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

Transcript

Ex 6.5,1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (i) f (๐ฅ) = (2๐ฅ โ 1)^2 + 3 f(๐ฅ)=(2๐ฅโ1)^2+3 Hence, Minimum value of (2๐ฅโ1)^2 = 0 Minimum value of (2๐ฅโ1^2 )+3 = 0 + 3 = 3 Also, there is no maximum value of ๐ฅ โด There is no maximum value of f(x) Ex 6.5,1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (i) f (๐ฅ) = (2๐ฅ โ 1)^2+3 Step 1: Finding fโ(x) f(๐ฅ)=(2๐ฅโ1)^2+3 fโ(๐ฅ)= 2(2๐ฅโ1) Step 2: Putting fโ(๐ฅ)=0 2(2๐ฅโ1)=0 2๐ฅ โ 1 = 0 2๐ฅ = 1 ๐ฅ = 1/2 Step 3: Thus, x = 1/2 is the minima Finding minimum value f(๐ฅ)=(2๐ฅโ1)^2+3 Putting ๐ฅ = 1/2 f(1/2)=(2 ร 1/2โ1)^2+3= (1โ1)^2+3= 3 โด Minimum value = 3 There is no maximum value Ex 6.5,1 (Method 3) Find the maximum and minimum values, if any, of the following functions given by (i) ๐ (๐ฅ)= (2๐ฅ โ 1)^2 + 3 Double Derivative Test f(๐ฅ)=(2๐ฅโ1)^2+3 fโ(๐ฅ)=2(2๐ฅโ1)^(2โ1) = 2(2๐ฅโ1) Putting fโ(๐ฅ)=0 2(2๐ฅโ1)=0 (2๐ฅโ1)=0 2๐ฅ = 0 + 1 ๐ฅ = 1/2 Now, finding fโโ(๐ฅ) fโ(๐ฅ)=2(2๐ฅโ1) fโ(๐ฅ) = 4๐ฅ โ 2 fโโ(๐ฅ)= 4 So, fโโ (1/2) = 4 Since fโโ (1/2) > 0 , ๐ฅ = 1/2 is point of local minima Putting ๐ฅ = 1/2 , we can calculate minimum value f(๐ฅ) = (2๐ฅโ1)^2+3 f(1/2)= (2(1/2)โ1)^2+3= (1โ1)^2+3= 3 Hence, Minimum value = 3 There is no Maximum value Ex 6.5,1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (ii) f (๐ฅ) = 9๐ฅ2+12๐ฅ+2 Step 1: Finding fโ(x) f (๐ฅ)=9๐ฅ2+12๐ฅ+2 Diff. w.r.t ๐ฅ fโ(๐ฅ)=18๐ฅ+12 fโ(๐ฅ)=6(3๐ฅ+2) Step 2: Putting fโ(๐ฅ)=0 6(3๐ฅ+2)=0 3๐ฅ+2=0 3๐ฅ=โ2 ๐ฅ=(โ2)/( 3) Step 3: Hence ๐ฅ=(โ2)/3 is point of minima of f(๐ฅ) Finding minimum value of f(๐ฅ) at ๐ฅ=(โ2)/3 f(๐ฅ)=9๐ฅ^2+12๐ฅ+2 Putting ๐ฅ=(โ2)/3 f(๐ฅ)=9((โ2)/3)^2+12((โ2)/3)+2=9(4/3)โ12(2/3)+2=โ2 Thus, Minimum value of f(๐)=โ๐ There is no maximum value Ex 6.5,1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (ii) f (๐ฅ) = 9๐ฅ2+12๐ฅ+2 Step 1: f (๐ฅ)=9๐ฅ2+12๐ฅ+2 Diff. w.r.t ๐ฅ fโ(๐ฅ)=๐(9๐ฅ^2 + 12๐ฅ + 2)/๐๐ฅ fโ(๐ฅ)=18๐ฅ+12 fโ(๐ฅ)=6(3๐ฅ+2) Step 2: Putting fโ(๐ฅ)=0 6(3๐ฅ+2)=0 3๐ฅ+2=0 3๐ฅ=โ2 ๐ฅ=(โ2)/3 Step 3: Finding fโโ(๐ฅ) fโ(๐ฅ)= 6(3๐ฅ+2) Again diff w.r.t ๐ฅ fโโ(๐ฅ)=๐(6(3๐ฅ+2))/๐๐ฅ fโโ(๐ฅ)=6 ๐(3๐ฅ + 2)/๐๐ฅ fโโ(๐ฅ)=6(3+0) fโโ(๐ฅ)=6(3) fโโ(๐ฅ)=18 So, fโโ((โ2)/3)=18 Since fโโ(๐ฅ)>0 is for ๐ฅ=(โ2)/3 ๐ฅ=(โ2)/3 is point of local minima Step 4: Putting ๐ฅ=(โ2)/3 we can calculate minimum value of f(๐ฅ) f (๐ฅ)=9๐ฅ2+12๐ฅ+2 f ((โ2)/3)=9((โ2)/3)^2+12((โ2)/3)+2 =9(4/9)+12((โ2)/3)+2 =4โ8+2 =โ2 Hence minimum value = โ2 There is no maximum value Ex 6.5,1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (iii) ๐ (๐ฅ) = โ (๐ฅ โ 1)^2+10 f(๐ฅ)=โ(๐ฅโ1)^2+10 Step 1: Diff w.r.t ๐ฅ fโ(๐ฅ)=๐(โ(๐ฅโ1)^2+10)/๐๐ฅ fโ(๐ฅ) = โ2(๐ฅโ1)(๐(๐ฅโ1)/๐๐ฅ)+0 fโ(๐ฅ) = โ2(๐ฅโ1)(1โ0) + 0 fโ(๐ฅ)=โ2(๐ฅโ1) Step 2: Putting fโ(๐ฅ)=0 โ2(๐ฅโ1)=0 (๐ฅโ1)=0 ๐ฅ=1 Step 3: Hence, ๐ฅ=1 is point of Maxima & No point of Minima Thus, f(๐ฅ) has maximum value at ๐ฅ=1 f(๐ฅ)=โ(๐ฅโ1)^2+10 Putting ๐ฅ=1 f(1)=โ(1โ1)^2+10 = 0 + 10 = 10 Maximum value of f(๐ฅ) is 10 There is no minimum value of f(๐) Ex 6.5,1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (iii) ๐ (๐ฅ) = โ (๐ฅ โ 1)^2+10 f(๐ฅ)=โ(๐ฅโ1)^2+10 Step 1: Diff w.r.t ๐ฅ fโ(๐ฅ)=๐(โ(๐ฅโ1)^2+10)/๐๐ฅ fโ(๐ฅ) = โ2(๐ฅโ1)(๐(๐ฅโ1)/๐๐ฅ)+0 fโ(๐ฅ) = โ2(๐ฅโ1)(1โ0) + 0 fโ(๐ฅ)=โ2(๐ฅโ1) Step 2: Putting fโ(๐ฅ)=0 โ2(๐ฅโ1)=0 (๐ฅโ1)=0 ๐ฅ=1 Step 3: Finding fโโ(๐ฅ) fโ(๐ฅ)=โ2(๐ฅโ1) Again diff w.r.t ๐ฅ fโโ(๐ฅ)=๐(โ2(๐ฅ โ 1))/๐๐ฅ =โ2 ๐(๐ฅ โ 1)/๐๐ฅ =โ2(1โ0) =โ2 Since fโโ(๐ฅ) < 0 for ๐ฅ=1 Hence f(๐ฅ) has Maximum value at ๐ฅ=1 Finding maximum value of f(๐ฅ) f(๐ฅ)=โ(๐ฅโ1)^2+10 Putting ๐ฅ=1 f(๐ฅ) =โ(1โ1)^2+10 = 0 + 10 = 10 Maximum value of f(๐) is 10 There is no minimum value of f(๐) Ex 6.5,1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (iv) f(๐ฅ) = ๐ฅ3 + 1 f(๐ฅ)=๐ฅ^3+1 Step 1: Finding fโ(x) fโ(๐ฅ)=๐(๐ฅ^3+1)/๐๐ฅ =3๐ฅ^2 Step 2: Putting fโ(๐ฅ)=0 3๐ฅ^2=0 ๐ฅ^2=0 ๐ฅ=0 Step 3: โ Therefore by first derivate test, the point ๐ฅ=0 is Neither a point of local maxima nor a point of local Minima Hence ๐=๐ is point of inflexion Hence, there is no minimum or maximum value Ex 6.5,1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (iv) f(๐ฅ) = ๐ฅ3 + 1 f(๐ฅ)=๐ฅ^3+1 Step 1: Finding fโ(x) fโ(๐ฅ)=๐(๐ฅ^3+1)/๐๐ฅ =3๐ฅ^2 Step 2: Putting fโ(๐ฅ)=0 3๐ฅ^2=0 ๐ฅ^2=0 ๐ฅ=0 Step 3: Finding fโโ(x) fโ(x) = 3x2 fโโ(x) = 6x Finding fโโ(x) at x = 0 fโโ(0) = 6 ร 0 = 0 Since fโโ(x) = 0 at x = 0 โ the point ๐ฅ=0 is Neither a point of local maxima nor a point of local Minima Hence ๐=๐ is point of inflexion Hence, there is no minimum or maximum value

Ex 6.5