{"id":50,"date":"2021-03-23T03:12:36","date_gmt":"2021-03-22T21:42:36","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=50"},"modified":"2021-03-23T16:27:46","modified_gmt":"2021-03-23T10:57:46","slug":"samacheer-kalvi-10th-maths-chapter-1-ex-1-4","status":"publish","type":"post","link":"https:\/\/samacheerkalviguru.com\/samacheer-kalvi-10th-maths-chapter-1-ex-1-4\/","title":{"rendered":"Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Ex 1.4"},"content":{"rendered":"

You can Download Samacheer Kalvi 10th Maths Book Solutions<\/a> Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.<\/p>\n

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Ex 1.4<\/h2>\n

Exercise 1.4 Class 10 Maths Samacheer Question 1.<\/strong>
\nDetermine whether the graph given below represent functions. Give reason for your answers concerning each graph.
\n\"Exercise
\nSolution:
\n\"10th
\n(i) It is not a function. The graph meets the vertical line at more than one points.
\n(ii) It is a function as the curve meets the vertical line at only one point.
\n(iii) It is not a function as it meets the vertical line at more than one points.
\n(iv) It is a function as it meets the vertical line at only one point.<\/p>\n

10th Maths Exercise 1.4 Samacheer Kalvi Question 2.<\/strong>
\nLet f :A \u2192 B be a function defined by f(x) = \\(\\frac{x}{2}\\)\u00a0\u2013 1, Where A = {2, 4, 6, 10, 12},
\nB = {0, 1, 2, 4, 5, 9}. Represent f by
\n(i) set of ordered pairs;
\n(ii) a table;
\n(iii) an arrow diagram;
\n(iv) a graph
\nSolution:
\nf: A \u2192 B
\nA = {2, 4, 6, 10, 12}, B = {0, 1, 2, 4, 5, 9}
\n\"Ex
\n(i) Set of ordered pairs
\n= {(2, 0), (4, 1), (6, 2), (10, 4), (12, 5)}
\n(ii) a table
\n\"10th
\n(iii) an arrow diagram;
\n\"10th<\/p>\n

Ex 1.4 Class 10 Samacheer Question 3.<\/strong>
\nRepresent the function f = {(1, 2),(2, 2),(3, 2), (4,3), (5,4)} through
\n(i) an arrow diagram
\n(ii) a table form
\n(iii) a graph
\nSolution:
\nf = {(1, 2), (2, 2), (3, 2), (4, 3), (5, 4)}
\n(i) An arrow diagram.
\n\"10th<\/p>\n

10th Maths Exercise 1.4 Question 4.<\/strong>
\nShow that the function f : N \u2192 N defined by f{x) = 2x \u2013 1 is one \u2013 one but not onto.
\nSolution:
\nf: N \u2192 N
\nf(x) = 2x \u2013 1
\nN = {1, 2, 3, 4, 5,\u2026}
\nf(1) = 2(1) \u2013 1 = 1
\nf(2) = 2(2) \u2013 1 = 3
\nf(3) = 2(3) \u2013 1 = 5
\nf(4) = 2(4) \u2013 1 = 7
\nf(5) = 2(5) \u2013 1 = 9
\n\"10th
\nIn the figure, for different elements in x, there are different images in f(x).
\nHence f : N \u2192 N is a one-one function.
\nA function f: N \u2192 N is said to be onto function if the range of f is equal to the co-domain of f
\nRange = {1, 3, 5, 7, 9,\u2026}
\nCo-domain = {1, 2, 3,..}
\nBut here the range is not equal to co-domain. Therefore it is one-one but not onto function.<\/p>\n

10th Maths 1.4 Exercise Question 5.<\/strong>
\nShow that the function f: N \u2192 N defined by f (m) = m2<\/sup>\u00a0+ m + 3 is one \u2013 one function.
\nSolution:
\nf: N \u2192 N
\nf(m) = m2<\/sup>\u00a0+ m + 3
\nN = {1, 2, 3, 4, 5\u2026..}m \u2208 N
\nf{m) = m2<\/sup>\u00a0+ m + 3
\nf(1) = 12<\/sup>\u00a0+ 1 + 3 = 5
\nf(2) = 22<\/sup>\u00a0+ 2 + 3 = 9
\nf(3) = 32<\/sup>\u00a0+ 3 + 3 = 15
\nf(4) = 42<\/sup>\u00a0+ 4 + 3 = 23
\n\"Samacheer
\nIn the figure, for different elements in the (X) domain, there are different images in f(x). Hence f: N \u2192 N is a one to one but not onto function as the range of f is not equal to co-domain.
\nHence it is proved.<\/p>\n

10th Maths Ex 1.4 Question 6.<\/strong>
\nLet A = {1,2, 3, 4} and B = N .
\nLet f: A \u2192 B be defined by f(x) = x3<\/sup> then,
\n(i) find the range of f
\n(ii) identify the type of function
\nAnswer:
\nA = {1,2, 3,4}
\nB = {1,2, 3, 4, 5,….}
\nf(x) = x3<\/sup>
\nf(1) = 13<\/sup> = 1
\nf(2) = 23<\/sup> = 8
\nf(3) = 33<\/sup> = 27
\nf(4) = 43<\/sup> = 64
\n(i) Range = {1,8, 27, 64}
\n(ii) one -one and into function.<\/p>\n

10th Maths Exercise 1.4 Answers Question 7.<\/strong>
\nIn each of the following cases state whether the function is bijective or not. Justify your answer.
\n(i) f: R \u2192 R defined by f(x) = 2x + 1
\n(ii) f: R \u2192 R defined by f(x) = 3 \u2013 4x2<\/sup>
\nSolution:
\n(i) f : R \u2192 R
\nf(x) = 2x + 1
\nf(1) = 2(1) + 1 = 3
\nf(2) = 2(2) + 1 = 5
\nf(-1) = 2(-1) + 1 = -1
\nf(0) = 2(0) + 1 = 1
\nIt is a bijective function. Distinct elements of A have distinct images in B and every element in B has a pre-image in A.
\n(ii) f: R \u2192 R; f(x) = 3 \u2013 4x2<\/sup>
\nf(1) = 3 \u2013 4(12<\/sup>) = 3 \u2013 4 = -1
\nf(2) = 3 \u2013 4(22<\/sup>) = 3 \u2013 16 = -13
\nf(-1) = 3 \u2013 4(-1)2<\/sup>\u00a0= 3 \u2013 4 = -1
\nIt is not bijective function since it is not one-one<\/p>\n

Samacheer Kalvi 10th Maths Book Graph Solution Question 8.<\/strong>
\nLet A = {-1, 1} and B = {0, 2}. If the function f: A \u2192 B defined by f(x) = ax + b is an onto function? Find a and b.
\nSolution:
\nA= {-1, 1},B = {0, 2}
\nf: A \u2192 B, f(x) = ax + b
\nf(-1) = a(-1) + b = -a + b
\nf(1) = a(1) + b = a + b
\nSince f(x) is onto, f(-1) = 0
\n\u21d2 -a + b = 0 \u2026(1)
\n& f(1) = 2
\n\u21d2 a + b = 2 \u2026(2)
\n-a + b = 0
\n\"10th<\/p>\n

10th Maths Exercise 1.4 In Tamil Question 9.<\/strong>
\nIf the function f is defined by
\n\"Samacheer
\n(i) f(3)
\n(ii) f(0)
\n(iii) f(-1.5)
\n(iv) f(2) + f(-2)
\nSolution:
\n(i) f(3) \u21d2 f(x) = x + 2 \u21d2 3 + 2 = 5
\n(ii) f(0) \u21d2 2
\n(iii) f (- 1.5) = x \u2013 1
\n= -1.5 \u2013 1 = -2.5
\n(iv) f(2) + f(-2)
\nf(2) = 2 + 2 = 4\u00a0 \u00a0 \u00a0[\u2235 f(x) = x + 2]
\nf(-2) = -2 – 1 = -3\u00a0 \u00a0 [\u2235 f(x) = x – 1]
\nf(2) + f(-2) = 4 – 3 = 1<\/p>\n

Samacheer Kalvi 10th Maths Exercise 1.4 Question 10.<\/strong>
\nA function f: [-5,9] \u2192 R is defined as follows:
\n\"10th
\nSolution:
\nf : [-5, 9] \u2192 R
\n(i) f(-3) + f(2)
\nf(-3) = 6x + 1 = 6(-3) + 1 = -17
\nf(2) = 5 \u00d7 2 \u2013 1 = 5(22<\/sup>) \u2013 1 = 19
\n\u2234 f(-3) + f(2) = -17 + 19 = 2<\/p>\n

(ii) f(7) \u2013 f(1)
\nf(7) = 3x \u2013 4 = 3(7) \u2013 4 = 17
\nf(1) = 6x + 1 = 6(1) + 1 = 7
\nf(7) \u2013 f(1) = 17 \u2013 7 = 10<\/p>\n

(iii) 2f(4) + f(8)
\nf(4) = 5x2<\/sup>\u00a0\u2013 1 = 5 \u00d7 42<\/sup>\u00a0\u2013 1 = 79
\nf(8) = 3x \u2013 4 = 3 \u00d7 8 \u2013 4 = 20
\n\u2234 2f(4) + f(8) = 2 \u00d7 79 + 20 = 178
\n\"Samacheer<\/p>\n

10th Maths 1.4 Question 11<\/strong>
\nThe distance S an object travels under the influence of gravity in time t seconds is 1 2 given by S(t) = \\(\\frac { 1 }{ 3 } \\)gt2<\/sup> + at + b, where, (g is the acceleration due to gravity), a, b are constants. Check if the function S(t) is one-one.
\nAnswer:
\nS(t) = \\(\\frac { 1 }{ 2 } \\)gt2<\/sup> + at + b
\nLet the time be 1, 2, 3 …. n seconds
\nS(1) = \\(\\frac { 1 }{ 2 } \\)g(1)2<\/sup> + a(1) + b
\n= \\(\\frac { g }{ 2 } \\) + a + b
\nS(2) = \\(\\frac { 1 }{ 2 } \\) g(2)2<\/sup> + a(2) + b
\n= \\(\\frac { 4g }{ 2 } \\) + 2a + b
\n= 2g + 2a + b
\nS(3) = \\(\\frac { 1 }{ 2 } \\) g(3)2<\/sup> + a(3) + 6
\n= \\(\\frac { 9 }{ 2 } \\) g + 3a + b
\nFor every different value of t, there will be different distance.
\n\u2234 It is a one-one function.<\/p>\n

Samacheer Kalvi 10th Maths Book Graph Solutions Question 12.<\/strong>
\nThe function \u2018t\u2019 which maps temperature in Celsius (C) into temperature in Fahrenheit (F) is defined by t(C)= F where F = \\(\\frac{9}{5}\\) C + 32 . Find,
\n(i) t(0)
\n(ii) t(28)
\n(iii) t(-10)
\n(iv) the value of C when t(C) = 212
\n(v) the temperature when the Celsius value is equal to the Farenheit value.
\nSolution:
\n\"10th
\n\"10th<\/p>\n","protected":false},"excerpt":{"rendered":"

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations. Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Ex 1.4 Exercise 1.4 Class 10 Maths Samacheer Question 1. Determine whether the graph given below …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/50"}],"collection":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/comments?post=50"}],"version-history":[{"count":0,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/50\/revisions"}],"wp:attachment":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/media?parent=50"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/categories?post=50"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/tags?post=50"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}