# Ex 6.2,12

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 6.2,12 Which of the following functions are strictly decreasing on 0 ,𝜋2? (A) cos 𝑥 Let f𝑥 = cos 𝑥 Step 1: Finding f’𝑥 f’𝑥 = – sin 𝑥 Hence f’𝑥 < 0 for 𝑥 ∈ 0 , 𝜋2 ⇒ f𝑥 is strictly decreasing on 𝟎 , 𝝅𝟐 Ex 6.2,12 Which of the following functions are strictly decreasing on 0,𝜋2 ? (B) cos 2𝑥 Let f𝑥 = cos 2𝑥 Step 1: Finding f’𝑥 f’𝑥 = cos2𝑥′ 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 , 𝜋2 0 < 𝑥 < 𝜋2 0 . 2 < 2𝑥 < 𝜋2 × 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 , 𝜋2 ⇒ f’𝑥 < 0 for 𝑥 ∈ 0 , 𝜋2 ⇒ f 𝑥 is strictly decreasing for 𝒙 ∈ 𝟎 , 𝝅𝟐 Ex 6.2,12 Which of the following functions are strictly decreasing on 0,𝜋2 ? (C) cos 3𝑥 Let f𝑥 = cos 3𝑥 as 𝑥 ∈ 0 ,𝜋2 Step 1: Finding f’𝑥. f’𝑥 = cos3𝑥′ 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 = 𝜋3 , 2𝜋3 As 𝑥 ∈ 0 , 𝜋2 So, 𝑥 = 𝜋3 Step 3: Since 𝑥 ∈ 0 , 𝜋2. so we start number line from 0 , 𝜋2 The point x = 𝜋3 divide the interval 0 , 𝜋2 into 2 disjoint interval. i.e. 0 , 𝜋3 & 𝜋3 , 𝜋2 Step 4: Checking sign of f’𝑥 Case 1: For 𝒙 ∈ 𝟎 , 𝝅𝟑 Now, 0 < 𝑥 < 𝜋3 3 × 0 < 3𝑥 < 3𝜋3 0 < 3𝑥 < π So, when 𝑥 ∈ 0 , 𝜋3 , 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 , 𝜋3 – sin 3𝑥 < 0 for 𝑥 ∈ 0 ,𝜋3 ⇒ f’𝑥 < 0 for 𝑥 ∈ 0 , 𝜋3 ⇒ f𝑥 is strictly decreasing for 𝑥 ∈ 0 ,𝜋3 Case 2: For 𝒙 ∈ 𝝅𝟑, 𝝅𝟐 Since 𝜋3 < 𝑥 < 𝜋2 3 × 𝜋3 < 3𝑥 < 𝜋2 × 3 π < 3𝑥 < 3𝜋2 We know that sin θ < 0 in 3rd quadrant sin θ < 0 for θ ∈ 𝜋, 3𝜋2 sin 3x < 0 for 3x ∈ 𝜋, 3𝜋2 sin 3x < 0 for x ∈ 𝜋3, 3𝜋2 × 3 sin 3𝑥 < 0 for 𝑥 ∈ 𝜋3, 𝜋2 –sin 3𝑥 > 0 for 𝑥 ∈ 𝜋3, 𝜋2 – 3sin 3𝑥 > 0 for 𝑥 ∈ 𝜋3, 𝜋2 f’𝑥 > 0 for 𝑥 ∈ 𝜋3, 𝜋2 Hence f𝑥 is strictly increasing for 𝑥 ∈ 𝜋3, 𝜋2 Thus f𝑥 is neither decreasing nor increasing on 𝒙 ∈ 𝟎 , 𝝅𝟐 Ex 6.2,12 Which of the following functions are strictly decreasing on 0,𝜋2 ? (D) tan 𝑥 Let f𝑥 = tan 𝑥 So, f’𝑥 = sec2 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 , 𝜋2 ∴ f𝑥 is strictly increasing function Hence 𝑨 & 𝑩 is strictly decreasing on 0 , 𝜋2

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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.