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

Last updated at Dec. 16, 2024 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

Share on WhatsApp

Transcript

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

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo