Ex 6.3

Chapter 6 Class 12 Application of Derivatives
Serial order wise

### Transcript

Ex 6.3, 12 Find the maximum and minimum values of ๐ฅ + sin 2๐ฅ on [0, 2ฯ ] Let f(๐ฅ)=๐ฅ + sin 2๐ฅ Finding fโ(๐) ๐โ(๐ฅ)=๐(๐ฅ + sin 2๐ฅ)/๐๐ฅ =1+2 cosโก2๐ฅ Putting fโ(๐)=๐ 1 + 2 cos 2๐ฅ=0 2 cos 2๐ฅ=โ1 cos 2๐ฅ=(โ1)/2 cos 2๐ฅ=cosโกใ2๐/3ใ General solution for cos 2๐ฅ is 2๐ฅ=2๐๐ยฑ2๐/3 ๐ฅ=(2๐๐ ยฑ 2๐/3)/2 ๐ฅ= nฯ ยฑ ๐/3 Putting ๐=0 ๐ฅ=0(๐)ยฑ๐/3 =ยฑ๐/3 " " We know that ๐๐๐  60ยฐ=1/2 And cos is negative in 2nd & 3rd quadrant ๐ = 180 โ 60 = 120 = 120 ร ๐/180 = 2๐/3 So, ๐ฅ=(โ๐)/3,๐/3 Since Given ๐ฅ โ[0 , 2๐] โด ๐ฅ=๐/3 only Putting ๐=1 ๐ฅ=(1)๐ยฑ๐/3 =๐ยฑ๐/3 =(3๐ + ๐)/3 , (3๐ โ ๐)/3 =4๐/3 , 2๐/3 Putting ๐=2 ๐ฅ=2(๐)ยฑ๐/3 ๐ฅ=2๐โ๐/3 & 2๐+๐/3 ๐ฅ=(6๐ โ ๐)/3 & (6๐ + ๐)/3 ๐ฅ=5๐/3 & 7๐/3 So, ๐ฅ=5๐/3 only Also, We are given interval ๐ฅ โ[0 , 2๐] Hence , calculating f(๐ฅ) at ๐ฅ=0 , ๐/3 , 2๐/3 , 4๐/3 , 5๐/3 , 2๐ Hence, f(๐ฅ) is Maximum at ๐ฅ=2๐ Maximum value of f(๐)=๐๐ & f(๐ฅ) is Minimum at ๐ฅ=0 Minimum value of f(๐)=๐