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

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  1. Chapter 5 Class 12 Continuity and Differentiability
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Ex 5.1, 17 Find the relationship between a and b so that the function f defined by ๐‘“ ๐‘ฅ๏ทฏ= ๐‘Ž๐‘ฅ+1, ๐‘–๐‘“ ๐‘ฅโ‰ค3๏ทฎ&๐‘๐‘ฅ+3, ๐‘–๐‘“ ๐‘ฅ>3๏ทฏ๏ทฏ is continuous at x = 3. Given ๐‘“ ๐‘ฅ๏ทฏ= ๐‘Ž๐‘ฅ+1, ๐‘–๐‘“ ๐‘ฅโ‰ค3๏ทฎ&๐‘๐‘ฅ+3, ๐‘–๐‘“ ๐‘ฅ>3๏ทฏ๏ทฏ Given function is continuous at x = 3 f is continuous at x = 3 if L.H.L = R.H.L = ๐‘“ 3๏ทฏ i.e. if lim๏ทฎxโ†’ 3๏ทฎโˆ’๏ทฏ๏ทฏ ๐‘“ ๐‘ฅ๏ทฏ = lim๏ทฎxโ†’ 3๏ทฎ+๏ทฏ๏ทฏ ๐‘“ ๐‘ฅ๏ทฏ = ๐‘“ 1๏ทฏ And ๐‘“ ๐‘ฅ๏ทฏ=ax+1 So, ๐‘“ 3๏ทฏ=3a +1 Now, lim๏ทฎxโ†’ 3๏ทฎโˆ’๏ทฏ๏ทฏ ๐‘“ ๐‘ฅ๏ทฏ = lim๏ทฎxโ†’ 3๏ทฎ+๏ทฏ๏ทฏ ๐‘“ ๐‘ฅ๏ทฏ = ๐‘“ 1๏ทฏ 3๐‘Ž + 1 = 3๐‘ + 3 = 3๐‘Ž + 1 โ‡’ 3๐‘Ž + 1 = 3๐‘ + 3 โ‡’ 3๐‘Žโˆ’3b=3โˆ’1 โ‡’ 3๐‘Ž โˆ’3๐‘=2 โ‡’ 3 ๐‘Žโˆ’๐‘๏ทฏ=2 โ‡’ ๐‘Žโˆ’๐‘= 2๏ทฎ3๏ทฏ โ‡’ ๐‘Ž=๐‘+ 2๏ทฎ3๏ทฏ Thus , for any arbitrary value of b. We can find value of a.

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.