Example 12 - Find intervals where f(x) = sin 3x is decreasing

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Example 12 - Chapter 6 Class 12 Application of Derivatives - Part 2

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Example 12 - Chapter 6 Class 12 Application of Derivatives - Part 3 Example 12 - Chapter 6 Class 12 Application of Derivatives - Part 4 Example 12 - Chapter 6 Class 12 Application of Derivatives - Part 5 Example 12 - Chapter 6 Class 12 Application of Derivatives - Part 6 Example 12 - Chapter 6 Class 12 Application of Derivatives - Part 7 Example 12 - Chapter 6 Class 12 Application of Derivatives - Part 8

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

Transcript

Example 12 Find intervals in which the function given by f (x) = sin 3x, x, ∈ [0, πœ‹/2] is (a) increasing (b) decreasing. f(π‘₯) = sin 3π‘₯ where π‘₯ ∈ [0 ,πœ‹/2] Finding f’(x) f’(π‘₯) = 𝑑(sin⁑3π‘₯ )/𝑑π‘₯ f’(π‘₯) = cos 3π‘₯ Γ— 3 f’(𝒙) = 3. cos 3𝒙 Putting f’(𝒙) = 0 3 cos 3π‘₯ = 0 cos 3π‘₯ = 0 We know that cos ΞΈ = 0 When ΞΈ = πœ‹/2 & 3πœ‹/2 So, for cos 3𝒙 = 0 3π‘₯ = πœ‹/2 & 3π‘₯ = 3πœ‹/2 π‘₯ = πœ‹/(2 Γ—3) & π‘₯ = 3πœ‹/(2 Γ— 3) 𝒙 = 𝝅/πŸ” & 𝒙 = 𝝅/𝟐 Since π‘₯ = πœ‹/6 ∈ [𝟎 ,𝝅/𝟐] & π‘₯ = πœ‹/2 ∈ [𝟎,𝝅/𝟐] ∴ Both values of π‘₯ are valid Plotting points on number line So, point π‘₯ = πœ‹/6 divides the interval into two disjoint intervals [0 ,πœ‹/6) and (πœ‹/6, πœ‹/2] Checking sign of f’(𝒙) f’(π‘₯) = 3. cos 3π‘₯ Case 1: For 𝒙 ∈ (𝟎 ,𝝅/πŸ”) 0<π‘₯<πœ‹/6 3 Γ— 0<3π‘₯<3πœ‹/6 𝟎<πŸ‘π’™<𝝅/𝟐 So when π‘₯ ∈ (0 ,πœ‹/6), then 3π‘₯ ∈ (0 , πœ‹/2) We know that cos 𝜽>𝟎 for 𝜽 ∈ (𝟎 , 𝝅/𝟐) cos 3x >0 for 3x ∈ (0 , πœ‹/2) cos 3x >0 for x ∈ (0 , πœ‹/6) 3 cos 3x >0 for x ∈ (0 , πœ‹/6) 𝒇′(𝒙)>𝟎 for x ∈ (0 , πœ‹/6) Since f’(0) = 3 and f’(𝝅/πŸ”) = 0 Therefore, f’(x) β‰₯ 0 for π‘₯ ∈ [0 , πœ‹/6] Thus, f(x) is increasing for π‘₯ ∈ [0 , πœ‹/6] Case 2: For 𝒙 ∈ (𝝅/πŸ”, 𝝅/𝟐) πœ‹/6<π‘₯<πœ‹/2 3 Γ— πœ‹/6<3π‘₯<3πœ‹/2 𝝅/𝟐<πŸ‘π’™<πŸ‘π…/𝟐 So when π‘₯ ∈(πœ‹/6 , πœ‹/2), then 3π‘₯ ∈ (πœ‹/2 , 3πœ‹/2) We know that, cos πœƒ<0 for πœƒ ∈ (πœ‹/2 , 3πœ‹/2) cos 3π‘₯<0 for 3π‘₯ ∈ (πœ‹/2 , 3πœ‹/2) cos 3π‘₯<0 for π‘₯ ∈ (πœ‹/6 , πœ‹/2) 3 cos 3π‘₯<0 for π‘₯ ∈ (πœ‹/6 , πœ‹/2) fβ€˜(x) <𝟎 for π‘₯ ∈ (πœ‹/6 , πœ‹/2) Since f’(𝝅/πŸ”) = 0 and f’(𝝅/𝟐) = 0 Therefore f’(x) ≀ 0 for π‘₯ ∈ [πœ‹/6,πœ‹/2] Thus, f(x) is decreasing for π‘₯ ∈ [πœ‹/6,πœ‹/2] (As cos πœƒ is negative in 2nd and 3rd quadrant) Thus, f(x) is increasing for 𝒙 ∈ [𝟎 , 𝝅/πŸ”] & f(x) is decreasing for 𝒙 ∈ [𝝅/πŸ” , 𝝅/𝟐]

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.