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Ex 5.1

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Last updated at March 22, 2023 by Teachoo

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) = 𝑥2 – sin x + 5 continuous at x = π? f (x) = 𝑥2 – sin x + 5 We need to check continuity at 𝑥=𝜋 We know that A function f is continuous at 𝑥=𝜋 i.e. limx→𝜋 𝑓 𝑥=𝑓 𝜋 Thus limx→𝜋 𝑓 𝑥=𝑓 𝑐 ⇒ f is continuous at 𝒙=𝝅 Ex 5.1, 20 (Method 2) Is the function defined by f (x) = 𝑥2 – sin x + 5 continuous at x = π? f (x) = 𝑥2 – sin x + 5 Let 𝑝 𝑥= 𝑥2 , 𝑞 𝑥=sin x & 𝑟 𝑥 = 5 𝑝 𝑥 = 𝑥2 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 = 𝑥2 – sin x + 5 is continuous at all real numbers. ⇒ f is continuous at 𝒙=𝝅