Ex 5.1, 29 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at Dec. 16, 2024 by

Ex 5.1, 29 - Find k. f(x) = {kx + 1, 3x - 5 is continuous at x=5

Ex 5.1, 29 - Chapter 5 Class 12 Continuity and Differentiability - Part 2

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Ex 5.1, 29 Find the values of k so that the function f is continuous at the indicated point 𝑓(π‘₯)={β–ˆ(π‘˜π‘₯+1, 𝑖𝑓 π‘₯≀5@3π‘₯βˆ’5, 𝑖𝑓 π‘₯>5)─ at x = 5 Given that function is continuous at π‘₯ =5 𝑓 is continuous at π‘₯ =5 if L.H.L = R.H.L = 𝑓(5) i.e. lim┬(xβ†’5^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’5^+ ) " " 𝑓(π‘₯)= 𝑓(5) LHL at x β†’ 5 (π‘™π‘–π‘š)┬(π‘₯β†’5^βˆ’ ) f(x) = (π‘™π‘–π‘š)┬(β„Žβ†’0) f(5 βˆ’ h) = lim┬(hβ†’0) k(5 – h) + 1 = k(5 βˆ’ 0) + 1 = 5k + 1 RHL at x β†’ 5 (π‘™π‘–π‘š)┬(π‘₯β†’5^+ ) f(x) = (π‘™π‘–π‘š)┬(β„Žβ†’0) f(5 + h) = lim┬(hβ†’0) 3(5 + h) βˆ’ 5 = 3(5 + 0) βˆ’ 5 = 3 Γ— 5 βˆ’ 5 = 15 βˆ’ 5 = 10 Since L.H.L = R.H.L 5k + 1 = 10 5k = 10 βˆ’ 1 5k = 9 π’Œ= 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