Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12


Last updated at Jan. 7, 2020 by Teachoo
Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12
Transcript
Ex 6.3, 9 Find the point on the curve ๐ฆ=๐ฅ^3โ11๐ฅ+5 at which the tangent is ๐ฆ=๐ฅ โ11. Equation of Curve is ๐ฆ=๐ฅ^3โ11๐ฅ+5 We know that Slope of tangent is ๐๐ฆ/๐๐ฅ ๐๐ฆ/๐๐ฅ=๐(๐ฅ^3 โ 11๐ฅ + 5)/๐๐ฅ ๐๐ฆ/๐๐ฅ=ใ3๐ฅใ^2โ11 Also, Given tangent is ๐ฆ=๐ฅโ12 Comparing with ๐ฆ=๐๐ฅ+๐ , when m is the Slope Slope of tangent =1 From (1) and (2) ๐๐ฆ/๐๐ฅ=1 3๐ฅ^2โ11=1 3๐ฅ^2=1+11 3๐ฅ^2=12 ๐ฅ^2=12/3 ๐ฅ^2=4 ๐ฅ=ยฑ2 When ๐=๐ ๐ฆ=(2)^3โ11(2)+5 ๐ฆ=8โ22+5 ๐ฆ=โ 9 So, Point is (2 , โ9) When ๐=โ๐ ๐ฆ=(โ2)^3โ11(โ2)+5 ๐ฆ=โ 8+22+5 ๐ฆ=19 So, Point is (โ2 , 19) Hence , Points (2 , โ9) & (โ2 , 19) But (โ2, 19) does not satisfy line y = x โ 11 As 19 โ โ2 โ 11 โด Only point is (2, โ9)
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