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Ex 6.2, 12 - Which functions are strictly decreasing on - Ex 6.2

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
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Ex 6.2,12 Which of the following functions are strictly decreasing on ﷐0 ,﷐𝜋﷮2﷯﷯? (A) cos 𝑥 Let f﷐𝑥﷯ = cos 𝑥 Step 1: Finding f’﷐𝑥﷯ f’﷐𝑥﷯ = – sin 𝑥 Hence f’﷐𝑥﷯ < 0 for 𝑥 ∈ ﷐0 , ﷐𝜋﷮2﷯﷯ ⇒ f﷐𝑥﷯ is strictly decreasing on ﷐𝟎 , ﷐𝝅﷮𝟐﷯﷯ Ex 6.2,12 Which of the following functions are strictly decreasing on ﷐0,﷐𝜋﷮2﷯﷯ ? (B) cos 2𝑥 Let f﷐𝑥﷯ = cos 2𝑥 Step 1: Finding f’﷐𝑥﷯ f’﷐𝑥﷯ = ﷐﷐cos﷮2𝑥﷯﷯′ f’﷐𝑥﷯ = – 2 sin 2𝑥 Step 2: Putting f’﷐𝑥﷯ = 0 f’﷐𝑥﷯ = 0 – 2 sin 2𝑥 = 0 sin 2𝑥 = 0 Step 3: As 𝑥 ∈ ﷐0 , ﷐𝜋﷮2﷯﷯ 0 < 𝑥 < ﷐𝜋﷮2﷯ 0 . 2 < 2𝑥 < ﷐𝜋﷮2﷯ × 2 0 < 2𝑥 < π We know that sin θ > 0 for θ ∈ ﷐0 , 𝜋﷯ sin 2𝑥 > 0 for 2𝑥 ∈ ﷐0 , 𝜋﷯ –sin 2𝑥 < 0 for 2𝑥 ∈ ﷐0 , 𝜋﷯ – 2 sin 2𝑥 < 0 for 𝑥 ∈ ﷐0 , ﷐𝜋﷮2﷯﷯ ⇒ f’﷐𝑥﷯ < 0 for 𝑥 ∈ ﷐0 , ﷐𝜋﷮2﷯﷯ ⇒ f ﷐𝑥﷯ is strictly decreasing for 𝒙 ∈ ﷐𝟎 , ﷐𝝅﷮𝟐﷯﷯ Ex 6.2,12 Which of the following functions are strictly decreasing on ﷐0,﷐𝜋﷮2﷯﷯ ? (C) cos 3𝑥 Let f﷐𝑥﷯ = cos 3𝑥 as 𝑥 ∈ ﷐0 ,﷐𝜋﷮2﷯﷯ Step 1: Finding f’﷐𝑥﷯. f’﷐𝑥﷯ = ﷐﷐﷐cos﷮3𝑥﷯﷯﷮′﷯ f’﷐𝑥﷯ = –3 sin 3𝑥 Step 2: f’﷐𝑥﷯ = 0 –3 sin 3𝑥 = 0 sin 3𝑥 = 0 As sin θ = 0 if θ =0 , π , 2π , 3π ⇒ 3𝑥 = π , 2π 𝑥 = ﷐𝜋﷮3﷯ , ﷐2𝜋﷮3﷯ As 𝑥 ∈ ﷐0 , ﷐𝜋﷮2﷯﷯ So, 𝑥 = ﷐𝜋﷮3﷯ Step 3: Since 𝑥 ∈ ﷐0 , ﷐𝜋﷮2﷯﷯. so we start number line from ﷐0 , ﷐𝜋﷮2﷯﷯ The point x = ﷐𝜋﷮3﷯ divide the interval ﷐0 , ﷐𝜋﷮2﷯﷯ into 2 disjoint interval. i.e. ﷐0 , ﷐𝜋﷮3﷯﷯ & ﷐﷐𝜋﷮3﷯ , ﷐𝜋﷮2﷯﷯ Step 4: Checking sign of f’﷐𝑥﷯ Case 1: For 𝒙 ∈ ﷐𝟎 , ﷐𝝅﷮𝟑﷯﷯ Now, 0 < 𝑥 < ﷐𝜋﷮3﷯ 3 × 0 < 3𝑥 < ﷐3𝜋﷮3﷯ 0 < 3𝑥 < π So, when 𝑥 ∈ ﷐0 , ﷐𝜋﷮3﷯﷯ , 3𝑥 ∈ ﷐0 , 𝜋﷯ We know that sinθ > 0 for θ ∈ ﷐0 , 𝜋﷯ sin 3𝑥 > 0 for 3𝑥 ∈ ﷐0 , 𝜋﷯ sin 3𝑥 > 0 for 𝑥 ∈ ﷐0 , ﷐𝜋﷮3﷯﷯ – sin 3𝑥 < 0 for 𝑥 ∈ ﷐0 ,﷐𝜋﷮3﷯﷯ ⇒ f’﷐𝑥﷯ < 0 for 𝑥 ∈ ﷐0 , ﷐𝜋﷮3﷯﷯ ⇒ f﷐𝑥﷯ is strictly decreasing for 𝑥 ∈ ﷐0 ,﷐𝜋﷮3﷯﷯ Case 2: For 𝒙 ∈ ﷐﷐𝝅﷮𝟑﷯, ﷐𝝅﷮𝟐﷯﷯ Since ﷐𝜋﷮3﷯ < 𝑥 < ﷐𝜋﷮2﷯ 3 × ﷐𝜋﷮3﷯ < 3𝑥 < ﷐𝜋﷮2﷯ × 3 π < 3𝑥 < ﷐3𝜋﷮2﷯ We know that sin θ < 0 in 3rd quadrant sin θ < 0 for θ ∈ ﷐𝜋, ﷐3𝜋﷮2﷯﷯ sin 3x < 0 for 3x ∈ ﷐𝜋, ﷐3𝜋﷮2﷯﷯ sin 3x < 0 for x ∈ ﷐﷐𝜋﷮3﷯, ﷐3𝜋﷮2 × 3﷯﷯ sin 3𝑥 < 0 for 𝑥 ∈ ﷐﷐𝜋﷮3﷯, ﷐𝜋﷮2﷯﷯ –sin 3𝑥 > 0 for 𝑥 ∈ ﷐﷐𝜋﷮3﷯, ﷐𝜋﷮2﷯﷯ – 3sin 3𝑥 > 0 for 𝑥 ∈ ﷐﷐𝜋﷮3﷯, ﷐𝜋﷮2﷯﷯ f’﷐𝑥﷯ > 0 for 𝑥 ∈ ﷐﷐𝜋﷮3﷯, ﷐𝜋﷮2﷯﷯ Hence f﷐𝑥﷯ is strictly increasing for 𝑥 ∈ ﷐﷐𝜋﷮3﷯, ﷐𝜋﷮2﷯﷯ Thus f﷐𝑥﷯ 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﷮2﷯﷯ ? (D) tan 𝑥 Let f﷐𝑥﷯ = tan 𝑥 So, f’﷐𝑥﷯ = ﷐sec﷮2﷯ 𝑥 As square of any number is always positive So, f’﷐𝑥﷯ > 0 for any value of 𝑥 Hence f’﷐𝑥﷯ > 0 for x ∈ ﷐0 , ﷐𝜋﷮2﷯﷯ ∴ f﷐𝑥﷯ is strictly increasing function Hence ﷐𝑨﷯ & ﷐𝑩﷯ is strictly decreasing on ﷐0 , ﷐𝜋﷮2﷯﷯

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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