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

Transcript

Ex 5.1, 23 Find all points of discontinuity of f, where 𝑓(π‘₯)={β–ˆ(sin⁑π‘₯/π‘₯, 𝑖𝑓 π‘₯<0@&π‘₯+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 For x < 0, f(x) = sin⁑π‘₯/π‘₯ Since sin x and x are continuous So, sin⁑π‘₯/π‘₯ 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) sin⁑〖 (βˆ’β„Ž)γ€—/((βˆ’β„Ž)) = 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) β„Ž+1 = 0 + 1 = 1 & 𝑓(0) = π‘₯+1 = 0+1 = 1 Hence, L.H.L = R.H.L = 𝑓(0) ∴ f is continuous at x=0 Case 3 : When x > 0 For x > 0, f(x) = x + 1 Since this a polynomial It is continuous ∴ f(x) is continuous for x > 1 Hence, there is no point of discontinuity

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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.