1. Chapter 6 Class 12 Application of Derivatives
2. Serial order wise
3. Ex 6.5

Transcript

Ex 6.5,2 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (π) π (π₯)=|π₯ + 2| β1 π (π₯)=|π₯ + 2| β1 Minimum value of |π₯ + 2|=0 Minimum value of f(π₯)= minimum value of |π₯ + 2| β1 =0β1 =β1 Hence minimum value of f(π)=βπ 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 (π) π (π₯)=|π₯ + 2| β1 π (π₯)=|π₯ + 2| β1 f(π₯)= {β((π₯+2)β1" w" βπππ π₯β₯β2@"β" (π₯+2)β1 π€βπππ " " π₯<β2" " )β€ f(π₯)= {β(π₯+1" , " π€βπππ π₯β₯β2@βπ₯β3" , where " π₯<β2)β€ We check sign of fβ(π₯) at π₯=β2 For π<βπ f(π₯)=βπ₯β3 fβ(π₯)=β1 β΄ fβ(π₯)<0 for any value of π₯<β2 For x > β2 f(π₯) =π₯+1 fβ(π₯) =1 β΄ fβ(π₯)=1>0 for any value of π₯>β2 Thus, at x = β2, fβ(x) changes sign from negative to positive β π₯=β2 is point of minima of f(π₯) Finding minimum value of f(π₯) Putting π₯=β2 f(π₯) =|π₯ + 2| β1 f(2)=|β2+ 2| β1 = 0 β 1 = β1 Minimum value of f(π₯) is β1 at π₯=β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) π(π₯)= β | π₯ +1|+3 f(π₯)= β | π₯ +1|+3 We know that | π₯ +1|β₯0 So, β| π₯ +1|β€0 Maximum value of g(π₯) = maximum value of β | 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) π(π₯)= β |π₯+1|+3 g(π₯)= β|π₯+1|+3 g(π₯)= {β("β" (π₯+1)+3" " π₯β₯β1@"β" (β(π₯+1))+3" " π₯<β1)β€ g(π₯)= {β(βπ₯+2" " π₯β₯β1@π₯+4" " π₯<β1)β€ We check sign of fβ(π₯) at π₯=β1 For π<βπ f(π₯)=π₯+4 fβ(π₯)=1 β΄ fβ(π₯)>0 for any value of π₯<β1 For x > β1 f(π₯) =βπ₯+2 fβ(π₯) =β1 β΄ fβ(π₯)=<0 for any value of π₯>β1 Thus, at x = β1, fβ(x) changes sign from positive to negative β π₯=β1 is point of maxima of f(π₯) Finding maximum value of f(π₯) Putting π₯=β1 f(π₯)=β | π₯ +1|+3 f(π₯)=β | π₯ +1|+3 f(β1) =β |β1 +1|+3 =β |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(π₯)= sin β‘(2π₯)+ 5 β(π₯)= sin β‘(2π₯)+ 5 We know that β1 β€ sin ΞΈ β€ 1 β1 β€ sin 2π₯ β€ 1 Adding 5 both sides β1 + 5 β€ sin2π₯ + 5 β€ 1 + 5 4 β€ sin 2π₯ + 5 β€ 6 4 β€ f(π₯)β€6 Hence Maximum value of f(π)=π & Minimum value of f(π)=π Ex 6.5,2 Find the maximum and minimum values, if any, of the following functions given by (iv) π (π₯)=|sinβ‘4π₯+3| π (π₯)=| sinβ‘4π₯+3| We know that β1 β€ sin ΞΈ β€ 1 So, β1 β€ sin 4π₯ β€ 1 Adding 3 both sides β1 + 3 β€ sin 4π₯ + 3 β€ 1 + 3 2 β€ sin 4π₯ +3 β€ 4 Taking modulus |2| β€ | sinβ‘4π₯+3| β€ |4| 2 β€ | sinβ‘4π₯+3| β€ |4| 2 β€ f(π₯)β€4 Hence Maximum value of f(π) is 4 & Minimum value of f(π) is 2 Ex 6.5,2 Find the maximum and minimum values, if any, of the following functions given by (v) β(π₯)=π₯ + 1 , π₯ β (β1 , 1) Drawing graph of f(π₯)=π₯+1 β(π₯) have Maximum value of point closest to x = 1 & Minimum value of point closest to x = β1 but its not possible to locate such points Thus the given function has neither the maximum value nor minimum value

Ex 6.5