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

Transcript

Ex 6.5,2 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (𝑖) 𝑓 (π‘₯)=|π‘₯ + 2| –1 𝑓 (π‘₯)=|π‘₯ + 2| –1 Minimum value of |π‘₯ + 2|=0 Minimum value of f(π‘₯)= minimum value of |π‘₯ + 2| –1 =0βˆ’1 =βˆ’1 Hence minimum value of f(𝒙)=βˆ’πŸ And there is no maximum value of f(x) Ex 6.5,2 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (𝑖) 𝑓 (π‘₯)=|π‘₯ + 2| –1 𝑓 (π‘₯)=|π‘₯ + 2| –1 f(π‘₯)= {β–ˆ((π‘₯+2)βˆ’1" w" β„Žπ‘’π‘Ÿπ‘’ π‘₯β‰₯βˆ’2@"–" (π‘₯+2)βˆ’1 π‘€β„Žπ‘’π‘Ÿπ‘’ " " π‘₯<βˆ’2" " )─ f(π‘₯)= {β–ˆ(π‘₯+1" , " π‘€β„Žπ‘’π‘Ÿπ‘’ π‘₯β‰₯βˆ’2@βˆ’π‘₯βˆ’3" , where " π‘₯<βˆ’2)─ We check sign of f’(π‘₯) at π‘₯=βˆ’2 For 𝒙<βˆ’πŸ f(π‘₯)=βˆ’π‘₯βˆ’3 f’(π‘₯)=βˆ’1 ∴ f’(π‘₯)<0 for any value of π‘₯<βˆ’2 For x > –2 f(π‘₯) =π‘₯+1 f’(π‘₯) =1 ∴ f’(π‘₯)=1>0 for any value of π‘₯>βˆ’2 Thus, at x = –2, f’(x) changes sign from negative to positive β‡’ π‘₯=βˆ’2 is point of minima of f(π‘₯) Finding minimum value of f(π‘₯) Putting π‘₯=βˆ’2 f(π‘₯) =|π‘₯ + 2| –1 f(2)=|βˆ’2+ 2| –1 = 0 – 1 = βˆ’1 Minimum value of f(π‘₯) is –1 at π‘₯=βˆ’2 f(π‘₯) has no maximum value Ex 6.5,2 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (ii) 𝑓(π‘₯)= – | π‘₯ +1|+3 f(π‘₯)= – | π‘₯ +1|+3 We know that | π‘₯ +1|β‰₯0 So, βˆ’| π‘₯ +1|≀0 Maximum value of g(π‘₯) = maximum value of – | π‘₯ +1|+3 = 0 + 3 = 3 Hence maximum value of f(x) is 3 And there is no minimum value of f(x) Ex 6.5,2 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (ii) 𝑓(π‘₯)= – |π‘₯+1|+3 g(π‘₯)= –|π‘₯+1|+3 g(π‘₯)= {β–ˆ("–" (π‘₯+1)+3" " π‘₯β‰₯βˆ’1@"–" (βˆ’(π‘₯+1))+3" " π‘₯<βˆ’1)─ g(π‘₯)= {β–ˆ(βˆ’π‘₯+2" " π‘₯β‰₯βˆ’1@π‘₯+4" " π‘₯<βˆ’1)─ We check sign of f’(π‘₯) at π‘₯=βˆ’1 For 𝒙<βˆ’πŸ f(π‘₯)=π‘₯+4 f’(π‘₯)=1 ∴ f’(π‘₯)>0 for any value of π‘₯<βˆ’1 For x > –1 f(π‘₯) =βˆ’π‘₯+2 f’(π‘₯) =βˆ’1 ∴ f’(π‘₯)=<0 for any value of π‘₯>βˆ’1 Thus, at x = –1, f’(x) changes sign from positive to negative β‡’ π‘₯=βˆ’1 is point of maxima of f(π‘₯) Finding maximum value of f(π‘₯) Putting π‘₯=βˆ’1 f(π‘₯)=βˆ’ | π‘₯ +1|+3 f(π‘₯)=βˆ’ | π‘₯ +1|+3 f(βˆ’1) =βˆ’ |βˆ’1 +1|+3 =βˆ’ |0|+3 = 3 Hence Maximum value is 3 And there is no minimum value of f(x) Ex 6.5,2 Find the maximum and minimum values, if any, of the following functions given by (iii) β„Ž(π‘₯)= sin ⁑(2π‘₯)+ 5 β„Ž(π‘₯)= sin ⁑(2π‘₯)+ 5 We know that –1 ≀ sin ΞΈ ≀ 1 –1 ≀ sin 2π‘₯ ≀ 1 Adding 5 both sides –1 + 5 ≀ sin2π‘₯ + 5 ≀ 1 + 5 4 ≀ sin 2π‘₯ + 5 ≀ 6 4 ≀ f(π‘₯)≀6 Hence Maximum value of f(𝒙)=πŸ” & Minimum value of f(𝒙)=πŸ’ Ex 6.5,2 Find the maximum and minimum values, if any, of the following functions given by (iv) 𝑓 (π‘₯)=|sin⁑4π‘₯+3| 𝑓 (π‘₯)=| sin⁑4π‘₯+3| We know that –1 ≀ sin ΞΈ ≀ 1 So, –1 ≀ sin 4π‘₯ ≀ 1 Adding 3 both sides –1 + 3 ≀ sin 4π‘₯ + 3 ≀ 1 + 3 2 ≀ sin 4π‘₯ +3 ≀ 4 Taking modulus |2| ≀ | sin⁑4π‘₯+3| ≀ |4| 2 ≀ | sin⁑4π‘₯+3| ≀ |4| 2 ≀ f(π‘₯)≀4 Hence Maximum value of f(𝒙) is 4 & Minimum value of f(𝒙) is 2 Ex 6.5,2 Find the maximum and minimum values, if any, of the following functions given by (v) β„Ž(π‘₯)=π‘₯ + 1 , π‘₯ ∈ (βˆ’1 , 1) Drawing graph of f(π‘₯)=π‘₯+1 β„Ž(π‘₯) have Maximum value of point closest to x = 1 & Minimum value of point closest to x = –1 but its not possible to locate such points Thus the given function has neither the maximum value nor minimum value

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.