     1. Chapter 6 Class 12 Application of Derivatives
2. Serial order wise
3. Ex 6.2

Transcript

Ex 6.2, 9 Prove that 𝑦 = (4 sin⁡𝜃)/((2 +〖 cos〗⁡〖𝜃)〗 ) – θ is an increasing function of θ in[0,𝜋/2] . 𝑦 = (4 sin⁡𝜃)/((2 + cos⁡𝜃 ) )−𝜃 We need to prove that function is increasing function of θ in (0 , 𝜋/2) i.e. we need to prove that 𝑑𝑦/(𝑑𝜃 ) > 0 for θ ∈ [0 , 𝜋/2] We have 𝑦 = (4 sin⁡𝜃)/(2 + cos⁡𝜃 ) – θ 𝑑𝑦/(𝑑𝜃 ) = 𝑑/𝑑𝜃 ((4 sin⁡𝜃)/(2 + cos⁡𝜃 ) −𝜃) 𝑑𝑦/(𝑑𝜃 ) = 𝑑/𝑑𝜃 ((4 sin⁡𝜃)/(2 +cos⁡𝜃 ) )− 𝑑𝜃/𝑑𝜃 𝑑𝑦/(𝑑𝜃 ) = 𝑑/𝑑𝜃 ((4 sin⁡𝜃)/(2 +cos⁡𝜃 ) )− 1 𝑑𝑦/𝑑𝑥 = ((4 sin⁡𝜃 )^′ (2 + cos⁡𝜃 ) − (2 + cos⁡𝜃 )^′ (4 sin⁡𝜃 ))/(2 + cos⁡𝜃 )2 −1 𝑑𝑦/𝑑𝑥 = (4 cos⁡𝜃 (2 + cos⁡𝜃 ) − (0 − sin⁡𝜃 )(4 sin⁡𝜃 ))/(2 + cos⁡𝜃 )2 −1 𝑑𝑦/𝑑𝑥 = (8 cos⁡〖𝜃 + 4 (𝑐𝑜𝑠2𝜃 + 𝑠𝑖𝑛2𝜃)〗)/(2 + cos⁡𝜃 )2 −1 𝑑𝑦/𝑑𝑥 = (8 cos⁡〖𝜃 + 4(1)〗)/(2 + cos⁡𝜃 )2 −1 Using quotient rule as (𝑢/𝑣)^′= (𝑢^′ 𝑣 − 𝑣^′ 𝑢)/𝑣2 Where u = 4 sin⁡𝜃 & 𝑣= 2+cos⁡𝜃 𝑑𝑦/𝑑𝑥 = (8 cos⁡〖𝜃 + 4〗)/(2 + cos⁡𝜃 )2 −1 𝑑𝑦/𝑑𝑥 = (8 cos⁡〖𝜃 + 4〗 − (2 + cos⁡𝜃 )2)/(2 + cos⁡𝜃 )2 𝑑𝑦/𝑑𝑥 = (8 cos⁡〖𝜃 + 4〗 − (4 + 𝑐𝑜𝑠2 𝜃 + 4 𝑐𝑜𝑠𝜃))/(2 + cos⁡𝜃 )2 𝑑𝑦/𝑑𝑥 = (8 cos⁡〖𝜃 + 4〗 − 4 − 𝑐𝑜𝑠2𝜃 − 4 𝑐𝑜𝑠𝜃)/(2 + cos⁡𝜃 )2 𝑑𝑦/𝑑𝑥 = (8 cos⁡〖𝜃 − 4 cos⁡𝜃 + 4 − 4 − 𝑐𝑜𝑠2𝜃〗)/(2 + cos⁡𝜃 )2 𝑑𝑦/𝑑𝑥 = (4 cos 𝜃 + 0 − 𝑐𝑜𝑠2𝜃)/(2 + cos⁡𝜃 )2 𝑑𝑦/𝑑𝑥 = (4 cos⁡〖𝜃 − 𝑐𝑜𝑠2 𝜃〗)/(2 + cos⁡𝜃 )2 𝑑𝑦/𝑑𝑥 = (cos⁡𝜃 (4 − cos⁡𝜃 ))/(2 + cos⁡𝜃 )2 For the function to be increasing 𝑑𝑦/𝑑𝜃 = (cos⁡𝜃 (4 − cos⁡𝜃 ))/(2 + cos⁡𝜃 )2 > 0 Since denominator is a square, We need to show that cos θ (4−cos⁡𝜃 ) > 0 for θ ∈ [0 , 𝜋/2] Now, 0 ≤ cos θ ≤ 1 Multiplying by –1 –1 ≤ –cos θ ≤ 0 Adding 4 both sides –1 + 4 ≤ – cos θ + 4 < 0 + 4 3 ≤ 4 – cos θ < 4 Thus, (4 – cos θ) is positive And cos θ is also positive Hence cos θ (4 – cos2 θ) is also positive ⇒ cos θ (4 – cos2 θ) > 0 for θ ∈ [0 , 𝜋/2 ] Hence 𝑦 = (4 sin⁡𝜃)/(2 +cos⁡𝜃 ) – θ is an increasing function for θ ∈ [𝟎 , 𝝅/𝟐]

Ex 6.2 