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

Transcript

Ex 5.1, 16 Discuss the continuity of the function f, where f is defined by 𝑓(π‘₯)={β–ˆ(βˆ’&2, 𝑖𝑓 π‘₯β‰€βˆ’1@2&π‘₯, 𝑖𝑓 βˆ’1<π‘₯≀1@2, 𝑖𝑓 π‘₯>1 )─ Since we need to find continuity at of the function We check continuity for different values of x When x < βˆ’1 When x = βˆ’1 When βˆ’1 < x < 1 When x = 1 When x > 1 Case 1 : When x < βˆ’1 For x < βˆ’1, f(x) = βˆ’2 Since this constant It is continuous ∴ f(x) is continuous for x < βˆ’1 Case 2 : When x = βˆ’1 f(x) is continuous at π‘₯ =βˆ’1 if L.H.L = R.H.L = 𝑓(βˆ’1) if lim┬(xβ†’γ€–βˆ’1γ€—^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’γ€–βˆ’1γ€—^+ ) " " 𝑓(π‘₯)= 𝑓(βˆ’1) Since there are two different functions on the left & right of βˆ’1, we take LHL & RHL . LHL at x β†’ βˆ’1 lim┬(xβ†’γ€–βˆ’1γ€—^βˆ’ ) f(x) = lim┬(hβ†’0) f(βˆ’1 βˆ’ h) = lim┬(hβ†’0) 2(βˆ’β„Ž) = 2 Γ— 0 = 0 RHL at x β†’ 0 lim┬(xβ†’γ€–βˆ’1γ€—^+ ) f(x) = lim┬(hβ†’0) f(βˆ’1 + h) = lim┬(hβ†’0) 0 = 0 & 𝑓(βˆ’1) = βˆ’2 Hence, L.H.L = R.H.L = 𝑓(βˆ’1) ∴ f is continuous at x=βˆ’1 Case 3 : When βˆ’1 < x < 1 For βˆ’1 < x < 1 f(x) = 2x Since this a polynomial It is continuous ∴ f(x) is continuous for βˆ’1 < x < 1 Case 4 : 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 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) 0 = 0 RHL at x β†’ 1 lim┬(xβ†’1^+ ) f(x) = lim┬(hβ†’0) f(1 + h) = lim┬(hβ†’0) 4(1+β„Ž) = 4(1 + 0) = 4 & 𝑓(1) = 2(1) = 2 Hence, L.H.L = R.H.L = 𝑓(1) ∴ f is continuous at x=1 Case 5: When x > 1 For x > 1, f(x) = 2 Since this constant It is continuous ∴ f(x) is continuous for x > 1 Thus, f(x) is continuous for all real numbers, i.e. f is continuous for all x ∈ R lim┬(xβ†’γ€–βˆ’1γ€—^βˆ’ ) 𝑓(π‘₯) = lim┬(xβ†’βˆ’1) (βˆ’2) = βˆ’2 lim┬(xβ†’γ€–βˆ’1γ€—^+ ) 𝑓(π‘₯) = lim┬(xβ†’βˆ’1) 2π‘₯ Putting x = βˆ’ 1 = 2(βˆ’1) = βˆ’ 2 And 𝑓(βˆ’2) = βˆ’2 Thus L.H.L = R.H.L = 𝑓(βˆ’2) = βˆ’2 Hence 𝒇(𝒙) is continuous at x = βˆ’πŸ Case 1 At x = βˆ’1 A function is continuous at x = βˆ’ 1 if A function is continuous at x = 1 if if L.H.L = R.H.L = 𝑓(1) i.e. lim┬(xβ†’1^βˆ’ ) 𝑓(π‘₯) = lim┬(xβ†’1^+ ) 𝑓(π‘₯) = 𝑓(1) & 𝑓(1) = 2(1) = 2 Thus L.H.L = R.H.L = 𝑓(βˆ’2) Hence 𝒇(𝒙) is continuous at x = 1 lim┬(xβ†’1^βˆ’ ) 𝑓(π‘₯) = lim┬(xβ†’1^βˆ’ ) 2π‘₯ Putting x = 1 = 2(1) = 2 lim┬(xβ†’1^+ ) 𝑓(π‘₯) = lim┬(xβ†’1^+ ) 2 = 2 Case 3 For π‘₯\<βˆ’1 𝑓(π‘₯) = βˆ’2 Thus, 𝑓(π‘₯) is a constant function . & Every constant function is continuous for all real number. Hence 𝒇(𝒙) is continuous at 𝒙\<βˆ’πŸ Case 4 For π‘₯>1 𝑓(π‘₯) = 2 Thus, 𝑓(π‘₯) is a constant function . & Every constant function is continuous for all real number. Hence 𝒇(𝒙) is continuous at 𝒙>𝟏 Case 5 For βˆ’1≀π‘₯≀1 𝑓(π‘₯) = 2π‘₯ So, f(x) is a polynomial & Every polynomial is continuous. β‡’ 𝒇(𝒙) is continuous at βˆ’πŸ<π’™β‰€πŸ Thus, f(x) is continuous for all real numbers, i.e. f is continuous for all x ∈ 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 9 years. He provides courses for Maths and Science at Teachoo.