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

Transcript

Ex 5.1, 34 Find all the points of discontinuity of f defined by 𝑓(π‘₯)= |π‘₯| – |π‘₯+1|. Given 𝑓(π‘₯)= |π‘₯| – |π‘₯+1|. Here, we have 2 critical points x = 0 and x + 1 = 0 i.e. x = 0, and x = βˆ’1 So, our intervals will be When π‘₯β‰€βˆ’1 When βˆ’1<π‘₯<0 When π‘₯β‰₯0 When π’™β‰€βˆ’πŸ 𝑓(π‘₯)= |π‘₯| – |π‘₯+1|. Here, both will be negative 𝑓(π‘₯)=(βˆ’π‘₯) –(βˆ’(π‘₯+1)) 𝑓(π‘₯)=βˆ’π‘₯+(π‘₯+1) " " 𝑓(π‘₯)=1 When βˆ’πŸ<π’™β‰€πŸŽ 𝑓(π‘₯)= |π‘₯| – |π‘₯+1|. Here, x will be negative, but (x + 1) will be positive 𝑓(π‘₯)=(βˆ’π‘₯) –(π‘₯+1) 𝑓(π‘₯)=βˆ’π‘₯βˆ’π‘₯βˆ’1 " " 𝑓(π‘₯)=βˆ’2π‘₯βˆ’1 |π‘₯| = {β–ˆ(π‘₯, π‘₯ β‰₯0@βˆ’π‘₯, π‘₯<0)─ |π‘₯+1| = {β–ˆ((π‘₯+1) , π‘₯+1β‰₯0@βˆ’(π‘₯+1) π‘₯+1<0)─ = {β–ˆ((π‘₯+1) , π‘₯β‰₯βˆ’1@βˆ’(π‘₯+1) π‘₯<1)─ When 𝒙β‰₯𝟎 𝑓(π‘₯)= |π‘₯| – |π‘₯+1|. Here, both will be positive 𝑓(π‘₯)=π‘₯ –(π‘₯+1) 𝑓(π‘₯)=π‘₯βˆ’π‘₯βˆ’1 " " 𝑓(π‘₯)=βˆ’1 Thus, our function becomes 𝑓(π‘₯)={β–ˆ(1 𝑖𝑓 π‘₯β‰€βˆ’1@βˆ’2π‘₯βˆ’1 𝑖𝑓 βˆ’1<π‘₯<0@βˆ’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 < βˆ’1 When x = βˆ’1 When βˆ’1 < x < 0 When x = 0 When x > 0 Checking continuity Case 1 : When x < βˆ’1 For x < βˆ’1, f(x) = 1 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) 1 = 1 RHL at x β†’ 0 lim┬(xβ†’γ€–βˆ’1γ€—^+ ) f(x) = lim┬(hβ†’0) f(βˆ’1 + h) = lim┬(hβ†’0) (βˆ’2(βˆ’1+β„Ž))βˆ’1 = lim┬(hβ†’0) (2βˆ’2β„Ž)βˆ’1 = (2 βˆ’ 2(0)) βˆ’ 1 = 2 βˆ’ 0 βˆ’ 1 = 1 & 𝑓(βˆ’1) = 1 Hence, L.H.L = R.H.L = 𝑓(βˆ’1) ∴ f is continuous at x=βˆ’1 Case 3 : When βˆ’1 < x < 0 For βˆ’1 < x < 0 f(x) = βˆ’2x βˆ’ 1 Since this a polynomial It is continuous ∴ f(x) is continuous for βˆ’1 < x < 0 Case 4 : 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) βˆ’2(βˆ’β„Ž)βˆ’1 = lim┬(hβ†’0) 2β„Žβˆ’1 = 2(0) βˆ’ 1 = βˆ’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) βˆ’1 = βˆ’1 & 𝑓(0) = βˆ’1 Hence, L.H.L = R.H.L = 𝑓(0) ∴ f is continuous at x=0 Case 5: When x > 0 For x > 0, f(x) = βˆ’1 Since this constant It is continuous ∴ f(x) is continuous for x > 0 Since there is no point of discontinuity Therefore, 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.