Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12

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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise

Transcript

Ex 5.1, 1 Prove that the function 𝑓 (π‘₯) = 5π‘₯ – 3 is continuous at π‘₯ = 0, at π‘₯ = –3 and at π‘₯ = 5 Given 𝑓(π‘₯)= 5π‘₯ –3 At 𝒙=𝟎 f(x) is continuous at x = 0 if (π₯𝐒𝐦)┬(π±β†’πŸŽ) 𝒇(𝒙) = 𝒇(𝟎) (π₯𝐒𝐦)┬(π±β†’πŸŽ) 𝒇(𝒙) "= " lim┬(xβ†’0) " "(5π‘₯βˆ’3) Putting x = 0 = 5(0) βˆ’ 3 = βˆ’3 𝒇(𝟎) = 5(0) βˆ’ 3 = 0 βˆ’ 3 = βˆ’3 Since L.H.S = R.H.S Hence, f is continuous at 𝒙 = 𝟎 At x = βˆ’3 f(x) is continuous at x = βˆ’3 if ( lim)┬(xβ†’βˆ’3) 𝑓(π‘₯)= 𝑓(βˆ’3) Since, L.H.S = R.H.S Hence, f is continuous at 𝒙 =βˆ’3 (π₯𝐒𝐦)┬(π±β†’πŸ‘) 𝒇(𝒙) "= " lim┬(xβ†’3) " "(5π‘₯βˆ’3) Putting x = βˆ’3 = 5(βˆ’3) βˆ’ 3 = βˆ’18 𝒇(βˆ’πŸ‘) = 5(βˆ’3) βˆ’ 3 = βˆ’15 βˆ’ 3 = βˆ’18 At 𝒙 =πŸ“ f(x) is continuous at x = 5 if ( lim)┬(xβ†’5) 𝑓(π‘₯)= 𝑓(5) Since, L.H.S = R.H.S Hence, f is continuous at x = 5 Thus, the function is continuous at x = 0, x = βˆ’3, x = 5 (π’π’Šπ’Ž)┬(π’™β†’πŸ“) 𝒇(𝒙) "= " (π‘™π‘–π‘š)┬(π‘₯β†’5) " "(5π‘₯βˆ’3) Putting x = 5 = 5(5) βˆ’ 3 = 22 𝒇(πŸ“) = 5(5) βˆ’ 3 = 25 βˆ’ 3 = 22

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.