# Ex 5.1, 14 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 5.1, 14 Discuss the continuity of the function f, where f is defined by 𝑓 𝑥= 3, 𝑖𝑓 0≤𝑥≤14, 𝑖𝑓 1<𝑥<3 5, 𝑖𝑓 3≤𝑥≤10 𝑓 𝑥= 3, 𝑖𝑓 0≤𝑥≤14, 𝑖𝑓 1<𝑥<3 5, 𝑖𝑓 3≤𝑥≤10 Case 1:- At x = 1 A function is continuous at x = 1 if L.H.L = R.H.L = 𝑓 1 i.e. limx→ 1− 𝑓 𝑥 = limx→ 1+ = 𝑓 1 Thus L.H.L ≠ R.H.L ⇒ limx→ 1− 𝑓 𝑥 ≠ limx→ 1+ = 𝑓 𝑥 Hence 𝒇 𝒙 is not continuous as 𝒙=𝟏 Case 2:- At x = 3 f is continuous at x = 3 if L.H.L = R.H.L = 𝑓 3 i.e. limx→ 3− 𝑓 𝑥 = limx→ 3+𝑓 𝑥 = 𝑓 3 Since L.H.L ≠ R.H.L Hence 𝒇 𝒙 is not continuous as 𝒙=𝟑 Case 3:- 0≤𝑥<1 𝑓 𝑥 = 3 3 can be written as 𝑓 𝑥 = 3. 𝑥0 So, f(x) is a polynomial We know that every polynomial function is continuous for every real number Therefore 𝒇 𝒙 = 3 is continuous at 𝟎≤𝒙≤𝟏 . Case 4:- 1<𝑥<3 𝑓 𝑥 = 4 4 can be written as 𝑓 𝑥 = 4. 𝑥0 So, f(x) is a polynomial We know that every polynomial function is continuous for every real number Therefore 𝒇 𝒙 = 3 is continuous at 𝟏<𝒙<𝟑 . Case 4:- 3 <𝑥≤ 10 𝑓 𝑥 = 5 5 can be written as 𝑓 𝑥 = 5. 𝑥0 So, f(x) is a polynomial We know that every polynomial function is continuous for every real number Therefore 𝒇 𝒙 = 3 is continuous at 3 <𝒙≤ 10 Hence points of discontinuity are x = 1 & x = 3 Thus, f is continuous for all x ∈ R − 𝟏,𝟑

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.