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Example 13 - Find intervals where f(x) = sin x + cos x is - Find intervals of increasing/decreasing

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
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Example 13 Find the intervals in which the function f given by f ﷐𝑥﷯=﷐sin﷮𝑥﷯+﷐cos﷮𝑥﷯ , 0 ≤ 𝑥 ≤ 2𝜋 is strictly increasing or strictly decreasing. f﷐𝑥﷯ = sin 𝑥 + cos 𝑥 0 ≤ x ≤ 2π Step 1: Finding f’﷐𝑥﷯ f﷐𝑥﷯ = sin x + cos 𝑥 f’﷐𝑥﷯ = ﷐𝑑 ﷮𝑑𝑥﷯ (sin 𝑥 + cos 𝑥) f’﷐𝑥﷯ = ﷐𝑑﷐﷐sin﷮𝑥﷯﷯﷮𝑑𝑥﷯ + ﷐𝑑﷐﷐cos﷮𝑥﷯﷯﷮𝑑𝑥﷯ f’﷐𝑥﷯ = cos 𝑥 + ﷐−𝑠𝑖𝑛𝑥﷯ f’﷐𝑥﷯ = cos⁡𝑥 – sin⁡𝑥 Step 2: Putting f’﷐𝑥﷯ = 0 cos 𝑥 − sin 𝑥 = 0 cos 𝑥 = sin 𝑥 ∴𝑥=﷐𝜋﷮4﷯ ,﷐5𝜋﷮4﷯ 𝑎𝑠 0 ≤ 𝑥 ≤ 2𝜋 Step 3: Plotting points Points 𝑥=﷐𝜋﷮4﷯ ,﷐5𝜋﷮4﷯ divide interval ﷐0 , 2𝜋﷯ into 3 disjoint intervals ﷐0 , ﷐𝜋﷮4﷯﷯, ﷐﷐𝜋﷮4﷯,﷐5𝜋﷮4﷯﷯, ﷐﷐5𝜋﷮4﷯ , 2𝜋﷯ Step 4: Checking sign of f’﷐𝑥﷯ = cos 𝑥 – sin 𝑥 Case 1 When 𝑥 ∈ ﷐0 , ﷐𝜋﷮4﷯﷯ as 0 ≤ 𝑥 < ﷐𝜋﷮4﷯ Thus, f’﷐𝑥﷯ > 0 for 𝑥 ∈ ﷐0 , ﷐𝜋﷮4﷯﷯ Case 2 When 𝑥 ∈ ﷐﷐𝜋﷮4﷯,﷐5𝜋﷮4﷯﷯ As ﷐𝜋﷮4﷯ < x < ﷐5𝜋﷮4﷯ Let us find value of f’(x) at any value of 𝑥 lies between ﷐𝜋﷮4﷯,﷐5𝜋﷮4﷯ f’﷐𝑥﷯ < 0 for 𝑥 ∈ ﷐﷐𝜋﷮4﷯,﷐5𝜋﷮4﷯﷯ Case 3 When 𝑥 ∈ ﷐﷐5𝜋﷮4﷯ , 2𝜋﷯ as ﷐5𝜋﷮4﷯ < 𝑥 ≤ 2𝜋 At 𝑥 = 2π f’﷐𝑥﷯ = cos 𝑥 – sin 𝑥 f’﷐2𝜋﷯ = cos 2𝜋 – sin 2𝜋 = cos ﷐𝜋+𝜋﷯ – 0 = – cos π – 0 = − ﷐−1﷯ −0 = 1 > 0 So, f’﷐𝑥﷯ > 0 at 𝑥 = 2π Hence f’(x) > 0 for 𝑥 ∈ ﷐﷐5𝜋﷮4﷯ , 2𝜋﷯ Thus, f is strictly increasing intervals ﷐𝟎 , ﷐𝝅﷮𝟒﷯﷯& ﷐﷐𝟓𝝅﷮𝟒﷯ , 𝟐𝝅﷯ f is strictly increasing intervals ﷐﷐𝝅﷮𝟒﷯ , ﷐𝟓𝝅﷮𝟒﷯﷯

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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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