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Ex 5.1 ,6 - Find all points of discontinuity of f(x) = {2x + 3

Ex 5.1 ,6 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.1 ,6 - Chapter 5 Class 12 Continuity and Differentiability - Part 3 Ex 5.1 ,6 - Chapter 5 Class 12 Continuity and Differentiability - Part 4

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Ex 5.1, 6 Find all points of discontinuity of f, where f is defined by 𝑓(π‘₯)={β–ˆ(2π‘₯+3, 𝑖𝑓 π‘₯≀[email protected]&2π‘₯βˆ’3, 𝑖𝑓 π‘₯>2)─ 𝑓(π‘₯)={β–ˆ(2π‘₯+3, 𝑖𝑓 π‘₯≀[email protected]&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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.