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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise


Ex 6.3, 9 Find the point on the curve ๐‘ฆ=๐‘ฅ^3โˆ’11๐‘ฅ+5 at which the tangent is ๐‘ฆ=๐‘ฅ โˆ’11. Equation of Curve is ๐‘ฆ=๐‘ฅ^3โˆ’11๐‘ฅ+5 We know that Slope of tangent is ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=๐‘‘(๐‘ฅ^3 โˆ’ 11๐‘ฅ + 5)/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=ใ€–3๐‘ฅใ€—^2โˆ’11 Also, Given tangent is ๐‘ฆ=๐‘ฅโˆ’12 Comparing with ๐‘ฆ=๐‘š๐‘ฅ+๐‘ , when m is the Slope Slope of tangent =1 From (1) and (2) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=1 3๐‘ฅ^2โˆ’11=1 3๐‘ฅ^2=1+11 3๐‘ฅ^2=12 ๐‘ฅ^2=12/3 ๐‘ฅ^2=4 ๐‘ฅ=ยฑ2 When ๐’™=๐Ÿ ๐‘ฆ=(2)^3โˆ’11(2)+5 ๐‘ฆ=8โˆ’22+5 ๐‘ฆ=โˆ’ 9 So, Point is (2 , โˆ’9) When ๐’™=โˆ’๐Ÿ ๐‘ฆ=(โˆ’2)^3โˆ’11(โˆ’2)+5 ๐‘ฆ=โˆ’ 8+22+5 ๐‘ฆ=19 So, Point is (โˆ’2 , 19) Hence , Points (2 , โˆ’9) & (โˆ’2 , 19) But (โ€“2, 19) does not satisfy line y = x โ€“ 11 As 19 โ‰  โ€“2 โ€“ 11 โˆด Only point is (2, โ€“9)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.