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

Misc 21 The line ๐ฆ=๐๐ฅ+1 is a tangent to the curve ๐ฆ^2=4๐ฅ if the value of ๐ is (A) 1 (B) 2 (C) 3 (D) 1/2 Let (โ , ๐) be the point at which tangent is to be taken & Given Equation of tangent ๐ฆ=๐๐ฅ+1 & Curve is ๐ฆ^2=4๐ฅ We know that Slope of tangent to the Curve is ๐๐ฆ/๐๐ฅ ๐ฆ^2=4๐ฅ Differentiating w.r.t. ๐ฅ ๐(๐ฆ^2 )/๐๐ฅ=๐(4๐ฅ)/๐๐ฅ ๐(๐ฆ^2 )/๐๐ฅ ร ๐๐ฆ/๐๐ฆ=4 ๐(๐ฆ^2 )/๐๐ฆ ร ๐๐ฆ/๐๐ฅ=4 2๐ฆ ร๐๐ฆ/๐๐ฅ=4 ๐๐ฆ/๐๐ฅ=2/๐ฆ Since tangent to be taken from (โ , ๐) So, Slope of tangent at (โ , ๐) is โ ๐๐ฆ/๐๐ฅโค|_((โ , ๐) )=2/๐ But Given Equation of tangent ๐ฆ=๐๐ฅ+1 Hence Slope of tangent = m โ ๐ =2/๐ Now, Point (โ , ๐) lie on the Curve ๐ฆ^2=4๐ฅ โ(โ , ๐) will Satisfies the Equation of Curve Putting ๐ฅ=โ , ๐ฆ=๐ in equation โ ๐^2=4โ โฆ(1) โฆ(2) Also, Point (โ , ๐) lie on the tangent โ (โ , ๐) will satisfies the Equation of tangent ๐ฆ=๐๐ฅ+1 Putting ๐ฅ=โ , ๐ฆ=๐ in equation ๐=๐โ+1 From (1) Putting Value of ๐=2/๐ ๐=2/๐ รโ+1 ๐=(2โ + ๐)/๐ ๐^2=2โ+๐ ๐^2โ๐=2โ Now Our Equation are ๐^2=4โ โฆ(3) โฆ(2) ๐^2โ๐=2โ Solving (2) ๐^2=4โ ๐^2=2(2โ) Putting 2โ=๐^2โ๐ from (3) ๐^2=2(๐^2โ๐) ๐^2=2๐^2โ2๐ 0=2๐^2โ2๐โ๐^2 2๐^2โ๐^2โ2๐=0 ๐^2โ2๐=0 ๐(๐โ2)=0 Hence ๐=0, 2 When ๐=๐ From (1) ๐=2/๐ ๐=2/0 ๐=โ When ๐=๐ From (1) ๐=2/๐ ๐=2/2 ๐=1 Since ๐=โ not in the Option So, m = 1 Hence Correct Answer is (A)

Miscellaneous

Misc 1
Important
Not in Syllabus - CBSE Exams 2021

Misc 2 Important

Misc 3 Important Not in Syllabus - CBSE Exams 2021

Misc 4

Misc 5 Important

Misc 6 Important

Misc 7

Misc 8 Important

Misc 9 Important

Misc 10

Misc 11 Important

Misc 12 Important

Misc 13 Important

Misc 14 Important

Misc 15 Important

Misc 16 Important

Misc 17 Important

Misc 18 Important

Misc. 19 Not in Syllabus - CBSE Exams 2021

Misc 20 Important

Misc 21 Important You are here

Misc 22

Misc. 23 Important

Misc 24 Important

Chapter 6 Class 12 Application of Derivatives

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About the Author

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.