Ex 5.1, 9 - Find all points of discontinuity - Chapter 5 Class 12

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Ex 5.1, 9 - Chapter 5 Class 12 Continuity and Differentiability - Part 2

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Ex 5.1, 9 - Chapter 5 Class 12 Continuity and Differentiability - Part 3 Ex 5.1, 9 - Chapter 5 Class 12 Continuity and Differentiability - Part 4

  1. Class 12
  2. Important Questions for exams Class 12

Transcript

Ex 5.1, 9 Find all points of discontinuity of f, where f is defined by 𝑓(π‘₯)={β–ˆ(π‘₯/|π‘₯| , 𝑖𝑓 π‘₯<0@&βˆ’1 , 𝑖𝑓 π‘₯β‰₯ 0)─ Since we need to find continuity at of the function We check continuity for different values of x When x = 0 When x > 0 When x < 0 Case 1 : When x = 0 f(x) is continuous at π‘₯ =0 if L.H.L = R.H.L = 𝑓(0) Since there are two different functions on the left & right of 0, we take LHL & RHL . if lim┬(xβ†’0^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’0^+ ) " " 𝑓(π‘₯)= 𝑓(0) And, f(0) = βˆ’1 LHL at x β†’ 0 lim┬(xβ†’0^βˆ’ ) f(x) = lim┬(hβ†’0) f(0 βˆ’ h) = lim┬(hβ†’0) f(βˆ’h) = lim┬(hβ†’0) (βˆ’β„Ž)/|βˆ’β„Ž| = lim┬(hβ†’0) (βˆ’β„Ž)/β„Ž = lim┬(hβ†’0) βˆ’1 = βˆ’1 RHL at x β†’ 0 lim┬(xβ†’0^+ ) f(x) = lim┬(hβ†’0) f(0 + h) = lim┬(hβ†’0) f(h) = lim┬(hβ†’0) βˆ’1 = βˆ’1 Hence, L.H.L = R.H.L = 𝑓(0) ∴ f is continuous at x = βˆ’3 Case 2 : When x < 0 For x < 0, f(x) = π‘₯/(|π‘₯|) f(x) = π‘₯/((βˆ’π‘₯)) f(x) = βˆ’1 Since this constant It is continuous ∴ f(x) is continuous for x < 0 (As x < 0, x is negative) Case 3 : When x > 0 For x > 0, f(x) = βˆ’1 Since this constant It is continuous ∴ f(x) is continuous for x > 0 ∴ f is continuous for all real numbers Thus, f is continuous for π’™βˆˆ R

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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.