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

Transcript

Ex 5.1, 14 Discuss the continuity of the function f, where f is defined by 𝑓(π‘₯)={β–ˆ(3, 𝑖𝑓 0≀π‘₯≀1@4, 𝑖𝑓 1<π‘₯<3@ 5, 𝑖𝑓 3≀π‘₯≀10)─ Since we need to find continuity at of the function We check continuity for different values of x When 0 ≀ x < 1 When x = 1 When 1 < x < 3 When x = 3 When 3 < x ≀ 10 Case 1 : When 0 ≀ x < 1 For 0 ≀ x < 1, f(x) = 3 Since this constant It is continuous ∴ f(x) is continuous for 0 ≀ x < 1 Case 2 : When x = 1 f(x) is continuous at π‘₯ =1 if L.H.L = R.H.L = 𝑓(1) if lim┬(xβ†’1^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’1^+ ) " " 𝑓(π‘₯)= 𝑓(1) Since L.H.L β‰  R.H.L f(x) is not continuous at x=1 Since there are two different functions on the left & right of 1, we take LHL & RHL . LHL at x β†’ 1 lim┬(xβ†’1^βˆ’ ) f(x) = lim┬(hβ†’0) f(1 βˆ’ h) = lim┬(hβ†’0) 3 = 3 RHL at x β†’ 1 lim┬(xβ†’1^+ ) f(x) = lim┬(hβ†’0) f(1 + h) = lim┬(hβ†’0) 4 = 4 Case 3 : When 1 < x < 3 For 1 < x < 3 f(x) = 4 Since this constant It is continuous ∴ f(x) is continuous for 1 < x < 3 Case 4 : When x = 3 f(x) is continuous at π‘₯ =3 if L.H.L = R.H.L = 𝑓(3) if lim┬(xβ†’3^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’3^+ ) " " 𝑓(π‘₯)= 𝑓(3) Since L.H.L β‰  R.H.L f(x) is not continuous at x=3 LHL at x β†’ 3 lim┬(xβ†’3^βˆ’ ) f(x) = lim┬(hβ†’0) f(3 βˆ’ h) = lim┬(hβ†’0) 4 = 4 RHL at x β†’ 3 lim┬(xβ†’3^+ ) f(x) = lim┬(hβ†’0) f(3 + h) = lim┬(hβ†’0) 5 = 5 Since there are two different functions on the left & right of 3, we take LHL & RHL . Case 5: When 3 < x ≀ 10 For 3 < x ≀ 10, f(x) = 5 Since this constant It is continuous ∴ f(x) is continuous for 3 < x ≀ 10 Hence points of discontinuity are x = 1 & x = 3 Thus, f is continuous for all x ∈ R βˆ’ {𝟏,πŸ‘}

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.