Ex 5.1, 30 - Find a and b such that f(x) = {5, ax + b, 21 - Checking continuity using LHL and RHL

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  1. Chapter 5 Class 12 Continuity and Differentiability
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Ex 5.1, 30 Find the values of a and b such that the function defined by ๐‘“ ๐‘ฅ๏ทฏ= 5, ๐‘–๐‘“ ๐‘ฅโ‰ค2๏ทฎ๐‘Ž๐‘ฅ+๐‘, ๐‘–๐‘“ 2<๐‘ฅ<10๏ทฎ21, ๐‘–๐‘“ ๐‘ฅโ‰ฅ10๏ทฏ๏ทฏ is a continuous function ๐‘“ ๐‘ฅ๏ทฏ= 5, ๐‘–๐‘“ ๐‘ฅโ‰ค2๏ทฎ๐‘Ž๐‘ฅ+๐‘, ๐‘–๐‘“ 2<๐‘ฅ<10๏ทฎ21, ๐‘–๐‘“ ๐‘ฅโ‰ฅ10๏ทฏ๏ทฏ At x = 2 A function is continuous at x = 2 if L.H.L = R.H.L = ๐‘“(2) i.e. lim๏ทฎxโ†’ 2๏ทฎโˆ’๏ทฏ๏ทฏ ๐‘“ ๐‘ฅ๏ทฏ= lim๏ทฎxโ†’ 2๏ทฎ+๏ทฏ๏ทฏ ๐‘“ ๐‘ฅ๏ทฏ= ๐‘“ 2๏ทฏ Since, LHL = RHL 2a + b = 5 At x = 10 ๐‘“ is continuous at x = 10 if L.H.L = R.H.L = ๐‘“ 10๏ทฏ i.e. lim๏ทฎxโ†’ 10๏ทฎโˆ’๏ทฏ๏ทฏ ๐‘“ ๐‘ฅ๏ทฏ= lim๏ทฎxโ†’ 10๏ทฎ+๏ทฏ๏ทฏ ๐‘“ ๐‘ฅ๏ทฏ= ๐‘“ 10๏ทฏ Since, L.H.L = R.H.L 10a + b = 21 Now our equation are 2a + b = 5 โ€ฆ(1) 10 a + b = 21 โ€ฆ(2) From (1) 2a + b = 5 b = 5 โˆ’ 2a Putting value of b in (2) 10 ๐‘Ž+ 5โˆ’2๐‘Ž๏ทฏ = 21 10 ๐‘Ž+5โˆ’2๐‘Ž = 21 8๐‘Ž = 21โˆ’5 8๐‘Ž = 16 ๐‘Ž = 16๏ทฎ8๏ทฏ ๐‘Ž = 2 Putting value of a in (1) 2๐‘Ž+๐‘=5 2 2๏ทฏ+๐‘=5 4+๐‘=5 ๐‘=5โˆ’4 ๐‘=1 Hence, a = 2 & b = 1 .

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