Ex 5.1, 17 - Find relationship a, b so that f(x) is continuous

Ex 5.1, 17 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.1, 17 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

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Ex 5.1, 17 Find the relationship between a and b so that the function f defined by 𝑓(π‘₯)={β–ˆ(π‘Žπ‘₯+1, 𝑖𝑓 π‘₯≀3@&𝑏π‘₯+3, 𝑖𝑓 π‘₯>3)─ is continuous at x = 3.Given function is continuous at x = 3 f(x) is continuous at π‘₯ =3 if L.H.L = R.H.L = 𝒇(πŸ‘) if lim┬(xβ†’3^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’3^+ ) " " 𝑓(π‘₯)= 𝑓(3) Since there are two different functions on the left & right of 3, we take LHL & RHL . LHL at x β†’ 3 lim┬(xβ†’3^βˆ’ ) f(x) = lim┬(hβ†’0) f(3 βˆ’ h) = lim┬(hβ†’0) π‘Ž(3βˆ’β„Ž)+1 = π‘Ž(3βˆ’0)+1 = 3a + 1 RHL at x β†’ 3 lim┬(xβ†’3^+ ) f(x) = lim┬(hβ†’0) f(3 + h) = lim┬(hβ†’0) 𝑏(3+β„Ž)+3 = b(3 + 0) + 3 = b + 3 And 𝑓(3)=π‘Žπ‘₯+1 𝒇(πŸ‘)=πŸ‘π’‚ +𝟏 Now, lim┬(xβ†’3^βˆ’ ) 𝑓(π‘₯) = lim┬(xβ†’3^+ ) 𝑓(π‘₯) = 𝑓(1) 3π‘Ž + 1 = 3𝑏 + 3 = 3π‘Ž + 1 Comparing values πŸ‘π’‚ + 𝟏 = πŸ‘π’ƒ + πŸ‘ 3π‘Žβˆ’3b=3βˆ’1 3π‘Ž βˆ’3𝑏=2 3(π‘Žβˆ’π‘)=2 π‘Žβˆ’π‘=2/3 𝒂=𝒃+ 𝟐/πŸ‘ Thus , for any value of b, We can find value of a

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.