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Ex 5.1, 18 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Last updated at April 13, 2021 by

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.