Ex 5.1, 27 - Chapter 5 Class 12 Continuity and Differentiability

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Ex 5.1, 27 Find k so that f(x) = {kx2 , 3 is continuous at x = 2

Ex 5.1, 27 - Chapter 5 Class 12 Continuity and Differentiability - Part 2

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Ex 5.1, 27 Find the values of k so that the function f is continuous at the indicated point 𝑓(π‘₯)={β–ˆ(π‘˜π‘₯2 , 𝑖𝑓 π‘₯≀2@3, 𝑖𝑓 π‘₯>2)─ at x = 2Given that function is continuous at π‘₯ = 2 𝑓 is continuous at π‘₯ = 2 if L.H.L = R.H.L = 𝑓(2) i.e. lim┬(xβ†’2^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’2^+ ) " " 𝑓(π‘₯)= 𝑓(2) LHL at x β†’ 2 lim┬(xβ†’2^βˆ’ ) f(x) = lim┬(hβ†’0) f(2 βˆ’ h) = lim┬(hβ†’0) γ€–π‘˜(2βˆ’β„Ž)γ€—^2 = γ€–π‘˜(2βˆ’0)γ€—^2 = γ€–π‘˜(2)γ€—^2 = πŸ’π’Œ RHL at x β†’ 2 lim┬(xβ†’2^+ ) f(x) = lim┬(hβ†’0) f(2 + h) = lim┬(hβ†’0) 3 = 3 RHL at x β†’ 2 lim┬(xβ†’2^+ ) f(x) = lim┬(hβ†’0) f(2 + h) = lim┬(hβ†’0) 3 = 3 RHL at x β†’ 2 lim┬(xβ†’2^+ ) f(x) = lim┬(hβ†’0) f(2 + h) = lim┬(hβ†’0) 3 = 3 RHL at x β†’ 2 lim┬(xβ†’2^+ ) f(x) = lim┬(hβ†’0) f(2 + h) = lim┬(hβ†’0) 3 = 3 Since LHL = RHL 4k = 3 k = πŸ‘/πŸ’

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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