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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) lim┬(xβ†’0) 𝑓(π‘₯) "= " 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 ∴ lim┬(xβ†’0) 𝑓(π‘₯) = 𝑓(0) 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 ∴ lim┬(xβ†’βˆ’3) 𝑓(π‘₯) = 𝑓(βˆ’3) Hence, f is continuous at 𝒙 =βˆ’3 lim┬(xβ†’3) 𝑓(π‘₯) "= " lim┬(xβ†’3) " "(5π‘₯βˆ’3) Putting x = βˆ’3 = 5(βˆ’3) βˆ’ 3 = βˆ’18 𝑓(βˆ’3) = 5(βˆ’3) βˆ’ 3 = βˆ’15 βˆ’ 3 = βˆ’18 Since, L.H.S = R.H.S ∴ lim┬(xβ†’βˆ’3) 𝑓(π‘₯) = 𝑓(βˆ’3) Hence, f is continuous at 𝒙 =βˆ’3 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 ∴ ( lim)┬(xβ†’5) 𝑓(π‘₯)= 𝑓(5) (π‘™π‘–π‘š)┬(π‘₯β†’5) 𝑓(π‘₯) "= " (π‘™π‘–π‘š)┬(π‘₯β†’5) " "(5π‘₯βˆ’3) Putting x = 5 = 5(5) βˆ’ 3 = 22 𝑓(5) = 5(5) βˆ’ 3 = 25 βˆ’ 3 = 22 Hence, f is continuous at 𝒙 =πŸ“ Thus the function is continuous at 𝒙 =𝟎, at 𝒙 =βˆ’πŸ‘ & at 𝒙 =πŸ“

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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.