Ex 6.2, 9 - Prove that y = 4 sin/2 + cos - theta is increasing

Ex 6.2,9 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.2,9 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.2,9 - Chapter 6 Class 12 Application of Derivatives - Part 4 Ex 6.2,9 - Chapter 6 Class 12 Application of Derivatives - Part 5

  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

Transcript

Ex 6.2, 9 Prove that 𝑦 = (4 sinβ‘πœƒ)/((2 +γ€– cosγ€—β‘γ€–πœƒ)γ€— ) – ΞΈ is an increasing function of ΞΈ in[0,πœ‹/2] . 𝑦 = (4 sinβ‘πœƒ)/((2 + cosβ‘πœƒ ) )βˆ’πœƒ We need to prove that function is increasing function of ΞΈ in (0 , πœ‹/2) i.e. we need to prove that π’…π’š/(π’…πœ½ ) > 0 for ΞΈ ∈ [𝟎 , 𝝅/𝟐] Differentiating w.r.t ΞΈ 𝑑𝑦/(π‘‘πœƒ ) = 𝑑/π‘‘πœƒ ((4 sinβ‘πœƒ)/(2 + cosβ‘πœƒ ) βˆ’πœƒ) 𝑑𝑦/(π‘‘πœƒ ) = 𝑑/π‘‘πœƒ ((4 sinβ‘πœƒ)/(2 +cosβ‘πœƒ ) )βˆ’ π‘‘πœƒ/π‘‘πœƒ 𝑑𝑦/(π‘‘πœƒ ) = 𝑑/π‘‘πœƒ ((4 sinβ‘πœƒ)/(2 +cosβ‘πœƒ ) )βˆ’ 1 𝑑𝑦/𝑑π‘₯ = ((πŸ’ π’”π’Šπ’β‘πœ½ )^β€² (𝟐 + π’„π’π’”β‘πœ½ ) βˆ’ (𝟐 + π’„π’π’”β‘πœ½ )^β€² (πŸ’ π’”π’Šπ’β‘πœ½ ))/(𝟐 + π’„π’π’”β‘πœ½ )𝟐 βˆ’1 𝑑𝑦/𝑑π‘₯ = (4 cosβ‘πœƒ (2 + cosβ‘πœƒ ) βˆ’ (0 βˆ’ sinβ‘πœƒ )(4 sinβ‘πœƒ ))/(2 + cosβ‘πœƒ )2 βˆ’1 𝑑𝑦/𝑑π‘₯ = (8 cosβ‘γ€–πœƒ + 4 (π’„π’π’”πŸπœ½ + π’”π’Šπ’πŸπœ½)γ€—)/(2 + cosβ‘πœƒ )2 βˆ’1 𝑑𝑦/𝑑π‘₯ = (8 cosβ‘γ€–πœƒ + 4(𝟏)γ€—)/(2 + cosβ‘πœƒ )2 βˆ’1 𝑑𝑦/𝑑π‘₯ = (8 cosβ‘γ€–πœƒ + 4γ€— βˆ’ (2 + cosβ‘πœƒ )2)/(2 + cosβ‘πœƒ )2 Using quotient rule as (𝑒/𝑣)^β€²= (𝑒^β€² 𝑣 βˆ’ 𝑣^β€² 𝑒)/𝑣2 Where u = 4 sinβ‘πœƒ & 𝑣= 2+cosβ‘πœƒ 𝑑𝑦/𝑑π‘₯ = (8 cosβ‘γ€–πœƒ + 4γ€— βˆ’ (4 + π‘π‘œπ‘ 2 πœƒ + 4 π‘π‘œπ‘ πœƒ))/(2 + cosβ‘πœƒ )2 𝑑𝑦/𝑑π‘₯ = (8 cosβ‘γ€–πœƒ + 4γ€— βˆ’ 4 βˆ’ π‘π‘œπ‘ 2πœƒ βˆ’ 4 π‘π‘œπ‘ πœƒ)/(2 + cosβ‘πœƒ )2 𝑑𝑦/𝑑π‘₯ = (8 cosβ‘γ€–πœƒ βˆ’ 4 cosβ‘πœƒ + 4 βˆ’ 4 βˆ’ π‘π‘œπ‘ 2πœƒγ€—)/(2 + cosβ‘πœƒ )2 𝑑𝑦/𝑑π‘₯ = (4 cos πœƒ + 0 βˆ’ π‘π‘œπ‘ 2πœƒ)/(2 + cosβ‘πœƒ )2 𝑑𝑦/𝑑π‘₯ = (4 cosβ‘γ€–πœƒ βˆ’ π‘π‘œπ‘ 2 πœƒγ€—)/(2 + cosβ‘πœƒ )2 π’…π’š/𝒅𝒙 = (π’„π’π’”β‘πœ½ (πŸ’ βˆ’ π’„π’π’”β‘πœ½ ))/(𝟐 + π’„π’π’”β‘πœ½ )𝟐 For the function to be increasing 𝑑𝑦/π‘‘πœƒ > 0 (𝐜𝐨𝐬⁑𝜽 (πŸ’ βˆ’ π’„π’π’”β‘πœ½ ))/(𝟐 + π’„π’π’”β‘πœ½ )𝟐 > 0 Since denominator is a square, We need to show that cos ΞΈ (πŸ’βˆ’πœπ¨π¬β‘πœ½ ) > 0 for ΞΈ ∈ [0 , πœ‹/2] Now, 0 ≀ cos ΞΈ ≀ 1 Multiplying by –1 –1 ≀ –cos ΞΈ ≀ 0 Adding 4 both sides –1 + 4 ≀ – cos ΞΈ + 4 < 0 + 4 3 ≀ 4 – cos ΞΈ < 4 Thus, (4 – cos ΞΈ) is positive And, cos ΞΈ is also positive Hence, cos ΞΈ (4 – cos2 ΞΈ) is also positive ∴ cos ΞΈ (4 – cos2 ΞΈ) > 0 for ΞΈ ∈ [0 , πœ‹/2 ] Hence 𝑦 = (4 sinβ‘πœƒ)/(2 +cosβ‘πœƒ ) – ΞΈ is an increasing function for ΞΈ ∈ [𝟎 , 𝝅/𝟐]

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.