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Ex 5.1, 29 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

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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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