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

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

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

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.