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  1. Chapter 5 Class 12 Continuity and Differentiability
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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‘₯≀2@&2π‘₯βˆ’3, 𝑖𝑓 π‘₯>2)─ We have, 𝑓(π‘₯)={β–ˆ(2π‘₯+3, 𝑖𝑓 π‘₯≀2@&2π‘₯βˆ’3, 𝑖𝑓 π‘₯>2)─ Case 1 At π‘₯ =2 f is continuous at x = 2 if L.H.L = R.H.L = 𝑓(2) i.e. lim┬(xβ†’2^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’2^+ ) 𝑓(π‘₯)=𝑓(2) L.H.L lim┬(xβ†’2^βˆ’ ) 𝑓(π‘₯) = lim┬(xβ†’2^βˆ’ ) 2π‘₯+3 Putting π‘₯ =2 = 2(2) + 3 = 4 + 3 = 7 R.H.L lim┬(xβ†’2^+ ) 𝑓(π‘₯) = lim┬(xβ†’2^+ ) 2π‘₯βˆ’3 Putting π‘₯ =2 =2(2) βˆ’ 3 = 4 βˆ’ 3 = 1 Since, L.H.L β‰  R.H.L ∴ f is not continuous at x=2. Case 2 At π‘₯ =𝑐 where c < 2 𝑓(π‘₯)= 2π‘₯+3 (As π‘₯ =𝑐, where c < 2) f is continuous at x=c if lim┬(x→𝑐) 𝑓(π‘₯)=𝑓(𝑐) L.H.L lim┬(x→𝑐) 𝑓(π‘₯) = lim┬(x→𝑐) 2π‘₯+3 Putting π‘₯=𝑐 = 2𝑐+3 R.H.L 𝑓(𝑐)= 2𝑐+3 Hence, lim┬(x→𝑐) 𝑓(π‘₯)=𝑓(𝑐) ∴ f is continuous at x=c where c<2 Thus, f is continuous at all real number less than 2. Case 3 At π‘₯ =𝑐 where c > 2 ∴ 𝑓(π‘₯)= 2 π‘₯βˆ’3 (As π‘₯ =𝑐, c > 2) f is continuous at x=c if lim┬(x→𝑐) 𝑓(π‘₯)=𝑓(𝑐) L.H.L lim┬(x→𝑐) 𝑓(π‘₯) = lim┬(x→𝑐) 2 π‘₯βˆ’3 Putting π‘₯=𝑐 = 2 (𝑐)βˆ’3 = 2 π‘βˆ’3 R.H.L 𝑓(π‘₯)= 2π‘₯βˆ’3 𝑓(𝑐)=2π‘βˆ’3 Hence, lim┬(x→𝑐) 𝑓(π‘₯)=𝑓(𝑐) β‡’ f is continuous at x=c where c>2 β‡’ f is continuous at all real number greater than 2 Hence, only x=2 is point is discontinuity. β‡’ 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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.