Ex 5.1 ,3 - Chapter 5 Class 12 Continuity and Differentiability - Part 8

Advertisement

Ex 5.1 ,3 - Chapter 5 Class 12 Continuity and Differentiability - Part 9

Advertisement

Ex 5.1 ,3 - Chapter 5 Class 12 Continuity and Differentiability - Part 10 Ex 5.1 ,3 - Chapter 5 Class 12 Continuity and Differentiability - Part 11

  1. Chapter 5 Class 12 Continuity and Differentiability (Term 1)
  2. Serial order wise

Transcript

Ex 5.1, 3 Examine the following functions for continuity. (d) f (x) = |x – 5| f(x) = |π‘₯βˆ’5| = {β–ˆ((π‘₯βˆ’5), π‘₯βˆ’5β‰₯0@βˆ’(π‘₯βˆ’5), π‘₯βˆ’5<0)─ = {β–ˆ((π‘₯βˆ’5), π‘₯β‰₯5@βˆ’(π‘₯βˆ’5), π‘₯<5)─ Since we need to find continuity at of the function We check continuity for different values of x When x = 5 When x < 5 When x > 5 Case 1 : When x = 5 f(x) is continuous at π‘₯ = 5 if L.H.L = R.H.L = 𝑓(5) if lim┬(xβ†’5^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’5^+ ) " " 𝑓(π‘₯)= 𝑓(5) Since there are two different functions on the left & right of 5, we take LHL & RHL . LHL at x β†’ 5 lim┬(xβ†’5^βˆ’ ) f(x) = lim┬(hβ†’0) f(5 βˆ’ h) = lim┬(hβ†’0) |(5βˆ’β„Ž)βˆ’5| = lim┬(hβ†’0) |βˆ’β„Ž| = lim┬(hβ†’0) β„Ž = 0 RHL at x β†’ 5 lim┬(xβ†’5^+ ) f(x) = lim┬(hβ†’0) f(5 + h) = lim┬(hβ†’0) |(5+β„Ž)βˆ’5| = lim┬(hβ†’0) |β„Ž| = lim┬(hβ†’0) β„Ž = 0 & 𝒇(πŸ“) = |π‘₯βˆ’5| = |5βˆ’5| = 0 Hence, L.H.L = R.H.L = 𝑓(5) ∴ f is continuous at x = 5 Case 2 : When x < 5 For x < 5, f(x) = βˆ’ (x βˆ’ 5) Since this a polynomial It is continuous ∴ f(x) is continuous for x < 5 Case 3 : When x > 5 For x > 5, f(x) = (x βˆ’ 5) Since this a polynomial It is continuous ∴ f(x) is continuous for x > 5 Hence, 𝑓(π‘₯)= |π‘₯βˆ’5| is continuous at all points. i.e. f is continuous at 𝒙 ∈ R.

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.