1. Chapter 5 Class 12 Continuity and Differentiability
2. Serial order wise

Transcript

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๐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๐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๏ทฎโ๏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๏ทฎโ๏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 โ 2a Putting value of b in (2) 10 ๐+ 5โ2๐๏ทฏ = 21 10 ๐+5โ2๐ = 21 8๐ = 21โ5 8๐ = 16 ๐ = 16๏ทฎ8๏ทฏ ๐ = 2 Putting value of a in (1) 2๐+๐=5 2 2๏ทฏ+๐=5 4+๐=5 ๐=5โ4 ๐=1 Hence, a = 2 & b = 1 .