### Sets And Functions Class 10th Mathematics Tamilnadu Board Solution

##### Question 1.State whether each of the following arrow diagrams define a function or not. Justify your answer. Answer:If P and Q are two relation then the two most important conditions for the relation to be treated as a function are as follows,● Each element of P must have a unique image in Q.● Each element of P has to be mapped with only one element in QIf either of the above conditions is not satisfied, then the relation can’t be consider as a function. (i) The figure shows mapping from P to Q, the element "C" in P does not have any image in Q. Since, the above given relation does not meet the conditions for relation and functions mentioned above so it is not a function. (ii) In the above mapping from L to m, each element of L has an image in M and the number of image for each element is one. Since, the above given relation satisfies all the conditions of relation and function mentioned above, so it is a function.Question 2.For the given function F = { (1, 3), (2, 5), (4, 7), (5, 9), (3, 1) }, write the domain and range.Answer:We know that for any two non-empty sets A and B. A function f from A to B may be defined as a rule that assigns each element of set A to tha of set B in a unique way (∀ x ∈ A and y ∈ B).We denote y = f(x) which means mean y is a function of x. so the set A is called the domain of the function and set B is called the co-domain of the function. y is called as the image of x under f and x is called a preimage of y. Range may be defined as the set of all image and it’s a subset of co-domain.We can represent the function in set as described belowf = {(a,b) / a ∈ A and b ∈ B}Where A is the domain and B is the pre-domain.So as per data given,F = {(1, 3), (2, 5), (4, 7), (5, 9), (3, 1)}HereDomain = {1, 2, 3, 4, 5}Range = {1, 3, 5, 7, 9}Question 3.Let A = { 10, 11, 12, 13, 14 }; B = { 0, 1, 2, 3, 5 } and fi : A → B , i = 1,2,3.State the type of function for the following (give reason):(i) f1 = { (10, 1), (11, 2), (12, 3), (13, 5), (14, 3) }(ii) f2 = { (10, 1), (11, 1), (12, 1), (13, 1), (14, 1) }(iii) f3 = { (10, 0), (11, 1), (12, 2), (13, 3), (14, 5) }Answer:There are several kind of function such as,One-one function:If f : A → B is a function then its treated as an one to one function if no element of B is associated with more than one element of A . A one-one function is also called an injective function.Onto function:A function f:A→B is called as an onto function if every element in B has a pre-image in A.One-one and onto function:A function f:A→B is called as an One-one and onto function if f maps all distinct elements of A with all distinct images in B and all element in B is an image of some element in A.Constant function:A function f:A→B is called as Constant function if all distinct elements of A has a single image in B.Here we are givenA = { 10, 11, 12, 13, 14 }B = { 0, 1, 2, 3, 5 }fi:A → B , where i = 1,2,3.(i) Here both 12 and 14 have same image i.e. 3 and there is no preimage of 0 of B in A, So it’s clear that the function given is neither an one to one nor onto.(ii) Here it’s clear that for all element of A there is only one image in B i.e. 1.we can also write that for f2:A→ B, f2(x) = 1, ∀ x ∈ 1. So the above function is a constant function.(iii) Here range of f3 = {0,1,2,3,4,5} = BIt’s an one-one and onto function as all element in A have a distinct and non-repetitive image in BQuestion 4.If X = {1, 2, 3, 4, 5 }, Y = { 1, 3, 5, 7, 9 } determine which of the following relations from X to Y are functions? Give reason for your answer. If it is a function, state its type.(i) R1 = {(x, y)| y = x + 2, x ∈ X , y ∈ Y }(ii) R2 = {(1, 1), (2, 1), (3, 3), (4, 3), (5, 5)}(iii) R3 = {(1, 1), (1, 3), (3, 5), (3, 7), (5, 7)}(iv) R4 = {(1, 3), (2, 5), (4, 7), (5, 9), (3, 1)}Answer:Here we are givenX = {1, 2, 3, 4, 5}, Y = { 1, 3, 5, 7, 9 }(i) R1 = {(x, y)| y = x + 2, x ∈ X , y ∈ Y }So using the value we can write the function in set formR1 = {(1,3),(3,5),(5,7)}From the above form it’s clear that as the elements of X doesn’t have a unique image in Y so it’s not a function.(ii) R2 = {(1, 1), (2, 1), (3, 3), (4, 3), (5, 5)}Here R2 is a function as all the element of X has a distinct image in Y. As 1 and 2 of X is related to 1 of Y and 3, 4 of X is related to 3 of Y so this function is onto function.(iii) R3 = {(1, 1), (1, 3), (3, 5), (3, 7), (5, 7)}Here in this relation 1 of X is related to 1 and 3 of Y which clearly contradict the definition of a function which says that every elements of a domain should have at most one image. So it’s clear that it’s not a function.(iv) R4 = {(1, 3), (2, 5), (4, 7), (5, 9), (3, 1)}The above expression is a onto and one-one function as all the elements of X has a unique image in Y and no two elements of X have same image in Y. these functions are also termed as bijective function.Question 5.If R = {(a, - 2), (- 5, b), (8, c), (d, - 1)} represents the identity function, find the values of a, b, c and d.Answer:Identity function is a function given by f(x) = x.Given function: R = {(a, - 2), (- 5, b), (8, c), (d, - 1)}This means,f(a) = -2f(-5) = bf(8) = cf(d) = -1Now, from the definition, f(x) = xtherefore,a = -2-5 = b, b = -58 = c, c = 8d = -1So a = -2, b = -5, c = 8 and d = -1Question 6.A = {–2, –1, 1, 2 } and . Write down the range of f. Is f a function from A to A?Answer:From the elements of A we can write the function in set orderSo f = Here from the above expression the range of the function is given asRange of f = As the element of the range are not in the set A so clearly it’s not a function from A to A.Question 7.Let f = {(2, 7), (3, 4), (7, 9), (–1, 6), (0, 2), (5, 3) } be a function from A = { –1, 0, 2, 3, 5, 7 } to B = { 2, 3, 4, 6, 7, 9}.Answer:Here we are given thatf = {(2, 7), (3, 4), (7, 9), (–1, 6), (0, 2), (5, 3)}A = { –1, 0, 2, 3, 5, 7 }B = { 2, 3, 4, 6, 7, 9 } where f:A→ BHere from the above expression the range of the function can be given asRange of f = {7, 4, 9, 6, 2, 3} = BSo it clear that it is an onto function, again the mapping shows that every element of A is mapped to an unique element of B and no two elements of A is linked with a single element of B so it’s a one-one function too.Hence its clear that the expression is a one-one and onto functionQuestion 8.Write the pre-images of 2 and 3 in the functionf = { (12, 2), (13, 3), (15, 3), (14, 2), (17, 17) }.Answer: So from the image we can see thatPre-image of 2 are 12 and 14Pre-image of 3 are 13 and 15Question 9.The following table represents a function from A = { 5, 6, 8, 10 } to B = { 19, 15, 9, 11 } where f = 2x - 1. Find the values of a and b. Answer:It’s given that f(x) = 2x-1So f(5) = 2 × 5 -1= 10-1= 9∴ a = 5Similarly f(8) = 2 × 8 -1= 16-1= 15∴ b = 15So the values of a = 9 and b = 15Question 10.Let A = {5, 6, 7, 8 }; B = { –11, 4, 7, –10,–7, –9,–13 } and f = {( x,y) : y = 3 - 2x, x ∈ A, y ∈ B }|(i) Write down the elements of f. (ii) What is the co-domain?(iii) What is the range? (iv) Identify the type of function.Answer:Here we are givenA = {5, 6, 7, 8 }B = {–11, 4, 7, –10,–7, –9,–13 }f = {( x,y) : y = 3 - 2x, x A, y B }so we can findf = {( x,y) : y = 3 - 2x, x A, y B }f(5) = 3- 2×5 = 3-10 = -7f(6) = 3 – 2×6 = 3-12 = -9f(7) = 3 – 2×7 = 3-14 = -11f(8) = 3 – 2×8 = 3- 16 = -13(i) the elements of f areFrom the above expression we get thatf = {(5,-7),(6,-9),(7,-11),(8,-13)}(ii) co-domainThe co-domain is B and it can be given asCodomain B = {-11,4, 7,-10,-7,-9,-13}(iii) RangeRange may be defined as the image of A in B present in the function.Range of f = {-7,-9,-11,-13}(iv) type of function From the above data and the diagram it’s clear that the given function is a one-one function as all the element A has a unique and single image in B. it’s not a onto function as the range is not same as the co-domain .Question 11.State whether the following graphs represent a function. Give reason for your answer.  Answer:(i) The above graph represents a function as we know that a graph represents a function only when the curve in the graph cuts the vertical axis only at a single point A.Here the curve cuts the vertical line at point X1 so it’s a function.(ii) The above graph represents a function as a vertical line cuts the curve at X1 point.(iii) The above graoh doesn’t represents a function as the vertical line and the curve intersects ech other at two points X1 and X2.(iv) The above figure doesn’t represent a function as the vertical line cuts the curve at three different points,.(v) The graph is are presentation of a function as the straight line is cut at a single point X2 by the vertical line.Question 12.Represent the function f = { (–1, 2), (– 3, 1), (–5, 6), (– 4, 3) } as(i) a table (ii) an arrow diagramAnswer:We are givenf = { (–1, 2), (– 3, 1), (–5, 6), (– 4, 3) }(i) a tableRepresenting the given data in tabular form (ii) an arrow diagramwe are givenA = {-1, -3, -5, -4}B = {2, 1, 6, 3} Question 13.Let A = { 6, 9, 15, 18, 21 }; B = { 1, 2, 4, 5, 6 } and f : A → B be defined by . Represent f by(i) an arrow diagram(ii) a set of ordered pairs(iii) a table(iv) a graph.Answer:Given A = { 6, 9, 15, 18, 21 }B = { 1, 2, 4, 5, 6 } and f : A BAnd We can find the domain     (i) an arrow diagram (ii) a set of ordered pairsFrom the given data and function value we can write the function as set of order pair as followsf(x) = {(6,1),(9,2),(15,4),(18,5),(21,6)}(iii) a tableFrom the data we can draw the tabular representation taking X and f(X) (iv) a graph.We can represent the function values in graph as followsf(x) = {(6,1),(9,2),(15,4),(18,5),(21,6)} Question 14.Let A = {4, 6, 8, 10 } and B = { 3, 4, 5, 6, 7 }. If f : A → B is defined by then represent f by(i) an arrow diagram(ii) a set of ordered pairs(iii) a table.Answer:given A = {4, 6, 8, 10 }B = { 3, 4, 5, 6, 7 }.f : A B is defined by so we can find the domain    (i) an arrow diagram (ii) a set of ordered pairsBasing on the available data we can write the function as asset of order pair as followsf(x) = {(4,3),(6,4),(8,5),(10,6)}(iii) a table.Using the data we have we can draw the tabel Question 15.A function is defined as follows Find (i) f(5) + f(6) (ii) f(1)- f(-3)(iii) f(-2h)- f(4) (iv) Answer:We are given,f: [-3,7) → ℝAnd the function given is (i) f(5) + f(6)Here it is clear that 5 and 6 lies between 4 and 7 so we have to use the function,f(x) = 2x - 3; 4 < x < 7So,f(5) = 2×5 - 3 = 10 - 3 = 7f(6) = 2×6 - 1 = 12 - 3 = 9∴ f(5) + f(6) = 7 + 9 = 16(ii) f(1) - f(-3)here its clear that 1 and -3 lies between the range of -3 ≤ x < 2 so we use the functionf(x) = 4x2 - 1⇒ f(1) = 4(1)2-1 = 4-1 = 3⇒ f(-3) = 4(-3)2-1 = 4×9 - 1 = 35So f(1) - f(-3)= 3 - 35= -32 (iii) f(-2) - f(4)As, x = -2f(x) = 4x2 -1; -3 ≤ x <2⇒ f(-2) = 4×(-2)2-1= 4×4 - 1= 16 - 1= 15For, x = 4f(x) = 3x-2; 2 ≤ x ≤ 4⇒ f(4) = 3 × 4 - 2= 12 - 2= 10⇒ f(-2)- f(4) = 15-10 = 5(iv) Here the entire values ranges from -1 to 6 so the range will lie between -3≤x<2, 2≤ x<4 and 4<x<7Sof(3) = 3x - 2; 2≤ x<4= 3 × 3 – 2= 9 - 2 = 7f(-1) = 4x2 - 1: -3≤x<2= 4(-1)2-1= 4-1 = 3f(6) = 2x - 3; 4<x<7= 2 × 6-3= 12-3 = 9⇒ 2f(6) = 2 × 9 = 18f(1) = 4x2 - 1: -3≤x<2= 4(1)2 - 1= 4 - 1 = 3So the function    Question 16.A function is defined as follows (i) 2f(-4) + 3f(2)(ii) f (- 7) - f (- 3)(iii) Answer:We are given,f: [-7,6) → ℝ[-7,6) = {x ∈ A: -7≤x<6}= {-7, -6, -5, -4,-3, -2, -1, 0, 1, 2, 3, 4, 5}And the function given are (i) 2f(-4) + 3f(2)Here the function lies between -5<x<2Sof(x) = x + 5f(-4) = -4 + 5= 1f(2) = 2 + 5= 7∴ 2f(-4) = 2 × 1 = 2∴ 3f(2) = 3× 7 = 21So 2f(-4) + 3f(2)= 2 + 21 = 23(ii) f (- 7) - f (- 3)here the function lies between -7≤x<-5 and -5≤ x < 2so f(x) = x2 + 2x + 1⇒ f(-7) = (-7)2 + 2×(-7) + 1= 49-14 + 1= 36Similarly f(x) = x + 5⇒ f(-3) = -3 + 5= 2∴ f (- 7) - f (- 3)= 36 -2= 34(iii) .For the above statement the function lies between-7≤ x <-5 for f(-6)-5≤ x<2 for for f(-3) and f(1)2<x<6 for f(4)at f(x) = x2 + 2x + 1 ; -7≤ x <-5⇒ f(-6) = (-6)2 + 2(-6) + 1= 36-12 + 1 = 25At f(x) = x + 5; -5≤ x<2 for⇒ f(-3) = -3 + 5 = 2⇒ 4f(-3) = 4 × 2 = 8Similarly,f(1) = 1 + 5 = 6⇒ 3f(1) = 3 × 6 = 18At f(x) = x-1; 2<x<6⇒ f(4) = 4-1 = 3⇒ 2f(4) = 2× 3 = 6So the function given   PDF FILE TO YOUR EMAIL IMMEDIATELY PURCHASE NOTES & PAPER SOLUTION. @ Rs. 50/- each (GST extra)

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