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Ex 5.1, 27 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Last updated at March 16, 2023 by

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‘₯≀[email protected], 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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