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 6.5, 8 At what points in the interval [0, 2Ο ], does the function sin 2π₯ attain its maximum value? Let f(π₯)=sinβ‘2π₯, π₯ β [0 , 2π] Finding fβ(π) fβ(π₯)=π(sinβ‘2π₯ )/ππ₯ fβ(π₯)=2 cosβ‘2π₯ Putting fβ(π)=π 2cos 2π₯=0 cos 2π₯=0 cos 2π₯=cosβ‘γπ/2γ If cos π₯=cosβ‘π¦ Then general solution of π₯=2ππΒ±π¦ 2π₯=2ππ+π/2 π₯=(2π+1) π/4 Putting n = 0 π₯=(0+1) π/4= π/4 Putting n = 1 π₯=(2+1) π/4= 3π/4 Putting n = 2 π₯=(2(2)+1) π/4= 5π/4 Putting n = 3 π₯=(2(3)+1) π/4= 7π/4 Putting n = 4 π₯=(2(4)+1) π/4= 9π/4 Putting n = 5 π₯=(2(5)+1) π/4 =11π/4>2π Since π₯ β [0 , 2π] So, Critical Point are π₯ = 0 , π/4 , 3π/4 , 5π/4 , 7π/4 & 2Ο Finding value of f(x) at critical points i.e. π₯ = 0 , π/4 , 3π/4 , 5π/4 , 7π/4 & 2Ο π₯ = 5π/4 f(5π/4)= sin 2(5π/4) = sin(5π/2) = sin(2π+ π/2) = sin (π/2) = 1 (sin (2Ο + ΞΈ) = sin ΞΈ) f(7π/4)= sin 2(7π/4) = sin(7π/2) = sin(3π+ π/2) = β sin (π/2) = β 1 (sin (Ο + ΞΈ) = β sin ΞΈ) f(2π" " )=sin 2(2π" " ) = sin 4Ο = 0 Hence Maximum Value of f(π)=π at π = π /π , ππ /π

Ex 6.5

Ex 6.5,1
Important

Ex 6.5,2 Important

Ex 6.5,3

Ex 6.5,4

Ex 6.5,5 Important

Ex 6.5,6

Ex 6.5,7 Important

Ex 6.5,8 You are here

Ex 6.5,9 Important

Ex 6.5,10

Ex 6.5,11 Important

Ex 6.5,12 Important

Ex 6.5,13

Ex 6.5,14 Important

Ex 6.5,15 Important

Ex 6.5,16

Ex 6.5,17

Ex 6.5,18 Important

Ex 6.5,19 Important

Ex 6.5, 20 Important

Ex 6.5,21

Ex 6.5,22 Important

Ex 6.5,23 Important

Ex 6.5,24 Important

Ex 6.5,25 Important

Ex 6.5,26 Important

Ex 6.5, 27

Ex 6.5,28 Important

Ex 6.5,29

Chapter 6 Class 12 Application of Derivatives

Serial order wise

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.