Ex 5.1, 20 - Ex 5.1

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Ex 5.1, 20 Is the function defined by f (x) = 𝑥^2 – sin x + 5 continuous at x = π? f (x) = 𝑥^2 – sin x + 5 Let 𝑝(𝑥)=𝑥^2 , 𝑞(𝑥)="sin x " & 𝑟(𝑥) = 5 𝒑(𝒙) = 𝑥^2 is continuous as it is a polynomial 𝒒(𝒙)" = sin x" is continuous at all real numbers 𝒓(𝒙) = 5 is continuous as it is a constant function By Algebra of continuous functions, If 𝑝(𝑥)" ", 𝑞(𝑥) & 𝑟(𝑥) all are continuous at all real numbers then 𝒇(𝒙)= 𝒑(𝒙)−𝒒(𝒙)+𝒓(𝒙) is continuous at "all real numbers" ∴ 𝑓(𝑥) = 𝑥^2 " – sin x + 5" is continuous at all real numbers. Thus, 𝑓(𝑥) is continuous at 𝒙=𝝅 Ex 5.1, 20 (Method 1) Is the function defined by f (x) = 𝑥﷮2﷯ – sin x + 5 continuous at x = π? f (x) = 𝑥﷮2﷯ – sin x + 5 We need to check continuity at 𝑥=𝜋 We know that A function f is continuous at 𝑥=𝜋 i.e. lim﷮x→𝜋﷯ 𝑓 𝑥﷯=𝑓 𝜋﷯ Thus lim﷮x→𝜋﷯ 𝑓 𝑥﷯=𝑓 𝑐﷯ ⇒ f is continuous at 𝒙=𝝅 Ex 5.1, 20 (Method 2) Is the function defined by f (x) = 𝑥﷮2﷯ – sin x + 5 continuous at x = π? f (x) = 𝑥﷮2﷯ – sin x + 5 Let 𝑝 𝑥﷯= 𝑥﷮2﷯ , 𝑞 𝑥﷯=sin x & 𝑟 𝑥﷯ = 5 𝑝 𝑥﷯ = 𝑥﷮2﷯ is continuous as it is a polynomial 𝑞 𝑥﷯=sin x is continuous at all real numbers 𝑟 𝑥﷯ = 5 is continuous as it is a constant function By Algebra of continuous function, If 𝑝 𝑥﷯ , 𝑞 𝑥﷯ & 𝑟 𝑥﷯ all are continuous at all real numbers then 𝑓 𝑥﷯= 𝑝 𝑥﷯−𝑞 𝑥﷯+𝑟 𝑥﷯ is continuous at all real numbers ∴ 𝑓 𝑥﷯ = 𝑥﷮2﷯ – sin x + 5 is continuous at all real numbers. ⇒ f is continuous at 𝒙=𝝅

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.