# Ex 6.2,12 - Chapter 6 Class 12 Application of Derivatives

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 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

Chapter 6 Class 12 Application of Derivatives

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.