Ex 5.1, 16 - Discuss continuity of f(x) = { -2, 2x, 2 - Chapter 5 - Ex 5.1

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  1. Chapter 5 Class 12 Continuity and Differentiability
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Ex 5.1, 16 Discuss the continuity of the function f, where f is defined by ๐‘“ ๐‘ฅ๏ทฏ= โˆ’&2, ๐‘–๐‘“ ๐‘ฅโ‰คโˆ’1๏ทฎ2&๐‘ฅ, ๐‘–๐‘“ โˆ’1โ‰ค๐‘ฅโ‰ค1๏ทฎ2, ๐‘–๐‘“ ๐‘ฅ>1 ๏ทฏ๏ทฏ Case 1:- At x = โˆ’1 A function is continuous at x = โˆ’ 1 if L.H.L = R.H.L = ๐‘“ โˆ’1๏ทฏ i.e. lim๏ทฎxโ†’ โˆ’1๏ทฎโˆ’๏ทฏ๏ทฏ ๐‘“ ๐‘ฅ๏ทฏ = lim๏ทฎxโ†’ โˆ’1๏ทฎ+๏ทฏ๏ทฏ ๐‘“ ๐‘ฅ๏ทฏ = ๐‘“ โˆ’1๏ทฏ And ๐‘“ โˆ’2๏ทฏ = โˆ’2 Thus L.H.L = R.H.L = ๐‘“ โˆ’2๏ทฏ = โˆ’2 Hence ๐’‡ ๐’™๏ทฏ is continuous at x = โˆ’๐Ÿ Case 1:- At x = โˆ’1 A function is continuous at x = โˆ’ 1 if A function is continuous at x = 1 if if L.H.L = R.H.L = ๐‘“ 1๏ทฏ i.e. lim๏ทฎxโ†’ 1๏ทฎโˆ’๏ทฏ๏ทฏ ๐‘“ ๐‘ฅ๏ทฏ = lim๏ทฎxโ†’ 1๏ทฎ+๏ทฏ๏ทฏ ๐‘“ ๐‘ฅ๏ทฏ = ๐‘“ 1๏ทฏ & ๐‘“ 1๏ทฏ = 2 1๏ทฏ = 2 Thus L.H.L = R.H.L = ๐‘“ โˆ’2๏ทฏ Hence ๐’‡ ๐’™๏ทฏ is continuous at x = 1 Case 3:- For ๐‘ฅ<โˆ’1 ๐‘“ ๐‘ฅ๏ทฏ = โˆ’2 Thus, ๐‘“ ๐‘ฅ๏ทฏ is a constant function . & Every constant function is continuous for all real number. Hence ๐’‡ ๐’™๏ทฏ is continuous at ๐’™<โˆ’๐Ÿ Case 4:- For ๐‘ฅ>1 ๐‘“ ๐‘ฅ๏ทฏ = 2 Thus, ๐‘“ ๐‘ฅ๏ทฏ is a constant function . & Every constant function is continuous for all real number. Hence ๐’‡ ๐’™๏ทฏ is continuous at ๐’™>๐Ÿ Case 5:- For โˆ’1โ‰ค๐‘ฅโ‰ค1 ๐‘“ ๐‘ฅ๏ทฏ = 2๐‘ฅ So, f(x) is a polynomial & Every polynomial is continuous. โ‡’ ๐’‡ ๐’™๏ทฏ is continuous at โˆ’๐Ÿ<๐’™โ‰ค๐Ÿ Thus, f(x) is continuous for all real numbers, i.e. f is continuous for all x โˆˆ R

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