Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12

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

Transcript

Ex 5.1, 6 Find all points of discontinuity of f, where f is defined by 𝑓(π‘₯)={β–ˆ(2π‘₯+3, 𝑖𝑓 π‘₯≀2@&2π‘₯βˆ’3, 𝑖𝑓 π‘₯>2)─ 𝑓(π‘₯)={β–ˆ(2π‘₯+3, 𝑖𝑓 π‘₯≀2@&2π‘₯βˆ’3, 𝑖𝑓 π‘₯>2)─ Since we need to find continuity at of the function We check continuity for different values of x When x = 2 When x < 2 When x > 2 Case 1 : When x = 2 f(x) is continuous at π‘₯ =2 if L.H.L = R.H.L = 𝑓(2) if lim┬(xβ†’2^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’2^+ ) " " 𝑓(π‘₯)= 𝑓(2) Since there are two different functions on the left & right of 2, we take LHL & RHL . LHL at x β†’ 2 lim┬(xβ†’2^βˆ’ ) f(x) = lim┬(hβ†’0) f(2 βˆ’ h) = lim┬(hβ†’0) (2(2βˆ’β„Ž)+3) = lim┬(hβ†’0) (4βˆ’2β„Ž+3) = lim┬(hβ†’0) (7βˆ’2β„Ž) = 7 βˆ’ 0 = 7 RHL at x β†’ 2 lim┬(xβ†’2^+ ) f(x) = lim┬(hβ†’0) f(2 + h) = lim┬(hβ†’0) (2(2+β„Ž)βˆ’3) = lim┬(hβ†’0) (4βˆ’2β„Žβˆ’3) = lim┬(hβ†’0) (1βˆ’2β„Ž) = 1 βˆ’ 0 = 1 Since L.H.L β‰  R.H.L f(x) is not continuous at x=2 Case 2 : When x < 2 For x < 2, f(x) = 2x + 3 Since this a polynomial It is continuous ∴ f(x) is continuous for x < 2 Case 3 : When x > 2 For x > 2, f(x) = 2x βˆ’ 3 Since this a polynomial It is continuous ∴ f(x) is continuous for x > 2 Hence, only x=2 is point is discontinuity. f is continuous at all real numbers except 2. Thus, f is continuous for 𝐱 ∈ R βˆ’ {2}

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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.