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Ex 5.1, 18 - For what value of is f(x) continuous at x = 0, 1

Ex 5.1, 18 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.1, 18 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

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Ex 5.1, 18 For what value of Ξ» is the function defined by 𝑓(π‘₯)={β–ˆ("Ξ»" (π‘₯^2βˆ’2π‘₯), 𝑖𝑓 π‘₯≀0@&4π‘₯+1, 𝑖𝑓 π‘₯>0)─ continuous at x = 0? What about continuity at x = 1? At x = 0 f(x) is continuous at π‘₯ =0 if L.H.L = R.H.L = 𝑓(0) if 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 3, we take LHL & RHL . LHL at x β†’ 0 lim┬(xβ†’3^βˆ’ ) f(x) = lim┬(hβ†’0) f(0 βˆ’ h) = lim┬(hβ†’0) f(βˆ’h) = lim┬(hβ†’0) "Ξ»" (γ€–(βˆ’β„Ž)γ€—^2βˆ’2(βˆ’β„Ž)) = "Ξ»" (02+2(0)) = "Ξ» (0)" = 0 RHL at x β†’ 3 lim┬(xβ†’3^+ ) f(x) = lim┬(hβ†’0) f(0 + h) = lim┬(hβ†’0) f(h) = lim┬(hβ†’0) 4β„Ž+1 = 4 Γ— 0 + 1 = 0 + 1 = 1 Since L.H.L β‰  R.H.L ∴ f(x) is not continuous at x = 0. So, for any value of "Ξ»"βˆˆπ‘, f is discontinuous at x = 0 When x = 1 For x > 1, f(x) = 4x + 1 Since this a polynomial It is continuous ∴ f(x) is continuous for x = 1 Thus, we can write that for any value of "Ξ»"βˆˆπ‘, f is continuous at x = 1

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.