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We will learn how to link pairs of objects from two sets and then introduce relations b/w the two objects in the pair. AXB = { (a,b): a ε A, b ε B} for non-empty sets A,B otherwise AXB=φ, this is the set of all ordered pairs of elements from A and B. n(AXB)=n(A)Xn(B) AXBXC = {(a,b,c):aεA,bεB,cεC} is called set of all ordered triplets for non-empty sets. n(AXBXC) =n(A)Xn(B)Xn(C) AX(BXC)=(AXB)XC=AXBXC. AXB,BXA are not equal (not commutative).
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Let P,Q be two sets then a relation R from P to Q is a subset of PXQ. Example: If A={a,b,c,d},B={p,q,r,s},then which of the following are relations (i) R1={(a,p),(b,r),(c,s)} (ii) R2 ={(q,b),(c,s),(d,r)} Solution: R1is a relation because R1is the subset of AXB but R2 is the not the relation of AXB because (q,b)εR2 but not in AXB.(R2 is not a subset of AXB) Total number of relations that can be defined from a set A to B is the number of possible subsets of AXB. If n(A)=P , n(B)=q , then total number of relations is 2pq OR total number of subsets of AXB. Example: If R is relation on a finite set having n elements, then the number of relations on A is Answer 2nxn Inverse Function is a relation from B to A defined by { (b,a) :(a,b)εR}
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State the domain and range of the following relation. Is the relation a function? {(2, –3), (4, 6), (3, –1), (6, 6), (2, 3)} they gave me two points with the same x-value: (2, –3) and (2, 3). Since x = 2 gives me two possible destinations, then this relation is not a function. domain: {2, 3, 4, 6} range: {–3, –1, 3, 6}RelationsA relation is a set of inputs and outputs, oftenwritten as ordered pairs (input, output). We canalso represent a relation as a mapping diagramor a graph. For example, the relation can berepresented as: * If output has one more unmapped element say 5 then output becomes codomain then Range is the subset of codomain Range ={-2,1,2,4} and Codomain ={-2,1,2.4,5}
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A relation f from a set A to B is said to be a function if every element of set A has unique image in set B. Or f : A→ B s.t. f(a) = b for all aεA , b is image of a and a is pre image of b under f. Types of Functions and their Graphs Signum functionModulus functionf(x) = |x|
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Reciprocal function f(x) = [x], x εR is called the smallest integer function or theDomain = R – {0} ceiling function.Range = R – {0} Domain=R , Range =Z. Square root function f : R → R (x > 0) such that Domain =Range is set of all positive real numbers.
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State the domain and range of the following relation. Is the relation a function? {(–3, 5), (–2, 5), (–1, 5), (0, 5), (1, 5), (2, 5)} domain: {–3, –2, –1, 0, 1, 2} range: {5}this relation is indeed a function. ARROW DIGRAMS are given below : State in each case ,whether it is a function or not ?
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f : R → R defined by f(x) = x2 Domain = R , Range = *0,∞) Graph of f : R → R such that f(x) = |x-1| Graph of f : R → R such that f(x) = |x+1| Draw a graph of the function defined by f(x) = Draw a graph of f : R → R defined by f(x) = |x-2|Ques. If f : R → R Be defined by f(x) = x2 +2x+1 then find (i) f(-1) x f(1) ,is f(-1)+f(1)=f(0) (ii) f(2) x f(3) , is f(2) x f(3)=f(6)Q.1 Find the domain of f(x) = [Hint R-{2,6} ]
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Ques. Find the domain and Range of following functions: (i) f(x) = (ii) f(x) = { ( x, ) : xεR} Sol. (i) Domain = R , Range = [-1/2 , 1/2+ , y ≠ 0 (How) Y= it implies yx2 – x + y =0 ,if y=0 then x=0 If y ≠ 0 for x to be real 1 – 4y2 ≥ 0(D≥0) (ii) Domain = R – {-1,1} , Range = (-∞,0) U [1,∞) ={ y:y<0 or y≥1} (How) 1 –x2 = 1/y it implies x = ± , ≥0 for x to be real.Find the domain and Range of the function f(x) = Sol. Domain = R , Range = [1/ 3 , 1] (How) -1 ≤ sin3x ≤ 1 it implies 1 ≥ sin3x ≥ -1 then 3 ≥ 2- sin3x ≥ 1 1/3 ≤ y 1 by using inequality result ,where y = f(x).
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A relation R in a set A is called (i) reflexive, if (a, a) ∈ R, for every a ∈ A, (ii) symmetric, if (a, b) ∈ R implies that (b, a) ∈ R, for all a, b ∈ A. (iii) transitive, if (a, b) ∈ R and (b, c) ∈ R implies that (a, c) ∈ R, for all a, b, c ∈ A.Example : Let T be the set of all triangles in a plane with R a relation in T given byR = {(T1, T2) : T1 is congruent to T2}. Show that R is an equivalence relation.R is reflexive, since every triangle is congruent to itself. Further,(T1, T2) ∈ R⇒ T1 is congruent to T2 ⇒ T2 is congruent to T1 ⇒ (T2, T1) ∈ R. Hence,R issymmetric. Moreover, (T1, T2), (T2, T3) ∈ R ⇒ T1 is congruent to T2 and T2congruent to T3 ⇒ T1 is congruent to T3 ⇒ (T1, T3) ∈ R. Therefore, R is anequivalence relation.Example: Let L be the set of all lines in a plane and R be the relation in L definedasR = {(L1, L2) : L1 is perpendicular to L2}. Show that R is symmetric but neitherreflexive nor transitive or in other words we can say it is not an equivalencerelation. 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. It is not transitive and cannot be equivalence relation also.
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QUESTION BANK WITH HINTS OF RELATIONS AND FUNCTIONS ↓ BY:--INDU THAKUR
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Q.1 The relation R is defined on the set N as R={ (x,x+5) : x <4 ; x,yε N} Write R in roster form .Write its domain & range.Q.2 If AXB={(p,q),(p,r),(m,q),(m,r)},find A&B.Q.3 If A={x,y,z} and B={1,2}.Find the number of relations from A to B.Q.4 Examine the relation :R={{(2,1),(3,1),(4,1)} and state whether it is a function or not?Q.5 Let A={1,2,3,4,5,6,7,8,9,10}.Define a relation R from A to A by R={(x,y):2x – y=0,where x,y εA}. Write down its domain ,co-domain & range.Q.6 Let a relation R1 on the set of real numbers be defined as (a,b)ε R1↔1+ab>0 for all a,bεR. Show that R1 is reflexive & symmetric but not transitive.Q.7 If a={1,2,3,4,6}.Let R be the relation on A defined by {(a,b):a,bεA, b is divisible by a} Write R in roster form, domain & range of R.Q.8 If R is a relation from set A={11,12,13} to set B={8,10,12} defined by y=x-3, then write R-1. Q.9 Let R be the relation on the set N of natural numbers defined by R={(a,b):a+3b=12,a,bεN} Find R, domain of R & range of R.Q.10 If A={1,2,3},B={1,4,6,9} and R is a relation from A to B defined by ‘x is greater than y’.Find the range of R.
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SOLUTIONS: Ans.1 R={(1,6),(2,7),(3,8)} domain={1,2,3},range={6,7,8} Ans .3 22x3 =26 Ans.5 R={(1,2),(2,4),(3,6),(4,8),(5,10)} , Domain= {1,2,3,4,5} range ={2,4,6,8,10} ,co-domain = AAns.6 Reflexive (a,a)εR1 [1+a²>0] & symmetric also asab=ba. (1,1/2)εR1,(1/2,-1)ε R1Ans.7 {(1,1),(1,2)……(1,6),(2,4)….(4,4),(6,6),(3,3),(3,6)} , domain=range={1,2,3,4,6}. Ans.8 {8,11),(10.13)} Ans.9 R={(9,1),(6,2),3,3)} Ans 10 {1} BY:--INDU THAKUR
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. Q.1 Find the domain and range of the function f defined by f(x)= Q.2 Let f(x)=x2 and g(x)=2x+1 be two real functions .Find (f – g)(x),(fg)(x),(f/g)(x) Q.3 If f={(1,1),(2,3),(o,-1),(-1,-3)} be a linear function from Z into Z . Find f(x). Q.4 If f(x) = x2 , find Q.5 Let f(x)= find f(-1) , f(3). Q.6 If f(x)= x2 – 1/x2 , then find the value of : f(x) +f(1/x) Q.7 Let A= {-2,-1,0.1,2} and f:A→ Z given by f(x) =x2 – 2x – 3,find range of f and also pre-image of 6,-3,5. Q.8 Find the domain and range of real valued function f(x)= Q.9 Y=f(x)= , then show that x=f(y). Q.10 If f(x) = , show that f[f(x)].
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SOLUTIONS : Ans.1 f(x) is defined real function if 9-x2>0 ,so domain=(-3,3) and x2=(9 - 1/y2 ) ,x is defined when 9 – 1/y2 ≥0,so range=*1/3,∞)Ans.2 (f – g)(x)=x2-2x-1, (fg)(x)=2x3+x2 and (f/g)(x)= ,x≠-1/2.Ans.3 Let f(x)=ax+b be a linear function. If (0,-1),(1,1)εf then weCan get a=2,b=-1 so linear function becomes f(x)=2x-1Ans.4 = 2.2 Ans.5 f(-1)=2,f(3)=10. Ans.6 0 Ans.7 Range of f={0,5,-3,-4} and no pre-image of 6 ;0,2 are the pre-images of -3;-2 is thepre-image of 5. Ans.8 It is defined when x ≠4 otherwise it becomes 0/0 form(indeterminate form) Domain(f)=R-{4}, range (f)={-1} as f(x) =-1 [x-4=-(4-x)]