1. Class 12
2. Important Question for exams Class 12
3. Chapter 6 Class 12 Application of Derivatives

Transcript

Ex 6.3,7 Find points at which the tangent to the curve ๐ฆ=๐ฅ^3โ3๐ฅ^2โ9๐ฅ+7 is parallel to the x-axis Equation of Curve is ๐ฆ=๐ฅ^3โ3๐ฅ^2โ9๐ฅ+7 Differentiating w.r.t. ๐ฅ ๐๐ฆ/๐๐ฅ=๐(๐ฅ^3โ 3๐ฅ^2 โ 9๐ฅ + 7)/๐๐ฅ ๐๐ฆ/๐๐ฅ=3๐ฅ^2โ6๐ฅโ9+0 ๐๐ฆ/๐๐ฅ=3๐ฅ^2โ6๐ฅโ9 ๐๐ฆ/๐๐ฅ=3(๐ฅ^2โ2๐ฅโ3) ๐๐ฆ/๐๐ฅ=3(๐ฅ^2โ3๐ฅ+๐ฅโ3) ๐๐ฆ/๐๐ฅ=3(๐ฅ(๐ฅโ3)+1(๐ฅโ3)) ๐๐ฆ/๐๐ฅ=3(๐ฅ+1)(๐ฅโ3) Given tangent to the curve is parallel to the ๐ฅโ๐๐ฅ๐๐  i.e. the Slope of tangent = Slope of ๐ฅโ๐๐ฅ๐๐  ๐๐/๐๐=๐ 3(๐ฅ+1)(๐ฅโ3)=0 (๐ฅ+1)(๐ฅโ3)=0 Thus ๐ฅ=โ1 & ๐ฅ=3 When ๐=โ๐ ๐ฆ=๐ฅ^3โ3๐ฅ^2โ9๐ฅ+7 =(โ1)^3โ3(โ1)^2โ9(โ1)+7 =โ1โ3+9+7 =12 Point is (โ1 , 12) When ๐=๐ ๐ฆ=๐ฅ^3โ3๐ฅ^2โ9๐ฅ+7 =(3)^3โ3(3)^2โ9(3)+7 =27โ27โ27+7 =โ 20 Point is (3 , โ20) Hence , the tangent to the Curve is parallel to the ๐ฅโ๐๐ฅ๐๐  at (โ๐ , ๐๐) & (๐ , โ๐๐)

Chapter 6 Class 12 Application of Derivatives

Class 12
Important Question for exams Class 12