Ex 5.1, 20 - Chapter 5 Class 12 Continuity and Differentiability (Term 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