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Ex 5.1, 1 Class 12 - Prove f(x) = 5x - 3 is continuous at x = 0, -3, 5

Ex 5.1 ,1 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.1 ,1 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Ex 5.1 ,1 - Chapter 5 Class 12 Continuity and Differentiability - Part 4

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.