{"id":8395,"date":"2024-09-24T09:43:26","date_gmt":"2024-09-24T04:13:26","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=8395"},"modified":"2024-09-24T09:43:26","modified_gmt":"2024-09-24T04:13:26","slug":"ncert-solutions-class-11-maths-chapter-2-relations-and-functions-ex-2-3","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-11-maths-chapter-2-relations-and-functions-ex-2-3\/","title":{"rendered":"NCERT Solutions Class 11 Maths Chapter 2 Relations and Functions – Ex 2.3"},"content":{"rendered":"

NCERT Solutions Class 11 Maths\u00a0<\/span><\/strong><\/span><\/h2>\n

The NCERT Solutions in English Language for Class 11 Mathematics Chapter – 2 Relations and Functions <\/strong>Exercise 2.3 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n

Chapter 2 (Relations and Functions)\u00a0<\/span><\/strong><\/span><\/h2>\n\n\n\n
Exercise – 2.3<\/span><\/strong><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

1. Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.
\n<\/strong>(i) <\/strong>{(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}
\n(ii) <\/strong>{(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
\n(iii) <\/strong>{(1, 3), (1, 5), (2, 5)} <\/span><\/p>\n

Solution –\u00a0 <\/strong><\/span><\/p>\n

(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}
\n<\/strong>Here, each element in domain is having unique\/distinct image. So, the given relation is a function.<\/span>
\nDomain = {2, 5, 8, 11, 14, 17}<\/span>
\nRange of the function = {1}<\/span><\/p>\n

(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
\n<\/strong>Here, each element in domain is having unique\/distinct image. So, the given relation is a function.<\/span>
\nDomain = {2, 5, 8, 11, 14, 17}<\/span>
\nRange of the function = {1}<\/span><\/p>\n

(iii) {(1, 3), (1, 5), (2, 5)}
\n<\/strong>This relation is not a function since an element 1 corresponds to two elements\/images i.e, 3 and 5.<\/span>
\nHence, this relation is not a function.<\/span><\/p>\n

2. Find the domain and range of the following real function:
\n<\/strong>(i)\u00a0f<\/em>(x<\/em>) = \u2013|x<\/em>|
\n<\/strong>Solution –\u00a0 \u00a0<\/strong><\/span><\/p>\n

(ii) f(x) = \u221a(9 \u2013 x2<\/sup>)
\n<\/strong>Solution –\u00a0 \u00a0<\/strong><\/span><\/p>\n

(i)\u00a0<\/strong>Given, f<\/em>(x<\/em>) = \u2013|x<\/em>|,\u00a0x<\/em> \u2208 R
\nWe know that, \u00a0|x| =<\/span><\/p>\n

\n\n\n\n\n
x<\/span><\/td>\nif x >= 0<\/span><\/td>\n<\/tr>\n
-x<\/span><\/td>\nif x < 0<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/figure>\n

Here f(x) = -x =<\/span><\/p>\n

\n\n\n\n\n
-x<\/span><\/td>\nx >= 0<\/span><\/td>\n<\/tr>\n
x<\/span><\/td>\nx < 0<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/figure>\n

As f(x) is defined for x \u2208 R, the domain of f is R. <\/span>
\nIt is also seen that the range of f(x) = \u2013|x| is all real numbers except positive real numbers. <\/span>
\nTherefore, the range of f is given by (\u2013\u221e, 0].<\/span><\/p>\n

(ii)\u00a0f(x) = \u221a(9 \u2013 x2<\/sup>)
\n<\/strong>As \u221a(9 \u2013 x2<\/sup>) is defined for all real numbers that are greater than or equal to \u20133 and less than or equal to 3, for 9 \u2013 x2<\/sup> \u2265 0. <\/span>
\n|x| <=3 <\/span>
\nSo, the domain of f(x) is {x: \u20133 \u2264 x \u2264 3} or Domain of f = [\u20133, 3]. <\/span>
\nFor any value of x in the range [\u20133, 3], the value of f(x) will lie between 0 and 3. <\/span>
\nTherefore, the range of f(x) is {x: 0 \u2264 x \u2264 3} or we can say Range of f = [0, 3].<\/span><\/p>\n

3. A function\u00a0f<\/em>\u00a0is defined by\u00a0f<\/em>(x<\/em>) = 2x<\/em> \u2013 5. Write down the values of
\n<\/strong>(i)\u00a0f<\/em>(0), (ii)\u00a0f<\/em>(7), (iii)\u00a0f<\/em>(\u20133) <\/strong><\/span><\/p>\n

Solution –\u00a0 <\/strong>Given, Function,\u00a0f<\/em>(x<\/em>) = 2x<\/em> \u2013 5
\nTherefore, <\/span><\/p>\n

(i)\u00a0<\/strong>f<\/em>(0) = 2 \u00d7 0 \u2013 5 = 0 \u2013 5 = \u20135<\/span><\/p>\n

(ii)<\/strong>\u00a0f<\/em>(7) = 2 \u00d7 7 \u2013 5 = 14 \u2013 5 = 9<\/span><\/p>\n

(iii)\u00a0<\/strong>f<\/em>(\u20133) = 2 \u00d7 (\u20133) \u2013 5 = \u2013 6 \u2013 5 = \u201311 <\/span><\/p>\n

4. The function \u2018t<\/em>\u2019, which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by\u00a0<\/strong>t(C) = 9C\/5 + 32.
\n<\/strong>Find (i) t<\/em>\u00a0(0) (ii)\u00a0t<\/em>\u00a0(28) (iii)\u00a0t<\/em>\u00a0(\u201310) (iv) The value of C, when\u00a0t<\/em>(C) = 212 <\/strong><\/span><\/p>\n

Solution – <\/strong>Here in ques , it is given that : <\/span>
\nt(C) = 9C\/5 + 32\u00a0 <\/span>
\nSo, <\/span>
\n(i)<\/strong> t(0) = 9(0) \/ 5 + 32\u00a0 <\/span>
\n= 0 + 32 <\/span>
\n= 32<\/span><\/p>\n

(ii)<\/strong> t(28) = 9(28) \/ 5 + 32\u00a0 <\/span>
\nTaking LCM and solving, <\/span>
\n= ( 252 +160 ) \/ 5\u00a0 <\/span>
\n= 412 \/ 5<\/span><\/p>\n

(iii)<\/strong> t(-10) = 9(-10) \/ 5 + 32 <\/span>
\n= -18 + 32 <\/span>
\n= 14<\/span><\/p>\n

(iv)<\/strong> Here , in this ques we have to find the value of C. <\/span>
\nGiven that , t(C) = 212, <\/span>
\n9C \/ 5 + 32 = 212<\/span>
\n9C \/ 5 = 180<\/span>
\n9C = 180 \u00d7 5<\/span>
\nC = (180 \u00d7 5)\/9 <\/span>
\nC = 100 <\/span>
\nThe value of C is 100. <\/span><\/p>\n

5. Find the range of each of the following functions:
\n<\/strong>(i)\u00a0<\/strong>f<\/em>(x<\/em>) = 2 \u2013 3x<\/em>,\u00a0x<\/em>\u00a0\u2208\u00a0R,\u00a0x<\/em> > 0\u00a0
\n<\/strong>(ii)\u00a0<\/strong>f<\/em>(x<\/em>) =\u00a0x<\/em>2<\/sup>\u00a0+ 2,\u00a0x<\/em> is a real number\u00a0
\n<\/strong>(iii)<\/strong>\u00a0f<\/em>(x<\/em>) =\u00a0x<\/em>,\u00a0x<\/em> is a real number\u00a0 <\/span><\/p>\n

Solution –\u00a0 <\/strong><\/span><\/p>\n

(i) <\/strong>Given f (x) = 2 \u2013 3x, x \u2208 R, x > 0 <\/span>
\nThe values of f (x) for various values of\u00a0real numbers\u00a0x > 0 can be written in the tabular form as:<\/span><\/p>\n\n\n\n\n
x<\/span><\/td>\n0.01<\/span><\/td>\n0.1<\/span><\/td>\n0.9<\/span><\/td>\n1<\/span><\/td>\n2<\/span><\/td>\n2.5<\/span><\/td>\n4<\/span><\/td>\n5<\/span><\/td>\n….<\/span><\/td>\n<\/tr>\n
f ( x)<\/span><\/td>\n1.97<\/span><\/td>\n1.7<\/span><\/td>\n– 0.7<\/span><\/td>\n– 1<\/span><\/td>\n– 4<\/span><\/td>\n– 5.5<\/span><\/td>\n– 10<\/span><\/td>\n– 13<\/span><\/td>\n….<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Thus, it can be clearly observed that the range of f is the set of all real numbers less than 2. that is\u00a0range of <\/span>
\nf = (- \u221e, 2)<\/span><\/p>\n

(ii)<\/strong> Given, f<\/em>(x<\/em>) =\u00a0x<\/em>2<\/sup>\u00a0+ 2,\u00a0x<\/em> is a real number
\nThe values of f(x) for various values of real numbers x can be written in the tabular form as:<\/span><\/p>\n\n\n\n\n
x<\/span><\/td>\n0<\/span><\/td>\n\u00b1 0.3<\/span><\/td>\n\u00b1 0.8<\/span><\/td>\n\u00b1 1<\/span><\/td>\n\u00b1 2<\/span><\/td>\n\u00b1 3<\/span><\/td>\n….<\/span><\/td>\n<\/tr>\n
f ( x)<\/span><\/td>\n2<\/span><\/td>\n2.09<\/span><\/td>\n2.64<\/span><\/td>\n3<\/span><\/td>\n6<\/span><\/td>\n11<\/span><\/td>\n….<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Thus, it can be clearly observed that the range of f is the set of all real numbers greater than 2. that is\u00a0range of <\/span>
\nf = [2, \u221e]<\/span><\/p>\n

(iii)<\/strong> Given, f<\/em>(x<\/em>) =\u00a0x, x<\/em> is a real number
\nIt is clear that the range of f is the set of all real numbers. Therefore, the Range of f = R.\u00a0<\/span><\/p>\n

Go Back To Chapters<\/i><\/b><\/a><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"

NCERT Solutions Class 11 Maths\u00a0 The NCERT Solutions in English Language for Class 11 Mathematics Chapter – 2 Relations and Functions Exercise 2.3 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter 2 (Relations and Functions)\u00a0 Exercise – 2.3 1. Which of the following relations are functions? Give […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1674],"tags":[1675,1679,1687,1690,1678,1680,1681],"class_list":["post-8395","post","type-post","status-publish","format-standard","hentry","category-class-11-maths","tag-class-11-ncert-mathematics-solutions","tag-class-11-ncert-solutions","tag-ncert-class-11-mathematics-chapter-2-relations-and-functions-solutions","tag-ncert-class-11-mathematics-exercise-2-3-solutions","tag-ncert-class-11-mathematics-solutions","tag-ncert-solutions-class-11-mathematics","tag-ncert-solutions-class-11-maths"],"yoast_head":"\nNCERT Class 11 Maths Solutions - Exercise 2.3 | TheExamPillar NCERT<\/title>\n<meta name=\"description\" content=\"NCERT Solutions Class 11 Maths\u00a0The NCERT Solutions in English Language for Class 11 Mathematics Chapter - 2 Relations and Functions Exercise 2.3 has been provided here to help the students in solving the questions from this exercise.\u00a0Chapter 2 (Relations and Functions)\u00a0\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-11-maths-chapter-2-relations-and-functions-ex-2-3\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions Class 11 Maths Chapter 2 Relations and Functions - 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