Last updated at April 13, 2021 by Teachoo

Transcript

Ex 5.1, 18 For what value of Ξ» is the function defined by π(π₯)={β("Ξ»" (π₯^2β2π₯), ππ π₯β€0@&4π₯+1, ππ π₯>0)β€ 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 = π(0) if if limβ¬(xβ0^β ) π(π₯) = limβ¬(xβ0^+ ) π(π₯) = π(0) Since there are two different functions on the left & right of 3, we take LHL & RHL . LHL at x β 0 limβ¬(xβ3^β ) f(x) = limβ¬(hβ0) f(0 β h) = limβ¬(hβ0) f(βh) = limβ¬(hβ0) "Ξ»" (γ(ββ)γ^2β2(ββ)) = "Ξ»" (02+2(0)) = "Ξ» (0)" = 0 RHL at x β 3 limβ¬(xβ3^+ ) f(x) = limβ¬(hβ0) f(0 + h) = limβ¬(hβ0) f(h) = limβ¬(hβ0) 4β+1 = 4 Γ 0 + 1 = 0 + 1 = 1 Since L.H.L β R.H.L β΄ f(x) is not continuous at x = 0. So, for any value of "Ξ»"βπ, f is discontinuous at x = 0 When x = 1 For x > 1, f(x) = 4x + 1 Since this a polynomial It is continuous β΄ f(x) is continuous for x = 1 Thus, we can write that for any value of "Ξ»"βπ, f is continuous at x = 1

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.