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 ๐๏ทฏ ๐ (๐ฅ)=|๐ฅ + 2| โ1 ๐ (๐ฅ)=|๐ฅ + 2| โ1 Minimum value of ๐ฅ + 2๏ทฏ=0 Minimum value of f ๐ฅ๏ทฏ= minimum value of |๐ฅ + 2| โ1 =0โ1 =โ1 Hence minimum value of f ๐๏ทฏ=โ๐ Ex 6.5,2 (Method 2) Find the maximum and minimum values, if any, of the following functions given by ๐๏ทฏ ๐ (๐ฅ)=|๐ฅ + 2| โ1 ๐ (๐ฅ)=|๐ฅ + 2| โ1 f ๐ฅ๏ทฏ= ๐ฅ+2๏ทฏโ1 wโ๐๐๐ ๐ฅโฅโ2๏ทฎโ ๐ฅ+2๏ทฏโ1 ๐คโ๐๐๐ ๐ฅ<โ2 ๏ทฏ๏ทฏ f ๐ฅ๏ทฏ= ๐ฅ+1 , ๐คโ๐๐๐ ๐ฅโฅโ2๏ทฎโ๐ฅโ3 , where ๐ฅ<โ2๏ทฏ๏ทฏ We check sign of fโ ๐ฅ๏ทฏ at ๐ฅ=โ2 Thus, at x = โ2, fโ(x) changes sign from negative to positive โ ๐ฅ=โ2 is point of minima of f ๐ฅ๏ทฏ Finding minimum value of f ๐ฅ๏ทฏ Putting ๐ฅ=โ2 f ๐ฅ๏ทฏ =|๐ฅ + 2| โ1 f 2๏ทฏ=|โ2+ 2| โ1 = 0 โ 1 = โ1 Minimum value of f ๐ฅ๏ทฏ is โ1 at ๐ฅ=โ2 f ๐ฅ๏ทฏ 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) ๐(๐ฅ)= โ | ๐ฅ +1|+3 f(๐ฅ)= โ | ๐ฅ +1|+3 We know that ๐ฅ +1๏ทฏโฅ0 So, โ ๐ฅ +1๏ทฏโค0 Maximum value of g ๐ฅ๏ทฏ = maximum value of โ | ๐ฅ +1|+3 = 0 + 3 = 3 Hence maximum value of f(x) is 3 Ex 6.5,2 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (ii) ๐(๐ฅ)= โ |๐ฅ+1|+3 g(๐ฅ)= โ|๐ฅ+1|+3 g(๐ฅ)= โ ๐ฅ+1๏ทฏ+3 ๐ฅโฅโ1๏ทฎโ โ ๐ฅ+1๏ทฏ๏ทฏ+3 ๐ฅ<โ1๏ทฏ๏ทฏ g(๐ฅ)= โ๐ฅ+2 ๐ฅโฅโ1๏ทฎ๐ฅ+4 ๐ฅ<โ1๏ทฏ๏ทฏ We check sign of fโ ๐ฅ๏ทฏ at ๐ฅ=โ1 Thus, at x = โ1, fโ(x) changes sign from positive to negative โ ๐ฅ=โ1 is point of maxima of f ๐ฅ๏ทฏ Finding maximum value of f ๐ฅ๏ทฏ Putting ๐ฅ=โ1 f ๐ฅ๏ทฏ=โ | ๐ฅ +1|+3 f ๐ฅ๏ทฏ=โ | ๐ฅ +1|+3 f โ1๏ทฏ =โ |โ1 +1|+3 =โ |0|+3 = 3 Hence Maximum value is 3 Ex 6.5,2 Find the maximum and minimum values, if any, of the following functions given by (iii) โ ๐ฅ๏ทฏ= sin โก(2๐ฅ)+ 5 โ ๐ฅ๏ทฏ= sin โก(2๐ฅ)+ 5 We know that โ1 โค sin ฮธ โค 1 โ1 โค sin 2๐ฅ โค 1 Adding 5 both sides โ1 + 5 โค sin2๐ฅ + 5 โค 1 + 5 4 โค sin 2๐ฅ + 5 โค 6 4 โค f ๐ฅ๏ทฏโค6 Hence Maximum value of f ๐๏ทฏ=๐ & Minimum value of f ๐๏ทฏ=๐ Ex 6.5,2 Find the maximum and minimum values, if any, of the following functions given by (iv) ๐ (๐ฅ)=|sinโก4๐ฅ+3| ๐ (๐ฅ)=| sinโก4๐ฅ+3| We know that โ1 โค sin ฮธ โค 1 So, โ1 โค sin 4๐ฅ โค 1 Adding 3 both sides โ1 + 3 โค sin 4๐ฅ + 3 โค 1 + 3 2 โค sin 4๐ฅ +3 โค 4 Taking modulus |2| โค | sinโก4๐ฅ+3| โค |4| 2 โค | sinโก4๐ฅ+3| โค |4| 2 โค f ๐ฅ๏ทฏโค4 Hence Maximum value of f ๐ฅ๏ทฏ is 4 & Minimum value of f ๐ฅ๏ทฏ is 2 Hence Maximum value of f ๐๏ทฏ is 4 & Minimum value of f ๐๏ทฏ is 2 Ex 6.5,2 Find the maximum and minimum values, if any, of the following functions given by (v) โ ๐ฅ๏ทฏ=๐ฅ + 1 , ๐ฅ โ โ1 , 1๏ทฏ Drawing graph of f ๐ฅ๏ทฏ=๐ฅ+1 โ ๐ฅ๏ทฏ have Maximum value of point closest to x = 1 & Minimum value of point closest to x = โ1 but its not possible to locate such points Thus the given function has neither the maximum value nor minimum value