Slide6.JPG

Slide7.JPG
Slide8.JPG Slide9.JPG Slide10.JPG Slide11.JPG Slide12.JPG Slide13.JPG Slide14.JPG Slide15.JPG

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

Transcript

Ex 6.2,12 Which of the following functions are strictly decreasing on (0 ,πœ‹/2)? (A) cos π‘₯ Let f(π‘₯) = cos π‘₯ Finding f’(π‘₯) f’(π‘₯) = – sin π‘₯ Hence, f’(π‘₯) < 0 for π‘₯ ∈ (0 , πœ‹/2) β‡’ f(π‘₯) is strictly decreasing on (𝟎 , 𝝅/𝟐) Note :- sin π‘₯ > 0 for π‘₯ ∈ (0 , πœ‹/2) So, – sin π‘₯ < for π‘₯ ∈ (0 , πœ‹/2) f’ (π‘₯)<0 π‘“π‘œπ‘Ÿ π‘₯ ∈ πœ‹/2 Ex 6.2,12 Which of the following functions are strictly decreasing on (0,πœ‹/2) ? (B) cos 2π‘₯ Let f(π‘₯) = cos 2π‘₯ Finding f’(𝒙) f’(π‘₯) = (cos⁑2π‘₯ )β€² f’(π‘₯) = – 2 sin 2π‘₯ Putting f’(𝒙) = 0 f’(π‘₯) = 0 – 2 sin 2π‘₯ = 0 sin 2π‘₯ = 0 As π‘₯ ∈ (0 , πœ‹/2) 0 < π‘₯ < πœ‹/2 0 Γ— 2 < 2π‘₯ < πœ‹/2 Γ— 2 0 < 2π‘₯ < Ο€ We know that sin ΞΈ > 0 for ΞΈ ∈ (0 , πœ‹) sin 2π‘₯ > 0 for 2π‘₯ ∈ (0 , πœ‹) –sin 2π‘₯ < 0 for 2π‘₯ ∈ (0 , πœ‹) – 2 sin 2π‘₯ < 0 for π‘₯ ∈ (0 , πœ‹/2) β‡’ f’(π‘₯) < 0 for π‘₯ ∈ (0 , πœ‹/2) β‡’ f (π‘₯) is strictly decreasing for 𝒙 ∈ (𝟎 , 𝝅/𝟐) Ex 6.2,12 Which of the following functions are strictly decreasing on (0,πœ‹/2) ? (C) cos 3π‘₯ Let f(π‘₯) = cos 3π‘₯ as π‘₯ ∈ (0 ,πœ‹/2) Finding f’(𝒙) f’(π‘₯) = (cos⁑3π‘₯ )^β€² f’(π‘₯) = –3 sin 3π‘₯ Putting f’(𝒙) = 0 –3 sin 3π‘₯ = 0 sin 3π‘₯ = 0 As sin ΞΈ = 0 if ΞΈ = 0 , Ο€ , 2Ο€ , 3Ο€ ∴ 3π‘₯ = Ο€ , 2Ο€ π‘₯ = πœ‹/3 , 2πœ‹/3 As π‘₯ ∈ (0 , πœ‹/2) So, π‘₯ = πœ‹/3 Since π‘₯ ∈ (0 , πœ‹/2), so we start number line from (0 , πœ‹/2) The point x = πœ‹/3 divide the interval (0 , πœ‹/2) into 2 disjoint interval. i.e. (0 , πœ‹/3) & (πœ‹/3 , πœ‹/2) Checking sign of f’(π‘₯) Case 1: For 𝒙 ∈ (𝟎 , 𝝅/πŸ‘) Now, 0 < π‘₯ < πœ‹/3 3 Γ— 0 < 3π‘₯ < 3πœ‹/3 0 < 3π‘₯ < Ο€ So, when π‘₯ ∈ (0 , πœ‹/3) , 3π‘₯ ∈ (0 , πœ‹) We know that sinΞΈ > 0 for ΞΈ ∈ (0 , πœ‹) sin 3π‘₯ > 0 for 3π‘₯ ∈ (0 , πœ‹) sin 3π‘₯ > 0 for π‘₯ ∈ (0 , πœ‹/3) – sin 3π‘₯ < 0 for π‘₯ ∈ (0 ,πœ‹/3) β‡’ f’(π‘₯) < 0 for π‘₯ ∈ (0 , πœ‹/3) β‡’ f(π‘₯) is strictly decreasing for π‘₯ ∈ (0 ,πœ‹/3) Case 2: For 𝒙 ∈ (𝝅/πŸ‘, 𝝅/𝟐) Since πœ‹/3 < π‘₯ < πœ‹/2 3 Γ— πœ‹/3 < 3π‘₯ < πœ‹/2 Γ— 3 Ο€ < 3π‘₯ < 3πœ‹/2 We know that sin ΞΈ < 0 in 3rd quadrant sin ΞΈ < 0 for ΞΈ ∈ (πœ‹, 3πœ‹/2) sin 3x < 0 for 3x ∈ (πœ‹, 3πœ‹/2) sin 3x < 0 for x ∈ (πœ‹/3, 3πœ‹/(2 Γ— 3)) sin 3π‘₯ < 0 for π‘₯ ∈ (πœ‹/3, πœ‹/2) –sin 3π‘₯ > 0 for π‘₯ ∈ (πœ‹/3, πœ‹/2) – 3sin 3π‘₯ > 0 for π‘₯ ∈ (πœ‹/3, πœ‹/2) f’(π‘₯) > 0 for π‘₯ ∈ (πœ‹/3, πœ‹/2) Hence f(π‘₯) is strictly increasing for π‘₯ ∈ (πœ‹/3, πœ‹/2) Thus f(π‘₯) is neither decreasing nor increasing on 𝒙 ∈ (𝟎 , 𝝅/𝟐) Ex 6.2, 12 Which of the following functions are strictly decreasing on (0,πœ‹/2) ? (D) tan π‘₯ Let f(π‘₯) = tan π‘₯ So, f’(π‘₯) = sec^2 π‘₯ As square of any number is always positive So, f’(π‘₯) > 0 for any value of π‘₯ Hence f’(π‘₯) > 0 for x ∈ (0 , πœ‹/2) ∴ f(π‘₯) is strictly increasing function Hence, (𝑨) & (𝑩) is strictly decreasing on (0 , πœ‹/2)

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.