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

Transcript

Ex 5.1, 15 Discuss the continuity of the function f, where f is defined by 𝑓(π‘₯)={ β–ˆ(2π‘₯, 𝑖𝑓 π‘₯<0@ 0, 𝑖𝑓 0≀π‘₯≀1@ 4π‘₯, 𝑖𝑓 π‘₯>1 )─ 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 0 < x < 1 When x = 1 When x > 1 Case 1 : When x < 0 For x < 0, f(x) = 2x Since this a polynomial It is continuous ∴ f(x) is continuous for x < 0 Case 2 : When x = 0 f(x) is continuous at π‘₯ =0 if L.H.L = R.H.L = 𝑓(0) if lim┬(xβ†’0^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’0^+ ) " " 𝑓(π‘₯)= 𝑓(0) Since there are two different functions on the left & right of 0, we take LHL & RHL . 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) 2(βˆ’β„Ž) = 2 Γ— 0 = 0 RHL at x β†’ 0 lim┬(xβ†’1^+ ) f(x) = lim┬(hβ†’0) f(0 + h) = lim┬(hβ†’0) f(h) = lim┬(hβ†’0) 0 = 0 & 𝑓(0) = 0 Hence, L.H.L = R.H.L = 𝑓(0) ∴ f is continuous at x=0 Case 3 : When 0 < x < 1 For 0 < x < 1 f(x) = 0 Since this constant It is continuous ∴ f(x) is continuous for 0 < 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 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) 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 Case 5: When x > 1 For x > 1, f(x) = 4x Since this a polynomial It is continuous ∴ f(x) is continuous for x > 1 Hence points of discontinuity are x = 1 Thus, f is continuous for all x ∈ R βˆ’ {𝟏}

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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.