Ex 6.3

Chapter 6 Class 12 Application of Derivatives
Serial order wise

### Transcript

Ex 6.3,1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (iii) đ (đĽ) = â(đĽ â 1)^2+10 f(đĽ)=â(đĽâ1)^2+10 Finding fâ (x) Diff w.r.t đĽ fâ(đĽ)=đ(â(đĽâ1)^2+10)/đđĽ fâ(đĽ) = â2(đĽâ1)(đ(đĽâ1)/đđĽ)+0 fâ(đĽ) = â2(đĽâ1)(1â0) + 0 fâ(đĽ)=â2(đĽâ1) Putting fâ(đ)=đ â2(đĽâ1)=0 (đĽâ1)=0 đĽ=1 Hence, đĽ=1 is point of Maxima & No point of Minima Thus, f(đĽ) has maximum value at đĽ=1 Putting x=1 in f(x) f(đĽ)=â(đĽâ1)^2+10 f(1)=â(1â1)^2+10 = 0 + 10 = 10 Maximum value of f(đĽ) is 10 There is no minimum value of f(đ) Ex 6.3,1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (iii) đ (đĽ) = â (đĽ â 1)^2+10 f(đĽ)=â(đĽâ1)^2+10 Finding fâ(x) Diff w.r.t đĽ fâ(đĽ)=đ(â(đĽâ1)^2+10)/đđĽ fâ(đĽ) = â2(đĽâ1)(đ(đĽâ1)/đđĽ)+0 fâ(đĽ) = â2(đĽâ1)(1â0) + 0 fâ(đĽ)=â2(đĽâ1) Putting fâ(đ)=đ â2(đĽâ1)=0 (đĽâ1)=0 đĽ=1 Finding fââ(đ) fâ(đĽ)=â2(đĽâ1) Again diff w.r.t đĽ fââ(đĽ)=đ(â2(đĽ â 1))/đđĽ =â2 đ(đĽ â 1)/đđĽ =â2(1â0) =â2 Since fââ(đĽ) < 0 for đĽ=1 Hence, f(đĽ) has Maximum value at đĽ=1 Finding maximum value of f(đ) f(đĽ)=â(đĽâ1)^2+10 Putting đĽ=1 f(đĽ) =â(1â1)^2+10 = 0 + 10 = 10 Maximum value of f(đ) is 10 There is no minimum value of f(đ)