# 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.

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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.