Ex 6.2,12 - Chapter 6 Class 12 Application of Derivatives (Term 1)

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Ex 6.2, 12 Which of the following functions are strictly decreasing on (0 ,𝜋/2)? (A) cos 𝑥f(𝑥) = cos 𝑥 f’(𝒙) = – sin 𝒙 Since, sin 𝑥 > 0 for 𝑥 ∈ (0 , 𝜋/2) So, –sin 𝒙 < 0 for 𝑥 ∈ (0 , 𝜋/2) ∴ f’(𝑥) < 0 for 𝑥 ∈ (0 , 𝜋/2) So, f is strictly decreasing in (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 Finding f’(𝒙) f’(𝑥) = (cos⁡2𝑥 )′ f’(𝑥) = –2 sin 2𝑥 Let 2𝑥 = θ ∴ f’(𝑥) = –2 sin θ When 0 < x < 𝜋/2 , then 0 < θ < 𝜋 Since sin θ > 0 for 0 < θ < 𝜋 Therefore, sin 2x > 0 for 0 < 2x < 𝜋 Multiplying −2 both sides −2 × sin 2x < −2 × 0 for 0 < 2x < 𝜋 −2 sin 2x < 0 for 0 < 2x < 𝜋 f’(x) < 0 for 0 < 2x < 𝜋 f’(x) < 0 for 0 < x < 𝝅/𝟐 Therefore, 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 Finding f’(𝒙) f’(𝑥) = (cos⁡3𝑥 )′ f’(𝑥) = –3 sin 3𝑥 Let 3𝑥 = θ ∴ f’(𝑥) = –3 sin θ When 0 < x < 𝜋/2 , then 0 < θ < 𝟑𝝅/𝟐 For 0 < θ < 𝟑𝝅/𝟐 sin θ is positive for 0 < θ < 𝜋 sin θ is negative for 0 < θ < 𝟑𝝅/𝟐 Thus, we can say that sin θ is neither positive nor negative for 0 < θ < 𝟑𝝅/𝟐 Putting θ = 3x sin 3x is neither positive nor negative for 0 < 3x < 𝟑𝝅/𝟐 −3 sin 3x is neither positive nor negative for 0 < 3x < 3𝜋/2 f’(x) is neither positive nor negative for 0 < x < 𝝅/𝟐 Thus, we can write that f(x) is neither increasing nor decreasing for 𝒙 ∈ (𝟎 , 𝝅/𝟐) Ex 6.2, 12 Which of the following functions are strictly decreasing on (0,𝜋/2) ? (D) tan 𝑥f(𝑥) = tan 𝑥 f’(𝒙) = sec2 𝒙 As square of any number is always positive So, f’(𝑥) > 0 for all values of 𝑥 ∴ f is strictly increasing on (0 , 𝜋/2)