# Ex 5.1, 11

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 5.1, 11 Find all points of discontinuity of f, where f is defined by ๐ ๐ฅ๏ทฏ= ๐ฅ3โ3, ๐๐ ๐ฅโค2๏ทฎ&๐ฅ2+1 , ๐๐ ๐ฅ>2๏ทฏ๏ทฏ We have, ๐ ๐ฅ๏ทฏ= ๐ฅ3โ3, ๐๐ ๐ฅโค2๏ทฎ&๐ฅ2+1 , ๐๐ ๐ฅ>2๏ทฏ๏ทฏ Case 1: At x = 2 f is continuous at x = 2 if L.H.L = R.H.L = ๐ 2๏ทฏ i.e. lim๏ทฎxโ 2๏ทฎโ๏ทฏ๏ทฏ ๐ ๐ฅ๏ทฏ = lim๏ทฎxโ 2๏ทฎ+๏ทฏ๏ทฏ ๐ ๐ฅ๏ทฏ = ๐ 2๏ทฏ ๐ ๐ฅ๏ทฏ = ๐ฅ3โ3 ๐ 2๏ทฏ = 2๏ทฏ3โ3 = 8 โ 3 = 5 Thus LHL = RHL = f (2) โ f is continuous at ๐=๐ Case 2 Let x = c , where c < 2 ๐ ๐ฅ๏ทฏ=๐ฅ3โ3 f is continuous at x = c if lim๏ทฎxโ๐๏ทฏ ๐ ๐ฅ๏ทฏ=๐(๐) Thus, lim๏ทฎxโ๐๏ทฏ ๐ ๐ฅ๏ทฏ=๐(๐) โ f is continuous for ๐ฅ =๐ less than 2. โ f is at continuous for all real numbers less than 2. Case 3 Let x = c (where c > 2) ๐ ๐ฅ๏ทฏ=๐ฅ2+1 f is continuous at x = c if lim๏ทฎxโ๐๏ทฏ ๐ ๐ฅ๏ทฏ=๐(๐) Thus lim๏ทฎxโ๐๏ทฏ ๐ ๐ฅ๏ทฏ=๐(๐) โ f is continuous at ๐ฅ =๐ ( c is greater than 2) โ f is continuous at all real numbers greater than 2. Hence, there is no point of discontinuity โ f is continuous at all real point. Thus, f is continuous for all ๐โ๐.

Ex 5.1 ,1

Ex 5.1 ,2

Ex 5.1 ,3

Ex 5.1 ,4

Ex 5.1 ,5

Ex 5.1 ,6

Ex 5.1 ,7

Ex 5.1 ,8

Ex 5.1, 9 Important

Ex 5.1, 10

Ex 5.1, 11 You are here

Ex 5.1, 12

Ex 5.1, 13 Important

Ex 5.1, 14

Ex 5.1, 15

Ex 5.1, 16 Important

Ex 5.1, 17

Ex 5.1, 18 Important

Ex 5.1, 19

Ex 5.1, 20

Ex 5.1, 21

Ex 5.1, 22

Ex 5.1, 23

Ex 5.1, 24

Ex 5.1, 25

Ex 5.1, 26

Ex 5.1, 27

Ex 5.1, 28 Important

Ex 5.1, 29

Ex 5.1, 30 Important

Ex 5.1, 31

Ex 5.1, 32

Ex 5.1, 33

Ex 5.1, 34 Important

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.