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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

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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]

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.