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

Last updated at Jan. 3, 2020 by Teachoo

Last updated at Jan. 3, 2020 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

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.