by Shakespere is a set, but the collection of interesting books written by Shakespere is not a set, since a book found interesting by one person may not be liked by another. Example : The set of all known planets in solar system, set of days in a week, set of all whole numbers, set of consonants in English alphabet etc. Choose the collection of objects, among the following, that are sets. The collection of all students of Aakash Institute. The collection of most talented Artists of India. The collection of bright students at IIT Kanpur. CHAPTER INCLUDES : Representation of Set Number System Types of Sets Subset, Superset and proper subsets Universal Set Intervals as subsets of R Venn diagrams Operations on Sets Complement of Set Algebra of Sets Ordered pairs C artesia n produc t of sets Relations Doma in & ra nge of relation Representation of relation Types of relation Equivalence relations Composition of relation Partition of set Congruence modulation The graph of a function Domain of the function Algebraic operation on functions Some standard functions and their graphs Kinds of Function Types of Mappings or Functions Composition of Function Existence of an inverse function The collection of all prime-ministers of India. The collection of all lucky numbers. The collection of Indian states. The collection of all tasty dishes. The collection of all secular nations. Solution : 1, 4, 6, 8 are sets as there is no ambiguity about their members. 2 and 3, 6 and 7 do not represent sets as there is no definite yardstick for being most talented, bright, lucky or tasty. Different people shall address to these terms differently. Set-Notations A set is usually denoted by capital letters A, B, C,... etc. whereas its members or elements or objects are denoted by lowercase letters such as a, b, c, etc. The greek symbol ∈ is used to denote the phrase 'belongs to'. Symbol ∈ is called membership relation. x ∈ A ⇒ 'x belongs to A' or 'x is an element of A' or 'x is a member of A' or 'x is an object of A'. x ∉ A ⇒ 'x does not belong to A' e.g. a ∈ set of English alphabet. 6 ∉ set of English alphabet. Representation of a Set A set can be represented by two methods Roster form or tabular form Set builder form or rule method. Roster or Tabular Form Here the elements of set are listed seperated by commas within braces or curly brackets {}. Here order of listing is immaterial and no element is repeated. For example, the set A of all single digit natural numbers is written as A = {1, 2, 3, 4, 5, 6, 7, 8, 9} or A = {1, 3, 5, 2, 6, 4, 9, 8, 7} (order is immaterial) Set-Builder Form: Here we choose a variable (say x), which represents each element of the set satisfying a particular property. Inside the bracket, x is followed by symbol: (or ; or vertical line '|' or oblique line '/' followed by the property or properties, possessed by each element of set. For example, the set A of all even integers less than 10 is written as A = {x : x is an even integer less than 10} = {x | x is an even integer less than 10} = {x ; x is an even integer less than 10} = {x / x is an even integer less than 10} The symbol following x is read as 'such that'. The roster form of A is written as A = {0, 2, 4, 6, 8} Note : '0' is an even integer Set builder form is also called, rule method, property method or symbolic method. Write each of the following into another form of set writing. A = {x / x ∈N and x ≤ 6} B = ⎧ 1 3 ⎨ ⎩ , 5 , 7 ⎫ 6 ⎭ (c) C = {1, – 1, i, – i} (d) D = {2, 4, 8, 16, 32} (e) E = {x : x2 – 5x + 6 = 0} (f) F = {x | x is a letter of word IITJEE} (g) G = {n3 – n2 : n ∈ N and 2 ≤ n ≤ 4} (h) H = {1, 8, 27, 64, ,10) Solution : (a) A = {1, 2, 3, 4, 5, 6} Each element is of the form 2n − 1 2n − 1 2n hence B = {x : x = 2n , n ∈ N, 'C' is a set of fourth roots of Hence C = {x : x4 = 1} 2, 4, 8, 16, 32 are clearly of the form 2n, where n is a natural number less than 6. so D = {x : x = 2n; n ∈N, n < 6} The roots of given equation must form the solution, hence E = {2, 3} No element has to be repeated, hence F = { I, T, J, E} (g) G = {4, 18, 48} (h) All the listed numbers are cube of natural numbers. So H = {x : x = n3, n ∈ N, n ≤ 10} Standard Notations for Sets of Numbers Set of all Symbol i.e. 1. Natural number N N = {1, 2, 3 } 2. Integers Z or I Z or I = {.......–3, –2, –1, 0, 1, 2, 3, } 3. (a) Positive integers Z+ Z+ = {1, 2, 3, } (b) Negative integers Z– Z– = {. –3, –2, –1} 4. Integers excluding 0 I0 I0 = { ± 1, ± 2, ± 3 } 5. Even integers E E = {0, ± 2, ± 4 } 6. Odd integers O O = { ± 1, ± 3, ± 5 } Set of all Symbol i.e. p Rational numbers Q Q = {x : x = q , p and q are integers, q ≠ 0} Non-zero rational numbers Q0 Q0= {x : x ∈ Q, x ≠ 0} Positive rational numbers Q+ Q+ = {x : x ∈ Q, x > 0} 10. Real numbers R Here all rational and irrational numbers are included Non-zero real numbers Positive real number R0 R+ R0 = {x : x ∈ R, x ≠ 0} R+ = {x : x ∈ R, x > 0} 13. Complex numbers C C = {a + ib; a, b ∈ R and i = − 1 } 14. Non-zero complex number C0 C0 = {x : x ∈ C, x ≠ 0} 15. Natural numbers less than or equal Nk Nk = {1, 2, 3, 4, k} to K, where K is positive integer 16. Whole numbers W W = {0, 1, 2, 3, } NUMBER SYSTEM Number System Real Numbers (R) Imaginary Numbers (3i, w, w2 etc.) Rational Numbers (Q) Irrational Numbers ( 2, Integers (I) or (Z) Fractions π etc.) Positive Integers (Z+) O Negative Integers (Z–) Natural Numbers (N) Types of Sets Null Set (or Empty Set or Void Set) A set which has no element. It is denoted by φ or {}. Examples. A = Set of odd numbers divisible by 2. B = Set of all omnipresent humans. C = Set of all negative natural numbers D = Set of all Greek letters in English alphabet. Singleton Set A set having a single element only e.g. { φ }, {0}, {2}, {a} etc. each is sigleton set or unit set. Examples : A = Set of present chief justice of India. B = {x : x2 = 1, x > 0} C = {x : x is the slope of all straight lines parallel to x-axis} Pair–Set A set having two elements only. e.g. {0, 1}, {± 1}, {x : x is a root of x2 – 5x + 6 = 0} Set of Sets A set S having all its elements as sets is called set of sets or a family of sets or a class of sets. e.g. {{1, 2}, {2, 3} {1, 2, 3}} is a set of sets as each member is a set itself. {{1, 2}, 7, {1, 7, 4}} is not a set of set as 7 is not a set. Finite and Infinite Set A set having finite number of elements in it is called finite set otherwise infinite set. The number of members in an infinite set are infinite i.e. can not be counted. The number of elements in a finite set A is called Cardinal number, n(A), of set A. Example : I, N, W, Q, R all are infinite sets Set of all IIT in India is a finite set. Set of all supreme court judges in India is finite set. Equivalent Sets : Two sets A and B are equivalent iff n(A) = n(B). ⎧2 e.g. A = {a, b, c, d, e} and B = ⎨ , ⎩ 3 , 4 , 5 4 5 6 6 ⎫ , ⎬ are equivalent sets as n(A) = n(B) = 5 ⎭ Equal Sets: Two sets A and B are equal if both have all the elements same i.e. A is a subset of B and B is also a subset of A. The order of elements is immaterial. A = B ⇔ A ⊆ B and B ⊆ A e.g. {a, b, c} = {a, c, b} . Match the following columns. Column (A) (a) A = {x : x is prime, x ∈ E} (i) Column (B) Finite set but not void set (b) B = {(x, y} : y2 = 4x, x, y ∈ R} (ii) Pair set (c) C = {(x, y} : y2 = 4x, x ∈ I, 0 < x < 5} (iii) Singleton set (d) D = Set of all Muslim Indian Prime Ministers (iv) Void set (e) E = Set of all Positive integers less than 3 (v) Infinite set Solution : Answer : (a)–(iii); (b)–(v); (c)–(i); (d)–(iv); (e)–(ii) Only prime even number is 2 hence A = {2} There are infinite number of points (x, y) lying on curve y2 = 4x. So B is infinite set. (c) C = {(1, +2), (1, –2), (2, 2 set. ), (2, –2 ), (3, 2 ), (3, – 2 ), (4, 4), (4, –4)}. n(c) = 8 hence finite As no Indian Prime Minister has been a Muslim so far so D = φ E = {1, 2} hence pair set. Find the only correct option among following. {x : x2 = 3, x ∈ Q} is a pair set. A = {x : x ∈ Z and x2 ≤ 3} and B = {x : x ∈ R and x2 – 3x + 2 = 0} are equal sets. {x : x is people of India speaking Hindi} is an infinite set The set of prime numbers less than 99 is finite set. Solution : Answer (d) (a) It give x = ± which are not rational numbers, hence φ . False. (b) A= {–1, 0,+1}; B = {1, 2}. A ≠ B. False It is a finite set clearly. False Clearly it is a finite set. True SUBSETS, SUPERSETS, PROPER SUBSETS A set 'A' is called a subset of set B if every member of set A also belongs to set B. (sign of subset is ⊆) A ⊆ B ⇔ [x ∈ A ⇒ x ∈ B] ⇒ (A is contained in B) Here set B is called superset of A. B ⊇ A A ⊆/ B is read as 'A is not a subset of B'. Example : A = {x : x ∈ N} and B = {x : x ∈ Z} So A ⊆ B as every natural number is also an integer. Example : A = {a, e, i, o, u}; B = {x : x is a letter of English alphabet} ∴ A ⊆ B Example : A = {a, e, i, o, u} and B = {e, o, A ⊆ B and B ⊆ A ⇒ A = B. Proper Subset A set A is said to be a proper subset of a set B if every element of Set A is an element of set B and set B has atleast one extra element which is not an element of A. Proper subset is denoted by A ⊂ B(read as "A is a proper subset of B") A ⊄ B (read as "A is not a proper subset of B") Example : A = {a, e, i, o, u}; B = {Set of all letters of English alphabet}. Clearly A ⊆ B and A ⊂ B. If cardinal number of a set is n then number of its subsets is 2n and its number of proper subsets is (2n – 1). Comparability of Sets Two sets A and B are called comparable if A ⊂ B or B ⊂ A or A = B, otherwise A and B are called incomparable. Example : {a, e, i} and {a, e, o} are incomparable. {a, e, i} and {a, e, i, o, u} are comparable. {a, e, i} and {e, i, a} are comparable. Power Set The set of all subsets of a set A is called power set of A and is denoted by P(A) or 2A. P(A) = {x : x ⊆ A} x ∈ P(A) ⇔ x ⊆ A φ ∈ P(A) and A ∈ P(A) Example : A = {1, 2, 3} P(A) = 2A = {φ, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}} n(A) = 3 so n(2A) = 23 = 8 A = {(x, y) : y2 = x, x ∈ R} and B = {(x, y) : y = following. A = B B ⊂ A and A ⊆ B A and B are comparable sets. A and B both are infinite sets. Solution: Answers : (c) and (d) , x ∈ R}. Choose the correct option/options among the Set A and Set B include all the points lying on the respective curves below. Clearly A ≠ B, B ⊂ A, so A and B are comparable; A ⊄ B and both are infinite sets as infinite number of points satisfy each. y y x x –y –y Prove that if A has n elements then its power set has 2n elements. Solution : Methods of selecting r things from n different things is given by n = n(n − 1)(n – 2) upto r terms , where r | r | r = 1.2.3 r. The number of subsets of A having no element = nCo The number of subsets of A having one element = n C1 = 1 i.e. φ The number of subsets of A having two elements = nC2 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - The number of subsets of A having n elements = n Cn Thus total number of subsets of A = nCo + n C1 + nC2 + + n Cn Universal Set = (1 + 1)n = 2n (by binomial theorem) Any set which is the superset of all the sets under consideration is called the universal set ( Ω or S or U). Choice of universal set is not unique, but once chosen it is fixed for that discussion. Example: Let A = {a, e, i}; B = {i, o, u}; C = {e, f} then U = {a, e, i, o, u, f} or U = {a, e, i, p, o, u, f, g} or U = Set of all English alphabet. Mark T/F against the each statement given below: Every set has at least one proper subset. If A is a finite, non-void set, having n proper subsets and m subsets then n – m ∈N. A = {φ, {φ}} then cardinal number of P(A) is 4. (d) a ⊆ {a,{b},{c}} (e) 'Set of all squares in a plane' is a subset of 'all rectangles in the same plane'. Solution : Answer : (a) F (b) F (c) T (d) F (e) T Void set { }, has no proper subset. n – m = – 1 ∉ N (c) P(A) = {φ, {φ}, {{φ}}, {φ,{φ}}} so n {P(A)} = 4. Since 'a' is not a set hence it cannot be a subset. Every subset is a set in itself. Each square is also a rectangle hence true. Match the following columns. A = letters of word 'ball' (i) P(A) ⊂ P(B) B = letters of word 'lab' A ⊂ B (ii) A and B are incomparable −1 A = {x : cosx > 2 and 0 ≤ x ≤ π} (iii) A = B 1 π B = {x : sinx > 2 and 3 ≤ x ≤ π} (d) A = {(x, y) : x2 + y2 ≤ 1, x, y ∈ R} (iv) A ⊃ B Solution : B = {(x, y) : 0 ≤ x ≤ 1 and y = 0} 2 Answer : a(iii); b(i); c-(ii); d(iv) A = {a, b, l}; B = {l, a, b} clearly A = B, A and B are comparable P(A) = P(B) so P(A) ⊄ P(B); A ⊃/ B. Hence answer (iii) only. If A ⊂ B then P(A) ⊂ P(B); A and B are comparable, A ≠ B, A ⊃/ B. Hence answer (i) only. A and B are as shown on number line. Clearly P(A) ⊄ P(B) π π O 6 3 π 2π 2 3 5π π 6 Set - B Set - A A and B are not comparable. A ≠ B and A ⊃/ B. Hence answers (ii) only. A is all points within on a circle of radius 1 and centre (0, 0). B contains only the points lying on x-axis within the circle such that 0 ≤ x ≤ Hence only correct answer is (iv). Intervals as Subsets of R 1 , so clearly B ⊂ A; B ≠ A, A and B are comparable P(A) ⊄ P(B). 2 Four type of subsets can be defined on R as given below. Let a, b ∈ R, such that a < b Open Interval (a, b) or] a, b [ = {x : a < x < b} = Set of all real numbers between a and b, not including a and b both. a b ] a, b[ or (a, b) Closed Interval [a, b] = {x : a ≤ x ≤ b} = Set of all real numbers between a and b as well as including a and b both. a [a, b] b Open-closed Interval (semi closed or semi open interval) (a, b] or ]a, b] = {x : a < x ≤ b} = Set of all real numbers between a and b, a not included but b included. a b (a, b] or ]a, b] Closed-open interval (semi closed or semi open interval) [a, b) or [a, b[ = {x : a ≤ x < b} = Set of all real numbers between a and b including a but excluding b. a b [a, b) or [a, b[ Some More Representations on Number Line Infinite open interval x > a x < b Infinite close interval b x ≤ b x ≥ a a (0, ∞) = R+ (– ∞, 0) = R– (– ∞, ∞) = R VENN DIAGRAMS Introduced by Euler (a Swiss mathematician) and named after John Venn. It is a pictorial representation of sets in which a set is represented by a circle or a closed geometrical figure inside universal set which is shown by a rectangle. Each element of a set is represented by a point within the circle representing that set. A = {a, b, c} A ⊂ B; A = {a, b}; B = {a, b, c, d}; e ∉ A; e ∉ B Various Operations on Sets Union of Sets : A ∪ B (read as 'A union B' or 'A cup B' or 'A join B') is a set consisting of all the elements which are either in A or in B or in both. A ∪ B = {x : x ∈ A, x ∈ B} A ∪ B A ∪ B A ∪ B Example: Q → Set of all rational numbers Q′ → Set of all irrational numbers R = Q ∪ Q′ Choose the correct options among the following for any sets A and B. P(A) ∪ P(B) may be equal to P(A) P(A) ∪ P(B) may be equal to P(A ∪ B) P(A) ∪ P(B) must be a subset of P(A ∪ B) P(A) ∪ P(B) must be equal to P(A ∪ B) Solution : Answer: (a), (b), (c) are true. If A ⊇ B then P(A) ⊇ P(B), hence option (a) is true. If A = B then A ∪ B = A = B and P(A) ∪ P(B) = P (A ∪ B), option (b) is true. Option (c) is always correct ∀ A and B. Option (d) is not correct when A ≠ B. e.q. A = {1, 2} and B = {1, 4} then A ∪ B = {1, 2, 4} P(A) = {φ, {1}, {2}, [1, 2]} P(B) = {φ, {1}, {4}, [1, 4]} P(A) ∪ P(B) = {φ, {1}, {2}, {4}, {1, 2,}, {1, 4}} P(A ∪ B) = {φ, {1}, {2}, {4}, {1, 2,}, {1, 4}, {1, 2, 4}} So P(A) ∪ P(B) ≠ P (A ∪ B). INTERSECTION OF TWO SETS A ∩ B (read as 'A intersection B' or 'A cap B' or 'A meet B') is defined as a set containing all the elements common to A and B. A ∩ B = {x : x ∈ A and x ∈ B} = {x : x ∈ A Λ x ∈ B} x ∈ A ∩ B ⇔ x ∈ A and x ∈ B x ∉ A ∩ B ⇔ x ∉ A or x ∉ B A ∩ B A ∩ B A ∩ B = φ A1 ∩ A2 ∩ A2 ............. ∩ An = A ∩ B = AB n ∩ Ai i=1 = {x : x ∈ Ai ∀ i} Intersection of finite number of finite sets will be a finite set Intersection of finite set with infinite set will be finite set Intersection of two or more infinite sets may or may not be finite A ∩ A = A; A ∩ φ = φ; A ∩ S = A; S ∩ φ = φ φ ∩ φ = φ; if A ⊇ B then A ∩ B = B; (A ∪ B) ∩ A = A; (A ∩ B) ∪ A = A Considering the wall-clock as shown in figure. Let S = Set of all points in area covered by second's hand in 12 hours. 12 M = Set of all points in area covered by minute's hand in 12 hours. 9 3 H = Set of all points in area covered by hour's hand in 12 hours. Then pick the correct statement among following. 6 (a) S ∪ M ∪ H = S (b) (S ∩ M) ∪ H = S (c) S ∩ M ∩ H = H (d) All are correct Solution : Answer : (a) and (c). H ⊂ M ⊂ S (since hours hand is smallest in length) ∴ option (b) is wrong, since S ∩ M = M and (S ∪ M) ∪ H = M ∪ H = M. Disjoint Sets Two sets A and B having no element in common are disjoint or mutually exclusive A ∩ B = φ ⇔ A and B are disjoint. A and B are disjoint Example: z+ and z– are disjoint Set of all boys and set of all girls are disjoint Set of Hindi alphabet and set of English alphabet are disjoint Set of years of birth of adults and set of years of birth of minors are disjoint Q and Q′ are disjoint A ∩ B ∩ C = φ; but A and B, B and C are not disjoint. So A, B and C are not pairwise disjoint. Difference of Two Sets A – B (read 'A minus B') or (relative complement of B in A) is the set of all elements of A which are not elements of B. A – B = {x : x ∈ A and x ∉ B} B – A = {x : x ∈ B and x ∉ A} A – B when B ⊂ A A – B B – A A – B, when A and B are disjoint A – B = A A – B = φ when A = B or A ⊂ B Symmetric Difference of Two Sets Denoted by A Δ B or A ⊕ B, (A direct sum B) A Δ B = (A – B) ∪ (B – A) = (A ∪ B) – (A ∩ B) A Δ B = A ∪ B when A and B are disjoint A Δ B = (A ∪ B) – (A ∩ B) A Δ B = (A – B) when B ⊆ A A Δ B = (B – A) when A ⊆ B Complement of a Set If 'A' be a set and U be the universal set such that A ⊂ U then complement of set A is denoted by A′ or AC or C (A) or U – A A' = AC = C(A) = U – A = {x : x if x ∈ A ⇔ x ∉ A' x ∈ A' ⇔ x ∉ A U – A, or AC or A' or C(A) U′ = φ; φ′ = U; A ∪ A′ = U, A ∩ A′ = φ If A = {5, 6, 7, 8}; B = {3, 9, 8, 10} and S = {1, 2, 3, 4 10} then, (a) (A ∪ B)′ = {1, 2, 3, 4} (b) (A ∩ B)′ = {8} (c) (A′ ∩ B)′ = (A ∩ B)′ (d) None of these is true Solution : Answer (d) A ∪ B = {3, 5, 6, 7, 8, 9, 10} hence (A ∪ B)′ = {1, 2, 4}, (a) is wrong. A ∩ B = {8} hence (A ∩ B)′ = {1, 2, 3, 4, 5, 6, 7, 9, 10}, (b) is wrong. (A′ ∩ B)′ = {1, 2, 4} ≠ (A ∩ B)′, (c) is wrong. Hence answer (d) is correct. If A = {(x, y) : y = ex; x ∈ R} U = {(x, y) : x, y ∈ R} B = {(x, y) : y = x; x ∈ R} C = {(x, y) : y = –x; x ∈ R} Choose the correct statement/s among the following : (a) (A ∩ B)′ = φ (b) (A ∩ B ∩ C)′ = φ (c) A – B = φ (d) A Δ B = A ∪ B Solution : Answer : (d) Set A, B and C are the points on the curves as shown in adjacent diagram. Clearly A so (A ∩ B)⁄ = U, option (a) is wrong. ∩ B = φ Similarly A ∩ B ∩ C = φ as there are no common points to all the three curves. ∴ (A ∩ B ∩ C)′ = U. Option (b) is wrong. From figure it is clear that A – B = A, since A and B are disjoint sets. Option (c) is wrong. A Δ B = (A – B) ∪ (B – A) = A ∪ B ALGEBRA OF SETS Idempotent Laws : For A ∪ A = A A ∩ A = A Identity laws : For any set A, we have A ∪ φ = A (b) A ∩ φ = φ A ∪ U = U A ∩ U = A Commutative laws : For any two sets A and B, we have A ∪ B = B ∪ A A ∩ B = B ∩ A Associative laws : For any three sets A, B and C, we have (a) (A ∪ B) ∪ C = A ∪ (B ∪ C) (b) (A ∩ B) ∩ C = A ∩ (B ∩ C) Distributive laws : For any three sets A, B and C, we have (a) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) (b) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) Demorgan's laws : For any three sets A, B and C, we have (a) (A ∪ B)′ = A′ ∩ B' (b) (A ∩ B)′ = A′ ∪ B′ (c) A – (B ∪ C) = (A – B) ∩ (A – C) (d) A – (B ∩ C) = (A – B) ∪ (A – C) (A')' = A (for any set A) P(A) ∩ P(B) = P(A ∩ B) P(A) ∪ P(B) ⊆ P(A ∪ B) Shade (A ∪ B) ∩ (A ∪ C) in the following diagrams (i) (ii) (iii) (iv) Solution: (i) (iii) (iv) SOME THEOREMS For any sets A, B and C, we have A – B = A ∩ B' A ∪ B = B ⇔ A ⊆ B A ∩ B = A ⇔ A ⊆ B 4. (A – B) ∪ B = A ∪ B 5. A – (B ∪ C) = (A – B) ∩ (A – C) A – (B ∩ C) = (A – B) ∪ (A – C) Some more operations on sets. Some Basic Results about Cardinal Numbers RELATIONS a R b means 'a is R-related to b' i.e. a is related to b under relation R. If (a, b) ∈ R; (a, b) is called ordered pair in the sense that a and b can't be interchanged as a ∈ A and b ∈ B. Ordered Pair : It is a pair of objects written in a particular order. Two members are written in a particular order separated by a comma and enclosed in parentheses. Hence in ordered pair (a, b) a is called the first component or the first element or the first co-ordinate and b the second. CARTESIAN PRODUCT Cartesian product of two sets A × B : For any two non empty sets A and B A × B = {(a, b) : a ∈ A and b ∈ B} It is a set of all ordered pairs such that in each ordered pair first element belongs to set A and second element belongs to set B. A × B is read as 'A cross B' or 'Product set of A and B' A × B = {(a, b) : a ∈ A ∧ b ∈ B} Thus (a, b) ∈ A × B ⇔ a ∈ A and b ∈ B. B × A = {(b, a) : b ∈ B ∧ a ∈ A} A × B ≠ B × A (not commutative) n(A × B) = n(A) n(B) and n(P(A × B)) = 2n(A) n(B) A = φ and B = φ ⇔ A × B = φ Cartesian product of n non empty sets A1, A2, An is a set of all n tuples (a1, a2, an) such that each ai∈ Ai, i = 1, 2 n. n A1 × A2 × × An = ∏A i i =1 A × A = A2 : R × R = R2 is a set of all points lying in the plane R × R × R = R3 represents set of all points in 3-D space. If at least one of A and B is infinite set then A × B is also infinite set, provided that other is non-empty set. Let A = {a, b}, B = {c, d}, C = {e, f} then n(A × B × C) = n(A). n(B). n(C) = 8 A × B × C = {(a, c, e), (a, c, f), (a, d, e), (a, d, f), (b, c, e), (b, c, f), (b, d, e), (b, d, f)} A1 × A2 × A3× A4 = {(1, 1, 1, 1), (2, 4, 8, 16), (3, 9, 27, 81), }. Find A1, A2, A3 and A4. Solution : Each ordered pair {x1, x2, x3, x4} is of the form {x, x2, x3, x4} Hence x1 ∈ A1 ⇒ A1 = {x : x ∈ N} = {1, 2, 3, 4, } x2 ∈ A2 ⇒ A2 = {x2 : x ∈ N} = {12, 22, 32, 42, } x3 ∈ A3 ⇒ A3 = {x3 : x ∈ N} = {13, 23, 33, 43, } x4 ∈ A4 ⇒ A4 = {x4 : x ∈ N} = {14, 24, 34, 44, } Key Results on Cartesian Product If A and B are two non-empty sets having n elements in common then (A × B) and (B × A) have n2 elements in common. If n(A) = 7, n(B) = 8 and n(A ∩ B) = 4, then match the following columns. (i) n(A ∪ B) (a) 56 (ii) n(A × B) (b) 16 (iii) n((B × A) × A) (c) 392 (iv) n((A × B) ∩(B × A)) (d) 96 (v) n((A × B) ∪ (B × A)) (e) 11 Solution : Answer : (i)–(e); (ii)–(a); (iii)–(c); (iv)–(b); (v)–(d) (i) n(A ∪ B) = n(A) + n(B) – n(A ∩ B) = 7 + 8 – 4 = 11 (ii) n(A × B) = n(A) n(B) = 7 × 8 = 56 = n(B × A) (iii) n((B × A)× A) = n(B × A). n(A) = 56 × 7 = 392 (iv) n((A × B) ∩ (B × A)) = (n((A ∩ B))2 = 42 = 16 (v) n((A × B) ∪ (B × A)) = n(A × B) + n(B × A) – n(A × B) ∩ (B × A) = 56 + 56 – 16 = 96 If A = {2, 4} and B ={3, 4, 5}, then (A ∩ B) × (A ∪ B) is (1) {(2, 2), (3, 4), (4, 2), (5, 4)} (2) {(2, 3), (4, 3), (4, 5)} (3) {(2, 4), (3, 4), (4, 4), (4, 5)} (4) {(4, 2), (4, 3), (4, 4), (4, 5)} Solution : Answer (4) A ∩ B = {4} and A ∪ B = {2, 3, 4, 5} ∴ (A ∩ B) × (A ∪ B) = {(4, 2), (4, 3), (4, 4), (4, 5)} Pictorial Representation of Cartesian Product of Two Sets : Arrow diagram : Let A = {1, 2, 3} B = (a, b) A × B ⇒ Lattice-Diagram : Axis OX represents elements of A and perpendicular axis OY represents set B. Each dot represents an ordered pair of A × B. Let A = (1, 2, 3) B = (1, 3) y 3 2 1 O 1 2 3 x RELATIONS For any two non-empty sets A and B, every subset of A × B defines a relation from A to B and every relation from A to B is a subset of A × B. a R b ⊆ A × B ∀ R If (a, b) ∈ R, then a R b is read If (a, b) ∉ R, then a R/ b is read Domain and Range of Relation Domain of R = Dom(R) = Set of first components of all the ordered pairs belonging to R. Range of R = Set of second components of all the ordered pairs belonging to R. Co-domain of R = B where R is a relation from A to B Range of R ⊆ Co-domain of R Dom(R) = {a ∈ A : (a, b) ∈ R for some b ∈ B} Range of R = {b ∈ B : (a, b) ∈ R for some a ∈ A} If R = A × B, then Dom(R) = A and Range of R = B Dom (φ) = φ ; Range of φ = φ Let A = {1, 3, 4, 5, 7} and B = {1, 4, 6, 7} and R be the relation 'is one less than' from A to B, then list the domain, range and co-domain sets of R. Solution : R = {(3, 4), (5, 6)} So, Dom(R) = {3, 5} Range of R = {4, 6} Co-domain of R = B = {1, 4, 6, 7} Clearly Range of R ⊆ co-domain of R. Representation of a Relation Let A = {–2, –1, 4} B = {1, 4, 9} A relation from A to B i.e. a R b is defined as a is less than b. This can be represented in the following ways. Roster form: R = {(–2, 1), (–2, 4), (–2, 9), (–1, 1), (–1, 4), (–1, 9), (4, 9)} Set builder notation: R = {(a, b): a ∈ A and b ∈ B, a is less than b} Arrow - diagram: A B Lattice-diagram : Tabular form: R 1 4 9 – 2 1 1 1 – 1 1 1 1 4 0 0 1 Let A = {1, 2, 3, 4}, B = {1, 2, 3, 10} R1 = {(1, 4), (2, 5), (3, 6), (4, 7)} R2 = {(2, 5), (3, 6), (4, 7), (5, 8)} R3 = {(1, 1), (2, 4), (3, 9)} Among R1, R2, R3, choose those, that represent a relation from A to B, and represent the relations in set- builder form. Solution : R1 ⊆ A × B; R2 ⊄ A × B, R3 ⊆ A × B Hence R2 is not a relation as (5, 8) ∉ A × B R1 = {(a, b) : a ∈ A , b ∈ B and a + 3 = b} R3 = {(a, b) : a ∈ A , b ∈ B and a2 = b} Inverse Relation Let R ⊆ A × B be a relation from A to B. The inverse relation of R (denoted by R–1) is a relation from B to A defined as R–1 = {(b, a) : (a, b) ∈ R} If (a, b) ∈ R, then (b, a) ∈ R–1, ∀ a ∈ A, b ∈ B. domain of R–1 = Range of R Range of R–1 = domain of R (R–1)–1 = R (1, 3) (2, 3) (3, 3) (1, 1) (2, 1) (3, 1) Identity Relation The identity relation on a set A is the set of ordered pairs belonging to A × A and is denoted by IA. IA = {(a, a) : a ∈ A} i.e. every element of A is related to only itself. R is an identity relation if (a, b) ∈ R iff a = b, a ∈ A, b ∈ A. I –1 = I A A Domain of IA= Range of IA = A 'is equal to' is an identity relation on set of Natural number (N) i.e. {(1, 1), (2, 2), (3, 3). } = IN Universal Relation If A be a set and R is the set A × A, then R is called universal relation in A. If R = A × A, then R is universal relation in A. Void Relation φ is called the empty or void relation if φ ⊂ A × A Types of Relations on a Set If A is a non-empty set, then a relation R on A is said to be Reflexive : If (a, a) ∈ R, ∀ a ∈ A. i.e. a R a, ∀ a ∈ A “is equal to”, “is a friend of”, “is parallel to”, are some of reflexive relations. Symmetric : If a R b ⇒ b R a, ∀ a, b ∈ A i.e. if (a, b) ∈ R ⇒ (b, a) ∈ R, ∀ a, b ∈ A “is a friend of”, “is parallel to”, “is equal to”, are some of symmetric relations. Anti-Symmetric : If a R b and b R a ⇒ a = b, ∀ a, b ∈ A (If R ∩ R–1 = Identity, then R is anti-symmetric) “is divisible by” is an anti symmetric relation. Transitive : If a R b and b R c ⇒ a R c, ∀ a, b, c ∈ A i.e. If (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R, ∀ a, b, c ∈ A “is parallel to”, “is equal to”, “is congruent to” are some of the transitive relation. Equivalence Relation A relation R on a non-empty set A is called an equivalence relation if and only if it is Reflexive, Symmetric as well as Transitive. "is parallel to", "is equal to", "is congruent to" "Identity relation" are some of the equivalence relations. Every identity relation is an equivalence relation but every equivalence relation need not to be identity relation. Check the following relations for being reflexive, symmetric, transitive and thus choose the equivalence relations if any. a R b if a ≤ b; a, b ∈ set of real numbers. a R b iff a < b; a, b ∈ N. a R b iff a − b > 1 ; a, b ∈ R. 2 a R b iff a divides b; a, b ∈ N. a R b iff (a – b) is divisble by n; a, b ∈ I, n is a fixed positive integer. Solution : Not reflexive, not symmetric but transitive Let a = –2 and b = 3; (–2, 3) ∈ R. Since − 2 ≤ 3 is true Since − 2 = 2 ≤/ –2 hence relation is not Reflexive Since 3 ≤ –2 is wrong hence relation is not symmetric Now Let a, b, c be three real numbers such that a ≤ b ⇒ b ≥ 0, so b ≤ c ⇒ b ≤ c a ≤ b and b ≤ c Hence a ≤ c is true so the given relation is transitive. Not reflexive, not symmetric but transitive. Since no natural number is less than itself hence not reflexive, If a < b then b < a is false. Hence not symmetric. If a < b then b < c clearly a < c. Hence transitive Not reflexive, symmetric, not transitive. a − a = 0 >/ 1 2 hence it is not reflexive. a − b > 1 ⇒ 2 b − a = a − b > 1 2 hence symmetric. Let a = 1, b = – 1 and c = 3 , a − b = 2 > 1 so (a,b)∈ R; b − c = 5 > 1 so (b, c) ∈ R 2 2 2 2 But a − c = 1− = 1 >/ 2 1 so(a,c) ∉ R . Hence R is not a transitive relation. Reflexive, not symmetric, transitive a Since a = 1 i.e. every number divides itself, hence R is reflexive. If a divides b then b does not divide a (unless (a = b) hence the relation is not symmetric (but anti- symmetric). If a divides b and b divides c then it is clear that a will divide c. Hence transitive. Relfexive, symmetric as well as transitive, hence it is an equivalence relation. ⎛ 0 = 0⎞ Since 0 is divisible by n ⎜ ⎝ ⎟ so given relation is reflexive ⎠ If a – b is divisible by n, then (b – a) will also be divisible by n. Hence, symmetric. If a – b = nI1 and b – c = nI2, where I1, I2 are integer. Then, a – c = (a – b) + (b – c) = n(I1 + I2) so a – c is also divisible by n, hence transitive. Ordered Relation R is an ordered relation if it is transitive but not equivalence relation. e.g. a R b iff a < b, a, b ∈ N is an ordered relation. e.g. R = {(1, 1), (1, 3), (1, 2), (2, 1), (2, 2) (2, 3)} is not reflexive, not symmetric and transitive, hence not an equivalence relation. so, R is an ordered relation. Partial Order Relation R is an partial order relation if it is reflexive, transitive and antisymmetric at the same time. a R b iff a divides b; a, b ∈ N is partial order relation since it is reflexive, transitive and anti-symmetric. COMPOSITION OF TWO RELATIONS If A, B, C are three sets such that R ⊆ A × B and S ⊆ B × C then SOR ⊆ A × C. SOR, ROS, (SOR)–1, S–1OR–1, S–1OR etc. are called compositions of two relations. (SOR)–1 = R–1OS–1 (R1OR2OR3.....ORn)–1 = Rn–1OR –1 O OR –1OR –1OR –1 Pictorially n–1 3 2 1 SOR Let R be a relation such that R = {(a, d), (c, g) (d, e), (d, f) (g, f)} then find (i) R–1OR–1 (ii) (ROR–1)–1 Solution : R–1OR–1 = (ROR)–1 and (ROR–1)–1 = ROR–1 Clearly R–1 = {(d, a), (g, c), (e, d), (f, d), (f, g)} Domain of R = {a, c, d, g}; Range of R = {d, g, e, f} Domain of R–1 = {d, g, e, f}; Range of R–1 = {a, c, d, f} (i) R = {(a, d), (c, g) (d, e), (d, f) (g, f)} R–1={(d, a), (g, c), (e, d), (f, d), (f, g)} R R (ii) ROR From above figure clearly ROR = {(a, e), (a, f), (c, f)} so, R–1 OR–1 = (ROR)–1 = {(e, a), (f, a), (f, c)} R–1 R ROR–1 Hence ROR–1 = {(d, d), (g, g), (e, e), (e, f), (f, e), (f, f)} ∴ (ROR–1)–1 = ROR–1 = {(d, d), (g, g), (e, e), (e, f), (f, e), (f, f)} Partition of a Set If A is a non-empty set, then a partition of A is a collection of non-empty pairwise disjoints subsets of A, such that union of collection of subsets is A. Example : If A1, A2, A3.......An are non empty subsets of A, then the set {A1, A2, A3 An} is called partition of A if A1 ∪ A2 ∪ A3 ∪ An = A and (ii) Ai ∩ Aj = φ ∀ i ≠ j (i, j = 1, 2, 3 n) Congruence Modulo m : Let m be a positive integer and x, y ∈ I, then x is said to be congruent to y modulo m, (x ≡ y (mod m) iff x – y is divisible by m, i.e. x ≡ y (mod m) if Example: x − y = λ, λ ∈ I+ or x − y = mλ, λ ∈I+ m 24 ≡ 8 (Mod 4) since 24 − 8 = 16 = 4 ∈ I+ 4 4 FUNCTION Let A and B are two non empty sets. A function f from set A to set B is a rule which associated each element of A to a unique element of B, denoted by f : A → B set A is called domain of function ′f′ set B is called co-domain of function ′f′ If element x of A corresponds to y(∈B) under the function f, then we say that y is the image of x and write f(x) = y. Which of the following given below is/are a function, from R to R? f(x) = x2 f(x) = f(x) = 3x + 4. Solution : Yes, because all element of domain (which is R) have images in co-domain (R). No, this is not a function because all negative number in a domain (R), do not have images in co-domain. i.e. f(–1) = f(–2) = (imaginary no.) (imaginary no.) Yes, because all real numbers in domain have images in co-domain. The Graph of a Function The graph of a function y = f (x) consists of all points (x, f(x)) in the Cartesian plane since by definition of a function, there is exactly one value of y for each x, it follows that no vertical line can intersect the graph of a function of x for twice or more. not a function of x a function of x a function of x Domain of the Function Domain of the function is set of all those real numbers (x) for which f(x) exists or f(x) is meaningful. f(x) ≠ ∞ or any imaginary no.) f(x) = 1 , f(x) exist, if x ≠ 0, so domain is R . x 0 f(x) = , f(x) exist, if x ≥ 2 so domain [2, ∞). Range of Function Set of all the images of elements in domain is called the range. Range = {f(x) : x ∈ domain} Algebraic Operation on Functions Given functions f and g, their sum f + g, difference f –g, and fg are defined on dom f ∩ dom g as: (f + g) (x) = f (x) + g (x), (f – g) (x) = f (x) – g (x) and (fg) (x) = f(x) g (x). Moreover f/g is defined on dom f ∩ {x ∈ dom g: g(x) ≠ 0} by (f/g) (x) = f(x)/g(x). If k is any real number and f is a function then kf is defined on the domain of f by (k f) (x) = k f(x). We have the following formulae for domains of functions dom (f ± g) = dom f ∩ dom g dom (f g) = dom f ∩ dom g dom (f /g) = dom f ∩ {x ∈ dom g: g (x) ≠ 0} dom = {x ∈ dom f; f (x) ≥ 0} Find the domain of the function f(x) = cos −1⎛ x − 2 ⎞ . ⎜ ⎟ ⎝ ⎠ Solution : f(x) exist if –1 ≤ 5 ≤ 1 ⇒ – 5 < | x | – 2 ≤ 5 ⇒ – 3 ≤ | x | ≤ 7 ⇒ | x | ≥ – 3 true ∀ x ∈ R or | x | ≤ 7 ⇒ x ∈ [–7, 7]. Find the domain and range of the function f(x) = Solution : x 2 + x + 1 x 2 − x + 1 . x2 – x + 1 ≠ 0 for any value of x (Θ b2 – 4ac < 0) so domain of f(x) is R Range Let f(x) = y x 2 + x + 1 = ⇒ x 2 − x + 1 y ⇒ x2 (1 – y) + x(1 + y) + (1 – y) = 0 But x is real so b2 – 4ac ≥ 0 ⇒ (1 + y)2 – 4 (1 – y)2 ≥ 0 ⇒ 3y2 – 10y + 3 ≤ 0 ⇒ (y – 3) (3y – 1) ≤ 0 ⇒ so range of f(x) ⎡ 1 ,3⎤ . ⎢⎣ 3 ⎥⎦ Find the domain of the function f(x), if f(x) = . Solution : f(x) = Now we now that f(x) exist if log0.5 x ≥ 0 x > 0 (because log x not defined for zero and negative numbers) log0.5 x ≥ 0 ⇒ x ≤ (0.5)° ⇒ x ≤ 1 = x ∈ (– ∞, 1] But x > 0 so x ∈ (0, 1]. Find the range of the function f(x) = Solution : sin(π[x + 1]) [tan−1(x 2 + x + 1)]2 + 5 . (where [⋅] denotes step-function) Here denominator ≠ 0 ∀ x ∈ R and [x + 1] = Z (due to step. function) sin π [x + 1] = 0 (because sin of integer multiple of π is always zero) so f(x) = sin x[x + 1] [tan−1[x 2 + x + 1]2 + 5 = 0 (for x ∈ R) f(x) = 0 ∀ x ∈ R Range of the function = {0} Find the domain and range of the function, f(x) = Solution : 6−x Cx−3 . f(x) = 6−x Cx−3 f(x) exist if 6 – x > 0 and (6 – x) ∈ N x – 3 ≥ 0 and x – 3 ∈ W 6 – x ≥ x – 3 so x < 6 ⇒ x ∈ {...–2, –1, 0, 1, 2, 3, 4, 5,} (1) x ≥ 3 ⇒ x ∈ {3, 4 5, 6, 7......} (2) 6 – x ≥ x – 3 ⇒ 2x ≤ 9 ⇒ x ≤ 4.5 so x ∈ {...–2, –1, 0, 1, 2, 3, 4} (3) so final value from (1), (2) and (3) is x ∈ {3, 4} so domain is {3, 4} Range, f(3) = 6−3 C3−3 = 3C0 = 1 f(4) = 2C1 = 2 so range is {1, 2}. SOME STANDARD FUNCTIONS AND THEIR GRAPHS Constant Function A function denoted by f(x) = C (where C ∈ R) is known as constant function Domain = R Range = C x Identity Function [I(x)] : A function which is associated to itself is known as identity function and denoted by I(x) = x Since x can take any value so domain of this function is R, corresponding value of I(x) is also R, so range is R Domain = R Range = R x Modulus Function : This is also known as absolute value function and denoted by f(x) = |x| ⎧ x , x ≥ 0 i.e. f(x) = ⎨− x , x < 0 Domain of this function is set of all real numbers because f(x) exists for all x ∈ R but |x| ≥ 0 so range f(x) Domain = R x Range = [0, ∞] or R+ ∪ {0} Properties of modulus function : (a) | x |n = | xn | (b) | xn | = xn , where n is even and n ∈ z (c) | x y | = | x | | y | (d) = , (y ≠ 0) (e) | | x | – | y | | ≤ | x + y | ≤ | x | + | y | Signum Function The function f(x), defined as f(x) = ⎨ x ; ⎩ ; x ≠ 0 x = 0 is called signum function. This signum function may also defined as ⎧ 1 ; ⎪ f(x) = ⎨ 0 ; x > 0 x = 0 ⎪− 1 ; Domain = R x < 0 Range = {–1, 0, 1} Greatest Integer Function This function is also known as step function or floor function denoted by f(x) = [x]. By [x] we mean greatest integer less then or equal to x. If n is an integer and x is any real number between n and n + 1 i.e. n ≤ x < n + 1, then [x] = n Thus [3.4] = 3, [3.99] = 3 [–4.99] = –5, [–4.001] = –5 [0.3] = 0, [–0.2] = –1 Domain of [x] is set of all real numbers because [x] exist ∀ x ∈ R But [x] is always integral number so range is set of all integers Z. Some Properties of [x] : (a) [x + k] = [x] + k, if k ∈ Z (b) [–x] = –[x] – 1 (c) [x] + [–x] = 0, x ∈ Z (d) [x] + [–x] = –1, x ∉ Z (e) [x] – [–x] = 2x, x ∈ Z (f) [x] – [–x] = 2[x] + 1, x ∉ Z (g) x – 1 < [x] ≤ x (h) ⎡ 1 ⎤ ⎡ 2 ⎤ ⎡ 3 ⎤ ⎡ n − 1⎤ .......... .... x = [nx] ⎣⎢ n ⎥⎦ ⎢⎣ n ⎥⎦ ⎢⎣ n ⎥⎦ ⎣⎢ n ⎥⎦ (i) [x + y] ≥ [x] + [y] (j) ⎡(x) ⎤ = ⎡ x ⎤  ⎢⎣ n ⎥⎦ ⎢⎣ n ⎥⎦ (k) [x] ≥ n ⇒ x ≥ n, n ∈ z (l) [x] > n ⇒ x ≥ n + 1, n ∈ z (m) [x] ≤ n ⇒ x < n + 1, n ∈ z [x] < n ⇒ x < n, n ∈ z Fractional Part Function : Function denoted by f(x) = {x}, known as fractional part function. Also defined as f(x) = x – [x] If x ∈ Z, then f(x) = 0 [i.e. f(2) = 2 – [2] = 0] If x ∉ Z, then f(x) lies between 0 to 1. i.e. x ∉ Z, 0 < f(x) < 1 [i.e. f(3.4) = 3.4 – [3.4] = 3.4 – 3 = 0.4] Find the value of ⎡ 1 + 1 ⎤ + ⎡ 1 +  2 ⎤ + ⎡ 1 +  3 ⎤ + .......... ⎡ 1 +  99 ⎤ ⎢⎣ 3 100 ⎥⎦ ⎢⎣3 100 ⎥⎦ ⎢⎣ 3 100 ⎥⎦ ⎢⎣3 100 ⎥⎦ Where [⋅] denote greatest integer function. Solution : Using properties (h) of step function Here x = 1 3 n = 100 so ⎡ 1 + 1 ⎤ + ⎡ 1 +  2 ⎤ + ⎡ 1 +  3 ⎤ + .......... ⎡ 1 +  99 ⎤ = ⎡ 1 × 100⎤ = 33 . ⎢⎣ 3 100 ⎥⎦ ⎢⎣ 3 100 ⎥⎦ ⎢⎣3 100 ⎥⎦ ⎢⎣ 3 100 ⎥⎦ ⎢⎣ 3 ⎥⎦ Write the equivalent function of the function f(x) = |x + 2| + |x – 3|. Solution : First we find the critical values (values of x where modulus function vanish) which is x = – 2, 3. If x < – 2, then f(x) = – (x + 2) – (x – 3) = –2x + 1 If –2 ≤ x < 3, then f(x) = x + 2 – (x – 3) = 5 If x ≥ 3, then f(x) = x + 2 + x – 3 = 2x – 1 ⎧− 2x + 1 ; x < −2 so f(x) = ⎪ 5 ; − 2 ≤ x < 3 . ⎪2x − 1 ; x ≥ 3 LOGARITHMIC FUNCTION If f : R+ → R, f(x) = loga x, then f(x) is known as logarithmic function Here f(x) exist if x > 0 and 0 < a < 1 or a > 1 (a ≠ 1) y f(x) loga x 0 x a > 1 Properties of logarithmic function loga m.n = log m + log n log m = log m – log n a n loga mn = nloga m (iv) log bp = p log b aq q a log b = logx b logx a = log xb.log ax logb .loga = 1 a b If loga f(x) = y ⇒ f(x) = (a)y ⎧f (x) ≥ g(x) if a > 1 If loga f(x) ≥ loga g(x) ⇒ ⎨ ⎩f (x) ≤ g(x) if 0 < a < 1 ⎧⎪f (x) ≥ (a)y a > 1 If loga f(x) ≥ y ⇒ ⎨⎪⎩ ≤ (a)y if 0 < a < 1 ⎧⎪f (x) ≤ (a)y a > 1 If loga f(x) ≤ y ⇒ Exponential Function ⎨ ⎩f (x) y 0 < a < 1 f(x) = ax is known as exponential function (a > 0) ax If 0 < a < 1 Domain = R Range = R+ f(x) (0, 1) x′ 0 x How many solutions are there for equation log4 (x – 1) = log2 (x – 3)? Solution : log4 (x – 1) = log2 (x – 3) ⇒ log 2 (x – 1) = log (x – 3) ⇒ 1 log (x – 1) = log (x – 3) 2 2 2 ⇒ log2 (x – 1)1/2 = log2 (x – 3) ⇒ (x – 1)1/2 = (x – 3) ⇒ x – 1 = x2 – 6x + 9 ⇒ (x – 2) (x – 5) = 0 x = 2, 5 But x – 1 > 0 and x – 3 > 0 x > 1 and x > 3 So only one solution x = 5 KINDS OF FUNCTION Polynomial Function The function f(x) = a0 + a1x + a2x2 + ........ + anxn where a0, a1, a2, ....... an ∈ R and n ∈ N is called a polynomial function of degree n. Rational Function A function defined by the quotient of two polynomial function is called rational function for Example : x 2 + 1 x 3 + x + 1 is a rational function. Irrational Function A function involving one or more radicals of polynomial is called a irrational function Example : 3 x 2 + + x 2, x 2 + 2x + 3 etc. Algebraic Function An algebraic function is one which consist of a finite number of terms involving power and roots of the variable x and simple operation, addition, subtraction, multiplication and division i.e. all rational, and irrational functions are algebraic functions. Transcendental Function All function which are not algebric are called transcendental function. Example : All trigometric function i.e. sin x, cos x etc. All exponential function, ex, log x, ax etc. Inverse trigonometric function sin–1 x, cos–1 x, etc. Explicit Function A function in which dependent variable (y) is expressed directly in terms of independent variable (say x) i.e. y = x3 + x2 + 1, y = Implicit Function x 2 + 3x + 5 , etc. x + 2 A function in which we can't express dependent variable in terms of independent variable. Example: x3 + y3 + 3xy = 0, note that we can't write y or x in terms of x, or y separately. Even or Odd Function Even function : If f(–x) = f(x) then f(x) is said to be even function. Example : f(x) = cos x is a even function [Θ f(–x) = cos (–x) = cos x = f(x)] Odd function : If f(–x) = –f(x) then f(x) is said to odd function. Example : If f(x) = x3 + tan3 x is a odd function because f(–x) = (–x)3 + [tan (–x)]3 = – x3 – tan3 x = – [x3 + tan3 x] = – f(x) so f(–x) = – f(x) Is a function f(x) = Solution : ex + e −x x. e x − e−x even? e– x + e x = e x + e−x Yes, f(x) is even, because f(–x) = (–x). e – x − ex so f(–x) = f(x). x e x − e −x = f(x) Show that f(x) is a odd function if f(x) = log (x3 + Solution : 1+ x 6 ) . f(x) = log (x3 + 1+ x 6 ) f(–x) = log ⎡(−x)3 + = log [–x3 + 1+ (−x)6 ⎤ ⎦ ] = log ⎡ ⎢− x 3 + ⎢⎣ ⎡ x 6 − 1− x 6 ⎤ = ⎡ − 1 ⎤ = log ⎢ ⎥ log⎢ ⎥ − x 3 − 1+ x 6 ⎥⎦ ⎢⎣ − x 3 − 1+ x 6 ⎥⎦ f(–x) = log 1 = −log⎡x 3 + 1+ x 6 ⎤ x 3 + ⎣⎢ ⎥⎦ so f(–x) = –f(x) so f(x) is odd function. Periodic Function A function 'f' defined on its domain is said to be periodic function if their exist a positive number T such that f(x + T) = f(x) ∀ x ∈ D. Also both x + T and x – T should belong to D. The least value of T, it exists is called, the period of the function. f(x) = sin x f(x) = sin (x + 2π) = sin (x + 4π) = sin (x + 6π) = .................. Here T = 2π, 4π, 6π ...................... Least value of T is 2π, so time period of sin x is 2π Some Standard Functions and their Period Function Period sin x 2π cos x 2π tan x π {x} 1 Some Special Point about Periodic Function If period of f(x) is 'T' then (i) Period of |f(x)| is T . 2 Period of [f(x)]n is T , if n is even number (n ∈ N) 2 Period of [f(x)]n is T, if n is odd number (n ∈ N) Period of f(ax) and f(ax + b) is . f ⎛ x ⎞ Period of ⎜ a ⎟ is |a|.T. ⎝ ⎠ If Period of f(x) and g(x) are same say 'T' then period of f(x) ± g(x) is given by (i) T (if f(x) and g(x) both are even). 2 (ii) T (if f(x) is any function except even). If period f(x) is T1 and g(x) is T2. Then period of f(x) ± g(x) is given by L.C.M. of T1 and T2 f (x) (same for g(x) ) Calculate the period of f(x) = Solution : 2π Period of sin 3x = 3 Period of cos 2x = 2π = π 2 So, Period of f(x) is L.C.M. of 2π , π = 2π 3 1 If f(x) = is a periodic function, then find its period. Solution : f(x) = = f(x) = |sin x + cos x| Now period of sin x + cos x is 2π = ⎛⎜Remember ⎝ = x ⎞⎟ ⎠ So, period of |sin x + cos x| is 2π = π 2 If f(x) = esin3 x + cos x is a periodic function. If yes, then calculate its period. Solution : For periodic function f(x + T) = f(x) Now f(x + T) = esin3 (x+T) + cos( x + T) If T = 2π 3 f(x + 2π) = esin (x+2π) + cos( x + 2π) = esin3 x + cos x = f (x ) f(x + 2π) = f(x); f(x) is periodic function having period 2π. Bounded and Unbounded Function f(x) is said to be bounded above, if there exists a fixed number say M such that f(x) is never greater then M for all value of x. Similarly it's bounded below if there exists a fixed number m (say) that f(x) is never less then m i.e. M ≤ f(x) ≤ m for all value of x. f(x) is said to be unbounded if one or both of the upper and lower (M and m) bounds of the function are infinite Example : f(x) = 3 + sin x is a bounded function because maximum and minimum value of sin x are +1 and –1 So, 2 ≤ f(x) ≤ 4 for all value of x. TYPES OF MAPPINGS OR FUNCTIONS One-one Function or Injective Function : A function is said to be one-one function if different element in a domain have different images in co-domain. Set A Set B (domain) (co-domain) Many–one Function A function f : A → B is said to be many one if more than one element in set A have same image in Set B. A B Into Function A function f: A → B is said to be into function if there exist at least one element in set B having no any pre-image in set B. A B b c d In fig set B (co-domain) there is no pre-image, for element d, in set A, so function is into function. Onto Function f : A → B, said to be onto function if every element in set B has a pre image in set A. Range of f = co-domain of f. Example of Onto function : log x, linear polynomials, are always onto function. Possible mappings are One-one and onto (bijective function) Many one and onto One-one and into Many one-into Example : If f : R → R, f(x) = x2 + 3x + 2 then f(x) is many one function. ⎛ 3 ⎞2 1 because f(x) = x2 + 3x + 2 = ⎜ x + 2 ⎟ − 4 ⎝ ⎠ ⎛ 3 ⎞2 1 f(–2) = ⎜− 2 + 2 ⎟ − = 0 4 ⎝ ⎠ ⎛ 3 ⎞2 1 f(–1) = ⎜− 1+ 2 ⎟ − = 0 4 ⎝ ⎠ So image of –2 and –1 are same ∴ f(x) is many one. Example : f(x) = 2x + sin x, is one-one because f′(x) = 2 + cos x, minimum value of cos x is –1. ∴ f′(x) > 0 for all x ∈ Domain = R Composition of Function Let f : A→ B and g : B → C then the composition of g and f is denoted by gof and is defined as gof : A→ C given by gof (x) = g(f(x)) Similarly fog is defined. Note that, gof is defined only if Range f ⊆ gof dom g and fog is defined only if Range ⊆ dom f. dom fog = {x ∈ dom g : g(x) ∈ dom f} A B C Let f(x) = x2 + 3 and g(x) = . Since dom g = [0, ∞), dom f = R we have fog (x) = f(g(x)) = f ( x ) = ( x )2 + 3 = x + 3 So dom fog = {x ∈[0, ∞) : g(x) ∈R} = [0, ∞) Let us now find gof, we have (gof) (x) = g(f(x)) = g(x2 +3) = , then dom gof = {x ∈R : f(x) ∈[0, ∞)} = R. Inverse Function Two functions f and g are inverse of each other if f (g(x)) = x for x ∈ dom g and g(f(x)) = x for x ∈dom f , i.e., gof =Idom f and fog = Idom g where Idom f is identity function on dom f and Idom g is identity function on dom g. We denote g by f–1 or f by g–1. To find the inverse of f, write down the equation y = f(x) and then solve x as a function of y. The resulting equation is x = f–1 (y). To find the inverse of f (x) = ex − e− x ex − e− x 2 e2x − 1 We write y = ⇒ 2y = 2 ex ⇒ e2 x 2yex − 1 = 0 ⇒ ex = 2 ⇒ ex y ± since ex ≥ 0 so ex = y + ⇒ x = log y + y 2 + 1⎞⎟ . Thus f–1(x) = log x + x 2 + 1⎞⎟ ⎠ The graph of f and f–1 are related to each other in the following way : If the point (x, y) lies on the graph of f then the point (y, x) lies on the graph of f–1 and vice versa. Thus the graph of f–1 is the reflection of the graph of f in the line y = x as below (since we know that y = log x and y = ex are inverse of each other). Existence of inverse function y = x A function need not have an inverse. e.g. the function f(x) = x2 has no inverse (where dom f = R). To have an inverse, a function must be both one-one and onto, i.e. bijective. Example 1 : SOLVED EXAMPLES Suppose A1, A2,........A30 are thirty sets each with five elements and B1, B2 Bn are n sets each with three 30 n elements such that ΥAi = ΥB j = S . If each element S belongs to exactly ten of the i =1 j =1 Ai ' s and exactly nine of the Bj's, then the value of n is (1) 15 (2) 135 (3) 45 (4) 90 Solution : Answer: (3) Total number of elements in all A's = 30 × 5 = 150 Total number of elements in all B's = 3n Let there are p elements in set S and each is repeated ten times in A's and nine times in B's exactly, so 10p = 150 and 9p = 3n ⇒ p = 15 and n = 45. Example 2 : In a pollution study of 1500 Indian rivers the following data were reported, 520 were polluted by sulphur compounds, 335 were polluted by phosphates, 425 were polluted by crude oil. 100 were polluted by both crude oil and sulphur compounds, 180 were polluted by both sulphur compounds and phosphates, 150 were polluted by both phosphates and crude oil and 28 were polluted by sulphur compounds, phosphates and crude oil. How many of rivers were polluted by atleast one of the three impurties? How many rivers were polluted by exactly one of the three impurties? Solution : Let U = Set of all Indian rivers studie for pollution n(u) = 1500 S = Set of all Indian rivers polluted with Sulphur compounds n(s) = 520 P = Set of all Indian rivers polluted with Phosphates compounds n(p) = 335 C = Set of all Indian rivers polluted with Crude oil n(c) = 425 Number of rivers polluted by at least one of three impurities = n(S ∪ P ∪ C) = n(s) + n(p) + n(c) – n(s ∩ p) – n(p ∩ c) – n(c ∩ s) + n(s ∩ p ∩ c) = 520 + 335 + 425 – 180 – 150 – 100 + 28 = 878. Number of rivers polluted by exactly one of the three impurities = n(s) + n(p) + n(c) – 2n(c ∩ p) – 2n(p ∩ c) – 2n(s ∩ c) + 3n(s ∩ p ∩ c) = 520 + 335 + 425 – 2×180 – 2×150 – 2×100 + 3×28 = 504. Example 3 : Let A and B have 3 and 6 elements respectively. The minimum number of elements in A ∪ B is (1) 3 (2) 6 (3) 9 (4) 18 Solution : Answer (2) Quite clearly n(A ∪ B) will be minimum when either A ⊂ B or B ⊂ A. So here in this problem A ⊂ B. Hence n(A ∪ B) = n(B) = 6. By Venn diagram Example 4 : min. n(A ∪ B) = 6 If A, B and C are sets, then prove that (A – B) ∩ (A – C) = A – (B ∪ C). Verify the above result by venn diagrams. Solution: x ∈ (A – B) ∩ (A – C) ⇔ x ∈(A – B) and x ∈ (A – C) ⇔ (x ∈ A and x ∈/ B) and (x ∈ A and x ∈/ C) ⇔ x ∈ A and x ∈/ B and x ∈/ C ⇔ x ∈ A and x ∈/ (B ∪ C) ⇔ x ∈ A – (B ∪ C) (A – B) ∩ (A – C) (ii) B ∪ C (iii) A – (B ∪ C) Clearly (i) = (iii), hence verified. Example 5 : A survey of 500 television watchers produced the following information: 285 watch foot ball, 195 watch hockey, 115 watch basket ball, 45 watch foot ball and basket ball, 70 watch foot ball and hockey, 50 watch hockey and basket ball, 50 do not watch any of three games. How many watch all the three games? How many watch exactly one of three games? Solution : Let x watch all the three. 170 + x + 70 – x + 45 – x + x + 50 – x + 75 + x + 20 + x = 500 – 50 430 + x = 450 x = 20 So 20 watch all the three games. Exactly one of the three watches = 170 + x + 75 + x + 20 + x = 265 + 3x = 265 + 60 = 325 the following is an empty set? {x : x is a real number and x2 – 1 = 0} {x : x is a real number and x2 + 1 = 0} {x : x is a real number and x2 – 9 = 0} {x : x is a real number and x2 = x + 2} Solution : Answer (b) (a) x2 – 1 = 0 ∴ x = ± 1 hence {1, – 1} (b) x2 + 1 = 0; x2 = – 1 ∴ x = ± i, not a real number hence φ (c) x2 – 9 = 0; x = ± 3 hence {– 3, 3} (d) x2 – x – 2 = 0; x = –1, 2 hence {– 1, 2} Clearly option (b) is empty set. Example 7 : Two finite sets A and B are having m and n elements respectively. The total number of subsets of the first set is 56 more than the total number of subsets of the second set. The value of m, n are (1) 7, 6 (2) 6, 3 (3) 5, 1 (4) 8, 7 Solution : Answer (2) Number of subsets of A = 2m Number of subsets of B = 2n Hence 2m – 2n = 56 (given) (i) Clearly option (b) i.e. m = 6, n = 3 satisfies e.g. (i) above. B be two non-empty subsets of set X such that A is not a subset of B, then (1) A is a subset of B' (2) B ⊆ A (3) A and B are disjoint (4) A and B' are non-disjoint Solution: Answer (4) Since A ⊆/ B so only two possibilities arise as follows A ⊆ B' Or A / B' B / A B / A A and B are disjoint A and B' are non-disjoint Hence option (4) is correct in both. A and B are non-disjoint A and B' are non-disjoint Example 9 : If n(∪) = 80; n(A) = 40; n(B) = 30 and n(A ∪ B)' = 15 then n(A ∩ B) is Solution : n(A ∪ B) = n(∪) – n(A ∪ B)' n(A ∪ B) = n(A) + n(B) – n(A ∩ B) 65 = 40 + 30 – n(A ∩ B) ∴ n(A ∩ B) = 70 – 65 = 5 Example 10 : A = { x + 1 x B = { x + 1 x : x ∈ R+} : x ∈ R–} Show the sets A and B on a real number line and find A ∩ B. Solution : x + 1 ≥ 2 when x ≥ 0 and x B x + 1 ≤ – 2 when x < 0. x A –2 0 2 Hence Clearly A ∩ B = φ, since A and B are disjoint sets. Example 11 : Let A = {(x, y) : ax = ay; a > 0 and a ≠ 1; a, x, y ∈ R} B ={(x, y) : xy = 1; x, y ∈ Ro} Choose the correct statements amongst the following. A ∩ B = B (2) A ∩ B = A (3) n(B) > n(A) (4) A and B are non-comparable Solution : Answers (4) Clearly ax = ay only if x = y. So A is set of all points on the line y = x. B is a set on all points on rectangular hyperbola as shown. None of the sets is a subset of other. Both are infinite sets. A ≠ B, A ⊆ B, B ⊆ A hence A and B are non-comparable. Let A = {θ : sin θ ≤ B = {θ : 0 ≤ cos θ ≤ 1 , 0 ≤ θ ≤ 2π} 2 1 , 0 ≤ θ ≤ 2π} 2 Shade the region A ∩ B on a circular number line for θ. Solution : π 2 π 3 5π π 6 6 π 0 or 2π Clearly A ∩ B is shown by cross grid ⎧θ : 3π ≤ θ ≤ 5π⎫  5π Set A 3π 3 2 Set B A ∩ B = ⎨ ⎬ ⎩ 2 ⎭ Example 13 : Let Z+ = Set of all positive integers and R = {(a, b) : 3a + b = 0; a ∈ Z+}. Is R a relation on Z+. Explain Solution : 3a + b = 0 ⇒ b = – 3a, a ∈ Z+ So b = –3, –6, –9........... R = {(1, –3), (2, –6), (3, –9). } Here (1, –3) ∉ Z+ × Z+, hence R is not a relation on Z+. Example 14 : Find the quadratic relation between the components of the ordered pairs of the relation R, where R ={(0, –5), (1, 0), (2, 11), (3, 28). } Solution : Let y = ax2 + bx + c be the quadratic relation required. Since (0, –5) ∈ R, ∴ – 5 = 0 + 0 + c ⇒ c = – 5 (1, 0) ∈ R, ∴ 0 = a + b – 5 ⇒ a + b = 5 (i) (2, 11) ∈ R, ∴ 11 = a.4 + b.2 – 5 ⇒ 4a + 2b = 16 (ii) From (i) and (ii) above we get a = 3, b = 2 Hence y = 3x2 + 2x – 5 is the required relation, clearly (3, 28) satisfies the relation found. Example 15 : Determine the domain and range of following relation R = {(x, y) : y = x − 2 , x, y ∈ Z, x ≤ 2} Solution: x ≤ 2 ⇒ –2 ≤ x ≤ 2, x ∈ Z i.e. x = –2, –1, 0, 1, 2 Corresponding values of y are y = − 2 − 2 = 4 i.e. (–2, 4) ∈ R y = − 1− 2 = 3 i.e. (–1, 3) ∈ R y = 0 − 2 = 2 i.e. (0, 2) ∈ R y = 1− 2 = 1 i.e. (1, 1) ∈ R y = 2 − 2 = 0 i.e. (2, 0) ∈ R R = {(–2, 4), (–1, 3), (0, 2), (1, 1), (2, 0)} Hence dom (R) = {–2, –1, 0, 1, 2} Range of R = {0, 1, 2, 3, 4} Example 16 : Let R be a relation on set of real numbers (R) defined by R = {(a, b) : a = b3 and a, b are real numbers} Find domain and range of R. Solution : a R b ⇒ b = a1/3 Since cube and cube roots of all real numbers whether positive, negative or zero are defined on a set of real numbers then Dom (R) = All real numbers Range of R = All real numbers Example 17 : Let R be a relation on set of Real numbers defined by R = {(x, y): y = x − 1 + x − 2 , 0 ≤ x ≤ 3} By drawing a graph between x and y, find the range of R. Solution : Clearly domain of R = {x : 0 When 0 ≤ x ≤ 1 y 1 ≤ x ≤ 2 y 2 ≤ x ≤ 3 y Now the graph can be plotted Q From the graph 1 ≤ y ≤ 3 So Range of R ∈ [1, 3] Example 18 : In column A below some relation sets are defined. Column B enlists the cardinal number of these sets. Match the two columns. x Solution : Answer : (a)(iii), (b)(i), (c)(iv), (d)(ii) (a) (x + y) (y + 2004) = –1 where x, y ∈ Z Only possible when either x + y = 1 and y + 2004 = – 1 ⇒ x = 2006, y = – 2005 or x + y = –1 and y + 2004 = 1 ⇒ x = 2002, y = – 2003 or R = {(2002, –2003), (2006, – 2005)} so n(R) = 2 9x2 + 4y2 = 36 here least value of x2 = 0 (i.e. x = 0) and maximum value of y2 = 9 (i.e. y = ±3) also least value of y2 = 0 (y = 0) hence maximum of x2 = 4 (x = ±2) So we get four solution (0, 3), (0, –3), (2, 0), (–2, 0). For x = ±1, y ∉ Z, similarly for y = ±1 or ±2 x ∉ Z. So cardinal number of R is 4. Possible x = 2, 3, 4, 5, 6 and possible y = 4, 5, 6 x and y are co-prime so (2, 5), (3, 4), (3, 5), (4, 5), (5, 4), (5, 6), (6, 5) are only 7 solutions are possible. So, n(R) = 7. For x = 1, 2 and 3 corresponding So 3 ordered pairs satisfy R. n(R) = 3. 1 are x 1, 1 1 2 and 1 3 Example 19 : A B (a) R = {(x, y) : (x, y) ∈ Z × Z, (x + y) (y + 2004) + 1 = 0} (i) 4 (b) R = {(x, y) : (x, y) ∈ Z × Z, 9x2 + 4y2 = 62 (ii) 3 (c) R = {(x, y) : (x, y) ∈ N × N, 2 ≤ x ≤ 6, 3 < y < 7, x & y are co-prime} (iii) 2 (d) R = {( x, 1 ) : 0 < x < 4, x ∈ N} (iv) 7 Let S = {1, 2, 3, 4, 5} and A = S × S. A relation R on A is defined as follows : “(a, b) R (c, d) iff ad = cb”. Show that R is an equivalence relation. Solution : Check for reflexivity: (a, b) R (a, b) iff ab = ba true Hence R is reflexive Check for being symmetric: (a, b) R (c, d) ⇒ ad = bc (c, d) R (a, b) ⇒ cb = da true Hence R is symmetric Check for transitivity: (a, b) R (c, d) ⇒ ad = bc (i) (c, d) R (e, f) ⇒ cf = de (ii) (a, b) R (e, f) ⇒ af = be (iii) Multiply (i) and (ii) we get (iii) Hence R is transitive So R is an equivalence relation. – y + is an irrational number.” Consider the above relation and find whether R is an equivalence relation or not, when x, y ∈ set of real numbers. x, y ∈ set of rational numbers. Solution : x, y ∈ real numbers: x – y + is always an irrational number ∀ x ∈ R. Hence given relation is reflexive. Let us choose x = 3 and y = 2 2 then x – y + = 2 an irrational number but y – x + = 0 a rational number Hence R is not symmetric. Let us choose x = , y = – and Z = 2 x – y + y – z + ∈ Q′ ∈ Q′ x – z + 3 = 0 ∉ Q′ Hence R is not transitive, when x, y ∈ real number set. Not an equivalent relations. x, y ∈ Q It is well known that x – x = 0 ∈ Q x – x + 3 = 0 + 3 ∈ Q′ Hence reflexive x – y and y – x both will be rational. Hence x – y + 3 as well as y – x + 3 is always irrational (Note: Sum of a rational and irrational is always irrational) Hence R is symmetric x – y + y – z + ∈ Q′ ∈ Q′ x – z + 3 ∈ Q′ Hence transitive. So when x, y ∈ Q, given relation is an equivalence relation. Example 21 : Consider the following relations, defined as R1 = {(a, b) : a divides b; a, b ∈ N} R2 = {(a, b) : a is parallel to b; a, b ∈ set of all lines in a plane} R3 = {(a, b) : a = b2; a, b ∈ R} R4 = {(a, b) : a is perpendicular to b; a, b ∈ set of all lines in a plane} Complete the following table with True (T) or False (F) Reflexive Symmetric Anti- Symmetric Transitive Equivalence Relation Ordered Relation Partial Order Relation R1 R2 R3 R4 Solution: R1 is reflexive as a divides a is true. R1 is not symmetric as if a divide b, b does not divide a if a and b are different. R1 is antisymmetric (when a and b are equal) R1 is transitive as if a divides b and b divides c, ⇒ a divides c. R1 is hence not an equivalence relation. R1 is ordered relation. R1 is partial ordered relation. R2 is clearly reflexive, symmetric, transitive hence equivalence relation, not antisymmetric, not ordered relation and not partial ordered relation. R3 a = a2 not always true hence not reflexive if a = b2 then b may not be equal to a2, not symmetric if a = b2 and b = c2 then a may not be equal to c2 not transitive. if a = b then only two pairs of real numbers i.e. (0, 0) and (1, 1) satisfy R3 Hence R3 is antisymmetric, not equivalence relation, not ordered relation and not partial order relation. R4 not reflexive, symmetric, not transitive, not equivalence relation, not antisymmetric (as x0 line is perpendicular to itself) not partial order relation and not ordered relation. so the solution is as follow. Reflexive Symmetric Anti- Symmetric Transitive Equivalence Relation Ordered Relation Partial Order Relation R1 T F T T F T T R2 T T F T T F F R3 F F T F F F F R4 F T F F F F F If f : (3, 4) → (2, 4), f(x) : Solution : ⎡ x ⎤ ⎣ 2 ⎥⎦ where [⋅] denote step function then find the f–1(x). f(x) = x − ⎡ x ⎤ ⎣ 2 ⎦ ⎡ x ⎤ Domain of this function is (3, 4) in this domain So function is f(x) = x – 1 or y = x – 1 ⇒ x = y + 1 On interchanging x and y we get y = x + 1 ∴ f–1(x) = x + 1. ⎢⎣ 2 ⎥⎦ = 1 Find the domain of the function, f (x) = sin ⎛ ⎞ ⎜ ⎟ + cos −1⎛ x + 1⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ Solution : Domain of sin–1 ⎛ x − 3 ⎞ say D and ⎜ ⎟ ⎝ ⎠ ⎛ x + 1⎞ Domain of cos–1 ⎜ ⎟ say D ⎜ ⎟ ⎝ ⎠ So Domain of f(x) is D1 ∩ D2 For D1 − 1 ≤ 5 ≤ 1 for D2 – 1 ≤ x + 1 ≤ 1 4 –5 ≤ |x| –3 ≤ 5 –4 ≤ |x| + 1 ≤ 4 –2 ≤ |x| ≤ 8 |x| ≥ – 2 and |x| ≤ 8 so x ∈ [–8, 8] so Domain of f(x) is [–8, 8] = [–3, 3]. Example 24 : If f : [–1, 1] → B and f(x) = sin–1 x, if f(x) is one-one onto function then find value of B. Solution : In one-one onto function range of f = co-domain of f (which is B) So range of function for x ∈ [–1, 1] is ⎡− π , π⎤ so B ∈ ⎡− π , π⎤ . ⎣⎢ 2 2 ⎥⎦ ⎣⎢ 2 2 ⎥⎦ If f ⎛ x + 1 ⎞ = x + 2 x 2 − 1 , then find the value of f(2). x 2 + 2 ⎝ ⎠ Solution : ⎛ x + 1 ⎞ x 2 − 1 f ⎜ x + 2 ⎟ = x 2 + 2 ....(1) ⎝ ⎠ Put x + 1 = y ⇒ x + 2 x = 1− 2y y − 1 Put these value in equation (1) ⎛ 1− 2y ⎞2 ⎜ ⎟ − 1 y − 1 So f(y) = ⎝ ⎠ ⎛ 1− 2y ⎞2 ⎜ ⎟ + 2 ⎝ y − 1 ⎠ f(y) = (1− 2y )2 − (y − 1)2 (1− 2y )2 + 2(y − 1)2 (1− 4)2 − (2 − 1)2 = 8 So value of f(2) = (1− 4)2 + 2(2 − 1)2 11 . Example 26 : Find the domain of the function f(x) = logx (x2 – 3x + 2). Solution : Logax exist if x > 0 and 0 < a < 1, a > 1 Now f(x) = logx (x2 – 3x + 2) x2 – 3x + 2 > 0 ⇒ (x – 1) (x – 2) > 0, x ∈ (– ∞, 1) ∪ (2, ∞) But x > 1 or 0 < x < 1 So common value of x is (0, 1) ∪ (2, ∞). ∴ Domain of f(x) ∈ (0, 1) ∪ (2, ∞) and + 1 ⎪⎩x ; x ≤ 0 ; x > 0 g(x) = ⎧⎪ x + 1 ⎩ ; x ≤ 1 . ; x > 1 Then find (f + g) (x) and draw its graph. Solution : ⎧− x − 1 ; x ≤ −1 f (x) = ⎪x + 1 ⎪ ⎩ ; − 1 < x ≤ 0 ; x > 0 ⎧− x + 1 ⎪x + 1 g(x ) = ⎪ ⎪⎩− x + 2 ; x < 0 ; 0 < x ≤ 1 ; 1 < x < 2 ; x ≥ 2 ⎧− x − 1− x + 1 ; x ≤ −1 ⎧− 2x ; x ≤ −1 ⎪x + 1− x + 1 ; − 1 < x ≤ 0 ⎪ ; − 1 < x ≤ 0 f (x) + g(x)⎨x + x + 1 ⎪x + x − 2 ; 0 < x ≤ 1 ; 1 < x < 2 = ⎨2x + 1 ; ⎪2x − 2 ; 0 < x ≤ 1 1 < x < 2 ⎪⎩x − x + 2 ; x ≥ 2 ⎩⎪ 2 ; x ≥ 2 cos2x (sec2 x + 2 tanx) then find domain and range of f. Solution : f(sin2x) = 1 + 2sin x cosx = 1 + sin2x Let u = sin2x so f (u) = 1 + u since –1 ≤ sin 2x ≤ 1 so dom f = [–1, 1] Now – 1 ≤ u ≤ 1 ⇒ 0 ≤ 1 + u ≤ 2. Thus the range f = [0, 2] x 2 + 2x + c Show that the function f (x) = x 2 + 4x + 3c Solutions : attains any real value if 0 < c ≤1 Let m = x 2 + 2x + c 2 where m is an arbitrary real number, then (m–1) x2 + 2 (2m –1) x + c(3m–1) = 0, the argument x + 4x + 3c x must be a real number, hence (2m – 1)2 – (m – 1)(3m c – c) ≥ 0 > 0 or (4 – 3c)m2 – 4(c – 1)m – (c–1)≥0, but since m is a real number, this inequality in turn is valid under the conditions: (i) 4–3c > 0 (ii) 4(c–1)2 + (4–3c) 4(c–1) ≤ 0, Hence 0 ≤ c ≤ 1, but by hypothesis c ≠ 0 A function f : R → R, is defined by : αx 2 + 6x − 8 f (x) = α + 6x − 8x 2 Find the interval of values of α for which f is onto. Is the function one-one for α = 3 ? Solution : Let αx 2 + 6x − 8 m = ⇒ (α + 8m)x α + 6x − 8x 2 + 6(1− m)x − (8 + αm) = 0 Since x is real, 36 (1 – m)2 + 4(α + 8m)(8 + αm) ≥ 0 ⇒ (9 + 8α) m2 + (46 + α2) m + (9 + 8α) ≥ 0 f is onto if and only if the above relation hold for all m ∈ R. This will happen if 9 + 8α > 0 and (46 + α2)2 –4 (9 + 8α)2 ≤ 0 ⇒ 9 + 8α > 0 and (α2 + 16α + 64) (α2 – 16α +28) ≤ 0 ⇒ 9 + 8α > 0 and (α + 8)2 (α – 2) (α –14) ≤ 0 ⇒ α > –9/8 and 2 ≤ α ≤ 14. Thus 2 ≤ α ≤ 14 when α = 3, 3x 2 + 6x − 8 m = 3 + 6x − 8x 2 For m = 0, we get 3x2 + 6x – 8 = 0 ⇒ x = − 6 ± 36 + 96 6 = 1 (− 3 ± 3 33 ) Hence f is not one–one when α = 3 is not periodic Solution : Suppose that f(x) = sin is periodic with period T. Then, f (x + T ) = sin = f (x) = sin ; ∀x ≥ 0 ⇒ 2cos sin = 0 2 2 ⇒ cos = 0 2 or sin = 0 2 π ⇒ + 2 (2n + 1) = (2n + 1)π, n ∈ I ⎣⎢ 2 ⎥⎦ or − = 2nπ, n ∈ I The above equalities gives T as function of x. But for f(x) to be periodic T should be constant i.e. independent of x. Hence f(x) can not be periodic. Find the period of Solution : cos x + cot x + cos x 2 22 cot x 23 + cos x 2n −1 cot x 2n Since the period of cos ax(a > 0) is 2π/a and the period of cot ax (a > 0) is π/a, the periods of cosx, 2 n–1 2 n–1 x x n cosx/2 , ....., cos x/2 are 2π, 2 (2π) ......., 2 (2π) and the period of cot 2 , , cot 2n are 2π, ...., 2 π. Hence the period of the given function is L.C.M of (2π, 23π, ..., 2nπ) = 2nπ. ❑ ❑ ❑