Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Chapter 5 Class 12 Continuity and Differentiability
Ex 5.1, 13
Ex 5.1, 16
Ex 5.1, 18 Important You are here
Ex 5.1, 28 Important
Ex 5.1, 30 Important
Ex 5.1, 34 Important
Ex 5.2, 5
Ex 5.2, 9 Important
Ex 5.2, 10 Important
Ex 5.3, 10 Important
Ex 5.3, 14
Example 29 Important
Example 30 Important
Ex 5.5,6 Important
Ex 5.5, 7 Important
Ex 5.5, 11 Important
Ex 5.5, 16 Important
Ex 5.6, 7 Important
Ex 5.6, 11 Important
Example 38 Important
Ex 5.7, 14 Important
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 5 Important Deleted for CBSE Board 2024 Exams
Example 39 (i)
Example 40 (i)
Example 42 Important
Misc 6 Important
Misc 15 Important
Misc 16 Important
Misc 22 Important
Chapter 5 Class 12 Continuity and Differentiability
Last updated at May 29, 2023 by Teachoo
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